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Generalizations and analogues of the Nesbitts inequality
Fuhua Wei and Shanhe Wu
Department of Mathematics and Computer Science, Longyan
University, Longyan, Fujian 364012, P. R. China
E-mail: [email protected] Author
Abstract: The Nesbitts inequality is generalized by introducing
exponent and weight parameters. SeveralNesbitt-type inequalities
for n variables are provided. Finally, two analogous forms of
Nesbitts inequalityare given.Keywords: Nesbitts inequality;
Cauchy-Schwarz inequality; Chebyshevs inequality; power mean
inequal-ity; generalization; analogue
2000 Mathematics Subject Classification: 26D15
1 Introduction
The Nesbitts inequality states that ifx, y, z are positive real
numbers, then
x
y+z+
y
z+x+
z
x+y 3
2, (1)
the equality occurs if and only if the three variables are equal
([1], see also [2]).It is well known that this cyclic sum
inequality has many applications in the proof of fractional
inequalities.
In this paper we shall establish some generalizations and
analogous forms of the Nesbitts inequality.
2 Generalizations of the Nesbitts inequality
Theorem 1. Let x, y, z, k be positive real numbers. Then
x
ky+z+
y
kz+x+
z
kx+y 3
1 +k. (2)
Proof. By using the Cauchy-Schwarz inequality (see [3]), we
have
(kxy+zx +kyz +xy+kxz+yz)
x2
kxy+zx+
y2
kyz +xy+
z2
kxz+yz
(x+y+z)2.
Hence
x
ky+z+
y
kz+x+
z
kx+y (x+y+z)
2
(1 +k)(xy+yz +zx)
= x2 +y2 +z2 + 2xy+ 2yz + 2zx
(1 +k)(xy+yz +zx)
31 +k
.
The Theorem 1 is proved.
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Theorem 2. Let x1, x2, . . . , xn be positive real numbers, n 2.
Thenx1
x2+x3+ +xn+
x2x1+x3+x4+ +xn
+ + xnx1+x2+ +xn1
nn 1 . (3)
Proof. Lets = x1+x2+ +xn, one hasx1
x2+x3+ +xn + x2
x1+x3+x4+ +xn + + xn
x1+x2+ +xn1
= x1s x1
+ x2s x2
+ + xns xn
.
By symmetry, we may assume that x1 x2 xn, then
s x1 s x2 s xn, x1s x1
x2s x2
xns xn
.
Using the Chebyshevs inequality (see [3]) gives
x1s x1
(s x1) + x2s x2
(s x2) + + xns xn
(s xn)
1
n x1
s x1 + x2s x2 + +
xns xn
[(s x1) + (s x2) + + (s xn)],
or equivalently
x1s x1
+ x2s x2
+ + xns xn
nn 1 ,
this is exactly the required inequality.
Theorem 3. Let x1, x2, . . . , xn be positive real numbers, n 2,
k 1. Then
x1
x2+x3+ +xn
k+
x2
x1+x3+x4+ +xn
k+ +
xn
x1+x2+ +xn1
k n
(n 1)k . (4)
Proof. Using the power mean inequality and the inequality (3),
we have
x1
x2+x3+ +xn
k+
x2
x1+x3+x4+ +xn
k+ +
xn
x1+x2+ +xn1
k
n1k
ni=1
xis xi
k
n(n 1)k .
This completes the proof.
Theorem 4. Let x1, x2, . . . , xn be positive real numbers, and
let 1, r s > 0,n
i=1
xsi =p. Then
ni=1
xr
i
p xsi
n1
n
n 1
pn
( rs 1 )
. (5)
Proof. Using the power mean inequality (see [3]), we have
ni=1
xr
i
p xsi
n1
ni=1
xri
p xsi
.
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On the other hand, by symmetry, we may assume that x1 x2 xn,
then
xs1 xs2 xsn > 0, p xsn p xsn1 p xs1 > 0.Applying the
generalized Radons inequality (see [4-7])
ni=1
ai
bi n2
(
ni=1
ai)
/(
ni=1
bi)
(a1 a2 an > 0, bn bn1 b1 > 0, 1), we deduce that
ni=1
xri
p xsi
=
ni=1
(xsi
)r
s
p xsi
n2 rs (
ni=1
xsi)
r
s
ni=1
(p xsi
)=
n
n 1p
n
rs 1
,
Therefore
n
i=1
xri
p xsi
n1
n
i=1xri
p xsi
n1
n
n 1
p
n( r
s 1 )
.
