Research ArticleExtensions and Applications of Power-TypeAczeacutel-VasiT-PeIariTrsquos Inequalities

Kun Chen12 Fei Yan 1 and Hong-Ying Yue3

1North China Electric Power University Baoding Hebei Province 071003 China2Changde Power Supply of Hunan Electric Power Company Changde Hunan Province 415000 China3College of Science and Technology North China Electric Power University Baoding Hebei Province 071051 China

Correspondence should be addressed to Fei Yan yan feincepueducn

Received 15 May 2019 Accepted 24 June 2019 Published 7 July 2019

Academic Editor Peter Dabnichki

Copyright copy 2019 Kun Chen et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In this paper we enrich and develop power-type Aczel-Vasic-Pecaricrsquos inequalities First of all we give some new versions oftheorems and corollaries about Aczel-Vasic-Pecaricrsquos inequalities by quoting some lemmasMoreover in combinationwithHolderrsquosinequality we give some applications of the new version of Aczel-Vasic-Pecaricrsquos inequality and give its proof process

1 Introduction

In 1956 Aczel [1] discovered the Aczelrsquos inequality

eorem 1 Let n be a positive integer and let 120582119904 ]119904 (119904 =1 2 119899) 119899 ge 2 be positive numbers such that 12058221minussum119899119904=2 1205822119904 ge0 ]21 minus sum119899119904=2 ]2119904 ge 0 Then

(12058221 minus 119899sum119904=2

1205822119904)(]21 minus 119899sum119904=2

]2119904) le (1205821]1 minus 119899sum119904=2

120582119904]119904)2 (1)

It is matter of common observation that Aczelrsquos inequality(1) is of great significance in the theory of functional equa-tions in non-Euclidean geometry meanwhile many authorsincluding Bellman [2] Hu et al[3] Tian [4 5] Tian and Ha[6] Tian and Sun [7] Tian and Wu [8] Tian and Wang [9]Tian and Zhou [10] Wu [11] and Wu and Debnath [12 13]pay more attention to this inequality and its refinements

In 1959 Popoviciu [14] gave a generalization of Aczelrsquosinequality as follows

eorem 2 Let 119901 gt 1 119902 gt 1 and 1119901 + 1119902 = 1 andlet 120582119904 ]119904 (119904 = 1 2 119899) be positive numbers such that 1205821199011 minussum119899119904=2 120582119901119904 gt 0 and ]1199011 minus sum119899119904=2 ]119901119904 gt 0 Then

(1205821199011 minus 119899sum119904=2

120582119901119904)1119901 (]1199021 minus 119899sum

119904=2

]119902119904)1119902 le 1205821]1 minus 119899sum

119904=2

120582119904]119904 (2)

In 1979 Vasic and Pecaric [15] proved the followingextension of inequality (2)

eorem 3 Let 119901119905 gt 0 (119905 = 1 2 119898) and sum119898119905=1 1119901119905 ge 1and let 120582119904119905 (119904 = 1 2 119899 119905 = 1 2 119898) be positive integerssuch that 1205821199011199051119905 minus sum119899119904=2 120582119901119905119904119905 gt 0 Then

119898prod119905=1

(1205821199011199051119905 minus 119899sum119904=2

120582119901119905119904119905)1119901119905 le 119898prod

119905=1

1205821119905 minus 119899sum119904=2

119898prod119905=1

120582119904119905 (3)

Inequality (3) is known as Aczel- Vasic-Pecaricrsquos inequal-ity

In 2005 Wu and Debnath [13] generalized inequality (3)in the following form

eorem 4 Let 119901119905 and 120582119904119905 (119904 = 1 2 119899 119905 = 1 2 119898)be positive numbers such that 1205821199011199051119905 minus sum119899119904=2 120582119901119905119904119905 gt 0 for 119905 =1 2 119898 and let 120591 = minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

[1205821199011199051119905 minus 119899sum119904=2

120582119901119905119904119905]1119901119905 le 1198991minus120591 119898prod

119905=1

120582119905 minus 119899sum119904=2

119898prod119905=1

120582119905 (4)

Later Wu in [11] established the Aczel-Vasic-Pecaricinequality (3)

HindawiMathematical Problems in EngineeringVolume 2019 Article ID 1323474 8 pageshttpsdoiorg10115520191323474

2 Mathematical Problems in Engineering

eorem 5 Let 1199011 1199012 119901119898 gt 0 and 120588 = 11199011+11199012+sdot sdot sdot+1119901119898 and let 120582119904119905 gt 0 1205821199011199051119905 minus sum119899119904=2 120582119901119905119904119905 gt 0 (119904 = 1 2 119899 119905 =1 2 119898) and119898 ge 2 119899 ge 2 Then we have

119898prod119905=1

(1205821199011199051119905 minus 119899sum119904=2

120582119901119905119904119905)1119901119905

le 1198991minusmin1205881119898prod119905=1

120582119905 minus 119899sum119904=2

119898prod119905=1

120582119905

minus 119886 sum1le119905le119896le119898

( 119899sum119904=2

(1205821199011199051199041199051205821199011199051119905 minus1205821199011198961199041198961205821199011198961119896))2

(5)

where

119886 = prod119898119905=11205821119905120578

120578 = (119898 minus 1)max 1199011 1199012 119901119898 1198982 120588 ge 1119899120588minus11198982 (119898 minus 1)minus1max 1199011 1199012 119901119898 (1 minus 120588)minus1 120588 lt 1

(6)

Moreover in 2014 Tian [7] also presented an improve-ment of the Aczel-Vasic-Pecaric inequality (3)

eorem 6 Let 120582119903119905 gt 0 119901119905 gt 0 and 1205821199011199051119905 minus sum119899119903=2 120582119901119905119903119905 gt0 (119903 = 1 2 119899 119905 = 1 2 119898) let 119898 ge 2 and let 120588 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

(1205821199011199051119905 minus 119899sum119903=2

120582119901119905119903119905)1119901119905 le 1198991minus120588 1

minus 2119898 (119898 minus 1) sum1le119894le119905le119898

[ 119899sum119903=2

(1205821199011198941199031198941205821199011198941119894 minus1205821199011199051199031199051205821199011199051119905)]2

120574119898prod119905=1

1205821119905minus 119899sum119903=2

119898prod119905=1

120582119903119905 le 1198991minus120588 119898prod119905=1

1205821119905 minus 119899sum119903=2

119898prod119905=1

120582119903119905

(7)

where 120574 = 119898(2max1199011 1199012 119901119898)Recently the power mean has attracted of many

researcher ([16ndash24]) and many remarkable inequalities forthe power mean including Holder-type inequalities can befound in the literature [25ndash30] The purpose of the article isto establish some distinctive versions of Aczel-Vasic-Pecaricrsquosinequality (3) for power mean type As consequences severalintegral inequalities of the obtained results are given

2 Some Power Mean Types ofAczeacutel-VasiT-PeIariTrsquos Inequality

Lemma 7 (see [13]) Let 120582119904119905 gt 0 and 1119901119905 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) and let 120591 = minsum119898119905=1(1119901119905) 1Then

119899sum119904=1

119898prod119905=1

120582119904119905 le 1198991minus120591 119898prod119905=1

( 119899sum119904=1

120582119901119905119904119905)1119901119905 (8)

Like in [31] the power mean of order 119903 for 119899 positivenumbers 1205821 1205822 120582119899 is defined by

119872119899 (120582 119903) =(1119899119899sum119896=1

120582119903119896)1119903 119903 = 0

119899radic12058211205822 sdot sdot sdot 120582119899 119903 = 0(9)

where 120582 = (1205821 1205822 120582119899)Lemma 8 (see [32]) Let 1205821 1205822 120582119899 be a positive sequence119899 isin N If 119903 isin R then the function 119903 997891997888rarr 119872119899(120582 119903) is increasingfor fixed 1205821 1205822 120582119899Lemma 9 (see [31]) Let 120582119904 gt 0 (119904 = 1 2 119899) be positivenumbers and let 119901 gt 0 Then

119899sum119904=1

120582119901119904 le 1198991minusmin1199011 ( 119899sum119904=1

120582119904)119901 (10)

Putting 119901 ge 1 we get the following inequality119899sum119904=1

120582119901119904 le ( 119899sum119904=1

120582119904)119901 (11)

Let

120582119905 = (1205821119905 1205822119905 120582119899119905) (119905 = 1 2 119898) 119872119898 (120582119901119905 119903) = ( 1119898

119898sum119904=1

120582119901119903119904119905)1119903

119872119898119899 (120582119901119905 119903) = ( 1119899 minus 119898 + 1119899sum119904=119898

120582119901119903119904119905)1119903

119872120594119898119899 (120582119895 119903) = ( 1119899 minus 119898 + 1119899sum119894=119898

120582119903119904119905)120594119903

(12)

The main conclusions of this paper are as follows

eorem 10 Let 119903 gt 0 119901119905 gt 0 1198721199011199051 (120582119905 119903) minus 1198721199011199052119899(120582119905 119903) gt0 and prod119898119905=1120582119905 = (prod119898119905=11205821119905prod119898119905=11205822119905 prod119898119905=1120582119899119905) and let120591 = minsum119898119905=1(1119901119905) 1 and 120572 = 1 + sum119898119905=1min1119903 1119901119905 minussum119898119905=1(1119901119905) minus 119898119903 Then

119898prod119905=1

[1198721 (120582119901119905119905 119903) minus 1198721199011199052119899 (120582119905 119903)]1119901119905

le 1198991minus1205911198721 ( 119898prod119905=1

120582119905 1) minus (119899 minus 1)1205721198722119899 ( 119898prod119905=1

120582119905 1) (13)

Proof Simple computations lead to119898prod119905=1

[1198721 (120582119901119905119905 119903) minus 1198721199011199052119899 (120582119905 119903)]1119901119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

(14)

Mathematical Problems in Engineering 3

Since 119901119905 gt 0 1205821199011199051119905 minus ((1(119899 minus 1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 let

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903 = 120585119901119905119905

(120585119905 gt 0 119905 = 1 2 119898 minus 1) (15)

119898prod119905=1

1205821119905minus 119899120591minus1 (119899 minus 1)sum119898119905=1min11199031119901119905minussum

119898

119905=1(1119901119905)minus119898119903

119899sum119904=2

119898prod119905=1

120582119904119905= 119899120591minus1 119898prod

119905=1

120585119905(16)

Combining (15) and inequalities (10) and (8) we have

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 119898minus1prod119905=1

1205821119905 = 119898prod119905=1

[120585119901119905119905

+ ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905 ge 119898prod

119905=1

[120585119901119905119905

+ (119899 minus 1)min1199011199051199031minus119901119905119903minus1119899sum119904=2

120582119901119905119904119905]1119901119905 = 119898prod

119905=1

120585119901119905119905

+ 119899sum119904=2

((119899 minus 1)min11199031119901119905minus1119903minus1119901119905 120582119904119905)1199011199051119901119905

ge 119899120591minus1 119898prod119905=1

120585119905+ 119899sum119904=2

119898prod119905=1

((119899 minus 1)min11199031119901119905minus1119903minus1119901119905 120582119904119905)

= 119899120591minus1 [ 119898prod119905=1

120585119905 + (119899 minus 1)sum119898119905=1min11199031119901119905minussum119898

119905=1(1119901119905)minus119898119903

sdot 119899sum119904=2

119898prod119905=1

120582119904119905] = 119898prod119905=1

1205821119905

(17)

Therefore from (15) (16) and (17) we obtain

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 119898minus1prod119905=1

1205821119905 ge 119898prod119905=1

1205821119905 (18)

