R. M. M. - 23...Romanian Mathematical Society-Mehedinți Branch 2019 4 ROMANIAN MATHEMATICAL MAGAZINE NR. 23 GEOMETRIC INEQUALITIES VIA ONE ALGEBRAIC INEQUALITY By D.M. Bătinețu

ROMANIAN MATHEMATICAL SOCIETYMehedinți Branch

No. 23 - 1 September 2019

R. M. M. - 23ROMANIAN MATHEMATICAL

MAGAZINE

ISSN 2501-0099

Romanian Mathematical Society-Mehedinți Branch 2019

1 ROMANIAN MATHEMATICAL MAGAZINE NR. 23

ROMANIAN MATHEMATICAL

SOCIETY

Mehedinți Branch

ROMANIAN MATHEMATICAL MAGAZINE

R.M.M.

Nr.23-2019



ROMANIAN MATHEMATICAL

SOCIETY

Mehedinți Branch

DANIEL SITARU-ROMANIA EDITOR IN CHIEF

ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT ISSN 1584-4897

GHEORGHE CĂINICEANU-ROMANIA

EDITORIAL BOARD

DAN NĂNUȚI-ROMANIA

EMILIA RĂDUCAN-ROMANIA

MARIA UNGUREANU-ROMANIA

DANA PAPONIU-ROMANIA

GIMOIU IULIANA-ROMANIA

DRAGA TĂTUCU MARIANA-ROMANIA

CLAUDIA NĂNUȚI-ROMANIA

DAN NEDEIANU-ROMANIA

GABRIELA BONDOC-ROMANIA

OVIDIU TICUȘI-ROMANIA

ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO

DANIEL WISNIEWSKI-USA

EDITORIAL BOARD VALMIR KRASNICI-KOSOVO

ALEXANDER BOGOMOLNY-USA



CONTENT

Geometric inequalities via one algebraic inequality - D.M. Bătinețu – Giurgiu, Daniel Sitaru, Neculai Stanciu ......................................................................................................................................................4

Inequalities with concave functions – Daniel Sitaru, Claudia Nănuți..................................................8

Despre unele triunghiuri dreptunghice – Ștefan Marica ....................................................................12

O relație de egalitate între o sumă de cuburi de numere naturale consecutive și pătratul perfect al unui număr natural - Ștefan Marica.......................................................................................17

Two thousand nineteen - Bencze Mihály, Kovács Béla .......................................................................20

On certain algebraic inequalities - D.M. Bătinețu – Giurgiu, Daniel Sitaru, Neculai Stanciu ..22

Teorema lui Rolle şi teorema lui Pompeiu – Angela Nițoiu...............................................................24

About 1005 inequality in triangle – Marin Chirciu.................................................................................27



Monge’s parallelepiped – Marian Ursărescu ..........................................................................................35

Proposed problems………………………………………….………………………………………………………………...41

Romanian Mathematical Magazine-Spring Edition 2019………………………………………………….83

Romanian Mathematical Magazine-Summer Edition 2019………………………………………………89

Index of proposers and solvers RMM-23 Paper Magazine.…………………………………………….………96



GEOMETRIC INEQUALITIES VIA ONE ALGEBRAIC INEQUALITY

By D.M. Bătinețu – Giurgiu, Daniel Sitaru, Neculai Stanciu – Romania

Abstract. In this article we present some algebraic inequalities and some of their geometric

applications for general triangle.

Keywords: algebraic inequality, inequality in triangle.

MSC 2000: 51Mxx, 26D15.

The following inequality it’s well-known: for any holds:

(1)

with equality if and only if .

If , then:

(2)

with equality if and only if .

Indeed, if in (1) we take √

√

√

yields (2).

If , then:

( )( )

( )( )

( )( )

( )( ) (3)

Indeed, if in (2) we take we deduce

(3).

Lemma. For all , holds the following inequality:

( ) ( ) ( )( ) (4)

Proof. If in (3) we take we obtain that:

( )( )

( )( )

( )( )

( )

( )

( ) ( )

( ) ( ) ( ) ( )

Theorem 1. In any triangle , with usual notations, holds the following inequality:

( ) ( )( ) (5)



Proof. If in (4) we take , we obtain that:

( ) ( )

( ) ( ) (6)

Using the following identities:

(7)

(8)

(9)

and (6) we deduce:

( ) ( ) ( )

( ) ( )( )

( )( )


( ) ( )( ( )) (10)

Proof. If in (4) we take

, we obtain:

(

*

(

*

.

/ .

/ (11)

Using:

(12)

(13)

(14)

and by (11) we get:

( )

( )

( )

( ) ( )(( ) )

( )( ( ))


( ) ( ) (15)

Proof. If in (4) we take

, we get:



(

*

(

*

.

/ .

/

(16)

Taking into account by:

(17)

( )

(18)

( )

(19)

and (16) it follows that:

(( ) )

(( ) )

( )

( )

( )

( )( )

( )


( )( ) (20)

Proof. By (4) taking

, we obtain:

(

*

(

*

.

/ .

/

(21)

By well-known:

(22)

(23)

( )

(24)

and we deduce (21) we deduce that:

(( ) )

( )

( )

(( ) )( ( ))

( )( )




( ) ( )( ) (25)

Proof. Putting

, in (4) we have:

(

*

(

*

.

/ .

/

(26)

Using:

(27)

(29)

and then by (26) we get:

( )

( )

( )

( ) ( )( )

( )( )


( ) ( ) √ (30)

Proof. Setting in (4) we obtain that:

( ) ( ) ( ) ( ) (31)

By Ionescu – Weitzenböck’s inequality we have:

√ (I-W)

so (31) becomes:

( ) ( ) √ √ √


( ) (

) √ ( )

(32)

Proof. By (4) where we take yields that:

( ) (

)

( ) (

)

( )

( ), so, by (I-W) we obtain:

( ) (

)

( )

√



√ ( )

References

[1] Daniel Sitaru, Mihály Bencze, 699 Olympic Mathematical Challenges, Studis Publishing House, Iași, 2017.

[2] Daniel Sitaru, Analytical Phenomenon, Cartea Românescă Publishing House, Pitești, 2018.

[3] Daniel Sitaru, George Apostolopoulos: The Olympic Mathematical Marathon, Cartea Românescă Publishing House, Pitești, 2018.

[4] Daniel Sitaru, Contests Problems, Cartea Românescă Publishing House, Pitești, 2018. [5] Mihály Bencze, Daniel Sitaru, Quantum Mathematical Power, Studis Publishing House,

Iași, 2018. [6] Mihály Bencze, Daniel Sitaru, Marian Ursărescu, Olympic Mathematical Energy, Studis

Publishing House, Iași, 2018. [7] Romanian Mathematical Magazine, Interactive Journal, www.ssmrmh.ro

INEQUALITIES WITH CONCAVE FUNCTIONS

By Daniel Sitaru, Claudia Nănuți – Romania

Abstract. In this paper we prove a general inequality for concave functions and we construct

applications as consequences of it.

Theorem 1.

If , - is a concave function and then:

( )( ) ( )( ) ( )( ) (1)

Proof.

( ( )) ( ( )) ( ( ))

( ) ( ) ( )

, - , - , -



( ( ) ( ))( ) ( ( ) ( ))( ) ( ( ) ( ))( )

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

Equality holds for .

Theorem 2.

If , - is a concave function and then:

( )( ( ) ( )) ( )( ( ) ( )) (2)

Proof.

( ( )) ( ( )) ( ( )) ( ( ))

( ) ( ) ( ) ( )

, - , - , - , -

( ( ) ( ))( ) ( ( ) ( ))( )

( ( ) ( ))( ) ( ( ) ( ))( )

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( ) ( )( ) ( )( )

( )( ( ) ( )) ( )( ( ) ( ))

Equality holds for .

Application 1.

If , - is a concave function and

√

then:



( ) ( ) ( ) ( ) ( ) ( ) (3)

Proof.

We take in (1):

Inequality is a strictly because hence .

Application 2.

If , - is a concave function and √

√

then:

( )( ) ( )( ) ( )( ) (4)

Proof.

We take in (1):

Inequality is strictly because . Hence .

Application 3.

If , - is a concave function and

√

√

then:

( )( ( ) ( )) ( )( ( ) ( )) (5)

Proof.

We take in (2):

Inequality is strictly because . Hence:

Observation 1.

Function , - ( ) is concave because ( )

( )

. By (3):

( ) ( ) ( ) ( ) ( ) ( )

(( ) ( ) ( ) )

( ) ( ) ( )

(

*√

(√ )

(

*

√

By (4): ( ) ( ) ( ) ( ) ( ) ( )

(( ) ( ) ( ) )

( ) ( ) ( )



(√ )

(

*

√

√

(

*√

By (5): ( )( ( ) ( )) ( )( ( ) ( ))

4(

*

(

*

5

(

*

(

*

(

√

√

)

(

)

√

√

(√

)

4( )

5

√

√

Observation 2.

Function , - ( ) is concave because

( )

( )

( )

By (3):

( ) ( ) ( ) ( ) ( ) ( )

(√

* (

* (

* (√ )

(

√ * (

*

By (4):

( ) ( ) ( ) ( ) ( ) ( )

(

* (√ ) (

√ * (

*

(√

* (√

)

By (5):

( )( ( ) ( )) ( )( ( ) ( ))



( ) (

* ( ) (

*

(

*

(

√ √

√ ( )

)

(√

√ ) (

)

(

* 4

√ √

√ √ ( )5 (√

√ ) 4

( )

5

References

[1] Daniel Sitaru, Mihály Bencze, 699 Olympic Mathematical Challenges, Studis Publishing House, Iași, 2017.

[2] Daniel Sitaru, Analytical Phenomenon, Cartea Românescă Publishing House, Pitești, 2018.

[3] Daniel Sitaru, George Apostolopoulos: The Olympic Mathematical Marathon, Cartea Românescă Publishing House, Pitești, 2018.

[4] Daniel Sitaru, Contests Problems, Cartea Românescă Publishing House, Pitești, 2018. [5] Mihály Bencze, Daniel Sitaru, Quantum Mathematical Power, Studis Publishing House,



DESPRE UNELE TRIUNGHIURI DREPTUNGHICE

By Ștefan Marica – Romania

I. Să se afle mulțimea triunghiurilor dreptunghice știind că ariile și perimetrele lor sunt

exprimate prin relația:

– unde

√



. √ /

√

( ) ( )

( )

( )

(1)

În relația (1) luăm, pe rând, și aflăm lungimile laturile triunghiului

1)

unde * +

2)

unde * +

3)

unde * +

4)

unde * +



Obs: Dacă * + se obțin aceleași valori pentru și .

Relația:

( √ ) este simetrică în și .