The proof of Theorem 4 is complete.
In Theorem 4, choosing = 1, s = 1, n = 3, x1 = x, x2 = y, x3 =
z, we get
Theorem 5. Letx, y, z be positive real numbers, and let x+y+z =
p, r 1. Thenxr
y+z+
yr
z+x+
zr
x+y 3
2
p3
r1. (6)
In particular, whenr = 1, the inequality (6) becomes the
Nesbitts inequality (1).
3 Analogous forms of the Nesbitts inequality
Theorem 6. Letx, y, z be positive real numbers, Then x
x+y+
y
y+z+
z
z+x 3
2
2 (7)
Proof. Note that x
x+y+
y
y+z+
z
z+x
=
z+x
x
(x+y)(z+x)+
x+y
y
(y+z)(x+y)+
y+z
z
(z+x)(y+z).
By using the Cauchy-Schwarz inequality, we have
z+x
x
(x+y)(z+x)+
x+y
y
(y+z)(x+y)+
y+z
z
(z+x)(y+z)
2
(z+x+x+y+y+z)
x
(x+y) (z+x)+
y
(y+z) (x+y)+
z
(z+x) (y+z)
.
Thus, to prove the inequality (7), it suffices to show that
(x+y+z)
x
(x+y) (z+x)+
y
(y+z) (x+y)+
z
(z+x) (y+z)
9
4.
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Direct computation gives
(x+y+z)
x
(x+y) (z+x)+
y
(y+z) (x+y)+
z
(z+x) (y+z)
9
4
=
(x+y+z)[x(y+z) +y(z+x) +z(x+y)]
(x+y) (y+z) (z+x) 9
4
= 4(x+y+z)[x(y+z) +y(z+x) +z(x+y)] 9 (x+y) (y+z) (z+x)
4 (x+y) (y+z) (z+x)
= 8(x+y+z)(xy+yz +zx) 9 (x+y) (y+z) (z+x)
4 (x+y) (y+z) (z+x)
= 6xyz x2y x2z xy2 y2z xz2 yz2
4 (x+y) (y+z) (z+x)
0,
where the inequality sign is due to the arithmetic-geometric
means inequality. The Theorem 6 is thus proved.Theorem 7. Letx, y,
z be positive real numbers, 1/2, Then
x
x+y
+
yy+z
+
zz+x
3
2. (8)
Proof. It follows from the power mean inequality that x
x+y
+
y
y+z
+
z
z+x
312
x
x+y+
y
y+z+
z
z+x
2
312
32
2
= 32
.
The inequality (8) is proved.
Remark. The inequality (8) is the exponential generalization of
inequality (7). As a further generaliza-tion of inequality (7), we
put forward the following the following conjecture.
Conjecture. Let x1, x2, . . . , xn be positive real numbers, n
2, 1/2. Then x1
x1+x2
+
x2
x2+x3
+ +
xn1
xn1+xn
+
xn
xn+x1
n
2. (9)
Acknowledgements. The present investigation was supported, in
part, by the innovative experimentproject for university students
from Fujian Province Education Department of China under Grant
No.214,and, in part, by the innovative experiment project for
university students from Longyan University of China.
References
[1] A. M. Nesbitt, Problem 15114 , Educational Times, 3 (1903),
3738.
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[2] M. O. Drmbe, Inequalities - Ideas and Methods, Ed. Gil,
Zalau, 2003.
[3] D. S. Mitrinovi and P. M. Vasi, Analytic Inequalities,
Springer-Verlag, New York, 1970.
[4] Sh.-H. Wu, An exponential generalization of a Radon
inequality, J. Huaqiao Univ. Nat. Sci. Ed., 24 (1)(2003),
109112.
[5] Sh.-H. Wu, A result on extending Radons inequality and its
application, J. Guizhou Univ. Nat. Sci.Ed., 22 (1) (2004), 14.
[6] Sh.-H. Wu, A new generalization of the Radon inequality,
Math. Practice Theory, 35 (9) (2005), 134139.
[7] Sh.-H. Wu, A class of new Radon type inequalities and their
applications, Math. Practice Theory, 36(3) (2006), 217224.
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