Hence we obtain

120585119898 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 (19)

Therefore we have

119898prod119905=1

120585119905 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 119898minus1prod119905=1

120585119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

(20)

By using (16) we immediately obtain

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120572 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(21)

which leads to inequality (13) The proof of Theorem 10 isaccomplished

According to Theorem 10 we can get the followingcorollaries

Corollary 11 Let 119903 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and let120591 = minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(22)

Proof From inequality (21) we can get

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120572 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905le 1198991minus120591 119898prod

119905=1

1205821119905 minus (119899 minus 1)120572minus1 119899sum119904=2

119898prod119905=1

120582119904119905le 1198991minus120591 119898prod

119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905

(23)

The proof of Corollary 11 is accomplished

Corollary 12 Let 1 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119904119905)119901119905 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119904119905)119901119905]1119901119905

le 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(24)

4 Mathematical Problems in Engineering

If we put sum119898119905=1(1119901119905) = 1 in Corollary 11 the conclusionsthat we can draw from this are as follows

Corollary 13 Let 119903 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) = 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(25)

Let 119901119905 = 119903 (119905 = 1 2 119898) then fromTheorem 10 we get

Corollary 14 Let 119903 gt 0 1205821199031119905 minus (1(119899 minus 1))sum119899119904=2 120582119903119904119905 gt 0 (119904 =2 3 119899 119905 = 1 2 119898) and let 120591 = min119898119903 1 Then

119898prod119905=1

[1205821199031119905 minus 1119899 minus 1119899sum119904=2

120582119903119904119905]1119903

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)minus119898119903 119899sum119904=2

119898prod119905=1

120582119904119905(26)

Using the obtained inequality of inequality (26) we canget the following result of Wu [13]

Corollary 15 Let 119903 gt 0 and 1205951199031119905 minus sum119899119904=2 120595119903119904119905 gt 0 (119904 =2 3 119899 119905 = 1 2 119898) and let 120591 = min119898119903 1 Then

119898prod119905=1

(1205951199031119905 minus 119899sum119904=2

120595119903119904119905)1119903 le 1198991minus120591 119898prod

119905=1

1205951119905 minus 119899sum119904=2

119898prod119905=1

120595119904119905 (27)

eorem 16 Let 119903 gt 0 119901119905 gt 0 and 1198721(120582119901119905119905 119903) minus1198722119899(120582119901119905119905 119903) gt 0 let 120582119905 = (1205821119905 1205822119905 120582119899119905) and prod119898119905=1120582119905 =(prod119898119905=11205821119905prod119898119905=11205822119905 prod119898119905=1120582119899119905) and denote 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

[1198721 (120582119901119905119905 119903) minus 1198722119899 (120582119901119905119905 119903)]1119901119905

le 21minus1205911198721 ( 119898prod119905=1

120582119905 119903)

minus (119899 minus 1)(120591minussum119898119905=1(1119901119905))1199031198722119899 ( 119898prod119905=1

120582119905 119903) (28)

Proof For the hypotheses 119901119905 gt 0 and 1198721(120582119901119905119905 119903) minus1198722119899(120582119901119905119905 119903) gt 0 (119905 = 1 2 119898) we have

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903 gt 0 (29)

Denote

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903 = 120585119901119905119905

(120585119905 gt 0 119905 = 1 2 119898 minus 1) (30)

119898prod119905=1

1205821119905

minus 2120591minus1 (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

= 2120591minus1 119898prod119905=1

120585119905

(31)

By using inequality (8) simple computations lead to

119898prod119905=1

[120585119901119905119905 + ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905 ge 2120591minus1 [ 119898prod

119905=1

120585119905

+ 119898prod119905=1

( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903119901119905] = 2120591minus1 [ 119898prod

119905=1

120585119905

+ (119899 minus 1)minus(1119903)sum119898119905=1(1119901119905) 119898prod119905=1

( 119899sum119904=2

120582119903119901119905119904119905 )1119903119901119905]

ge 2120591minus1 [ 119898prod119905=1

120585119905

+ (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903]

(32)

Therefore from (30) (31) and (32) we obtain

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898

sdot 119898minus1prod119905=1

[120585119901119905119905 + ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905 = [120585119901119898119898

+ ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

1205821119905 ge 2120591minus1 [ 119898prod119905=1

120585119905

+ (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903]

= 119898prod119905=1

1205821119905

(33)

Hence

120585119898 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 (34)

Mathematical Problems in Engineering 5

Thus

119898prod119905=1

120585119905 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

120585119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

(35)

Combining (31) and inequality (35) we get

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

le 21minus120591 119898prod119905=1

1205821119905

minus (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(36)

which is equivalent to inequality (28) Successful proof ofTheorem 16 is as follows

Corollary 17 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) gt 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 119898prod119905=1

1205821119905

minus (119899 minus 1)(1minussum119898119905=1(1119901119905))119903 ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(37)

In particular putting 119903 = 1 in Corollary 17 we obtain anew version of the Aczel-Vasic-Pecaric inequality (3)

119898prod119905=1

[1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905]1119901119905

le 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(38)

Corollary 18 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(39)

Corollary 19 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let sum119898119905=1(1119901119905) le 1Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(40)

If we put 119903 = 1 inTheorem 16 then we have the followingcorollary

Corollary 20 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120591minussum119898119905=1(1119901119905)minus1 119899sum119904=2

119898prod119905=1

120582119904119905(41)

Using the substitutions 1205951119905 997888rarr 1205821119905 and (119899 minus 1)1119901119905120595119904119905 997888rarr120582119904119905 (119904 = 2 3 119899 119905 = 1 2 119898) in Corollary 20 weobtain the following refinement of the Aczel-Vasic-Pecaricinequality (3)

Corollary 21 Let 120595119904119905 119901119905 gt 0 and 1205951199011199051119905 minus sum119899119904=2 120595119901119905119904119905 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) and let 120591 = minsum119898119905=1(1119901119905) 1Then

119898prod119905=1

(1205951199011199051119905 minus 119899sum119904=2

120595119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205951119905 minus (119899 minus 1)120591minus1 119899sum119904=2

119898prod119905=1

120595119904119905(42)

If we put sum119898119905=1(1119901119905) ge 1 in Corollary 21 then we canderive inequality (3)

In particular putting 119898 = 2 1199011 = 119901 1199012 = 119902 1205821199041 = 120582119904and 1205821199042 = ]119904 (119904 = 1 2 119899) in Corollary 21 we obtain afresh improvement and promotion of nonequality (2)

Corollary 22 Let 120582119904 ]119904 gt 0 119901 119902 gt 0 and 120591 = min1119901 +1119902 1 and let 120582119901119904 minus sum119899119904=2 120582119901119904 gt 0 and ]119902119904 minus sum119899119904=2 ]119902119904 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) Then

(120582119901119904 minus nsum119904=2

120582119901119904)1119901 (]119902119904 minus 119899sum

119904=2

]119902119904)1119902

le 21minus1205911205821]1 minus (119899 minus 1)120591minus1 119899sum119904=2

120582119904]119904(43)

6 Mathematical Problems in Engineering

eorem 23 Let 120582119904119905 119901119905 gt 0 (119904 = 1 2 119899 119905 = 1 2 119898)and 1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 and let sum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905 (44)

Proof For sum119898119905=1(1119901119905) le 1 by inequality (28) we obtain119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(45)

We get this inequality (44) by adding 119903 997888rarr 0 on both sides ofinequality (45)

For the hypotheses sum119898119905=1(1119901119905) le 1 and1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 in Theorem 23 denote120577119905 = (prod119899119904=21205821199041199051205821119905)1(119899minus1) then we have 0 lt 120577119901119905119905 lt 120577119905 lt 1Thus we obtain the following corollaries

Corollary 24 Let 119901119905 gt 0 0 lt 120577119905 lt 1 (119905 = 1 2 119898) and letsum119898119905=1(1119901119905) lt 1 Then

119898prod119905=1

(1 minus 120577119901119905119905 )1119901119905 + 119898prod119905=1

120577119905 le 21minussum119898119905=1(1119901119905) (46)

Denote Θ(119903) = prod119898119905=1[1198721(120582119901119905 119903) minus 1198722119899(120582119901119905 119903)]1119901119905 byLemma 8 we realize that Θ(119903) is decreasing on (0 +infin)Combining Theorems 16 and 23 we obtain the followingcorollary

Corollary 25 Let 120582119904119905 gt 0 and 119901119905 gt 0 (119904 = 1 2 119899 119905 =1 2 119898) and let 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt 0 Ifsum119898119905=1(1119901119905) lt 1 then119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905

(47)

3 Applications

As iswell known analytic inequalities have various importantapplications in many branches of mathematics [33ndash37] Inthis section we will get some applications for this newinequality in Section 2

eorem 26 Let 120595119905 gt 0 119903 ge 119901119905 gt 0 sum119898119905=1(1119901119905) = 1 (119905 =1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119905 (119909)119889119909)119901119905119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119905 (119909) 119889119909)

119901119905119903]1119901119905

le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(48)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof For all positive integers 119899 the equidistance of thepartition of [0 1] is selected as

0 lt 1119899 lt sdot sdot sdot lt 119899 minus 1119899 lt 1119909119904 = 119904119899

119909119904 = 1119899 119904 = 0 1 119899 minus 1

(49)

Noting that 120595119901119905119905 minus (int10119891119903119905 (119909)119889119909)119901119905119903 gt 0 (119905 = 1 2 119898) we

have

120595119901119905119905 minus ( lim119899997888rarrinfin

119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0

(119905 = 1 2 119898) (50)

Therefore there is a positive integer119873 like that

120595119901119905119905 minus (119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0 (51)

for all 119899 gt 119873 and 119905 = 1 2 119898Moreover for any 119899 gt 119873 it follows fromCorollary 13 that

119898prod119905=1

[[120595119901119905119905 minus (1119899

119899minus1sum119904=0

119891119903119905 ( 119904119899))119901119905119903]

]1119901119905

le 119898prod119905=1

120595119905 minus 1119899119899minus1sum119904=0

119898prod119905=1

119891119905 ( 119904119899) (52)

We may find that 119891119905(119909) (119905 = 1 2 119898) are positive Rie-mann integrable functions on [0 1] we know thatprod119898119905=1119891119905(119909)and 119891119903119905 (119909) are also integrable on [0 1] Letting 119899 997888rarr infin onboth sides of inequality (52) we get the desired inequality(48) The proof of Theorem 26 is achieved

eorem 27 Let 120595119905 gt 0 and 119903 ge 119898 and let 119891119905(119909) (119905 =1 2 119898) be positive Riemann integrable functions on [0 1]such that 120595119903119905 minus int1

0119891119903119905 (119909)119889119909 gt 0 Then

119898prod119905=1

[120595119903119905 minus int10119891119903119905 (119909) 119889119909]

1119903 le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909 (53)

Mathematical Problems in Engineering 7

Proof Applying Corollary 14 and along the lines of the proofTheorem 26 Theorem 27 is simply given

eorem 28 Let 119903 gt 0 120595119905 gt 0 119901119905 gt 0 sum119898119905=1(1119901119905) le 1(119905 = 1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119901119905119905 (119909)119889119909)1119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119901119905119905 (119909) 119889119909)1119903]

1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus (int10

119898prod119905=1

119891119903119905 (119909) 119889119909)1119903

(54)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof Combining the proof Theorem 26 and Corollary 18 itis easy to get Theorem 28

Corollary 29 Let 119901119905 gt 0 sum119898119905=1(1119901119905) le 1 (119905 = 1 2 119898)and 120595119901119905119905 minus int1