Cele prezentate anterior se pot sintetiza astfel

Lat triunghi Aria

( )

Perimetru

( )

Relația

Pentru se obțin triunghiuri dreptunghice:

a) patru triunghiuri prin dublarea celor

b) al cincilea rezultă astfel:

, adică

( )( ) 2

2

II. Există triunghiuri oarecare la care se poate stabili o relație între ariile și perimetrele lor.

Să se afle mulțimea triunghiurilor despre care avem informațiile:

- aria perimetru ( )

- Lungimile laturilor sunt numere naturale care diferă între ele prin aceeași constantă .



√

√

( ) sau , -

Dacă notăm obținem ( )( )

În relația (1) dăm pe rând, lui valori din mulțimea numerelor naturale.

1) ( ) ( )

Singurul sistem care admite soluție în mulțimea numerelor naturale este:

2

Triunghiul este dreptunghic ( )

2) ( )( )

Următoarele trei sisteme au soluții în mulțimea numerelor naturale:

2

2

Triunghiul este oarecare ( )

2

2


2

2

Triunghiul este oarecare ( )

( )( )

și trebuie să fie de aceiași paritate

2

2


Cele prezentate până la se pot sintetiza astfel:

Laturile triunghiului Aria

( )

Perimetru

( )

Relația



. Pentru se aplică în continuare formula , -, adică

( )( ) și se obțin cinci triunghiuri.

De remarcat că în fiecare caz unul din triunghiuri este dreptunghic.

Tripletul de numere pitagorice ( ) generator de noi triplete.

Răspundem acestei probleme cu ajutorul următoarei metode de lucru:

Tripletul ( ) este soluție a relației: sau ( )( ) îl notăm

astfel: . Numerele le vom obține după cum urmează:

Rezultatele obținute le cuprindem în tabelul:

a) Tripletele noi sunt: ( ) ( ) ( ) ( ) etc.

b) Înmulțind tripletele de pe orizontală cu șirul numerelor naturale rezultă alte triplete.

c) Verificarea oricărui triplet se face ușor:



( ) ( ) ( )

O RELAȚIE DE EGALITATE ÎNTRE O SUMĂ DE CUBURI DE NUMERE NATURALE CONSECUTIVE ȘI PĂTRATUL PERFECT AL UNUI NUMĂR NATURAL

By Ștefan Marica – Romania

În suplimentul Gazetei Matematice din decembrie 2018 este publicată problema S:E 18357.

Găsiți șapte numere naturale consecutive cu proprietatea că suma cuburilor lor este un

pătrat perfect. In cele ce urmează vom arăta că există o mulțime de numere naturale cu

această proprietate. Vom folosi un raționament inductiv.

1) Trei numere naturale consecutive

( ) ( )

( )

Condiție . Două soluții:

Observație:

( ) ( ) ,( ) ( )-

Intr-adevăr, obținem ecuația:

2) Cinci numere naturale consecutive

( ) ( ) ( ) ( )

( )

Condiție . Două soluții:

Observație:

( ) ( ) ( ) ( )

,( ) ( ) ( ) ( )-

Într-adevăr rezultă ecuația: ( )

3) Șapte numere naturale consecutive:



( ) ( ) ( ) ( ) ( ) ( )

( )

Condiție: . Două soluții:

Observație:

( ) ( ) ( ) ( ) ( ) ( )

,( ) ( ) ( ) ( ) ( ) ( )-

Într-adevăr are loc relația: ( )

4) – numere naturale consecutive

( ) ( ) ( ) ( )

( ) ( )

Efectuând calcul se ajunge la:

( ) ( )

( ) ( )( )

( ) ( )( )

( ) , ( )-

Condiție ( ) ( )

( ) ( )

Observație:

( ) ( ) ( ) ( )

( ) ( )

,( ) ( ) ( ) ( ) ( ) ( )-

Calculul ne conduce la ecuația:

( ) ( )( ) ,( ) - sau

( ) ( )

5) Patru numere naturale consecutive

( ) ( ) ( )



Efectuând calculul se ajunge la

Două soluții:

Observație:

1) ( ) ( )

( ) ( ) ( ) } sunt pătrate perfecte pentru

2) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) } sunt pătrate perfecte

pentru

3) ( ) ( )

( ) ( ) ( ) } sunt pătrate perfecte pentru

Aplicații:

a) Să se scrie numărul ca o diferență de două pătrate perfecte.

( )

( ) ( )

6( )

7

6( )

7

b) Să se calculeze suma:

( )

( ) ( ) ( )

6( )

7

6( )

7

( ) ( )

c) Să se rezolve în mulțimea numerelor naturale ecuația:

( ) ( ) ( ) ( )

Soluție: Notăm . În noua necunoscută ecuația este:

( ) ( ) ( ) ( )

( )



Condiție

1)

2)

Temă:

1) Să se arate că se divide la

2) Să se scrie numărul ca o diferență de două pătrate perfecte.

TWO THOUSAND NINETEEN

By Bencze Mihály, Kovács Béla – Romania

2019 written with Roman numerals: MMXIX = MIX+MX 2019 is an odd natural number, product of two prime numbers: 2019 = 3 ∙ 673 It can be written with a single digit: 2019 = (1111 – 111 + 11) ∙ (1 + 1) – 1 – 1 – 1

2019 = 2222 – 222 + 22 – 2 – 2

2

2019 = 333 ∙ ( 3 + 3 ) + 3 ∙ 3 ∙ 3 – 3 – 3

2019 = ( 444 + 44 + 4 ∙ 4 ) ∙ 4 + 4 – 4

4

2019 = 555 ∙ 5 – 5 ∙ 5 ∙ 5 ∙ 5 – 5 ∙ 5 ∙ 5 – 5 – 5

5

2019 = ( 66 ∙6 – 66 + 6 ) ∙ 6 + 6 6 6

6

2019 = 7 ∙ 7 ∙ 7 ∙ 7 – 7 ∙ 7 ∙ 7 – 7 ∙ 7 + 7 + 7 7 7

7

2019 = 888 + ( 88 + 8 ∙ 8) ∙ 8 – 88 + 88

8

2019 = 999 + 999 + 9 + 9 + 9

Written using all digits at the same time: 2019 = 1 + 2345 – 6 ∙ 7 ∙ 8 + 9 2019 = 10 ∙ 9 ∙ 8 ∙ 7 : 6 : 5 ∙ 4 ∙ 3 + 2 + 1

Written as sum of the squares of three natural numbers: 2019 = 2 2 21 13 43

2019 = 2 2 25 25 37

2019 = 2 2 27 11 43

2019 = 2 2 27 17 41

2019 = 2 2 211 23 37

2019 = 2 2 213 13 41

2019 = 2 2 213 25 35

2019 = 2 2 217 19 37

2019 = 2 2 223 23 31 There are altogether 9 possibilities, from which in six cases we can use only prime numbers. Written as sum of the squares of four natural numbers:

2019 = 2 2 2 25 15 20 37



2019 = 2 2 2 27 9 17 40

2019 = 2 2 2 213 15 20 35

2019 = 2 2 2 213 21 25 28

2019 = 2 2 2 217 23 24 25

Written as sum of the squares of five natural numbers: 2019 = 2 2 2 2 213 15 20 21 28

2019 = 2 2 2 2 215 17 20 23 24

Written as a difference of two squares: 2019 = 2 21010 1009

2019 = 2 2338 335

as a Sum of powers of four prime factors: 2019 = 10 3 4 32 3 5 7

sum of biquadrats 2019 = 4 4 4 4 41 2 3 5 6

written with the powers of 2: 2019 = 11 5 2 02 2 2 2

2019 = 10 9 8 7 6 5 02 2 2 2 2 2 2 2

2019 = 10 9 8 7 6 5 4 3 2 02 2 2 2 2 2 2 2 2 2 2

written with other powers: 2019 = 7 4 4 23 3 3 3 + 3

2019 = 5 5 2 2 04 4 4 4 4 4

written with the powers of 5: 2019 = 4 4 4 3 2 05 5 5 5 5 5 5

As a term in a Pythagorean triple: 2019 = 2 21155 1656 = 3 ∙ 2 2385 552

witten as sum of terms of an arithmetic progression sum of consecutive odd natural numbers 2019 = 671 + 673 + 675 sum of consecutive natural numbers 2019 = 334 + 335 + 336 + 337 + 338 + 339 using Variations, Permutations and factorial, more shortly:

2019 = 3 18 3 3V P V = 8∙7∙6∙1∙2∙3 + 3 2019 =

8! 3!

5!

+

3!

2!

Using only two digits: 2019 = 8! 3!

(8 3)!

+ 3

Using descending digits and factorials:

2019 = 10! 9! 8! 7 7! 7! 6 6! 5! 5! 4! 3! 3! 2! 1!

written using factorials only: 2019 = 3 ∙ 6! – 5! – 4! + 2! + 1! written in another numeration system 2019 = 120312 = 157611 = 201910 = 26839 = 37438 = 56137 = 132036 2019 = 310345 = 1332034 = 22022103 = 11.111.100.0112 resolution in different based number systems

2019 = 17 ∙ 126 if we use base 11 2019 = 49 ∙ 51 if we use base 12 2019 = 39 ∙ 71 if we use base 13

Other interesting items:

2019 = 2 2 2 21 2 3 . . . 3028

1 2 3 . . . 3028

2019 = 1 1 1

. . . 20201 2 2 3 2019 2020

2019 = 2 2676 670

4

2019 =

3 3 3 3677 669 675 671

24



written with cubes: 3 3 3 3 3 320 25 42 10 15 32 = 30 ∙ 2019 3 3 3 3 3 316 24 47 2 10 33 = 42 ∙ 2019 3 3 3 3 3 328 33 48 2 7 22 = 78 ∙ 2019 3 3 3 3 3 340 41 42 6 7 8 = 102 ∙ 2019

or with numbers to 2nd power and 4th power 2 2 2 2 2 2 2 2 2 2 2 22 8 10 16 23 24 26 31 33 47 49 57 = 6 ∙ 2019 4 4 4 4 4 4 4 4 4 4 4 42 8 10 16 23 24 26 31 33 47 49 57 = 6 ∙ 20192 2 2 2 2 2 2 2 2 2 2 2 22 5 7 15 20 22 28 33 35 48 50 55 = 6 ∙ 2019 4 4 4 4 4 4 4 4 4 4 4 42 5 7 15 20 22 28 33 35 48 50 55 = 6 ∙ 20192 2 2 2 2 2 2 2 2 2 2 2 23 4 7 8 12 15 34 38 41 46 49 53 = 6 ∙ 2019 4 4 4 4 4 4 4 4 4 4 4 43 4 7 8 12 15 34 38 41 46 49 53 = 6 ∙ 20192

written with squared numbers

20192 = 2 2 2 2 2 2 22004 151 119 113 71 58 47

20192 = 2 2 2 2 2 2 21990 267 171 75 60 59 55

20192 = 2 2 2 2 2 2 21966 285 275 205 81 75 12

20192 = 2 2 2 2 2 2 21964 315 315 91 78 65 45

20192 = 2 2 2 2 2 2 21921 430 340 270 100 44 28

20192 = 2 2 2 2 2 2 2 2 2 21480 969 555 485 455 300 200 185 150 100

20192 = 2 2 2 2 2 2 2 2 2 21360 1311 433 360 280 241 153 126 119 32

20192 = 2 2 2 2 2 2 2 2 2 21344 815 740 600 555 455 370 175 100 75

20192 = 2 2 2 2 2 2 2 2 2 21200 977 782 575 552 425 408 240 167 49

20192 = 2 2 2 2 2 2 2 2 2 21104 991 850 600 575 408 391 336 167 47

or numbers to Higher powers 20192 = 11552 +16562 20193 = 6713 + 6733 + 6753 + 3 ∙ 1344 ∙ 1346 ∙ 1348 20195 = 6715 + 6735 + 6755 + 5 ∙ 1344 ∙ 1346 ∙ 1348 ∙ 2717578 20197 = 6717 + 6737 + 6757 + 7 ∙ 1344 ∙ 1346 ∙ 1348 ∙ 8000658788059

