R. M. M. - 28 · 2020. 3. 1. · claudia nĂnuȚi-romania neculai stanciu-romania floricĂ...

ROMANIAN MATHEMATICAL SOCIETYMehedinți Branch

SPRING EDITION 2021

R. M. M. - 28 ROMANIAN MATHEMATICAL

MAGAZINE

ISSN 2501-0099

Romanian Mathematical Society-Mehedinți Branch 2021

1 ROMANIAN MATHEMATICAL MAGAZINE NR. 28

ROMANIAN MATHEMATICAL

SOCIETY

Mehedinți Branch

ROMANIAN MATHEMATICAL MAGAZINE

R.M.M.

Nr.28-SPRING EDITION 2021



ROMANIAN MATHEMATICAL

SOCIETY

Mehedinți Branch

DANIEL SITARU-ROMANIA EDITOR IN CHIEF

ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT ISSN 1584-4897

GHEORGHE CĂINICEANU-ROMANIA

EDITORIAL BOARD

D.M.BĂTINEȚU-GIURGIU-ROMANIA

CLAUDIA NĂNUȚI-ROMANIA

NECULAI STANCIU-ROMANIA

FLORICĂ ANASTASE-ROMANIA

DAN NĂNUȚI-ROMANIA

IULIANA TRAȘCĂ-ROMANIA

EMILIA RĂDUCAN-ROMANIA

DRAGA TĂTUCU MARIANA-ROMANIA

DANA PAPONIU-ROMANIA

GIMOIU IULIANA-ROMANIA

DAN NEDEIANU-ROMANIA

OVIDIU TICUȘI-ROMANIA

DANIEL STRETCU-ROMANIA

MARIA UNGUREANU-ROMANIA

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

DANIEL WISNIEWSKI-USA

EDITORIAL BOARD VALMIR KRASNICI-KOSOVO



CONTENT

ABOUT BĂTINEȚU’S INEQUALITIES, PROBLEMS X.64, X.65, X.74-RMM 24-SPRING EDITION

2020-PAPER VARIANT – Marin Chirciu............................................................................................4

A SIMPLE PROOF FOR PADOA’S INEQUALITY– Florentin Vișescu.................…………...............10

PROBLEM 1131 TRIANGLE INEQUALITY RMM 2019 – Florentin Vișescu......…………...............11

THE EXTREMES OF THE SUMS OF ARCTANGENTS WITH RESTRICTIONS FOR ARGUMENTS

- Florentin Vișescu ………………………………………………………………………….…………………...……........13

TRIGONOMETRICALIZABLE QUINTIC EQUATION –Benny Lê Văn.................................………22 ABOUT NAGEL’S AND GERGONNE’S CEVIANS (II) - Bogdan Fuștei....……….………...…………….26

RMM 2019, NUMBER 18 – AUTUMN 2020,PROBLEM SP.263 - Marin Chirciu.....................38

SOME APPLICATIONS OF CHEBYSHEV INEQUALITY IN GEOMETRY - Florică Anastase.........40

STRUCTURI ALGEBRICE (V) - Vasile Buruiană ……………….………………………..…….…………………44

PROPOSED PROBLEMS……………………………………..……………………………………………………………...46

INDEX OF PROPOSERS AND SOLVERS RMM-25 PAPER MAGAZINE.……………………………………104



ABOUT BĂTINEȚU’S INEQUALITIES

PROBLEMS X.64, X.65, X.74-RMM 24-SPRING EDITION 2020-PAPER VARIANT

By Marin Chirciu – Romania

1) X.64. BĂTINEȚU’S INEQUALITY – 1

If then in the following relationship holds:

√

Proposed by D.M. Bătinețu – Giurgiu – Romania

Proof: We prove: Lemma:

2) If then in

∑

Proof:

∑

∑(

* ∑

∑

( )∑

∑

( )(∑ )

∑

.∑ /

∑ ∑ ∑ ∑ ∑

Let’s get back to the main problem.Using the Lemma, it suffices to prove that:

∑ √ ∑ √ .∑ /

( ) ( ) ( )

We’ve used the known identity in triangle ∑ . We distinguish the cases:

Case 1). If ( ) , the inequality is obvious.