0119891119901119905119905 (119909)119889119909 gt 0 Then119898prod119905=1

(120595119901119905119905 minus int10119891119901119905119905 (119909) 119889119909)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(55)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0A direct result from Theorem 28 is given by us Putting119898 = 2 1199011 = 119901 gt 0 1199012 = 119902 gt 0 1205951 = 1205821 1205952 = ]1 1198911 = 119891 and1198912 = 119892 in (54) we can get the following corollaries

Corollary 30 Let 119903 gt 0 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 11205821199011minus(int1

0119891119903119901(119909)119889119909)1119903 gt 0 and ]1199021minus(int1

0119892119903119902(119909)119889119909)1119903 gt 0Then

[1205821199011 minus (int10119891119903119901 (119909) 119889119909)1119903]

1119901

sdot []1199021 minus (int10119892119903119902 (119909) 119889119909)1119903]

1119902

le 21minus1119901minus11199021205821]1minus [int10(119891 (119909) 119892 (119909))119903 119889119909]1119903

(56)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0If we set 119903 = 1 in Corollary 30 then the following

inequality holds

Corollary 31 Let 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 1 1205821199011 minusint10119891119901(119909)119889119909 gt 0 and ]1199021 minus int1

0119892119902(119909)119889119909 gt 0 Then

(1205821199011 minus int10119891119901 (119909) 119889119909)1119901 (]1199021 minus int1

0119892119902 (119909) 119889119909)1119902

le 21minus1119901minus11199021205821]1 minus int10119891 (119909) 119892 (119909) 119889119909

(57)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Authorsrsquo Contributions

All authors contributed equally and significantly in writ-ing this article All authors read and approved the finalmanuscript

Acknowledgments

This work was supported by the Fundamental ResearchFunds for the Central Universities (no 2015ZD29) and theHigher School Science Research Funds of Hebei Province ofChina (no Z2015137)

References

[1] J Aczel ldquoSome general methods in the theory of functionalequations in one variable New applications of functionalequationsrdquo Uspekhi Matematicheskikh Nauk vol 11 no 3 pp3ndash68 1956 (Russian)

[2] R Bellman ldquoOn an inequality concerning an indefinite formrdquoTheAmericanMathematical Monthly vol 63 no 2 pp 108-1091956

[3] Z Hu and A Xu ldquoRefinements of Aczel and Bellmanrsquos inequal-itiesrdquo Computers amp Mathematics with Applications vol 59 no9 pp 3078ndash3083 2010

[4] J Tian ldquoA sharpened and generalized version of Aczel-Vasic-Pecaric inequality and its applicationrdquo Journal of Inequalitiesand Applications vol 2013 article no 497 2013

[5] J Tian ldquoReversed version of a generalized Aczelrsquos inequalityand its applicationrdquo Journal of Inequalities and Applications vol2012 article no 202 2012

[6] J Tian andM-HHa ldquoProperties and refinements of Aczel-typeinequalitiesrdquo Journal of Mathematical Inequalities vol 12 no 1pp 175ndash189 2018

[7] J Tian and Y Sun ldquoNew refinements of generalized Aczelinequalityrdquo Journal of Inequalities and Applications vol 2014article no 239 2014

[8] J Tian and S Wu ldquoNew refinements of generalized Aczelrsquosinequality and their applicationsrdquo Journal of MathematicalInequalities vol 10 no 1 pp 247ndash259 2016

[9] J Tian andWWang ldquoReversed versions of Aczel-type inequal-ity and Bellman-type inequalityrdquo Journal of MathematicalInequalities vol 9 no 2 pp 417ndash424 2015

[10] J-F Tian and Y-J Zhou ldquoNote on Aczel-type inequality andBellman-type inequalityrdquo Journal of Nonlinear Sciences andApplications vol 9 no 3 pp 1316ndash1322 2016

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

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Mathematical Problems in Engineering 3

Since 119901119905 gt 0 1205821199011199051119905 minus ((1(119899 minus 1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 let

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903 = 120585119901119905119905

(120585119905 gt 0 119905 = 1 2 119898 minus 1) (15)

119898prod119905=1

1205821119905minus 119899120591minus1 (119899 minus 1)sum119898119905=1min11199031119901119905minussum

119898

119905=1(1119901119905)minus119898119903

119899sum119904=2

119898prod119905=1

120582119904119905= 119899120591minus1 119898prod

119905=1

120585119905(16)

Combining (15) and inequalities (10) and (8) we have

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 119898minus1prod119905=1

1205821119905 = 119898prod119905=1

[120585119901119905119905

+ ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905 ge 119898prod

119905=1

[120585119901119905119905

+ (119899 minus 1)min1199011199051199031minus119901119905119903minus1119899sum119904=2

120582119901119905119904119905]1119901119905 = 119898prod

119905=1

120585119901119905119905

+ 119899sum119904=2

((119899 minus 1)min11199031119901119905minus1119903minus1119901119905 120582119904119905)1199011199051119901119905

ge 119899120591minus1 119898prod119905=1

120585119905+ 119899sum119904=2

119898prod119905=1

((119899 minus 1)min11199031119901119905minus1119903minus1119901119905 120582119904119905)

= 119899120591minus1 [ 119898prod119905=1

120585119905 + (119899 minus 1)sum119898119905=1min11199031119901119905minussum119898

119905=1(1119901119905)minus119898119903

sdot 119899sum119904=2

119898prod119905=1

120582119904119905] = 119898prod119905=1

1205821119905

(17)

Therefore from (15) (16) and (17) we obtain

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 119898minus1prod119905=1

1205821119905 ge 119898prod119905=1

1205821119905 (18)

Hence we obtain

120585119898 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 (19)

Therefore we have

119898prod119905=1

120585119905 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119898)119901119898119903]1119901119898 119898minus1prod119905=1

120585119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

(20)

By using (16) we immediately obtain

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120572 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(21)

which leads to inequality (13) The proof of Theorem 10 isaccomplished

According to Theorem 10 we can get the followingcorollaries

Corollary 11 Let 119903 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and let120591 = minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(22)

Proof From inequality (21) we can get

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120572 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905le 1198991minus120591 119898prod

119905=1

1205821119905 minus (119899 minus 1)120572minus1 119899sum119904=2

119898prod119905=1

120582119904119905le 1198991minus120591 119898prod

119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905

(23)

The proof of Corollary 11 is accomplished

Corollary 12 Let 1 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119904119905)119901119905 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119904119905)119901119905]1119901119905

le 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(24)

4 Mathematical Problems in Engineering

If we put sum119898119905=1(1119901119905) = 1 in Corollary 11 the conclusionsthat we can draw from this are as follows

Corollary 13 Let 119903 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) = 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(25)

Let 119901119905 = 119903 (119905 = 1 2 119898) then fromTheorem 10 we get

Corollary 14 Let 119903 gt 0 1205821199031119905 minus (1(119899 minus 1))sum119899119904=2 120582119903119904119905 gt 0 (119904 =2 3 119899 119905 = 1 2 119898) and let 120591 = min119898119903 1 Then

119898prod119905=1

[1205821199031119905 minus 1119899 minus 1119899sum119904=2

120582119903119904119905]1119903

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)minus119898119903 119899sum119904=2

119898prod119905=1

120582119904119905(26)

Using the obtained inequality of inequality (26) we canget the following result of Wu [13]

Corollary 15 Let 119903 gt 0 and 1205951199031119905 minus sum119899119904=2 120595119903119904119905 gt 0 (119904 =2 3 119899 119905 = 1 2 119898) and let 120591 = min119898119903 1 Then

119898prod119905=1

(1205951199031119905 minus 119899sum119904=2

120595119903119904119905)1119903 le 1198991minus120591 119898prod

119905=1

1205951119905 minus 119899sum119904=2

119898prod119905=1

120595119904119905 (27)

eorem 16 Let 119903 gt 0 119901119905 gt 0 and 1198721(120582119901119905119905 119903) minus1198722119899(120582119901119905119905 119903) gt 0 let 120582119905 = (1205821119905 1205822119905 120582119899119905) and prod119898119905=1120582119905 =(prod119898119905=11205821119905prod119898119905=11205822119905 prod119898119905=1120582119899119905) and denote 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

[1198721 (120582119901119905119905 119903) minus 1198722119899 (120582119901119905119905 119903)]1119901119905

le 21minus1205911198721 ( 119898prod119905=1

120582119905 119903)

minus (119899 minus 1)(120591minussum119898119905=1(1119901119905))1199031198722119899 ( 119898prod119905=1

120582119905 119903) (28)

Proof For the hypotheses 119901119905 gt 0 and 1198721(120582119901119905119905 119903) minus1198722119899(120582119901119905119905 119903) gt 0 (119905 = 1 2 119898) we have

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903 gt 0 (29)

Denote

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903 = 120585119901119905119905

(120585119905 gt 0 119905 = 1 2 119898 minus 1) (30)

119898prod119905=1

1205821119905

minus 2120591minus1 (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

= 2120591minus1 119898prod119905=1

120585119905

(31)

By using inequality (8) simple computations lead to

119898prod119905=1

[120585119901119905119905 + ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905 ge 2120591minus1 [ 119898prod

119905=1

120585119905

+ 119898prod119905=1

( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903119901119905] = 2120591minus1 [ 119898prod

119905=1

120585119905

+ (119899 minus 1)minus(1119903)sum119898119905=1(1119901119905) 119898prod119905=1

( 119899sum119904=2

120582119903119901119905119904119905 )1119903119901119905]

ge 2120591minus1 [ 119898prod119905=1

120585119905

+ (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903]

(32)

Therefore from (30) (31) and (32) we obtain

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898

sdot 119898minus1prod119905=1

[120585119901119905119905 + ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905 = [120585119901119898119898

+ ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

1205821119905 ge 2120591minus1 [ 119898prod119905=1

120585119905

+ (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903]

= 119898prod119905=1

1205821119905

(33)

Hence

120585119898 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 (34)

Mathematical Problems in Engineering 5

Thus

119898prod119905=1

120585119905 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

120585119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

(35)

Combining (31) and inequality (35) we get

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

le 21minus120591 119898prod119905=1

1205821119905

minus (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(36)

which is equivalent to inequality (28) Successful proof ofTheorem 16 is as follows

Corollary 17 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) gt 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 119898prod119905=1

1205821119905

minus (119899 minus 1)(1minussum119898119905=1(1119901119905))119903 ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(37)

In particular putting 119903 = 1 in Corollary 17 we obtain anew version of the Aczel-Vasic-Pecaric inequality (3)

119898prod119905=1

[1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905]1119901119905

le 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(38)

Corollary 18 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(39)

Corollary 19 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let sum119898119905=1(1119901119905) le 1Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(40)

If we put 119903 = 1 inTheorem 16 then we have the followingcorollary

Corollary 20 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120591minussum119898119905=1(1119901119905)minus1 119899sum119904=2

119898prod119905=1

120582119904119905(41)

Using the substitutions 1205951119905 997888rarr 1205821119905 and (119899 minus 1)1119901119905120595119904119905 997888rarr120582119904119905 (119904 = 2 3 119899 119905 = 1 2 119898) in Corollary 20 weobtain the following refinement of the Aczel-Vasic-Pecaricinequality (3)

Corollary 21 Let 120595119904119905 119901119905 gt 0 and 1205951199011199051119905 minus sum119899119904=2 120595119901119905119904119905 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) and let 120591 = minsum119898119905=1(1119901119905) 1Then