The 2019501 is a very large number, but this is the lowest power with 2019 as the last four digits, 2019! is a huge number whose last 502 digits are zero

TEOREMA LUI ROLLE ŞI TEOREMA LUI POMPEIU By Angela Niţoiu – Romania

Definiţie: Fie . O funcţie , - cu proprietăţile:

(1) f continuă pe , - şi

(2) f derivabilă pe ( )

se numeşte funcţie Rolle pe , -

Teorema lui Rolle: Fie , - o funcţie Rolle pe , - cu proprietatea că ( ) ( )

(3) Atunci există ( ) astfel încât ( )



Consecinţe:

1) Dacă, în particular, ( ) ( ) , teorema lui Rolle afirmă că între două rădăcini ale

unei funcţii derivabile, există cel puţin o rădăcină a derivatei sale.

2) Între două rădăcini consecutive ale derivatei există cel mult o rădăcină a funcţiei.

Observaţii:

1. Fiecare din condiţiile teoremei lui Rolle este fundamentală, în sensul că, renunţând la una

din cele trei condiţii nu mai rezultă concluzia.

Într-adevăr:

, - , ( ) îndeplineşte doar condiţiile (1) şi (2), dar nu şi (3):

( ) ( ) Se observă că ( ) , -.

, - ( ) { , )

îndeplineşte numai condiţiile (2) şi (3),

dar nu şi (1), nefiind continuă în . Evident, ( ) ( )

, - ( ) | | îndeplineşte numai condiţiile (1) şi (3), dar nu şi

(2), nefiind derivabilă în Evident, ( ) ( ) * +

2. Ipotezele teoremei lui Rolle sunt suficiente, dar nu şi necesare pentru ca derivata să aibă

cel puţin o rădăcină. Există funcţii care nu satisfac niciuna din condiţii pe un anumit interval

pe care totuşi, derivata se anulează.

Astfel, pentru ( ) { , -

, - avem ( )

Problemă rezolvată

Fie ̅̅ ̅̅̅ Să se demonstreze că există ( ) astfel încât:

∑( )

Soluţie:

Fie funcţia , -

( ) ∑

( )

Evident, f este derivabilă pe , - şi avem

( ) ( ) ∑(

*



deci, conform teoremei lui Rolle, există ( ) astfel încât ( ) Dar

( ) ∑ ( ) deci, ( ) ∑ (

)

Teorema lui POMPEIU: Fie funcţia Rolle , - şi , - Atunci există un punct

( ) astfel încât ( ) ( )

( ) ( )

Demonstraţie: Se consideră funcţia , - ( ) ( )

. Vom determina λ astfel

încât ( ) ( ) Se obţine ( ) ( )

Aplicând teorema lui Rolle, rezultă că există

( ) astfel încât ( ) , adică ( ) ( )

Interpretarea geometrică a teoremei lui Pompeiu

Dreapta AB, unde ( ( )) ( ( )) întâlneşte axa Oy în punctul ( ), unde

( ) ( )

Conform teoremei lui Pompeiu, există ( ) astfel încât tangenta în

punctul ( ( )) la graficul funcţiei f, întâlneşte axa Oy în punctul M.

Bibliografie:

1. M. Nicolescu, N. Dinculeanu, S. Marcus- Analiză matematică, EDP Bucureşti, 1980

2. M. Ganga- Elemente de analiză matematică- manual pentru clasa a XI-a, Mathpress, 2009

ON CERTAIN ALGEBRAIC INEQUALITIES

By D.M. Bătinețu-Giurgiu, Daniel Sitaru, Neculai Stanciu – Romania

Abstract. In this article we present some algebraic inequalities and some of their

generalizations and refinements.

1. New refinements for Nesbitt’s inequality



The inequality of Nesbitt (see [13]), i.e.

(N)

was refined by Marian Dincă (see [12]) as follows:

( )

( )

√ ( )

( )

(1)

In this paper we give, we prove that:

( )

( )( )

√ ( )

( )( )

( ) (2)

Indeed, by Harald Bergström’s inequality, we have that:

( )

( )( ) (3)

Since, ( ) ( )

by (3) yields that: ( )

( )( )

(4)

where for , we obtain (N)

In [1] it is shown that: ( )

√ ( )

(5)

and then (3) becomes:

( )

( )( )

√ ( )

( )( ) (6)

Taking into account QM-AM the inequality, i.e: √

(7)

which is equivalent with: √ ( ) √ ( )

(8)

Then, by above we get: ( )

( )( )

√ ( )

( )( )

i.e. the inequalities (2)

From (2) taking we deduce (1)

Remark. Many generalizations of Nesbitt’s inequality you will find if you check the

references of this paper.

2. New generalizations of Stergioiu’s inequality

In [15] were published the statements: Let with . Prove that:

( ) ( ) ( ) (9)

Babis Stergioiu, Athena, Greece

Next, we prove that: If and , then:



( )(( ) ( ) ( ) ) (10)

from which putting we obtain (9). Indeed, let:

∑

and we are setting

, where with .

Then, by the inequality of Harald Bergström we deduce that:

∑

(∑ )

∑ ( )

(∑ )

( )( ) ∑ (11)

Since: , by (11) and AM-GM inequality we get:

(∑ )

( )( )

. √( )( )( )

/

( )( )

( )( ),

and taking

yields that:

( ).

/

( )

( )(( ) ( ) ( ) ), and we are done.

Other generalization: ∑

( )

( ) (( ) ( ) ( ) ) .

Proof. Taking

, by the inequality of J. Radon and AM-GM inequality we

obtain that: ∑( )

( ) (∑

)

(∑ ( ) )

(∑

)

( ) ( )

(√( ) ( ) ( ) )

( ) ( )

( ) ( ) and if we take

, yields that:

( ) .

/

( )

( ) (( ) ( ) ( ) )

( ) (( ) ( ) ( ) ) , and the proof is complete.

References [1] D.M. Bătinețu-Giurgiu, Neculai Stanciu, An expansion and a refinement of Nesbitt’s

inequality, Rm. Sărat Simpozion, 2011. [2] D.M. Bătinețu-Giurgiu, Neculai Stanciu, New generalizations of Nesbitt’s inequality, Rm.

Sărat Simpozion 2011. [3] D.M. Bătinețu-Giurgiu, Mihály Bencze, Neculai Stanciu, New generalizations and new

approaches for Nesbitt’s inequality, Brașov Simpozion, 2011.



[4] D.M. Bătinețu-Giurgiu, Neculai Stanciu, An expansion and a refinement of Nesbitt, RMT, no. 1/2012.

[5] D.M. Bătinețu-Giurgiu, Neculai Stanciu, Another four demonstrations of the problem L:155 from Sclipirea minții, no. VII 2011, Sclipirea Minții, no. IX, 2012, pp. 6-8.

[6] D.M. Bătinețu -Giurgiu, Neculai Stanciu, Nesbitt’s inequality, Didactica Matematică, no. 1/2012.

[7] D.M. Bătinețu -Giurgiu, Neculai Stanciu, Problem 11634, The American Mathematical Monthly, Vol. 119, March 2012, pp. 248.

[9] D.M. Bătinețu -Giurgiu, Neculai Stanciu, Escolar de la Olimpiada Iberoamericana Mathematical Magazine, 2012.

[10] D.M. Bătinețu-Giurgiu, Neculai Stanciu, Generalizations of some remarkable inequalities, The Teaching of Mathematics, 2013, Vol. XVI, pp. 1-5.

[11] D.M. Bătinețu-Giurgiu, Neculai Stanciu, A new generalization of Nesbitt’s inequality, Journal of Science and Arts – Year 13, No. 3(24), pp. 255-260, 2013.

[12] Dincă Marian – An improvement of Nesbitt’s inequality, Minus Magazine, No. 1/2009, pp. 9-10.

[13] Titu Zvonaru, Neculai Stanciu, Six solutions for the problem L:155 from Sclipirea Minții, no. VII, 2011, Sclipirea Minții, no. VIII, 2011, pp. 9 – 10.

[15] Problem O.IX. 71 – RMT, No. 1 – 2005, pp. 48. [16] Daniel Sitaru, Mihály Bencze, 699 Olympic Mathematical Challenges, Studis Publishing

House, Iași, 2017. [17]. Daniel Sitaru, Analytical Phenomenon, Cartea Românescă Publishing House, Pitești,

2018. [18] Daniel Sitaru, George Apostolopoulos: The Olympic Mathematical Marathon, Cartea

Românescă Publishing House, Pitești, 2018. [19] Daniel Sitaru, Contests Problems, Cartea Românescă Publishing House, Pitești, 2018. [20] Mihály Bencze, Daniel Sitaru, Quantum Mathematical Power, Studis Publishing House,



ABOUT 1005 INEQUALITY IN TRIANGLE

ROMANIAN MATHEMATICAL MAGAZINE 2018 By Marin Chirciu – Romania

1) In

∑√

√

Proposed by Bogdan Fustei – Romania

Solution:

Using

the inequality can be equivalently transformed:



∑√

√ ∑√ ( ) , which follows from means inequality:

∑√ ( ) ∑ ( )

∑

Equality holds if and only if the triangle is equilateral.

Remark:

Let’s find an inequality having an opposite sense:

2) In

∑√

√

Proposed by Marin Chirciu – Romania

Solution:

Using

the inequality is equivalently transformed:

∑√

√ ∑√ ( )

, which follows from means inequality (GM-HM):

∑√ ( ) ∑

∑ ( )

( ) ∑

( )

Next, we use the following lemma:

Lemma

3) In the following inequality holds:

∑ ( )

Proof:

∑ ( )

∑ ( )( )( )

∏( )

( )

( )

( )

which follows from the identities:

∑ ( ) ( )( ) ( ) and

∏( ) ( )

It remains to prove that:

( )

( ) ( )



We distinguish the cases:

Case 1). If , the inequality is obvious.