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

( ) ( ), which follows from Blundon-Gerretsen’s inequality

( )

( ). It remains to prove that:

( ) ( )

( )( )

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

Equality holds if and only if the triangle is equilateral.



3) X.65. BĂTINEȚU INEQUALITY – 2

If then in the following relationship holds:

Proposed by D.M. Bătinețu – Giurgiu – Romania

Solution: We prove:

Lemma:

4) If then in :

∑

Proof:

∑

∑(

* ∑

∑

( )∑

∑

( )(∑ )

∑

.∑ /

∑ ∑ ∑ ∑ ∑

Let’s get back to the main problem: Using the Lemma, it suffices to prove that:

∑ ∑ ( ) ( )

( ) ( )

We’ve used the known identity in triangle ∑ ( ) ( )

We distinguish the cases:



( ) ( ), which follows from Blundon-Gerretsen’s inequality

( )


( ) ( )

( )( )



5) X.74. If then in the following relationship holds:



4

5 ( )

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

Solution:We prove: Lemma:

6) In the following relationship holds:

F. Goldner, 1949

Proof.

Using ∑ , ( ) ( ) -, we have

, ( ) ( ) -

( ) ( ) .




( ) ( ), which follows from Blundon-Gerretsen inequality

( )


( ) ( )

( )( )



Let’s get back to the main problem:

We have

, which follows from

, with equality for

and the analogs.

Using the Lemma and the above inequality we obtain the conclusion.

Equality holds if and only if the triangle is equilateral and .

Remark: In the same way we can propose:

7) If then in the following relationship holds:

( )

( )

( ) √

Proposed by Marin Chirciu – Romania

Solution: We prove: Lemma:



8) If then in :

( )

( )

( ) ∑

Proof:

∑

( ) ∑(

* ( )

∑

( ) ∑( ) ( )∑

( )

∑( )

( )( ∑ )

∑( ) .∑ /

∑( )

∑ ∑ ∑ ∑ ∑ ∑ ∑


∑ √ √

( )

( ) ( )

We’ve used the known identity in triangle ∑ . We distinguish the cases:



( ) ( ), which follows from Blundon-Gerretsen inequality

( )


( ) ( )

( )( ( ))

( )( ) , obvious from Euler’s inequality

. Equality holds if and only if the triangle is equilateral.


( )

( )

( )



10) If then in :

( )

( )

( ) ∑( ) ( )



Proof.

∑

( ) ∑(

* ( )

∑

( ) ∑( ) ( )∑

( )

∑( )

( )(∑( ) )

∑( ) .∑( ) /

∑( )

∑( ) ∑( ) ( ) ∑( ) ∑( ) ( )


∑( ) ( ) ∑( ) ( )

( ) ( )

( ) ( )

We’ve used the known identity in triangle

∑( ) ( ) ( ) ( )




( ) ( ), which follows from Blundon-Gerretsen

( )


( ) ( )

( )( ( ))

( )( ) , obvious from Euler’s inequality

. Equality if and only if the triangle is equilateral.




12) If then in :

∑



Proof.

∑

∑(

* ∑

∑

( ) ∑

∑

( ) (∑ )

∑

.∑ /

∑ ∑ ∑ ∑ ∑

Let’s get to the main problem:

Using the Lemma, it suffices to prove that:

∑ (Euler’s inequality)





14) If then in :

∑

Proof.

∑

∑(

*

∑

( )∑

∑

( )(∑ )

∑

.∑ /

∑

∑

∑ ∑

∑

Let’s get back to the main problem. Using the Lemma, it suffices to prove that:

∑ ∑ ( ) (Euler)



√





16) If then in :

∑

Proof.

∑

∑(

*

∑

∑

( ) ∑

∑

( )(∑ )

∑

.∑ /

∑ ∑

∑ ∑ ∑

Let’s get back to the main problem:

Using the Lemma, it suffices to prove that:

∑ √ ∑ √ √ √ (Mitrinovic)


Refferences:

Romanian Mathematical Magazine-Interactive Journal-www.ssmrmh.ro

A SIMPLE PROOF FOR PADOA’S INEQUALITY

By Florentin Vișescu-Romania

In the following relationship holds:

( )( )( )

Proof:

Considering ( ) ( ) ( )

( )

( )

( )

( )

( )

( )

concave

(

*

( ) ( ) ( )

( )

So, .