119898prod119905=1

(1205951199011199051119905 minus 119899sum119904=2

120595119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205951119905 minus (119899 minus 1)120591minus1 119899sum119904=2

119898prod119905=1

120595119904119905(42)

If we put sum119898119905=1(1119901119905) ge 1 in Corollary 21 then we canderive inequality (3)

In particular putting 119898 = 2 1199011 = 119901 1199012 = 119902 1205821199041 = 120582119904and 1205821199042 = ]119904 (119904 = 1 2 119899) in Corollary 21 we obtain afresh improvement and promotion of nonequality (2)

Corollary 22 Let 120582119904 ]119904 gt 0 119901 119902 gt 0 and 120591 = min1119901 +1119902 1 and let 120582119901119904 minus sum119899119904=2 120582119901119904 gt 0 and ]119902119904 minus sum119899119904=2 ]119902119904 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) Then

(120582119901119904 minus nsum119904=2

120582119901119904)1119901 (]119902119904 minus 119899sum

119904=2

]119902119904)1119902

le 21minus1205911205821]1 minus (119899 minus 1)120591minus1 119899sum119904=2

120582119904]119904(43)

6 Mathematical Problems in Engineering

eorem 23 Let 120582119904119905 119901119905 gt 0 (119904 = 1 2 119899 119905 = 1 2 119898)and 1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 and let sum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905 (44)

Proof For sum119898119905=1(1119901119905) le 1 by inequality (28) we obtain119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(45)

We get this inequality (44) by adding 119903 997888rarr 0 on both sides ofinequality (45)

For the hypotheses sum119898119905=1(1119901119905) le 1 and1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 in Theorem 23 denote120577119905 = (prod119899119904=21205821199041199051205821119905)1(119899minus1) then we have 0 lt 120577119901119905119905 lt 120577119905 lt 1Thus we obtain the following corollaries

Corollary 24 Let 119901119905 gt 0 0 lt 120577119905 lt 1 (119905 = 1 2 119898) and letsum119898119905=1(1119901119905) lt 1 Then

119898prod119905=1

(1 minus 120577119901119905119905 )1119901119905 + 119898prod119905=1

120577119905 le 21minussum119898119905=1(1119901119905) (46)

Denote Θ(119903) = prod119898119905=1[1198721(120582119901119905 119903) minus 1198722119899(120582119901119905 119903)]1119901119905 byLemma 8 we realize that Θ(119903) is decreasing on (0 +infin)Combining Theorems 16 and 23 we obtain the followingcorollary

Corollary 25 Let 120582119904119905 gt 0 and 119901119905 gt 0 (119904 = 1 2 119899 119905 =1 2 119898) and let 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt 0 Ifsum119898119905=1(1119901119905) lt 1 then119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905

(47)

3 Applications

As iswell known analytic inequalities have various importantapplications in many branches of mathematics [33ndash37] Inthis section we will get some applications for this newinequality in Section 2

eorem 26 Let 120595119905 gt 0 119903 ge 119901119905 gt 0 sum119898119905=1(1119901119905) = 1 (119905 =1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119905 (119909)119889119909)119901119905119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119905 (119909) 119889119909)

119901119905119903]1119901119905

le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(48)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof For all positive integers 119899 the equidistance of thepartition of [0 1] is selected as

0 lt 1119899 lt sdot sdot sdot lt 119899 minus 1119899 lt 1119909119904 = 119904119899

119909119904 = 1119899 119904 = 0 1 119899 minus 1

(49)

Noting that 120595119901119905119905 minus (int10119891119903119905 (119909)119889119909)119901119905119903 gt 0 (119905 = 1 2 119898) we

have

120595119901119905119905 minus ( lim119899997888rarrinfin

119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0

(119905 = 1 2 119898) (50)

Therefore there is a positive integer119873 like that

120595119901119905119905 minus (119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0 (51)

for all 119899 gt 119873 and 119905 = 1 2 119898Moreover for any 119899 gt 119873 it follows fromCorollary 13 that

119898prod119905=1

[[120595119901119905119905 minus (1119899

119899minus1sum119904=0

119891119903119905 ( 119904119899))119901119905119903]

]1119901119905

le 119898prod119905=1

120595119905 minus 1119899119899minus1sum119904=0

119898prod119905=1

119891119905 ( 119904119899) (52)

We may find that 119891119905(119909) (119905 = 1 2 119898) are positive Rie-mann integrable functions on [0 1] we know thatprod119898119905=1119891119905(119909)and 119891119903119905 (119909) are also integrable on [0 1] Letting 119899 997888rarr infin onboth sides of inequality (52) we get the desired inequality(48) The proof of Theorem 26 is achieved

eorem 27 Let 120595119905 gt 0 and 119903 ge 119898 and let 119891119905(119909) (119905 =1 2 119898) be positive Riemann integrable functions on [0 1]such that 120595119903119905 minus int1

0119891119903119905 (119909)119889119909 gt 0 Then

119898prod119905=1

[120595119903119905 minus int10119891119903119905 (119909) 119889119909]

1119903 le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909 (53)

Mathematical Problems in Engineering 7

Proof Applying Corollary 14 and along the lines of the proofTheorem 26 Theorem 27 is simply given

eorem 28 Let 119903 gt 0 120595119905 gt 0 119901119905 gt 0 sum119898119905=1(1119901119905) le 1(119905 = 1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119901119905119905 (119909)119889119909)1119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119901119905119905 (119909) 119889119909)1119903]

1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus (int10

119898prod119905=1

119891119903119905 (119909) 119889119909)1119903

(54)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof Combining the proof Theorem 26 and Corollary 18 itis easy to get Theorem 28

Corollary 29 Let 119901119905 gt 0 sum119898119905=1(1119901119905) le 1 (119905 = 1 2 119898)and 120595119901119905119905 minus int1

0119891119901119905119905 (119909)119889119909 gt 0 Then119898prod119905=1

(120595119901119905119905 minus int10119891119901119905119905 (119909) 119889119909)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(55)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0A direct result from Theorem 28 is given by us Putting119898 = 2 1199011 = 119901 gt 0 1199012 = 119902 gt 0 1205951 = 1205821 1205952 = ]1 1198911 = 119891 and1198912 = 119892 in (54) we can get the following corollaries

Corollary 30 Let 119903 gt 0 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 11205821199011minus(int1

0119891119903119901(119909)119889119909)1119903 gt 0 and ]1199021minus(int1

0119892119903119902(119909)119889119909)1119903 gt 0Then

[1205821199011 minus (int10119891119903119901 (119909) 119889119909)1119903]

1119901

sdot []1199021 minus (int10119892119903119902 (119909) 119889119909)1119903]

1119902

le 21minus1119901minus11199021205821]1minus [int10(119891 (119909) 119892 (119909))119903 119889119909]1119903

(56)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0If we set 119903 = 1 in Corollary 30 then the following

inequality holds

Corollary 31 Let 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 1 1205821199011 minusint10119891119901(119909)119889119909 gt 0 and ]1199021 minus int1

0119892119902(119909)119889119909 gt 0 Then

(1205821199011 minus int10119891119901 (119909) 119889119909)1119901 (]1199021 minus int1

0119892119902 (119909) 119889119909)1119902

le 21minus1119901minus11199021205821]1 minus int10119891 (119909) 119892 (119909) 119889119909

(57)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Authorsrsquo Contributions

All authors contributed equally and significantly in writ-ing this article All authors read and approved the finalmanuscript

Acknowledgments

This work was supported by the Fundamental ResearchFunds for the Central Universities (no 2015ZD29) and theHigher School Science Research Funds of Hebei Province ofChina (no Z2015137)

References

[1] J Aczel ldquoSome general methods in the theory of functionalequations in one variable New applications of functionalequationsrdquo Uspekhi Matematicheskikh Nauk vol 11 no 3 pp3ndash68 1956 (Russian)

[2] R Bellman ldquoOn an inequality concerning an indefinite formrdquoTheAmericanMathematical Monthly vol 63 no 2 pp 108-1091956

[3] Z Hu and A Xu ldquoRefinements of Aczel and Bellmanrsquos inequal-itiesrdquo Computers amp Mathematics with Applications vol 59 no9 pp 3078ndash3083 2010

[4] J Tian ldquoA sharpened and generalized version of Aczel-Vasic-Pecaric inequality and its applicationrdquo Journal of Inequalitiesand Applications vol 2013 article no 497 2013

[5] J Tian ldquoReversed version of a generalized Aczelrsquos inequalityand its applicationrdquo Journal of Inequalities and Applications vol2012 article no 202 2012

[6] J Tian andM-HHa ldquoProperties and refinements of Aczel-typeinequalitiesrdquo Journal of Mathematical Inequalities vol 12 no 1pp 175ndash189 2018

[7] J Tian and Y Sun ldquoNew refinements of generalized Aczelinequalityrdquo Journal of Inequalities and Applications vol 2014article no 239 2014

[8] J Tian and S Wu ldquoNew refinements of generalized Aczelrsquosinequality and their applicationsrdquo Journal of MathematicalInequalities vol 10 no 1 pp 247ndash259 2016

[9] J Tian andWWang ldquoReversed versions of Aczel-type inequal-ity and Bellman-type inequalityrdquo Journal of MathematicalInequalities vol 9 no 2 pp 417ndash424 2015

[10] J-F Tian and Y-J Zhou ldquoNote on Aczel-type inequality andBellman-type inequalityrdquo Journal of Nonlinear Sciences andApplications vol 9 no 3 pp 1316ndash1322 2016

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

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Submit your manuscripts atwwwhindawicom

4 Mathematical Problems in Engineering

If we put sum119898119905=1(1119901119905) = 1 in Corollary 11 the conclusionsthat we can draw from this are as follows

Corollary 13 Let 119903 ge 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119904119905)119901119905119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) = 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119904119905)119901119905119903]1119901119905

le 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(25)

Let 119901119905 = 119903 (119905 = 1 2 119898) then fromTheorem 10 we get

Corollary 14 Let 119903 gt 0 1205821199031119905 minus (1(119899 minus 1))sum119899119904=2 120582119903119904119905 gt 0 (119904 =2 3 119899 119905 = 1 2 119898) and let 120591 = min119898119903 1 Then

119898prod119905=1

[1205821199031119905 minus 1119899 minus 1119899sum119904=2

120582119903119904119905]1119903

le 1198991minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)minus119898119903 119899sum119904=2

119898prod119905=1

120582119904119905(26)

Using the obtained inequality of inequality (26) we canget the following result of Wu [13]

Corollary 15 Let 119903 gt 0 and 1205951199031119905 minus sum119899119904=2 120595119903119904119905 gt 0 (119904 =2 3 119899 119905 = 1 2 119898) and let 120591 = min119898119903 1 Then

119898prod119905=1

(1205951199031119905 minus 119899sum119904=2

120595119903119904119905)1119903 le 1198991minus120591 119898prod

119905=1

1205951119905 minus 119899sum119904=2

119898prod119905=1

120595119904119905 (27)

eorem 16 Let 119903 gt 0 119901119905 gt 0 and 1198721(120582119901119905119905 119903) minus1198722119899(120582119901119905119905 119903) gt 0 let 120582119905 = (1205821119905 1205822119905 120582119899119905) and prod119898119905=1120582119905 =(prod119898119905=11205821119905prod119898119905=11205822119905 prod119898119905=1120582119899119905) and denote 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

[1198721 (120582119901119905119905 119903) minus 1198722119899 (120582119901119905119905 119903)]1119901119905

le 21minus1205911198721 ( 119898prod119905=1

120582119905 119903)

minus (119899 minus 1)(120591minussum119898119905=1(1119901119905))1199031198722119899 ( 119898prod119905=1