Case 2). If , the inequality can be rewritten: ( ) ( )

which follows from Gerretsen’s inequality: .


( ) ( )( )

( )( ) , obviously from Euler’s inequality: .

Finally, using Lemma, the proposed inequality is true.


Remark:

We can write the double inequality:


√ ∑√

√

Solution:

See inequalities 1) and 2).


Remark:

For the sum ∑ ( )

we can write the double inequality:


∑

( )


Solution:

See Lemma and ∑ ( )

, which follows from ∑

( )

( )

It remains to prove that: ( )

true from Gerretsen’s inequality: and Euler’s inequality: .


Remark:

If we replace with , we propose:




√ ∑√ √


Solution:

The left-hand inequality:

Using

∏

, the means inequality, Euler’s inequality

and Mitrinovic’s inequality .

We obtain:

∑√ ∑√

√√

√

√ √

√ √ √

The right-hand inequality:

Using CBS inequality, ∑ and Euler’s inequality , we obtain:

(∑√ ) ∑( ) ∑ ( )

, wherefrom

∑√ √


ABOUT 1013 INEQUALITY IN TRIANGLE ROMANIAN MATHEMATICAL MAGAZINE 2018

By Marin Chirciu – Romania

1) In the following relationship holds:

∑√

√

Proposed by Mehmet Șahin – Ankara – Turkey

Solution:

With CBS inequality we obtain:

4∑√

5

∑( )∑

∑ ∑

.∑ /

( )



which follows from Leibniz inequality ∑ , wherefrom ∑√

√


Remark:

Let’s find an inequality of opposite sense:


∑√

√


Solution:

Using ( )

we obtain: ∑

√

∑

√( )

√ ∑


Lemma:


∑

Proof:

∑

∑

∑ ( )

∑


, true from Gerretsen’s inequality:

and Euler’s inequality: .

Finally, using Lemma, the proposed inequality is true.


Remark:



√ ∑

√

√

Solution:





Remark:

For the sum ∑



∑

( )

Solution:

See Lemma and ∑

( )



( )




Remark:

If we replace with we propose:


√ ∑

√

√


Solution:

The left-hand inequality:

Using ( )

we obtain: ∑

√

∑

√( )

√ ∑

.


Lemma


∑

Proof:

∑

∑

∑( )( )

∑

( )




( )



Finally using the Lemma, the propose inequality is true.


The right-hand inequality:

With CBS inequality we obtain:

4∑√

5

∑( )∑

∑ ∑

(

*

wherefrom ∑√

√

Above, we have used Leibniz inequality ∑ and the following inequality

∑

.

/ which follows from the identity ∑

( )

, Gerretsen’s

inequality: and Euler’s inequality: . Equality holds if and only

if the triangle is equilateral.

Remark:

For the sum ∑



∑

( )

Solution:

See Lemma and ∑

( )


( )

It remains to prove that: ( )

( )


and Euler’s inequality: . Equality holds if and only if the

triangle is equilateral.

ABOUT 1018 INEQUALITY IN TRIANGLE

ROMANIAN MATHEMATICAL MAGAZINE 2018 By Marin Chirciu – Romania




∑ ( )

√

Proposed by Daniel Sitaru – Romania

Solution:

We prove the following lemma:

Lemma

2) In

∑ ( )

( )

Proof:

We have ∑ ( )

( )


( )

∑ ( )( )( )

∏( ),

∑ ( ) ( )( ) ( ) and ∏( ) ( )

Back to the main problem.

Using the Lemma we prove that:

We obtain ∑ ( )

( )

, which follows from

Gerretsen’s inequality: and Euler’s inequality .

In order to prove ∑ ( )

√

, taking into account that ∑

( )

, it suffices to prove

that:

√

√

(Mitrinovic’s inequality).


Remark:

Let’s find an inequality having an opposite sense:

3) In

∑ ( )


which follows from Gerretsen’s inequality:


( )( )

obviously from Euler’s inequality .




Remark:


4) In

∑

( )

Solution:



MONGE’S PARALLELEPIPED By Marian Ursărescu – Romania

In this article, we will try to solve certain problems linked to tetrahedron’s geometry,

emphasizing a “special” parallelepiped associated to the tetrahedron.

The idea consists in solving the problem for a parallelepiped instead of solving the problem

for the tetrahedron, after we had first obtained a relation between certain properties

(characteristics) of the tetrahedron and certain properties (characteristics) of the

parallelepiped, by using this method, a difficult problem for a tetrahedron may become

easier for the parallelepiped.

For instance, let’s consider the following problem proposed at Laurențiu Duican math

competition: “Show that a tetrahedron is equifacial if and only if the treble volume of the

tetrahedron is equal to the product of the bimedians”.

(I remind that a tetrahedron is equifacial if the opposite edges are equal, and the bimedian is

the segment which links the middle points of two opposite edges.)

One method of solving this problem, although very long and laborious, consists of calculating

the lengths of the bimedians depending on the edges of the tetrahedron (using the theorem

of the median), and then, of expressing the volume of the tetrahedron depending on its

edges (sides).

We will see how this difficult problem can be solved in quite easy way.

Definition: Let be a certain tetrahedron. The parallelepiped limited by the planes built

through the edges of the tetrahedron and which are parallel to the opposite edges is called



parallelepiped circumscribed to the tetrahedron or Monge’s parallelepiped

associated to the tetrahedron .

Observation (remark): The edges of the tetrahedron represent the diagonals of the

faces in Monge’s parallelepiped.

Figure 1.

Properties:

1) The tetrahedron is equifacial (meaning with the opposite edges equal)

Monge’s parallelepiped associated to the tetrahedron is a rectangular

parallelepiped.

Demonstration.

‘ ’ Let be an equifacial (figure 1). Then , - , - is a rectangle; in the same

way the other faces of the parallelepiped are rectangles is a rectangular

parallelepiped.

2) The tetrahedron is regular Monge’s parallelepiped associated to the

tetrahedron is cube.

Demonstration: (obvious)

Theorem: The ratio between the volume of the tetrahedron and the volume of Monge’s

parallelepiped associated to the tetrahedron is

.

Demonstration: (figure 1)

We denote: the volume of the tetrahedron and the volume of Monge’s

parallelepiped.

{

( )

( )



(

( )

( )

( )

( )* (1)

But

(2)

and ( ) ( ) ( ) ( ) (3)

According to the relation (1), (2), (3)

Application 1.

If an equifacial tetrahedron with the edges has the volume

then the tetrahedron

is regular.

(Math Competition)

Proof. We will build Monge’s parallelepiped associated to the tetrahedron . The

tetrahedron being equifacial, then according to the first property Monge’s

parallelepiped is a rectangular parallelepiped.

We note and

Obviously

According to the third property

, but according

to the hypothesis

( )( )( )

( ) ( ) ( ) (1)

According to the inequality of the means, we have:

( ) ( ) ( ) (2)

From (1), (2) regular.

Application 2.

Show that a tetrahedron is equifacial the treble volume is equal to the product of the

bimedians.

(Math Competition)

Demonstration: (The bimedian links the middle points of two opposite edges.)



“ ” Let’s suppose that is euifacial Monge’s parallelepiped is a rectangular

parallelepiped.

We note: and .

According to the first application

(1)

The parallelepiped being rectangular the bimedians of the tetrahedron are equal to the

edges of the parallelepiped. (2)

From (1) and (2) the treble volume is equal to the product of the bimedians.

“ ” The treble volume of the tetrahedron being equal to the product of the bimedians

the volume of the parallelepiped is equal to the product of the bimedians the

parallelepiped is rectangular equifacial.

Application 3.

Demonstrate that if in a tetrahedron the mutual perpendiculars of the opposite edges are

perpendicular two by two, then the tetrahedron is equifacial.

(Barrage for I.M.O.)

Demonstration: We build Monge’s parallelepiped associated to the tetrahedron. The mutual

perpendicular of the edges and is perpendicular on the plane ( ) and the

mutual perpendicular of the edges and is perpendicular on the plane ( ). But

the perpendicular on one another (according to the hypothesis) ( ) ( )

In the same way ( ) ( ) and ( ) ( ).

Consequently is a rectangular parallelepiped equifacial.

Bibliography:

1. The Euclidian plane and space, Dan Brânzei.

2. Mathematical Gazette Collection.



PROPOSED PROBLEMS

5-CLASS-STANDARD

V.1 Să se afle despre care avem informațiile:

1) – pătratul perfect al unui număr prim

2) – este pătrat perfect 3) – este pătrat perfect 4) – este pătrat perfect

Proposed by Ștefan Marica – Romania

V.2. Să se afle numerele opuse

și

știind că are loc relația:


V.3. Find all the triplets ( ) of nonzero natural numbers which satisfy the relationship:

Proposed by Daniel Sitaru, Neculai Stanciu – Romania

V.4. Există astfel încât se divide cu ?

Proposed by Daianu Mihaela, Popescu Delia –Romania

V.5. Care dintre numerele și este mai mare?

Proposed by Beldea Daniela, Ghiță Georgiana Alina – Romania V.6. Prove that:

a) Number it is not a perfect square. b) Number it is not a perfect square. c) Number it is not a perfect square.

Proposed by Viespescu Carina Maria - Romania

V.7. Prove that the equation 2 2 2016x y z has an infinity of natural solutions.

Proposed by Marin Chirciu –Romania

V.8. Prove that there is an infinity of triplets of natural numbers , ,x y z for which the

number 3 3 3x y z to be a perfect cube. Proposed by Marin Chirciu-Romania

V.9. a) Prove that there is an infinity of triplets natural numbers , ,x y z for which the

number 3 3 3x y z is perfect square.



b) Prove that there isn’t a triplet of natural numbers , ,x y z for which the number

6 6 6x y z to be perfect square. Proposed by Marin Chirciu-Romania

V.10. Prove that 401n can be written as a sum of three distinct nonzero perfect squares, for

any *nN . Proposed by Marin Chirciu-Romania

V.11. Prove that the number 2n abc cab c is divisible with 11. Proposed by Marin Chirciu-Romania

V.12. Prove that 401n can be written as a sum of three nonzero distinct perfect squares, for

any *nN . Proposed by Marin Chirciu-Romania

V.13. Find the smallest natural numbers abcde , respectively fghij such that:

fghijabcde2 and the two numbers together to use all the digits from 0 to 9.

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.14. Prove that:

4046133

2023066

120112011

2011...

122

2

111

1242424

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.15. Find the sum of the digits of the number: 2011102011 . Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.16. How many digit has the number 962 ?

Proposed by Titu Zvonaru, Neculai Stanciu-Romania

V.17. Find all the pairs ),( yx of natural numbers which satisfy the relationship:


V.18. Prove that among any 305 natural numbers, two by two coprime contained between 2

and 20112010 it exists at least a prime number. Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.19. Find all the natural numbers a for which it exists exactly 2014 natural numbers b which

verify the relationship: 52 b

a.