/

( ) ( ) ( )

, with

.

/

, ( ) ( ) ( )-

√( )( )( )



( )( )

( )( )

( )( )

( ) ( ) ( )

( )( )( )

( )( )( )

( )( )( )

ABOUT PROBLEM 1131 TRIANGLE INEQUALITY

RMM 2019


1) In :

Proposed by Bogdan Fuștei – Romania

Solution:

Using

, (L. Panaitopol, Romanian GM 11/1982), we obtain:

Using ∑

( )

( ) ( )

( ), it follows:

∑

∑

( )

( ) ( )

( )

( ) ( )

( )

, the last inequality is equivalent with:

( ) ( ) , which follows from Gerretsen’s inequality:

. It remains to prove that:



( )( ) ( )

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


Remark: Let’s find an inequality having an opposite sense:

2) In :

Marin Chirciu-Romania

Solution:

Using Tereshin’s inequality:

and ∑

( ) ( )

( ), we obtain:

∑

∑

( ) ( )

( )

( ) ( )

( )

, the last inequality is equivalent with:

( ) ( ) , which follows from Gerretsen’s inequality:

. It remains to prove that:

( )( ) ( )

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


Remark: We can write the double inequality:

3) In

Solution: See inequalities 1) and 2). Equality holds if and only if the triangle is equilateral.

Observation by editor: Inequality 3) can be written:

.

/

This is a new refinement of Euler’s inequality.

Refferences:




THE EXTREMES OF THE SUMS OF ARCTANGENTS WITH RESTRICTIONS FOR

ARGUMENTS

By Florentin Vișescu-Romania

ABSTRACT: In this article are developed identities and inequalities related to sums of

arctangents with restrictions for arguments.

1. Let be ( ), such that . Then:

.

Solution: As and so

( ).

We consider the function ( ) ( ) ( ).

Then ( )

( )

( )

( )( ( ) )

( )( )

( )( ( ) ).

Obviously ( ) has the root

( )

( ) ( ) ( ) ( )

( ) ( ) (1)

( ) ( ) (2)

.

/

.

/

(3)

Let be ( )

( ) .

( )

.

/ .

/

.

/

( .

/ *( )

.

/.

/

( .

/ *( )

( )( )

( .

/ *( )

.

( )

( )

.

/

.



So

, if , equality for

. Let be ( ) ( ) ( ) .

( )

( )

( )

( )( ( ) )

( )

( )( ( ) ).

( )

( )

( )

( )

Then , if .

Then

, if

2) Let ( ), such that . Then

1. ( ), √

2.

3. ( )

√

.

Solution: As ( )

and so,

.

We consider the function ( )

.

( )

.

/

( )

( )

.

/

( )

( ) ( )

( ) ( ) ( )( )

( ),( ) ( ) -

( ),( ) ( ) -



( ) ( ) ( )

( ),( ) ( ) -

( )( )

( ),( ) ( ) -

( )( )

( ),( ) ( ) -

( ). √ /. √ /

( ),( ) ( ) -.

a) If ( ) constant.

So ( )

( )

.

In conclusion, if ( ) with , then

.

b) If or ( ), then:

√

( )

( ) ( ) ( ) ( )

( )

(1)

( )

(2)

.√ / .√ / (√ )

√

.√ / √

√

.√ / .√ /

(√ ) (3)

Let be ( )

( ) .

( )

( )( )

( ) ( )

( )( )

( )( )

( )( ) ( ) √ .

√

( )

( )

( )

( )



So for ( ) and ( ) with

.

Let ( ) (√ ) ( ) .

( )

(√ ) 4

√ 5

(√ )

√

√

( √ )( ) . (√ )

/√

√ . (√ ) / ( )

√ √ √

√ √ √ ( )

√ . (√ ) / ( )

√ √

√ . (√ ) / ( )

( ) √ ( )

√ . (√ ) / ( )

( )( √ )

√ . (√ ) / ( )

( ) √

√

.