120582119905 119903) (28)

Proof For the hypotheses 119901119905 gt 0 and 1198721(120582119901119905119905 119903) minus1198722119899(120582119901119905119905 119903) gt 0 (119905 = 1 2 119898) we have

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903 gt 0 (29)

Denote

1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903 = 120585119901119905119905

(120585119905 gt 0 119905 = 1 2 119898 minus 1) (30)

119898prod119905=1

1205821119905

minus 2120591minus1 (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

= 2120591minus1 119898prod119905=1

120585119905

(31)

By using inequality (8) simple computations lead to

119898prod119905=1

[120585119901119905119905 + ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905 ge 2120591minus1 [ 119898prod

119905=1

120585119905

+ 119898prod119905=1

( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903119901119905] = 2120591minus1 [ 119898prod

119905=1

120585119905

+ (119899 minus 1)minus(1119903)sum119898119905=1(1119901119905) 119898prod119905=1

( 119899sum119904=2

120582119903119901119905119904119905 )1119903119901119905]

ge 2120591minus1 [ 119898prod119905=1

120585119905

+ (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903]

(32)

Therefore from (30) (31) and (32) we obtain

[120585119901119898119898 + ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898

sdot 119898minus1prod119905=1

[120585119901119905119905 + ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905 = [120585119901119898119898

+ ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

1205821119905 ge 2120591minus1 [ 119898prod119905=1

120585119905

+ (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903]

= 119898prod119905=1

1205821119905

(33)

Hence

120585119898 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 (34)

Mathematical Problems in Engineering 5

Thus

119898prod119905=1

120585119905 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

120585119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

(35)

Combining (31) and inequality (35) we get

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

le 21minus120591 119898prod119905=1

1205821119905

minus (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(36)

which is equivalent to inequality (28) Successful proof ofTheorem 16 is as follows

Corollary 17 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) gt 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 119898prod119905=1

1205821119905

minus (119899 minus 1)(1minussum119898119905=1(1119901119905))119903 ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(37)

In particular putting 119903 = 1 in Corollary 17 we obtain anew version of the Aczel-Vasic-Pecaric inequality (3)

119898prod119905=1

[1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905]1119901119905

le 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(38)

Corollary 18 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(39)

Corollary 19 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let sum119898119905=1(1119901119905) le 1Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(40)

If we put 119903 = 1 inTheorem 16 then we have the followingcorollary

Corollary 20 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120591minussum119898119905=1(1119901119905)minus1 119899sum119904=2

119898prod119905=1

120582119904119905(41)

Using the substitutions 1205951119905 997888rarr 1205821119905 and (119899 minus 1)1119901119905120595119904119905 997888rarr120582119904119905 (119904 = 2 3 119899 119905 = 1 2 119898) in Corollary 20 weobtain the following refinement of the Aczel-Vasic-Pecaricinequality (3)

Corollary 21 Let 120595119904119905 119901119905 gt 0 and 1205951199011199051119905 minus sum119899119904=2 120595119901119905119904119905 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) and let 120591 = minsum119898119905=1(1119901119905) 1Then

119898prod119905=1

(1205951199011199051119905 minus 119899sum119904=2

120595119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205951119905 minus (119899 minus 1)120591minus1 119899sum119904=2

119898prod119905=1

120595119904119905(42)

If we put sum119898119905=1(1119901119905) ge 1 in Corollary 21 then we canderive inequality (3)

In particular putting 119898 = 2 1199011 = 119901 1199012 = 119902 1205821199041 = 120582119904and 1205821199042 = ]119904 (119904 = 1 2 119899) in Corollary 21 we obtain afresh improvement and promotion of nonequality (2)

Corollary 22 Let 120582119904 ]119904 gt 0 119901 119902 gt 0 and 120591 = min1119901 +1119902 1 and let 120582119901119904 minus sum119899119904=2 120582119901119904 gt 0 and ]119902119904 minus sum119899119904=2 ]119902119904 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) Then

(120582119901119904 minus nsum119904=2

120582119901119904)1119901 (]119902119904 minus 119899sum

119904=2

]119902119904)1119902

le 21minus1205911205821]1 minus (119899 minus 1)120591minus1 119899sum119904=2

120582119904]119904(43)

6 Mathematical Problems in Engineering

eorem 23 Let 120582119904119905 119901119905 gt 0 (119904 = 1 2 119899 119905 = 1 2 119898)and 1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 and let sum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905 (44)

Proof For sum119898119905=1(1119901119905) le 1 by inequality (28) we obtain119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(45)

We get this inequality (44) by adding 119903 997888rarr 0 on both sides ofinequality (45)

For the hypotheses sum119898119905=1(1119901119905) le 1 and1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 in Theorem 23 denote120577119905 = (prod119899119904=21205821199041199051205821119905)1(119899minus1) then we have 0 lt 120577119901119905119905 lt 120577119905 lt 1Thus we obtain the following corollaries

Corollary 24 Let 119901119905 gt 0 0 lt 120577119905 lt 1 (119905 = 1 2 119898) and letsum119898119905=1(1119901119905) lt 1 Then

119898prod119905=1

(1 minus 120577119901119905119905 )1119901119905 + 119898prod119905=1

120577119905 le 21minussum119898119905=1(1119901119905) (46)

Denote Θ(119903) = prod119898119905=1[1198721(120582119901119905 119903) minus 1198722119899(120582119901119905 119903)]1119901119905 byLemma 8 we realize that Θ(119903) is decreasing on (0 +infin)Combining Theorems 16 and 23 we obtain the followingcorollary

Corollary 25 Let 120582119904119905 gt 0 and 119901119905 gt 0 (119904 = 1 2 119899 119905 =1 2 119898) and let 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt 0 Ifsum119898119905=1(1119901119905) lt 1 then119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905

(47)

3 Applications

As iswell known analytic inequalities have various importantapplications in many branches of mathematics [33ndash37] Inthis section we will get some applications for this newinequality in Section 2

eorem 26 Let 120595119905 gt 0 119903 ge 119901119905 gt 0 sum119898119905=1(1119901119905) = 1 (119905 =1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119905 (119909)119889119909)119901119905119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119905 (119909) 119889119909)

119901119905119903]1119901119905

le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(48)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof For all positive integers 119899 the equidistance of thepartition of [0 1] is selected as

0 lt 1119899 lt sdot sdot sdot lt 119899 minus 1119899 lt 1119909119904 = 119904119899

119909119904 = 1119899 119904 = 0 1 119899 minus 1

(49)

Noting that 120595119901119905119905 minus (int10119891119903119905 (119909)119889119909)119901119905119903 gt 0 (119905 = 1 2 119898) we

have

120595119901119905119905 minus ( lim119899997888rarrinfin

119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0

(119905 = 1 2 119898) (50)

Therefore there is a positive integer119873 like that

120595119901119905119905 minus (119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0 (51)

for all 119899 gt 119873 and 119905 = 1 2 119898Moreover for any 119899 gt 119873 it follows fromCorollary 13 that

119898prod119905=1

[[120595119901119905119905 minus (1119899

119899minus1sum119904=0

119891119903119905 ( 119904119899))119901119905119903]

]1119901119905

le 119898prod119905=1

120595119905 minus 1119899119899minus1sum119904=0

119898prod119905=1

119891119905 ( 119904119899) (52)

We may find that 119891119905(119909) (119905 = 1 2 119898) are positive Rie-mann integrable functions on [0 1] we know thatprod119898119905=1119891119905(119909)and 119891119903119905 (119909) are also integrable on [0 1] Letting 119899 997888rarr infin onboth sides of inequality (52) we get the desired inequality(48) The proof of Theorem 26 is achieved

eorem 27 Let 120595119905 gt 0 and 119903 ge 119898 and let 119891119905(119909) (119905 =1 2 119898) be positive Riemann integrable functions on [0 1]such that 120595119903119905 minus int1

0119891119903119905 (119909)119889119909 gt 0 Then

119898prod119905=1

[120595119903119905 minus int10119891119903119905 (119909) 119889119909]

1119903 le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909 (53)

Mathematical Problems in Engineering 7

Proof Applying Corollary 14 and along the lines of the proofTheorem 26 Theorem 27 is simply given

eorem 28 Let 119903 gt 0 120595119905 gt 0 119901119905 gt 0 sum119898119905=1(1119901119905) le 1(119905 = 1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119901119905119905 (119909)119889119909)1119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119901119905119905 (119909) 119889119909)1119903]

1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus (int10

119898prod119905=1

119891119903119905 (119909) 119889119909)1119903

(54)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof Combining the proof Theorem 26 and Corollary 18 itis easy to get Theorem 28

Corollary 29 Let 119901119905 gt 0 sum119898119905=1(1119901119905) le 1 (119905 = 1 2 119898)and 120595119901119905119905 minus int1

0119891119901119905119905 (119909)119889119909 gt 0 Then119898prod119905=1

(120595119901119905119905 minus int10119891119901119905119905 (119909) 119889119909)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(55)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0A direct result from Theorem 28 is given by us Putting119898 = 2 1199011 = 119901 gt 0 1199012 = 119902 gt 0 1205951 = 1205821 1205952 = ]1 1198911 = 119891 and1198912 = 119892 in (54) we can get the following corollaries

Corollary 30 Let 119903 gt 0 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 11205821199011minus(int1

0119891119903119901(119909)119889119909)1119903 gt 0 and ]1199021minus(int1

0119892119903119902(119909)119889119909)1119903 gt 0Then

[1205821199011 minus (int10119891119903119901 (119909) 119889119909)1119903]

1119901

sdot []1199021 minus (int10119892119903119902 (119909) 119889119909)1119903]

1119902

le 21minus1119901minus11199021205821]1minus [int10(119891 (119909) 119892 (119909))119903 119889119909]1119903

(56)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0If we set 119903 = 1 in Corollary 30 then the following

inequality holds

Corollary 31 Let 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 1 1205821199011 minusint10119891119901(119909)119889119909 gt 0 and ]1199021 minus int1

0119892119902(119909)119889119909 gt 0 Then

(1205821199011 minus int10119891119901 (119909) 119889119909)1119901 (]1199021 minus int1

0119892119902 (119909) 119889119909)1119902

le 21minus1119901minus11199021205821]1 minus int10119891 (119909) 119892 (119909) 119889119909

(57)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Authorsrsquo Contributions

All authors contributed equally and significantly in writ-ing this article All authors read and approved the finalmanuscript

Acknowledgments

This work was supported by the Fundamental ResearchFunds for the Central Universities (no 2015ZD29) and theHigher School Science Research Funds of Hebei Province ofChina (no Z2015137)

References

[1] J Aczel ldquoSome general methods in the theory of functionalequations in one variable New applications of functionalequationsrdquo Uspekhi Matematicheskikh Nauk vol 11 no 3 pp3ndash68 1956 (Russian)

[2] R Bellman ldquoOn an inequality concerning an indefinite formrdquoTheAmericanMathematical Monthly vol 63 no 2 pp 108-1091956

[3] Z Hu and A Xu ldquoRefinements of Aczel and Bellmanrsquos inequal-itiesrdquo Computers amp Mathematics with Applications vol 59 no9 pp 3078ndash3083 2010

[4] J Tian ldquoA sharpened and generalized version of Aczel-Vasic-Pecaric inequality and its applicationrdquo Journal of Inequalitiesand Applications vol 2013 article no 497 2013

[5] J Tian ldquoReversed version of a generalized Aczelrsquos inequalityand its applicationrdquo Journal of Inequalities and Applications vol2012 article no 202 2012