V.20. Find all the natural numbers a for which it exists exactly 2013 natural numbers b

which verify the relationship: 52 b

a.




V.21. Find all the natural numbers a for which it exists exactly 2011 natural numbers b

which verify the relationship: 52 b

a.


V.22. How many divisors of the number 201130 are not divisors of the number 201020 ?(write

the searched number in function of number 2012). Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.23. Find , for which .

Proposed by Dan Nedeianu – Romania

All solutions for proposed problems can be finded on the

http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.

6-CLASS-STANDARD

VI.1. Find such that is divisible simultaneous with and . Proposed by Gheorghe Calafeteanu – Romania

VI.2. Să se rezolve în mulțimea numerelor întregi ecuația:


VI.3. If such that

and

, then find

.

Proposed by Daniel Sitaru, Neculai Stanciu – Romania

VI.4. Fie . Arătați că: ( ) ( ) Proposed by Daianu Mihaela, Popescu Delia – Romania

VI.5. Fie numărul ⏟

. Arătați că numărul se divide cu , iar

numărul se divide cu .

Proposed by Daianu Mihaela, Popescu Delia – Romania



VI.6. Prove that if ,a b Z such that 29 16a b is a multiple of 13, then 6 7a b is also a

multiple of 13. Mutually is true? Proposed by Marin Chirciu –Romania VI.7. Let be ,x y prime given numbers and the sets

, , 1A xa a B ya b C xy c c N N N . Find the set A B C .


VI.8. Let be 0, 0x y such that2 2 2x y . Prove that

1

3

xy

xy x y

.


VI.9. If ,a bZ such that 25 12a b is a multiple of 11, then 2 3a b is multiple of 11.


VI.10. Prove that 17 divides 8 181 64n , where nN .


VI.11. Prove that 13 divides 8 121 8n , where nN .

Proposed by Marin Chirciu-Romania

VI.12. Solve in integers the equation: 12 2 3 1 0x xx x Proposed by Marin Chirciu-Romania

VI.13. Solve in real numbers the system of equations:

1

1

1

xy y

yz z

zx x


VI.14. Let be *nN , n given. Solve in integers the system of equations: 3 1

3 2

xy z n

x yz n


VI.15. We consider the nonzero natural numbers n and a . Prove that the following is a

fraction in its lowest terms. 1

2

2 2 1

n

n

a

a


VI.16. If a and b are real positive numbers and 1x x a x a , compute x x .




VI.17. Prove that if and are nonzero integers such that

and , then the numbers | | | | and | | cannot be simultaneous prime numbers.

Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania

VI.18. Prove that: ⏞

⏟

⏞

⏟

Proposed by Naren Bhandari-Nepal VI.19. a) Does it exist three different natural numbers having the properties that

and ?

b) Does it exist three different natural numbers having the properties that

and ? Proposed by Dan Nedeianu – Romania

All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical

Magazine-Interactive Journal.

7-CLASS-STANDARD

VII.1. Find if ( ) and ( )

Proposed by Gheorghe Calafeteanu – Romania

VII.2. Să se rezolve ecuațiile:

1) ( ) 2) ( ) ( ) 3) 4) ( ) , ( )-


VII.3. Să se afle două numere naturale consecutive cu proprietățile: ̅̅ ̅ ̅̅ ̅ (număr

prim); ̅̅ ̅ ̅̅ ̅ ( ) Proposed by Ștefan Marica – Romania



VII.4. Să se afle și din egalitatea


VII.5. Să se afle ̅̅ ̅ știind că ̅̅ ̅


VII.6. este un paralelogram cu . Bisectoarea unghiului intersectează

dreapta în punctul și diagonalele în punctul . Să se stabilească relația Proposed by Ștefan Marica – Romania VII.7. Să se rezolve, în mulțimea numerelor naturale, ecuația:

( ) ( ) Proposed by Ștefan Marica – Romania

VII.8. Să se afle și din egalitatea:


VII.9. Dacă ( ) și ( ) atunci:

( ) ( )

√


VII.10. În triunghiul obtuzunghic ( ̂ ) luăm , iar pe luăm

. În triunghiul ducem: – mediană; – bisectoare. * +. Dreapta intersectează în . Să se arate că are loc relația


VII.11 Să se afle astfel încât ̅̅ ̅ ̅̅ ̅ ( )


VII.12. Să se afle și numere naturale din sistemul {

pentru Proposed by Ștefan Marica – Romania VII.13. 1) Să se scrie numărul ca sumă de trei pătrate perfecte cel puțin în trei moduri

diferite. 2) Să se scrie numărul ca diferență de două pătrate perfecte.


VII.14. If then:

√

√

√

Proposed by Vasile Mircea Popa-Romania



VII.15. If then:


VII.16. If

then:

( )

( )

( )


VII.17. Prove that in any triangle with the sides the following inequality is true:

√ √ √ ( ) Proposed by Daniel Sitaru, Neculai Stanciu – Romania

VII.18. Arătați că există o infinitate de numere naturale astfel încât să

fie pătrat perfect. Proposed by Mirea Mihaela Mioara, Oprea George Paul –Romania

VII.19. Fie triunghiul ascuțitunghic ABC cu H ortocentrul său și M mijlocul laturii BC.

Proiecția lui H pe AM este N astfel încât AN = k ∙MN, . Arătați că:

i)

ii)

√

Proposed by Iancu Daniela, Tuțescu Lucian –Romania

VII.20. Fie ABCD si AEFG două pătrate astfel încât ( ) ( ). Pe semidreapta

(EB se consideră punctul M astfel încât ( ) * + Arătați că dacă E, G și H sunt coliniare atunci ME = MD.

Proposed by Mirea Mihaela Mioara, Ivanescu Ionut -Romania

VII.21. If then:

Proposed by Boris Colakovic-Serbie

VII.22. Suppose | √ | , -; Show that:

i) | |

ii) | | √ Proposed by Nguyen Van Canh-Vietnam

VII.23. Solve the equation , where .

Proposed by Carmen – Victorița Chirfot – Romania

VII.24. Solve the equation:



, where . Proposed by Carmen – Victorița Chirfot – Romania



8-CLASS-STANDARD

VIII.1. Let . Prove:

∑ ( )

( )

∑

( )

Proposed by Nguyen Van Nho-Vietnam VIII.2. If then:

(( ) ( ) ( ) )

Proposed by Daniel Sitaru– Romania

VIII.3. Find if:


VIII.4. If then: | | | | | |


VIII.5. Se dă expresia ( ) ( ) ( ) unde . Să se determine astfel

încât: 1) ( ) 2) ( ) ( – număr prim)

Proposed by Ștefan Marica – Romania VIII.6. Să se afle ̅̅ ̅̅ ̅̅ ;stiind că sunt soluții ale ecuației

Din cele 30 soluții să se determine cel puțin 3. 2) Ecuația unde se înmulțesc cu și se obține o ecuație ce se notează . Se cere soluția în .




VIII.7. Let be . Prove that: ( )

( )

Proposed by Marian Ursărescu-Romania

VIII.8. If – fixed then find such that:

, - , - , - , - , - - great integer function Proposed by Seyran Ibrahimov-Azerbaijan

VIII.9. If then:

Proposed by Seyran Ibrahimov-Azerbaijan VIII.10. Solve for real numbers:

, - , - , - , - , - - great integer function Proposed by Seyran Ibrahimov-Azerbaijan

VIII.11. If √ and , then find | |. Proposed by Daniel Sitaru, Neculai Stanciu – Romania


∑√

( )

∑ √

( )

Proposed by Nguyen Van Nho-Vietnam


( )

( )

( )



( )

( )

( )



(( )( )( ))

√( )( )( ) √




VIII.16. Find such that: ( )

Proposed by Mehmet Șahin-Turkey

VIII.17. Fie astfel încât | | | | și | | Arătați că:

( ) √ √ √

Proposed by Bețiu Anicuța Patricia, Tuțescu Lucian – Romania

VIII.18. Fie și . Arătați că: ( )

.

Proposed by Beldea Daniela, Ghita Georgiana Alina –Romania

VIII.19. Let be non-negative numbers such that: . Prove that: ( )

Proposed by Iuliana Trașcă – Romania

VIII.20. Let be non-negative numbers such that . Prove that:

Proposed by Iuliana Trașcă – Romania

VIII.21. The strictly, positive, real numbers verify: .

Prove that:

In which case do we have inequality? Proposed by Ovidiu Năstase, Carmen Năstase – Romania

VIII.22. If and

, then:

Proposed by Vasile Mircea Popa – Romania

VIII.23. If then:

( )( )( )

( )( )( )

( )( )( )

( )( )( )

Proposed by Nguyen Van Canh-Vietnam

VIII.24. GENERALIZATION FOR HUNG NGUYEN VIET’S INEQUALITY

If then:

( )√ Proposed by Daniel Sitaru – Romania

VIII.25. If , then:



( )

( )

( )

√


VIII.26. Solve the equation , where . Proposed by Carmen - Victorița Chirfot – Romania

VIII.27. In triangle let be the bisector and , - the point such that .

If prove that ( ) ( ). Proposed by Dan Nedeianu – Romania

VIII.28. Find the real numbers strictly positive, knowing that:

( )( )




9-CLASS-STANDARD

IX.1. Solve for real numbers:

( ) √ ( (| | | |)) (| | | |)

Proposed by Rovsen Pirguliyev -Azerbaijan IX.2. If then:

( ∏ ( )

)(∑

+

(∑

+(∑

+(∑

+


IX.3. If .

/ then:




IX.4. If then:

( ) 4√

5 ∑(

*


IX.5. In the following relationship holds:

(

* (

* (

*

.

/

Proposed by Marian Ursărescu – Romania

IX.6. Let . Prove:

√

Proposed by Nguyen Van Nho- Vietnam

IX.7. If then:

( ) ( ) ( )



( )( )( ) √ (√ √ )


IX.9. If .

/ then:

√ ( ) ( ) ( ) √ ( )( )( )



√ √


IX.11. In let be the isogonal conjugates of altitudes, ( )

( ) ( ). Prove that:





√

√

√


IX.13. In acute the following relationship holds:

4 ( )( )

( ) ( )( )

( ) ( )( )

( ) 5




IX.15. Prove that in any the following inequality holds:

, where is the area of Proposed by Marian Ursărescu – Romania

IX.16. Prove that in any the following inequality holds:

(

* (

* (

*

.