[6] J Tian andM-HHa ldquoProperties and refinements of Aczel-typeinequalitiesrdquo Journal of Mathematical Inequalities vol 12 no 1pp 175ndash189 2018

[7] J Tian and Y Sun ldquoNew refinements of generalized Aczelinequalityrdquo Journal of Inequalities and Applications vol 2014article no 239 2014

[8] J Tian and S Wu ldquoNew refinements of generalized Aczelrsquosinequality and their applicationsrdquo Journal of MathematicalInequalities vol 10 no 1 pp 247ndash259 2016

[9] J Tian andWWang ldquoReversed versions of Aczel-type inequal-ity and Bellman-type inequalityrdquo Journal of MathematicalInequalities vol 9 no 2 pp 417ndash424 2015

[10] J-F Tian and Y-J Zhou ldquoNote on Aczel-type inequality andBellman-type inequalityrdquo Journal of Nonlinear Sciences andApplications vol 9 no 3 pp 1316ndash1322 2016

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

Hindawiwwwhindawicom Volume 2018

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Mathematical Problems in Engineering 5

Thus

119898prod119905=1

120585119905 ge [1205821199011198981119898 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119898 )1119903]1119901119898 119898minus1prod119905=1

120585119905

= 119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

(35)

Combining (31) and inequality (35) we get

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119898119904119905 )1119903]1119901119905

le 21minus120591 119898prod119905=1

1205821119905

minus (119899 minus 1)(1119903)(120591minussum119898119905=1(1119901119905)) ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(36)

which is equivalent to inequality (28) Successful proof ofTheorem 16 is as follows

Corollary 17 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) gt 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 119898prod119905=1

1205821119905

minus (119899 minus 1)(1minussum119898119905=1(1119901119905))119903 ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(37)

In particular putting 119903 = 1 in Corollary 17 we obtain anew version of the Aczel-Vasic-Pecaric inequality (3)

119898prod119905=1

[1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905]1119901119905

le 119898prod119905=1

1205821119905 minus (119899 minus 1)minussum119898119905=1(1119901119905) 119899sum119904=2

119898prod119905=1

120582119904119905(38)

Corollary 18 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus ((1(119899 minus1))sum119899119904=2 120582119903119901119905119904119905 )1119903 gt 0 (119904 = 2 3 119899 119905 = 1 2 119898) and letsum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(39)

Corollary 19 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let sum119898119905=1(1119901119905) le 1Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119904119905(40)

If we put 119903 = 1 inTheorem 16 then we have the followingcorollary

Corollary 20 Let 120582119904119905 119901119905 gt 0 and 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt0 (119904 = 2 3 119899 119905 = 1 2 119898) and let 120591 =minsum119898119905=1(1119901119905) 1 Then

119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205821119905 minus (119899 minus 1)120591minussum119898119905=1(1119901119905)minus1 119899sum119904=2

119898prod119905=1

120582119904119905(41)

Using the substitutions 1205951119905 997888rarr 1205821119905 and (119899 minus 1)1119901119905120595119904119905 997888rarr120582119904119905 (119904 = 2 3 119899 119905 = 1 2 119898) in Corollary 20 weobtain the following refinement of the Aczel-Vasic-Pecaricinequality (3)

Corollary 21 Let 120595119904119905 119901119905 gt 0 and 1205951199011199051119905 minus sum119899119904=2 120595119901119905119904119905 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) and let 120591 = minsum119898119905=1(1119901119905) 1Then

119898prod119905=1

(1205951199011199051119905 minus 119899sum119904=2

120595119901119905119904119905)1119901119905

le 21minus120591 119898prod119905=1

1205951119905 minus (119899 minus 1)120591minus1 119899sum119904=2

119898prod119905=1

120595119904119905(42)

If we put sum119898119905=1(1119901119905) ge 1 in Corollary 21 then we canderive inequality (3)

In particular putting 119898 = 2 1199011 = 119901 1199012 = 119902 1205821199041 = 120582119904and 1205821199042 = ]119904 (119904 = 1 2 119899) in Corollary 21 we obtain afresh improvement and promotion of nonequality (2)

Corollary 22 Let 120582119904 ]119904 gt 0 119901 119902 gt 0 and 120591 = min1119901 +1119902 1 and let 120582119901119904 minus sum119899119904=2 120582119901119904 gt 0 and ]119902119904 minus sum119899119904=2 ]119902119904 gt 0 (119904 =1 2 119899 119905 = 1 2 119898) Then

(120582119901119904 minus nsum119904=2

120582119901119904)1119901 (]119902119904 minus 119899sum

119904=2

]119902119904)1119902

le 21minus1205911205821]1 minus (119899 minus 1)120591minus1 119899sum119904=2

120582119904]119904(43)

6 Mathematical Problems in Engineering

eorem 23 Let 120582119904119905 119901119905 gt 0 (119904 = 1 2 119899 119905 = 1 2 119898)and 1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 and let sum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905 (44)

Proof For sum119898119905=1(1119901119905) le 1 by inequality (28) we obtain119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(45)

We get this inequality (44) by adding 119903 997888rarr 0 on both sides ofinequality (45)

For the hypotheses sum119898119905=1(1119901119905) le 1 and1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 in Theorem 23 denote120577119905 = (prod119899119904=21205821199041199051205821119905)1(119899minus1) then we have 0 lt 120577119901119905119905 lt 120577119905 lt 1Thus we obtain the following corollaries

Corollary 24 Let 119901119905 gt 0 0 lt 120577119905 lt 1 (119905 = 1 2 119898) and letsum119898119905=1(1119901119905) lt 1 Then

119898prod119905=1

(1 minus 120577119901119905119905 )1119901119905 + 119898prod119905=1

120577119905 le 21minussum119898119905=1(1119901119905) (46)

Denote Θ(119903) = prod119898119905=1[1198721(120582119901119905 119903) minus 1198722119899(120582119901119905 119903)]1119901119905 byLemma 8 we realize that Θ(119903) is decreasing on (0 +infin)Combining Theorems 16 and 23 we obtain the followingcorollary

Corollary 25 Let 120582119904119905 gt 0 and 119901119905 gt 0 (119904 = 1 2 119899 119905 =1 2 119898) and let 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt 0 Ifsum119898119905=1(1119901119905) lt 1 then119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905

(47)

3 Applications

As iswell known analytic inequalities have various importantapplications in many branches of mathematics [33ndash37] Inthis section we will get some applications for this newinequality in Section 2

eorem 26 Let 120595119905 gt 0 119903 ge 119901119905 gt 0 sum119898119905=1(1119901119905) = 1 (119905 =1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119905 (119909)119889119909)119901119905119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119905 (119909) 119889119909)

119901119905119903]1119901119905

le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(48)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof For all positive integers 119899 the equidistance of thepartition of [0 1] is selected as

0 lt 1119899 lt sdot sdot sdot lt 119899 minus 1119899 lt 1119909119904 = 119904119899

119909119904 = 1119899 119904 = 0 1 119899 minus 1

(49)

Noting that 120595119901119905119905 minus (int10119891119903119905 (119909)119889119909)119901119905119903 gt 0 (119905 = 1 2 119898) we

have

120595119901119905119905 minus ( lim119899997888rarrinfin

119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0

(119905 = 1 2 119898) (50)

Therefore there is a positive integer119873 like that

120595119901119905119905 minus (119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0 (51)

for all 119899 gt 119873 and 119905 = 1 2 119898Moreover for any 119899 gt 119873 it follows fromCorollary 13 that

119898prod119905=1

[[120595119901119905119905 minus (1119899

119899minus1sum119904=0

119891119903119905 ( 119904119899))119901119905119903]

]1119901119905

le 119898prod119905=1

120595119905 minus 1119899119899minus1sum119904=0

119898prod119905=1

119891119905 ( 119904119899) (52)

We may find that 119891119905(119909) (119905 = 1 2 119898) are positive Rie-mann integrable functions on [0 1] we know thatprod119898119905=1119891119905(119909)and 119891119903119905 (119909) are also integrable on [0 1] Letting 119899 997888rarr infin onboth sides of inequality (52) we get the desired inequality(48) The proof of Theorem 26 is achieved

eorem 27 Let 120595119905 gt 0 and 119903 ge 119898 and let 119891119905(119909) (119905 =1 2 119898) be positive Riemann integrable functions on [0 1]such that 120595119903119905 minus int1

0119891119903119905 (119909)119889119909 gt 0 Then

119898prod119905=1

[120595119903119905 minus int10119891119903119905 (119909) 119889119909]

1119903 le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909 (53)

Mathematical Problems in Engineering 7

Proof Applying Corollary 14 and along the lines of the proofTheorem 26 Theorem 27 is simply given

eorem 28 Let 119903 gt 0 120595119905 gt 0 119901119905 gt 0 sum119898119905=1(1119901119905) le 1(119905 = 1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119901119905119905 (119909)119889119909)1119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119901119905119905 (119909) 119889119909)1119903]

1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus (int10

119898prod119905=1

119891119903119905 (119909) 119889119909)1119903

(54)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof Combining the proof Theorem 26 and Corollary 18 itis easy to get Theorem 28

Corollary 29 Let 119901119905 gt 0 sum119898119905=1(1119901119905) le 1 (119905 = 1 2 119898)and 120595119901119905119905 minus int1

0119891119901119905119905 (119909)119889119909 gt 0 Then119898prod119905=1

(120595119901119905119905 minus int10119891119901119905119905 (119909) 119889119909)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(55)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0A direct result from Theorem 28 is given by us Putting119898 = 2 1199011 = 119901 gt 0 1199012 = 119902 gt 0 1205951 = 1205821 1205952 = ]1 1198911 = 119891 and1198912 = 119892 in (54) we can get the following corollaries

Corollary 30 Let 119903 gt 0 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 11205821199011minus(int1

0119891119903119901(119909)119889119909)1119903 gt 0 and ]1199021minus(int1

0119892119903119902(119909)119889119909)1119903 gt 0Then

[1205821199011 minus (int10119891119903119901 (119909) 119889119909)1119903]

1119901

sdot []1199021 minus (int10119892119903119902 (119909) 119889119909)1119903]

1119902

le 21minus1119901minus11199021205821]1minus [int10(119891 (119909) 119892 (119909))119903 119889119909]1119903

(56)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0If we set 119903 = 1 in Corollary 30 then the following

inequality holds

Corollary 31 Let 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 1 1205821199011 minusint10119891119901(119909)119889119909 gt 0 and ]1199021 minus int1

0119892119902(119909)119889119909 gt 0 Then

(1205821199011 minus int10119891119901 (119909) 119889119909)1119901 (]1199021 minus int1

0119892119902 (119909) 119889119909)1119902

le 21minus1119901minus11199021205821]1 minus int10119891 (119909) 119892 (119909) 119889119909

(57)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Authorsrsquo Contributions

All authors contributed equally and significantly in writ-ing this article All authors read and approved the finalmanuscript

Acknowledgments

This work was supported by the Fundamental ResearchFunds for the Central Universities (no 2015ZD29) and theHigher School Science Research Funds of Hebei Province ofChina (no Z2015137)

References

[1] J Aczel ldquoSome general methods in the theory of functionalequations in one variable New applications of functionalequationsrdquo Uspekhi Matematicheskikh Nauk vol 11 no 3 pp3ndash68 1956 (Russian)

[2] R Bellman ldquoOn an inequality concerning an indefinite formrdquoTheAmericanMathematical Monthly vol 63 no 2 pp 108-1091956

[3] Z Hu and A Xu ldquoRefinements of Aczel and Bellmanrsquos inequal-itiesrdquo Computers amp Mathematics with Applications vol 59 no9 pp 3078ndash3083 2010

[4] J Tian ldquoA sharpened and generalized version of Aczel-Vasic-Pecaric inequality and its applicationrdquo Journal of Inequalitiesand Applications vol 2013 article no 497 2013

[5] J Tian ldquoReversed version of a generalized Aczelrsquos inequalityand its applicationrdquo Journal of Inequalities and Applications vol2012 article no 202 2012

[6] J Tian andM-HHa ldquoProperties and refinements of Aczel-typeinequalitiesrdquo Journal of Mathematical Inequalities vol 12 no 1pp 175ndash189 2018

[7] J Tian and Y Sun ldquoNew refinements of generalized Aczelinequalityrdquo Journal of Inequalities and Applications vol 2014article no 239 2014

[8] J Tian and S Wu ldquoNew refinements of generalized Aczelrsquosinequality and their applicationsrdquo Journal of MathematicalInequalities vol 10 no 1 pp 247ndash259 2016

[9] J Tian andWWang ldquoReversed versions of Aczel-type inequal-ity and Bellman-type inequalityrdquo Journal of MathematicalInequalities vol 9 no 2 pp 417ndash424 2015

[10] J-F Tian and Y-J Zhou ldquoNote on Aczel-type inequality andBellman-type inequalityrdquo Journal of Nonlinear Sciences andApplications vol 9 no 3 pp 1316ndash1322 2016

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

Hindawiwwwhindawicom Volume 2018

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Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

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Hindawiwwwhindawicom Volume 2018

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Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

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Hindawiwwwhindawicom Volume 2018

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Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

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Volume 2018

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Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

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AnalysisInternational Journal of

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Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

6 Mathematical Problems in Engineering

eorem 23 Let 120582119904119905 119901119905 gt 0 (119904 = 1 2 119899 119905 = 1 2 119898)and 1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 and let sum119898119905=1(1119901119905) le 1 Then

119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905 (44)

Proof For sum119898119905=1(1119901119905) le 1 by inequality (28) we obtain119898prod119905=1

[1205821199011199051119905 minus ( 1119899 minus 1119899sum119904=2

120582119903119901119905119904119905 )1119903]1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus ( 1119899 minus 1119899sum119904=2

119898prod119905=1

120582119903119904119905)1119903

(45)

We get this inequality (44) by adding 119903 997888rarr 0 on both sides ofinequality (45)

For the hypotheses sum119898119905=1(1119901119905) le 1 and1205821199011199051119905 minus prod119899119904=2120582119901119905(119899minus1)119904119905 gt 0 in Theorem 23 denote120577119905 = (prod119899119904=21205821199041199051205821119905)1(119899minus1) then we have 0 lt 120577119901119905119905 lt 120577119905 lt 1Thus we obtain the following corollaries

Corollary 24 Let 119901119905 gt 0 0 lt 120577119905 lt 1 (119905 = 1 2 119898) and letsum119898119905=1(1119901119905) lt 1 Then

119898prod119905=1

(1 minus 120577119901119905119905 )1119901119905 + 119898prod119905=1

120577119905 le 21minussum119898119905=1(1119901119905) (46)

Denote Θ(119903) = prod119898119905=1[1198721(120582119901119905 119903) minus 1198722119899(120582119901119905 119903)]1119901119905 byLemma 8 we realize that Θ(119903) is decreasing on (0 +infin)Combining Theorems 16 and 23 we obtain the followingcorollary

Corollary 25 Let 120582119904119905 gt 0 and 119901119905 gt 0 (119904 = 1 2 119899 119905 =1 2 119898) and let 1205821199011199051119905 minus (1(119899 minus 1))sum119899119904=2 120582119901119905119904119905 gt 0 Ifsum119898119905=1(1119901119905) lt 1 then119898prod119905=1

(1205821199011199051119905 minus 1119899 minus 1119899sum119904=2

120582119901119905119904119905)1119901119905

le 119898prod119905=1

(1205821199011199051119905 minus 119899prod119904=2

120582119901119905(119899minus1)119904119905 )1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

1205821119905 minus 119899prod119904=2

119898prod119905=1

1205821(119899minus1)119904119905

(47)

3 Applications

As iswell known analytic inequalities have various importantapplications in many branches of mathematics [33ndash37] Inthis section we will get some applications for this newinequality in Section 2

eorem 26 Let 120595119905 gt 0 119903 ge 119901119905 gt 0 sum119898119905=1(1119901119905) = 1 (119905 =1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119905 (119909)119889119909)119901119905119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119905 (119909) 119889119909)

119901119905119903]1119901119905

le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(48)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof For all positive integers 119899 the equidistance of thepartition of [0 1] is selected as

0 lt 1119899 lt sdot sdot sdot lt 119899 minus 1119899 lt 1119909119904 = 119904119899

119909119904 = 1119899 119904 = 0 1 119899 minus 1

(49)

Noting that 120595119901119905119905 minus (int10119891119903119905 (119909)119889119909)119901119905119903 gt 0 (119905 = 1 2 119898) we

have

120595119901119905119905 minus ( lim119899997888rarrinfin

119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0

(119905 = 1 2 119898) (50)

Therefore there is a positive integer119873 like that

120595119901119905119905 minus (119899minus1sum119904=0

119891119903119905 (0 + 119904119899) 1119899)119901119905119903 gt 0 (51)

for all 119899 gt 119873 and 119905 = 1 2 119898Moreover for any 119899 gt 119873 it follows fromCorollary 13 that

119898prod119905=1

[[120595119901119905119905 minus (1119899

119899minus1sum119904=0

119891119903119905 ( 119904119899))119901119905119903]

]1119901119905

le 119898prod119905=1

120595119905 minus 1119899119899minus1sum119904=0

119898prod119905=1

119891119905 ( 119904119899) (52)

We may find that 119891119905(119909) (119905 = 1 2 119898) are positive Rie-mann integrable functions on [0 1] we know thatprod119898119905=1119891119905(119909)and 119891119903119905 (119909) are also integrable on [0 1] Letting 119899 997888rarr infin onboth sides of inequality (52) we get the desired inequality(48) The proof of Theorem 26 is achieved

eorem 27 Let 120595119905 gt 0 and 119903 ge 119898 and let 119891119905(119909) (119905 =1 2 119898) be positive Riemann integrable functions on [0 1]such that 120595119903119905 minus int1

0119891119903119905 (119909)119889119909 gt 0 Then

119898prod119905=1

[120595119903119905 minus int10119891119903119905 (119909) 119889119909]

1119903 le 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909 (53)

Mathematical Problems in Engineering 7

Proof Applying Corollary 14 and along the lines of the proofTheorem 26 Theorem 27 is simply given

eorem 28 Let 119903 gt 0 120595119905 gt 0 119901119905 gt 0 sum119898119905=1(1119901119905) le 1(119905 = 1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119901119905119905 (119909)119889119909)1119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119901119905119905 (119909) 119889119909)1119903]

1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus (int10

119898prod119905=1

119891119903119905 (119909) 119889119909)1119903

(54)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof Combining the proof Theorem 26 and Corollary 18 itis easy to get Theorem 28

Corollary 29 Let 119901119905 gt 0 sum119898119905=1(1119901119905) le 1 (119905 = 1 2 119898)and 120595119901119905119905 minus int1

0119891119901119905119905 (119909)119889119909 gt 0 Then119898prod119905=1

(120595119901119905119905 minus int10119891119901119905119905 (119909) 119889119909)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(55)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0A direct result from Theorem 28 is given by us Putting119898 = 2 1199011 = 119901 gt 0 1199012 = 119902 gt 0 1205951 = 1205821 1205952 = ]1 1198911 = 119891 and1198912 = 119892 in (54) we can get the following corollaries

Corollary 30 Let 119903 gt 0 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 11205821199011minus(int1

0119891119903119901(119909)119889119909)1119903 gt 0 and ]1199021minus(int1

0119892119903119902(119909)119889119909)1119903 gt 0Then

[1205821199011 minus (int10119891119903119901 (119909) 119889119909)1119903]

1119901

sdot []1199021 minus (int10119892119903119902 (119909) 119889119909)1119903]

1119902

le 21minus1119901minus11199021205821]1minus [int10(119891 (119909) 119892 (119909))119903 119889119909]1119903

(56)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0If we set 119903 = 1 in Corollary 30 then the following

inequality holds

Corollary 31 Let 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 1 1205821199011 minusint10119891119901(119909)119889119909 gt 0 and ]1199021 minus int1

0119892119902(119909)119889119909 gt 0 Then

(1205821199011 minus int10119891119901 (119909) 119889119909)1119901 (]1199021 minus int1

0119892119902 (119909) 119889119909)1119902

le 21minus1119901minus11199021205821]1 minus int10119891 (119909) 119892 (119909) 119889119909

(57)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Authorsrsquo Contributions

All authors contributed equally and significantly in writ-ing this article All authors read and approved the finalmanuscript

Acknowledgments

This work was supported by the Fundamental ResearchFunds for the Central Universities (no 2015ZD29) and theHigher School Science Research Funds of Hebei Province ofChina (no Z2015137)

References

[1] J Aczel ldquoSome general methods in the theory of functionalequations in one variable New applications of functionalequationsrdquo Uspekhi Matematicheskikh Nauk vol 11 no 3 pp3ndash68 1956 (Russian)

[2] R Bellman ldquoOn an inequality concerning an indefinite formrdquoTheAmericanMathematical Monthly vol 63 no 2 pp 108-1091956

[3] Z Hu and A Xu ldquoRefinements of Aczel and Bellmanrsquos inequal-itiesrdquo Computers amp Mathematics with Applications vol 59 no9 pp 3078ndash3083 2010

[4] J Tian ldquoA sharpened and generalized version of Aczel-Vasic-Pecaric inequality and its applicationrdquo Journal of Inequalitiesand Applications vol 2013 article no 497 2013

[5] J Tian ldquoReversed version of a generalized Aczelrsquos inequalityand its applicationrdquo Journal of Inequalities and Applications vol2012 article no 202 2012

[6] J Tian andM-HHa ldquoProperties and refinements of Aczel-typeinequalitiesrdquo Journal of Mathematical Inequalities vol 12 no 1pp 175ndash189 2018

[7] J Tian and Y Sun ldquoNew refinements of generalized Aczelinequalityrdquo Journal of Inequalities and Applications vol 2014article no 239 2014

[8] J Tian and S Wu ldquoNew refinements of generalized Aczelrsquosinequality and their applicationsrdquo Journal of MathematicalInequalities vol 10 no 1 pp 247ndash259 2016

[9] J Tian andWWang ldquoReversed versions of Aczel-type inequal-ity and Bellman-type inequalityrdquo Journal of MathematicalInequalities vol 9 no 2 pp 417ndash424 2015

[10] J-F Tian and Y-J Zhou ldquoNote on Aczel-type inequality andBellman-type inequalityrdquo Journal of Nonlinear Sciences andApplications vol 9 no 3 pp 1316ndash1322 2016

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Mathematical Problems in Engineering 7

Proof Applying Corollary 14 and along the lines of the proofTheorem 26 Theorem 27 is simply given

eorem 28 Let 119903 gt 0 120595119905 gt 0 119901119905 gt 0 sum119898119905=1(1119901119905) le 1(119905 = 1 2 119898) and 120595119901119905119905 minus (int1

0119891119903119901119905119905 (119909)119889119909)1119903 gt 0 Then

119898prod119905=1

[120595119901119905119905 minus (int10119891119903119901119905119905 (119909) 119889119909)1119903]