/



( )( )

( )( )

Proposed by Bogdan Fustei-Romania


√

∑ √

√

.∑ ∑ /



√

√

√

√ (√

√

)

Proposed by Bogdan Fustei-Romania IX.20. In the following relationship holds:

(

*





( ) ( ) ∑



( )


√ ∑

( )( )

( )√



∑√

√

∑√



∑

∑

∑


∑

(

*


∑ √

√( )( )∑



∑ √ ∑√ ∑ √


√ 4

5





∑√

∑


∑

(

* √


IX.32. If

then:

Proposed by Seyran Ibrahimov-Azerbaijan

IX.33. Let and √ √ √ . Prove:

∑ √

( )

√ ∑ √

( )

Proposed by Nguyen Van Nho-Vietnam IX.34. Let * +. Prove:

∑( )( )

( )( ) ( )


IX.35. Let . Solve in :

√

√

√

Proposed by Nguyen Van Nho-Vietnam IX.36. Let . Prove:

(

*



Proposed by Rovsen Pirguliyev-Azerbaijan

IX.38. Solve for real numbers:

[

] [

] [

] , -





( )

√

Proposed by Mehmet Șahin-Turkey IX.40. In the following relationship holds:

Proposed by Mehmet Șahin-Turkey IX.41. In the following relationship holds:

( ) ( )

( )


IX.42. Determinați funcțiile cu proprietatea că:

( ( )) | | ( ) ( ).

Proposed by Mirea Mihaela Mioara, Ivanescu Ionut – Romania

IX.43. If in – incenter, – circumradii of then:

( )

Proposed by Mustafa Tarek –Egypt

IX.44. In right angled the following relationship holds:

( √ )

Proposed by Mustafa Tarek -Egypt


( ) ( ) ( ) √


IX.46. In – incenter, – contact triangle, – circumradii in

( ) ( ) ( ). Prove that:

( )( ) ( )( ) ( )( ) Proposed by Mustafa Tarek –Egypt

IX.47. In acute – orthocenter, – incentre, the following relationship holds:

√

√

√

√ ( )




IX.48. a) Let be . Prove that ( ) (

)

b) Let be . Prove that ( ) (

). Generalization. Proposed Ramona Nălbaru , Alecu Orlando – Romania

IX.49. Let be an acute triangle. Prove that:

√ (

* (

* (

*

Proposed by Vasile Mircea Popa – Romania IX.50. If then in the following relationship holds:

(

*

(

*

(

*

√

Proposed by D.M. Bătinețu – Giurgiu – Romania, Martin Lukarevski – Macedonia

IX.51. If then in the following relationship holds:

( )

( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania


√

Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți – Romania


√

Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți – Romania


( ) ( ) ( ) √ Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți – Romania




( )

( )

Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania


(

*

(

*

(

*

( )




IX.58. Find .

1 such that:

( )( )

( )( )( )

Proposed by Daniel Sitaru – Romania IX.59. In the following relationship holds:

( )


IX.60.In the following relationship holds:

∑√

√

∑



∑ ( ) ( )



( ) (and the analogs) Proposed by Bogdan Fustei – Romania


√

∑√

∑√



√

√

√

Proposed by Bogdan Fustei – Romania IX.65. In the following relationship holds:

(√ √ √ ) (

*



∑( )( )

Proposed by Bogdan Fustei – Romania IX.67. In the following relationship holds:



√

√

√

√


IX.68. Solve the equation: , where . Proposed by Carmen – Victorița Chirfot – Romania

IX.69. Solve the equation:

0

1

0

1, where , - is great integer function.




10-CLASS-STANDARD

X.1. If

then:

∑ ( )

( )

∑ ( )

( )


X.2. In the following relationship holds:

(

*



(

)

(

)

(

)

Proposed by Rovsen Pirguliyev – Azerbaijan

X.4. Solve for real numbers:



,.

/

.

/

(

*

√ ( )


X.5. If in – Brocard angle then:

( ) ( ) ( )

( )

√


X.6. If

then:

.√ √

/ .√ √

/ .√ √

/


X.7. If – fixed, then solve for real numbers:


X.8. Solve for real numbers:


X.9. In – excenters the following relationship holds:

( ) Proposed by Marian Ursărescu – Romania

X.10. Find the minimum of the function:

, ) ( )



∑

(

*√



.∑ ∑ /√ .∑ ∑ /√

Proposed by Bogdan Fustei – Romania X.13 In the following relationship holds:

√





√

Proposed by Bogdan Fustei-Romania X.15. In the following relationship holds:

√

∑√ ∑ √


∑ √

√

∑


∑



( )


∑

∑


(

*



∑

∑



∑(

*

(

*





∑ ∑

( )


∑ ( )

Proposed by Bogdan Fustei-Romania X.26. Prove that in any triangle:

∑ ( )


X.27. Prove that in any triangle the following inequality is true:

√

√

√

(

*

where Proposed by Marin Chirciu – Romania X.28. Prove that in any triangle the following inequality is true:

√

√

√

√

4

5

where Proposed by Marin Chirciu – Romania X.29. Prove that in any triangle the following triangle is true:

√

√

√

√

(

*

where Proposed by Marin Chirciu – Romania X.30. Prove that in any triangle the following inequality holds:

√

√

√

√

(

*

where Proposed by Marin Chirciu – Romania X.31. Prove that in any triangle the following inequality holds:

√(

*

√

√

√

where . Proposed by Marin Chirciu – Romania X.32. Prove that in any triangle the following inequality holds:



∑


X.33. Prove that in any triangle the following inequality holds:

∑



∑



∑



∑



∑



( ) ∑



( ) ∑


X.40. Solve for natural numbers: ( ) ( ) ( )




X.41. Solve for natural numbers:

a. ( ) ( ) ( ) b. c.

Proposed by Seyran Ibrahimov-Azerbaijan X.42. Solve in

√

√


X.43. Solve in ( ) ( )

. Proposed by Nguyen Van Nho-Vietnam


√ (

*


X.45. In rightangled the following relationship holds:

√


X.46. In acute – circumcentre, -circumradii. Circle with radii is

simultaneous tangent to and ( ). Analogous we construct circles – with radii . Prove that:

√

√

√

√



√

√

√

√

Proposed by Mehmet Sahin-Turkey

X.48. In the following relationship holds: ( ) ( )

( ) ( )

( )

( ) ( )

( )


X.49. Arătați că dacă numărul este număr prim ( ), atunci nu se poate scrie

sub forma , unde p este un număr prim( ), iar .

Proposed by Mirea Mihaela Mioara, Oprea Paul George - Romania



X.50. Rezolvați în R ecuația: √

Proposed by Bețiu Anicuța Patricia, Grigore Dan –Romania

X.51. In – Gergonne’s point, – Nagel’s point; – Gergonne’s

cevians, – Nagel’s cevians. Prove that:

Proposed by Mustafa Tarek -Egypt

X.52. In acute – orthocentre, – incentre the following relationship holds:

√

√

√

√ ( )


X.53. In – incentre, the following relationship holds:

( ) ( )


X.54. – cyclic orthodiagonal quadrilater, – sides,

– bimedians. If * + – circumradii then: ( )( )

( )( )( )( )


X.55. If in – incenter then:

( ) √ Proposed by Mustafa Tarek -Egypt

X.56. In acute the following relationship holds:

√ ∑

√

Proposed by Mustafa Tarek -Egypt X.57. In the following relationship holds:

Proposed by Mustafa Tarek –Egypt X.58. In the following relationship holds:

( )

( )

( )


X.59. In – excenters. In – altitudes,

– medians. Prove that:



(

) √( )


X.60. If then in the following relationship holds:

( )


X.61 If then in the following relationship holds:

∑ ( )

√



√



√



∑(

√ *

√ ∑(

*


X.65. In – incentre, – Lemoine’s point. Prove that:

( )( )( ) ( )

( )

Proposed by Adil Abdullayev-Azerbaijan

X.66. ∑

( )

∑

.

/

. Find in terms of .

Proposed by Madan Mastemind-India

X.67. Let be the lengths of the medians of a triangle with circumradius

and inradius . Let be the lengths of the sides of the triangle . Prove that

Proposed by George Apostolopoulos-Greece

X.68. Prove that in any triangle:



.

( ) /; – orthocenter

– Lemoine’s point Proposed by Adil Abdullayev-Azerbaijan X.69. Let be positive real numbers. Prove the inequality:

( )( ) .

/

Proposed by Boris Colakovic-Serbie X.70. In the following relationship holds:

√

√

√

√

(

*

Proposed by Bogdan Fustei – Romania X.71. In the following relationship holds:

√ ∑√

√

∑


√

∑√

∑√



∑( ) ( )



∑√

√


∑

√

( )

√


∑

√ (

*√





∑

√ (

*√


(√ √ √ ) (

*


∑ ( )


∑( )

( )


X.81. In any with the usual notations the following relationship holds:

( ) √


X.82. Let equilateral of side ( ) and are the altitudes from

to . Show that:

a) ∑ √ ( )

b) ∑ ( )

Proposed by Mihalcea Andrei Stefan-Romania

X.83. In acute with sides different in pairs, – altitudes,

– medians, – symedians. Prove that:



√






11-CLASS-STANDARD

XI.1. If then:

( ) ( )

Proposed by Rovsen Pirguliyev -Azerbaijan XI.2. Find:

4√

√

( √

√

)5

Proposed by Daniel Sitaru – Romania XI.3. Find:

(

∑(.

/ (

*+

+

Proposed by Daniel Sitaru – Romania XI.4

( ) (∑

( )

+(∑

( )

+(∑

( )

+

( ) (∑

+(∑

+(∑

+

Find: . ( )

( )/

Proposed by Daniel Sitaru – Romania XI.5. Prove that:

( ) ( ). Proposed by Dan Nedeianu – Romania

XI.6. Solve the equation , where

. Proposed by Carmen - Victorița Chirfot – Romania

XI.7. ( ) ∑ 0

1

, - - great integer function. Find:

( ( ) ∑

( )

+




XI.8. ( )

Find:

( )


XI.9. If ( ) . Find:

( )


XI.10. .

/ . Find:

(

√ *


XI.11.

√

. Find:

( )

Proposed by Marian Ursărescu – Romania XI.12. Find:

(

((∑

+

4

5

∑

,

)


XI.13. If ;

then find

4(

)(

)

( ) 5


XI.14. Let be the sequence: and

√

Find:


XI.15. Find the continuous functions: having the property:

( ) ( ) ( ) ( ) Proposed by Marian Ursărescu – Romania



XI.16. Find:

(

( )

+


XI.17. Solve the equation: ( ) (

+, where ( ) and


XI.18. Let be ( ) such that ( ) . Prove that:

. Proposed by Marian Ursărescu – Romania

XI.19. Let , put: ( )

( )

( )

( )

( ) . Find: .


XI.20. If .

/ then:


XI.21. Solve for real numbers:

|

|

√( ) .

/


XI.22. If .

/ then:

√

√


XI.23. If ( ) ( ) then in the following relationship holds:

√



Proposed by D.M. Bătinețu – Giurgiu – Romania





XI.26. √

. Find:

( √

)

Proposed by Marian Ursărescu – Romania XI.27. Solve for real numbers:

(

)( )(

)( ) (

)

.