1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus (int10

119898prod119905=1

119891119903119905 (119909) 119889119909)1119903

(54)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0Proof Combining the proof Theorem 26 and Corollary 18 itis easy to get Theorem 28

Corollary 29 Let 119901119905 gt 0 sum119898119905=1(1119901119905) le 1 (119905 = 1 2 119898)and 120595119901119905119905 minus int1

0119891119901119905119905 (119909)119889119909 gt 0 Then119898prod119905=1

(120595119901119905119905 minus int10119891119901119905119905 (119909) 119889119909)1119901119905

le 21minussum119898119905=1(1119901119905) 119898prod119905=1

120595119905 minus int10

119898prod119905=1

119891119905 (119909) 119889119909(55)

where 119891119905(119909) can be integrated on [0 1] and 119891119905(119909) gt 0A direct result from Theorem 28 is given by us Putting119898 = 2 1199011 = 119901 gt 0 1199012 = 119902 gt 0 1205951 = 1205821 1205952 = ]1 1198911 = 119891 and1198912 = 119892 in (54) we can get the following corollaries

Corollary 30 Let 119903 gt 0 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 11205821199011minus(int1

0119891119903119901(119909)119889119909)1119903 gt 0 and ]1199021minus(int1

0119892119903119902(119909)119889119909)1119903 gt 0Then

[1205821199011 minus (int10119891119903119901 (119909) 119889119909)1119903]

1119901

sdot []1199021 minus (int10119892119903119902 (119909) 119889119909)1119903]

1119902

le 21minus1119901minus11199021205821]1minus [int10(119891 (119909) 119892 (119909))119903 119889119909]1119903

(56)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0If we set 119903 = 1 in Corollary 30 then the following

inequality holds

Corollary 31 Let 1205821 ]1 gt 0 119901 119902 gt 0 1119901 + 1119902 le 1 1205821199011 minusint10119891119901(119909)119889119909 gt 0 and ]1199021 minus int1

0119892119902(119909)119889119909 gt 0 Then

(1205821199011 minus int10119891119901 (119909) 119889119909)1119901 (]1199021 minus int1

0119892119902 (119909) 119889119909)1119902

le 21minus1119901minus11199021205821]1 minus int10119891 (119909) 119892 (119909) 119889119909

(57)

where 119891(119909) 119892(119909) can be integrated on [0 1] and 119891(119909) gt0 119892(119909) gt 0Data Availability

The data used to support the findings of this study areincluded within the article

Conflicts of Interest

The authors declare that they have no conflicts of interest

Authorsrsquo Contributions

All authors contributed equally and significantly in writ-ing this article All authors read and approved the finalmanuscript

Acknowledgments

This work was supported by the Fundamental ResearchFunds for the Central Universities (no 2015ZD29) and theHigher School Science Research Funds of Hebei Province ofChina (no Z2015137)

References

[1] J Aczel ldquoSome general methods in the theory of functionalequations in one variable New applications of functionalequationsrdquo Uspekhi Matematicheskikh Nauk vol 11 no 3 pp3ndash68 1956 (Russian)

[2] R Bellman ldquoOn an inequality concerning an indefinite formrdquoTheAmericanMathematical Monthly vol 63 no 2 pp 108-1091956

[3] Z Hu and A Xu ldquoRefinements of Aczel and Bellmanrsquos inequal-itiesrdquo Computers amp Mathematics with Applications vol 59 no9 pp 3078ndash3083 2010

[4] J Tian ldquoA sharpened and generalized version of Aczel-Vasic-Pecaric inequality and its applicationrdquo Journal of Inequalitiesand Applications vol 2013 article no 497 2013

[5] J Tian ldquoReversed version of a generalized Aczelrsquos inequalityand its applicationrdquo Journal of Inequalities and Applications vol2012 article no 202 2012

[6] J Tian andM-HHa ldquoProperties and refinements of Aczel-typeinequalitiesrdquo Journal of Mathematical Inequalities vol 12 no 1pp 175ndash189 2018

[7] J Tian and Y Sun ldquoNew refinements of generalized Aczelinequalityrdquo Journal of Inequalities and Applications vol 2014article no 239 2014

[8] J Tian and S Wu ldquoNew refinements of generalized Aczelrsquosinequality and their applicationsrdquo Journal of MathematicalInequalities vol 10 no 1 pp 247ndash259 2016

[9] J Tian andWWang ldquoReversed versions of Aczel-type inequal-ity and Bellman-type inequalityrdquo Journal of MathematicalInequalities vol 9 no 2 pp 417ndash424 2015

[10] J-F Tian and Y-J Zhou ldquoNote on Aczel-type inequality andBellman-type inequalityrdquo Journal of Nonlinear Sciences andApplications vol 9 no 3 pp 1316ndash1322 2016

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

8 Mathematical Problems in Engineering

[11] S Wu ldquoSome improvements of Aczelrsquos inequality and Popovi-ciursquos inequalityrdquo Computers amp Mathematics with Applicationsvol 56 no 5 pp 1196ndash1205 2008

[12] S Wu and L Debnath ldquoA new generalization of Aczelrsquosinequality and its applications to an improvement of Bellmanrsquosinequalityrdquo Applied Mathematics Letters vol 21 no 6 pp 588ndash593 2008

[13] S Wu and L Debnath ldquoGeneralizations of Aczelrsquos inequalityand Popoviciursquos inequalityrdquo Indian Journal of Pure and AppliedMathematics vol 36 no 2 pp 49ndash62 2005

[14] T Popoviciu ldquoOn an inequalityrdquo Gazeta Mathematica si FizicaA vol 11 article no 64 pp 451ndash461 1959 (Romanian)

[15] P M Vasic and J E Pearic ldquoOn the Jensen inequality formonotone functionsrdquo Analele Universitatii din Timisoara SeriaMatematica-Informatica vol 17 no 1 pp 95ndash104 1979

[16] Y-M Chu and B-Y Long ldquoBounds of the neuman-sandormean using power and identric meansrdquo Abstract and AppliedAnalysis vol 2013 Article ID 832591 6 pages 2013

[17] Y-M Chu and W-F Xia ldquoTwo optimal double inequalitiesbetween power mean and logarithmic meanrdquo Computers ampMathematics with Applications vol 60 no 1 pp 83ndash89 2010

[18] B Long and Y Chu ldquoOptimal power mean bounds forthe weighted geometric mean of classical meansrdquo Journal ofInequalities and Applications vol 2010 Article ID 905679 6pages 2010

[19] G Wang X Zhang and Y Chu ldquoA power mean inequalityfor the Grotzsch ring functionrdquo Mathematical Inequalities ampApplications vol 14 no 4 pp 833ndash837 2011

[20] G Wang X Zhang and Y Chu ldquoA power mean inequalityinvolving the complete elliptic integralsrdquo Rocky Mountain Jour-nal of Mathematics vol 44 no 5 pp 1661ndash1667 2014

[21] M-K Wang Y-M Chu Y-F Qiu and S-L Qiu ldquoAn optimalpower mean inequality for the complete elliptic integralsrdquoApplied Mathematics Letters vol 24 no 6 pp 887ndash890 2011

[22] W Xia and Y Chu ldquoOptimal inequalities for the convex combi-nation of error functionrdquo Journal of Mathematical Inequalitiesvol 9 no 1 pp 85ndash99 2015

[23] W Xia W Janous and Y Chu ldquoThe optimal convex combi-nation bounds of arithmetic and harmonic means in terms ofpower meanrdquo Journal of Mathematical Inequalities vol 6 no 2pp 241ndash248 2012

[24] W-F Xia X-H Zhang G-D Wang and Y-M Chu ldquoSomeproperties for a class of symmetric functions with applicationsrdquoIndian Journal of Pure and Applied Mathematics vol 43 no 3pp 227ndash249 2012

[25] J-F Tian ldquoTriple diamond-alpha integral and holder-typeinequalitiesrdquo Journal of Inequalities and Applications vol 2018article no 111 2018

[26] J-F Tian and M-H Ha ldquoProperties of generalized sharpHolderrsquos inequalitiesrdquo Journal of Mathematical Inequalities vol11 no 2 pp 511ndash525 2017

[27] J-F Tian and M-H Ha ldquoExtensions of Holderrsquos inequality viapseudo-integralrdquo Mathematical Problems in Engineering vol2018 Article ID 4080619 5 pages 2018

[28] J-F Tian M-H Ha and C Wang ldquoImprovements of gener-alized Holderrsquos inequalities and their applicationsrdquo Journal ofMathematical Inequalities vol 12 no 2 pp 459ndash471 2018

[29] J Tian Y Zhu and W Cheung ldquoN-tuple diamond-alphaintegral and inequalities on time scalesrdquo Revista de la RealAcademia de Ciencias Exactas Fısicas y Naturales Serie AMatematicas vol 113 no 3 pp 2189ndash2200 2019

[30] Zh-H Yang and J Tian ldquoOptimal inequalities involving power-exponential mean arithmetic mean and geometric meanrdquoJournal of Mathematical Inequalities vol 11 no 4 pp 1169ndash11832017

[31] D S Mitrinovic and P M Vasic Analytic Inequalities SpringerNew York NY USA 1970

[32] G H Hardy J E Littlewood and G Polya InequalitiesCambridge University Press London UK 2nd edition 1952

[33] J Tian ldquoNew property of a generalized Holderrsquos inequality andits applicationsrdquo Information Sciences vol 288 pp 45ndash54 2014

[34] J Tian ldquoReversed version of a generalized sharp Holderrsquosinequality and its applicationsrdquo Information Sciences vol 201pp 61ndash69 2012

[35] Z Yang and J Tian ldquoMonotonicity rules for the ratio of twoLaplace transforms with applicationsrdquo Journal of MathematicalAnalysis and Applications vol 470 no 2 pp 821ndash845 2019

[36] Z Yang and J Tian ldquoA comparison theorem for two divideddifferences and applications to special functionsrdquo Journal ofMathematical Analysis and Applications vol 464 no 1 pp 580ndash595 2018

[37] Z Yang and J Tian ldquoA class of completely mixed monotonicfunctions involving the gamma function with applicationsrdquoProceedings of the American Mathematical Society vol 146 no11 pp 4707ndash4721 2018

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom

Hindawiwwwhindawicom Volume 2018

MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Mathematical Problems in Engineering

Applied MathematicsJournal of

Hindawiwwwhindawicom Volume 2018

Probability and StatisticsHindawiwwwhindawicom Volume 2018

Journal of

Hindawiwwwhindawicom Volume 2018

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawiwwwhindawicom Volume 2018

OptimizationJournal of

Hindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom Volume 2018

Engineering Mathematics

International Journal of

Hindawiwwwhindawicom Volume 2018

Operations ResearchAdvances in

Journal of

Hindawiwwwhindawicom Volume 2018

Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018

International Journal of Mathematics and Mathematical Sciences

Hindawiwwwhindawicom Volume 2018

Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom

The Scientific World Journal

Volume 2018

Hindawiwwwhindawicom Volume 2018Volume 2018

Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in

Nature and SocietyHindawiwwwhindawicom Volume 2018

Hindawiwwwhindawicom

Dierential EquationsInternational Journal of

Volume 2018

Hindawiwwwhindawicom Volume 2018

Decision SciencesAdvances in

Hindawiwwwhindawicom Volume 2018

AnalysisInternational Journal of

Hindawiwwwhindawicom Volume 2018

Stochastic AnalysisInternational Journal of

Submit your manuscripts atwwwhindawicom