/ .

/ .

/ .

/ (

*

Proposed by Jhoaw Carlos-Bolivia

XI.28. Let be given positive integers with and nonnegative real

numbers. Prove

∑( )

[∑.

/

]

.

/

(

*

(∑

+

Proposed by Le Khanh Sy-Vietnam

XI.29 Let , - , - – continuous. Prove that exist , -

such that: ( ) ( ) ( ) ( ) Proposed by Nguyen Van Canh-Vietnam

XI.30. Let be positive real numbers such that: . Prove that:

√ ( )

√ ( )

√ ( )

( )

Proposed by Hoang Le Nhat Tung – Vietnam XI.31. Let be positive real numbers such that . Find the minimum value

of:

√

√

√

√

√ √

Proposed by Hoang Le Nhat Tung – Vietnam XI.32. In the following relationship holds:

∑ √

(∑

) √ ∑

∑

Proposed by Mihalcea Andrei Stefan-Romania XI.33. Solve the equation:

( ) √

( *| | | | + *| | | | +)




XI.34. If

then:


XI.35. ||

|| |

| |

|. If

then:

| | | |

( )

Proposed by Daniel Sitaru – Romania XI.36. If then:

( ) ( ) ( ) ( ) Proposed by Daniel Sitaru – Romania

XI.37. If then:

(( ) .

/

* ( .

/

*

(( ) √ .

/

*

( )

( )


XI.38. Find:

40 √ ( )( )

1

[√ ]5 , - - great integer function


XI.39. ( )

( )( ) ( ) ( )( ) th derivative. Find:

4 ( )( )

5


√ ∑

. /

∑. /


XI.41. Find:

( ( ) ∑ (

[√ ] 0

√ √ √

1*

* , - function




XI.42. Find:

(

∑(.

/ (

*+

+


(∑( ( )

.

/+

+


XI.44. If then:

.

/ .

/ .

/

.

√ / .

√ / .

√ /

(

( )( )( )*


XI.45. In the following relationship holds:

√

√

√

√

√

√


XI.46. If .

/

√

then:

( ) ( )


XI.47. If 0

/ then: ( ) ( ) ( ) ( )


XI.48 Find:

( 4(

*

(

*

5)

Proposed by Daniel Sitaru – Romania XI.49. If then:

4

5

4

5

4

5




XI.50. GENERALIZATION OF MARIAN URSĂRESCU’s SEQUENCE

( ( )( ))

Find in terms of . ( )/

Proposed by Daniel Sitaru – Romania XI.51. Solve the equation:

, where Proposed by Carmen - Victorița Chirfot – Romania

XI.52. Solve the equation , where .

Proposed by Carmen - Victorița Chirfot – Romania



12-CLASS-STANDARD

XII.1. Find:

( ∫(√

√ )

)

Proposed by Abdul Mukhtar-Nigeria

XII.2. If

then:

(∫ (√ )

√

)

(

∫ (√ )

)

(∫ (√ )

√

)

(

∫ (√ )

)

Proposed by Daniel Sitaru – Romania XII.3. Find:

∫4( ) ( )

5




XII.4. Find:

∫

.

/


XII.5. Find:

∫( ( )) .

/


XII.6. Find: ∫ ( (* +) ( (* +) ) )

(* + is the fraction of the real number ) Proposed by Nguyen Van Nho-Vietnam

XII.7. If , - ( ) ( ) continuous on , - and ( ) ( ) then:

√ ∫√ ( ( ))


XII.8. Evaluate:

∫ 4 4

55

Proposed by Artan Ajredini-Serbie

XII.9. Find:

∫

(|

|*

Proposed by Artan Ajredini-Serbie

XII.10. .

/ . Prove that exists ( ) such that:

(∫

)(∫

) ( )

Proposed by Daniel Sitaru – Romania XII.11.

∫√


XII.12. . / .

/ .

/ ( ).

/ . Find:



√

Proposed by Daniel Sitaru– Romania XII.13. Find:

(

∫

)

Proposed by Daniel Sitaru– Romania XII.14. Find:

∏4

5


XII.15. Find:

√

∑

( ) ( )


XII.16. If ∫

Find the limit:

[

( )( )]

Proposed by Dimitris Kastriotis-Athens-Greece

XII.17. If then:

∫( )

4

5


XII.18.

(∑( .

/ .

/

.

/ .

/

)

∫ √( )( )

)

Proposed by Daniel Sitaru – Romania XII.19. Find:

∫( ) .

/






UNDERGRADUATE PROBLEMS

U.1. Find:

∫ ( ) ( ) ( ) ( )

( ) ( )

Proposed by Nader Al Homsi-Jordan

U.2. A simple & Unique Arithmetical Identity involving Triangular Number. Let ( ) ( )

then prove that:

∑ ( )

. ( ( ))/

(

* ∑

( )

. ( ( ))/

(

*

Proposed by K. Srinivasa Raghava – AIRMC – India U.3.

∫ ( )

Proposed by Abdul Mukhtar-Nigeria U.4. Find:

( )

(∫ 4 ( )

5

)

Proposed by Abdul Mukhtar-Nigeria U.5.

( ) ∫( ( ))

Find: . ( )

/

Proposed by Sagar Kumar – India



U.6.

( ) ∫ (( ) .

/

( ))

Prove that: ∫ .

/

Catalan’s constant

Proposed by K. Srinivasa Raghava – AIRMC – India U.7. Let,

∫ ( ) ( )

( ( ) ( ) )

then prove that

∫ ( )

( )

Proposed by K. Srinivasa Raghava – AIRMC – India

U.8. For a sufficiently large positive integer ‘ ’, we have:

∫ ( ) ( )

.

/ ∫

( ) .

/

Proposed by K. Srinivasa Raghava – AIRMC – India U.9. If

∫(√ ) √( )

then prove that:

( ) ( )

( ) ( ), where, √


U.10. Let,

( ) ∫ ( ) ( )

( ( ) ( ) )

then prove that

∫ ( )

( )


U.11. If then:

( ) ( ) √




U.12. If

∑

( ) ( )

( )

( )∑

( )

( ) ( )

then show that √ Proposed by K. Srinivasa Raghava – AIRMC – India

U.13. Prove this sharp inequality:

∫

( )

√


( ) ∏

( )

then prove that,

∫

( )

( )

( )

(

.

√ /

)


U.15. Let, ( ) ∑ ∑

( )

( )

and if, ∑ ( )

then prove that

4

√

5

, for any integer .

Proposed by K. Srinivasa Raghava – AIRMC – India U.16. Let for any

( ) ∑ ( )

.

/

( )

then prove that

∫( ( ) ( ))

( )

( √ )

√ ( √ )

Proposed by K. Srinivasa Raghava – AIRMC – India U.17. Show that

∑(

( )( )

( )( )( )

( )( )( )( )*

( ) ( )

( )

( √ )


U.18. Let,

( ) ∫ ( )

( )



then show that

∑ ( )

( )

( )

(√ ) .

/

√ .

/


U.19. Establish these sharp inequalities:

√

∑ .

/

( )

( )

(

√ * ∑ .

/

( )

( )


U.20. If then: ( )

( )

( )

( )

( )

( )

Proposed by Daniel Sitaru – Romania U.21. Find:

(∫ (√ ( ) )

)

∑.

/(

*

( )

∑

( ) .

/(

*

( )

√ (√ )

Proposed by Ekpo Samuel-Nigeria

U.22. If

∫

√

√

then prove that Proposed by K. Srinivasa Raghava – AIRMC – India

U.23. An intriguing logarithmic sequence proved by Arafat Rahman Akib

( ( ( ( ( ( ( ))))))) then we have

∫

where is Euler’s Constant Proposed by K. Srinivasa Raghava – AIRMC – India



U.24. If we define, ( ) ,

then show that:

∫ ∫ ( )


∫( ( ) ( ))

( )

( ) ∫( ( ) ( ))

( )

then show that:

(∫ ( ( )

( )+

,

.

/


( ) ∫

( ) ( )

then show that

∫ ( )

( )

( )

where is Catalan’s constant. Proposed by K. Srinivasa Raghava – AIRMC – India

U.27. If

∫(∫ ( )

)

.

/ .

/

then prove that: ( ) , where ( ) is Poly – Gamma function.

Proposed by K. Srinivasa Raghava – AIRMC – India U.28.

∫

( )

( )

.

/

where √√ √ ( √ )


U.29. Let ( ) ∫

√√( )

√

√

, then prove that:



( )

( )


( )

.∫ ( )

( )

/, then show that (√ ( )

*


U.31. For , let

( ) ∑

∑

∫ ( )

then prove that:

∑ ( )( )

( )

where is catalan constant and ( ), is first derivatives at and ( ) is always fixed at .

Proposed by Naren Bhandari-Nepal U.32. If

( ) ∫∫

∫∫∫

∫∫ ∫

⏟

then prove that

∑( ( ) )

Proposed by Naren Bhandari-Nepal

U.33. Let . Find:

∑( )

( )

* ∫ (∑.

/

( )

+

+

Proposed by Naren Bhandari-Nepal

U.34. Let ( ) .

√ /, then prove that:

∑ ∑| |( )

∑( )

( )

(

( )+

where ( )

( ) is nth derivatives

Proposed by Naren Bhandari-Nepal U.35. Find:



(

(∑(

( )( ) ( )*

+

)

Proposed by Naren Bhandari-Nepal U.36. Find:

( ) ∫

( )( )


U.37. Find:

∫ ( )

Proposed by Vasile Mircea Popa – Romania U.38. Find:

( ∫

)


U.39. Prove that:

√

(

√ *

Proposed by Vasile Mircea Popa – Romania U.40. Find:

( ) ∫

( )( )


U.41. Prove that for any acute triangle the following inequality holds:

(

*


U.42.

( ) ∫

( )

then ( ) ( ) ( ), where √




U.43. Prove that:

∫ (∫(

)

( ) )

√√ √

√


U.44. Prove that:

.

/ .

/ .

/

( ) trigamma function


U.45.Prove that:

.

/ (

*

( )(√ )

√

[√ ( )

( ) ]

Note: is the golden ration and . Proposed by Naren Bhandari-Nepal

All solutions for proposed problems can be finded on the

http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.

ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-SPRING 2019

PROBLEMS FOR JUNIORS

JP.166. If , ) and then:

( )

∑

Proposed by Nguyen Van Nho– Vietnam

JP.167. Let be a tetrahedron with and let be any point inside the triangle . Denote respectively by the distances from to faces ( ) ( ) ( ). Prove that:



(a)

.

(b)

(c)

Proposed by Nguyen Viet Hung – Vietnam

JP.168. Let be positive real numbers such that:

√

√

√

Prove that: . Proposed by Nguyen Viet Hung – Vietnam

JP.169. Let be positive real numbers such that: . Prove that:

√ ( )

√ ( )

√ ( )

Proposed by Hoang Le Nhat Tung – Vietnam

JP.170. Let be positive real numbers such that: . Find the minimum value of:

√ ( )

√ ( )

√ ( )


JP.171. Let be an acute triangle with perimeter . Prove that:


JP.172. Let be positive real numbers such that: . Prove the inequality:

√

√

√ √

( )


JP.173. Prove that in any triangle :

√

Proposed by Nguyen Viet Hung – Vietnam JP.174. Prove that in any triangle :

√ ( )


JP.175. Prove that in any acute triangle :



Proposed by Nguyen Viet Hung – Vietnam JP.176. If , then:

( )

√

.

/

Proposed by Rovsen Pirguliyev - Azebaijan

JP.177. If then:

( ) ∑√

(√ √ √ )


JP.178. If then: (√ )

√ √


JP.179. In acute the following relationship holds:


JP.180. If then: 8 √ ( √ )

√ ( )( √ )


PROBLEMS FOR SENIORS

SP.166. Let and . Find:

∫ (∏( )

+ ( { | })


SP.167. Let be positive real numbers such that: . Prove that:

√ ( )

√ ( )

√ ( )

( )


SP.168. Let be positive real numbers. Find the minimum possible value of:



√


SP.169. Prove that for all non-negative real numbers :

√

√

√


SP.170. Let be positive real numbers such that . Prove that:

√ √

√ √

√ √

√ √


SP.171. Let be positive real numbers such that: . Find the minimum value of:

√ ( )

√ ( )

√ ( )


SP.172. Prove that for any real numbers :

( )( )( )( ) Proposed by Nguyen Viet Hung – Vietnam

SP.173. Prove that for any positive real numbers :

√ √ √

√


SP.174. Prove that for any positive real numbers ( )( )( ) ( )




SP.176. Prove that if , ) ( ), then in any triangle , with the usual notations holds:

∑(

)

( )

( )

( )





∑(

)

(

)

( )

( )



∑(

)

(

)

( )

( )

Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania SP.179. If , ) then:

Proposed by Seyran Ibrahimov – Azerbaidian SP.180. If then:

√ √ √ √ √

Proposed by Seyran Ibrahimov - Azerbaidian


UP.166. Solve the equation in :

√ √

√


UP.167. Let be positive real numbers such that: . Find the maximum value of:

√

√

√


UP.168. Let be and ( ) ( ) ( )

( ) . Find:

√|

∑ ( )( )

|


UP.169. Let be the sequence and

√ .

Find:




UP.170. Find:

∫ ( )


UP.171. Find that in any acute-angled the following inequality holds:

(

*

(

*


UP.172. Let be ( ), invertible such that: ( ) . Prove that:


UP.173. Find:

√ ∑

. / ∑.

/


UP.174. If , - , ) integrable then:

∫∫∫ ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) (∫

( )

)


UP.175. In acute the following relationship holds:


UP.176. Let be positive real numbers such that: . Find the minimum value of:

√

.


UP.177. If then: ( )( )( )( )




UP.178. Let be .

/ .

/. Find: ( ) ( – exponential matrix)


UP.179. In the following relationship holds:

√

√

√

√

√

√


UP.180. If ( ) ( ) such that exists:

( )

( ) and exists

( ( ))

then find:

(( ( ))

(( ( ))

( ) )

( ( ))

)

Proposed by D.M. Bătinețu-Giurgiu; Neculai Stanciu – Romania



ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-SUMMER 2019

PROBLEMS FOR JUNIORS

JP.181. Let be positive real numbers such that . Find the minimum value of:

√

√

√


JP.182. Let be positive real numbers such that . Prove that:



( )( )( ) √ ( ).

Equality occurs if and only if? Proposed by Hoang Le Nhat Tung – Vietnam

JP.183. In the following relationship holds:

(

*

∑ (

*



∑

(

*

Proposed by Marin Chirciu – Romania JP.185. In the following relationship holds:

∑ (

*


JP.186. Solve for real numbers: { √ ( √ √ )

√ √ √


JP.187. There is a positive integer of 2018’s digits such that the sequence:

( ( )) ( ( )) . ( ( ))/ is an increasing arithmetic progression formed by prime

numbers? Obs.: ( ) denotes sum of the digits of .

Proposed by Pedro H.O. Pantoja – Brazil

JP.188. Let be positive real numbers such that: . Find then minimum value of:

( )

( )

( )

√ √ √


JP.189. Prove that:

√ √



( )

( )

( )




JP.191. Solve for real numbers:

{ √ ( )

√

√ √

Proposed by Hoang Le Nhat Tung – Vietnam JP.192. If then:

4

5 4

5 4

5




JP.194. In internal bisectors; ( ) ( ) – circumcenter. Prove that: collinears


JP.195. If then in the following relationship holds:

( )

( )

( )

Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania

PROBLEMS FOR SENIORS

SP.181. If then:

4

( )( )5 4

( )( )5 4

( )( )5


SP.182. If ( ) ( ) and ( ) ( ) ( ) , then:

∫ ( )

( )

∫ ( )

( )

Proposed by Shivam Sharma – India

SP.183. If then:



∫ ( )

∫ ( ) √

√


SP.184. Let be positive real numbers such that: . Prove that:

√ √ √ ( ). Find the minimum value of:

√ √ √




SP.186. Let and be the lengths of the medians of an acute triangle with inradius and circumradius . Prove that:

√

√

Proposed by George Apostolopoulos – Greece

SP. 187. Let be the lengths of sides in a triangle such that . Find the minimum value of:


SP. 188. In are exradii. Prove that:


SP. 189. Let be ( ) .

/

. Find:


SP. 190. Let be

– fixed. Find:

√




SP.191. Let be , - – continuous and ∫ ( )

. Prove that exists ( )

such that:

( ) ∫ ( )

( ) ∫ ( )


SP.192. Let be ( ) ( ) . Prove that: ( )


SP.193. If ( ) ( ) then: ( ) ( ) ( )

( ) Proposed by Daniel Sitaru – Romania

SP.194. Find the continuous functions ( ) having the property: ( ) ( ) ( ) ( ) – fixed.


SP.195. Find:

√

( )( ) ( )

( )



UP.181. If

then:

∫4

5

Proposed by Daniel Sitaru – Romania UP.182.

∫ ( )

{

}

where * + denotes the Fractional Part. Proposed by Shivam Sharma – India

UP.183. Let be three sequences of real numbers such that:



Find:

(

)(

)(

)

( )


UP.184. If

then:

∫(( ) ( ) )

( )

Proposed by Daniel Sitaru – Romania UP.185. Calculate the integral:

∫


UP.186. If ; then find:

( (

* *


UP.187. Find:

. √( )

/

(√ )

(. √( )

/

.√( )

/

*


UP.188. If – fixed then find in terms of :

. √( )

/

.√( )

/

.. √( )

/

(√ )

/

Proposed by D.M. Bătinețu-Giurgiu; Neculai Stanciu – Romania UP.189. Find:

. √( )

/

( √

)

. √( )

/ – fixed


UP.190. If

then find:

4( )

√

√

5




UP.191. Let be:

∑

0 √( )

1

, - great integer function. Find:

( ∑

+

Proposed by Daniel Sitaru – Romania UP.192. Find:

(∑(

∑ 2

3

)

( ))

* + , - , - - great integer function. Proposed by Daniel Sitaru – Romania

UP.193.

( ) ∏4

5

Find:

( ( ) (

( )*

( )

( ) (

( )*+


UP.194. Let be two real numbers with . Calculate the next limit:

∫ √( ) (

*

Proposed by Vasile Mircea Popa – Romania UP.195. Calculate the integral:

∫ ( )

It is required to express the integral value with the usual mathematical constants, without using values of special functions.






INDEX OF AUTHORS RMM-23

No. Name and surname No. Name and surname

1 Dan Sitaru - Romania 32 Nguyen Van Nho - Vietnam

2 D.M. Bătinețu – Giurgiu- Romania 33 Seyran Ibrahimov - Azerbaijan

3 Neculai Stanciu - Romania 34 Bețiu Anicuța Patricia- Romania

4 Claudia Nănuți- Romania 35 Mehmet Șahin - Turkey

5 Ștefan Marica - Romania 36 Babis Stergioiu - Greece

6 Bencze Mihály- Romania 37 Carmen Năstase - Romania

7 Kovács Béla - Romania 38 Rovsen Pirguliyev - Azerbaijan

8 Angela Nițoiu- Romania 39 Mustafa Tarek - Egypt

9 Marin Chirciu- Romania 40 Ramona Nălbaru - Romania

10 Marian Ursărescu - Romania 41 Alecu Orlando-Romania

11 Ovidiu Năstase- Romania 42 Martin Lukarevski-Macedonia

12 Bogdan Fustei - Romania 43 Naren Bhandari - Nepal

13 Iuliana Trașcă - Romania 44 Oprea Paul George - Romania

14 Daianu Mihaela- Romania 45 Boris Colakovic - Serbie

15 Popescu Delia- Romania 46 Adil Abdullayev-Azerbaijan

16 Beldea Daniela- Romania 47 Madan Mastemind - India

17 Ghita Georgiana Alina - Romania 48 George Apostolopoulos - Greece

18 Viespescu Carina Maria - Romania 49 Jhoaw Carlos-Bolivia

19 Titu Zvonaru- Romania 50 Le Khanh Sy - Vietnam

20 Dan Nedeianu - Romania 51 Hoang Le Nhat Tung - Vietnam

21 Gheorghe Calafeteanu - Romania 52 Nguyen Van Canh - Vietnam

22 Dan Nănuți - Romania 53 Abdul Mukhtar - Nigeria

23 Vasile Mircea Popa- Romania 54 Artan Ajredini - Serbie

24 Mirea Mihaela Mioara- Romania 55 Dimitris Kastriotis-Greece

25 Oprea George Paul - Romania 56 Nader Al Homsi - Jordan

26 Iancu Daniela- Romania 57 K. Srinivasa Raghava - India

27 Tuțescu Lucian - Romania 58 Ekpo Samuel - Nigeria

28 Ivanescu Ionut - Romania 59 Nguyen Viet Hung - Vietnam

29 Grigore Dan - Romania 60 Pedro H.O. Pantoja - Brazil

30 Mihalcea Andrei Stefan- Romania 61 Shivam Sharma - India

31 Carmen – Victorița Chirfot- Romania 62 Sagar Kumar – India

NOTĂ: Pentru a publica probleme propuse, articole și note matematice în RMM puteți trimite materialele pe mailul: [email protected] All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.

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Documents

R. M. M. - 23...Romanian Mathematical Society-Mehedinți Branch 2019 4 ROMANIAN MATHEMATICAL MAGAZINE NR. 23 GEOMETRIC INEQUALITIES VIA ONE ALGEBRAIC INEQUALITY By D.M. Bătinețu