√

( )

( )

4√

5 :√

√

;

√

. √

√

/

√

√

.

Hence, for ( ) and ( ) with

√

.

c) If or ( ), then analogous point b) we have:



√

( )

( )

(√ )

So, for ( ) and ( ) with

and

√

.

3) Let be ( ), such that . Then:

1. For .

√ 1 we have √

2. For .

√ √ / we have

3. For [ √ ) we have

√

Solution: As

and then

.

Consider the function ( ) , ( )

.

( )

( )( )

( ) ( )

( )( )

( )( )

( )( )

a) If , then ( ) ( )

.

b) If , then

√

( )

( )

√

:√

; √

√

√



Consider ( ) ( ) √

.

( )

√

√

√

√ .

/

√

√ √

√ √ √ √

( )( √ √ )

√ (√ ) √ (√ )

( )( √ √ )

(√ )(√ √ )

( )( √ √ ).

i) For

√ √

. It is not true.

√ √ √ √ √

. It is not true

( )

( )

( )

( )

.

ii) For ( )

√ √

. It is true

√ √ √ √ √

. It is true.

√

( )

( )

√

( )

(√

) √√

√

√



(

*

( )

c) If , then analogous b), we have:

√

( )

( )

√

Consider ( ) ( ) √

.

( ) (√ )(√ √ )

( )( √ √ )

1) For

√ √

. It is true.

√ √ √

. it is true.

√

( )

( )

√

( )

( )

(

*

(√

) √

√ √

√

ii) For ( )

√

, it is not true.

√ √ √

, it is not true.



( )

( )

For .

√ 1, according to b), ii) and c), ii)

√

.

For [ √ ), according b), i) and c), i)

√

. For .

√ √ /

) .

√ /, according to b), ii) and c), ii)

.

) [ √ ), according to b), i) and c), i)

.

4) Let be ( ), such that . Then:

.

Solution: As ( ) .

If and . So , false.

Then

and as ( ) namely

.

In this situation,

.

Consider the function .

/ ( )

.

( )

.

/

( )

( )

( ) ( )

( ),( ) ( ) - constant on .

/.

( ) ( )

.

So ( ) with .

5) Let be ( ), such that .



then

.

Solution: ( ) .

If or , false.

Then

. As ( )

or ( ) . As ( ) . Then

.

We consider the function .

/ ( )

.

( )

.

/

( ) ( )

( )

( ) ( )

( ) ( ) ( )

( ),( ) ( ) -

( ),( ) ( ) -

( ) ( )

( ),( ) ( ) -

( ) ( )

( ),( ) ( ) -

( ) it isn’t true or

, it is true.

( )

( ) ( ) ( ) ( )

( )

( )

(*)

( )

(*)

.

/

(**).

We consider ( ) ( )

.

( )

.

/

( )

( )

( ),( ) -

( )

( ),( ) -

( ) it isn’t true, , it is true.



( )

( )

( )

( )

( )

So ( )

or

, ( ) with

. We consider ( ) ( )

( )

.

/

( )

( )

( )

( ),( ) -

( ),( ) -.

( ) , it isn’t true. , it is true.

( )

( )

( )

( )

( )

So ( ) or

( )

with .

Refferences:


TRIGONOMETRICALIZABLE QUINTIC EQUATION

By Benny Lê Văn-Ho Chi Minh City-Vietnam

This note discusses on a specific class of quintic equations which are solvable thanks to the multiple-angle trigonometric transformation. Besides, the author finds an interesting fact,



namely “Matryoshka paradox” where quintic equations are more convenient to solve than an order of generalized quartic, resolvent cubic, and factoring quadratic equations. Initiation This note focuses on the specific reduced form of quintic equations:

Except for the constant coefficient, the above equation contains odd-ordered parameters. Hence, the methodology is based on the expanding formula:

Replacing (of which ), the equation becomes:

( ) ( ) ( ) Hence, we shall find such that:

( ) ( ) ( ) Accordingly, we get:

{

{

4

5

In other words, the above mentioned quintic equation is solvable thanks to this method if there exists such that and . Amazingly, may be negative and even complex numbers. Illustrated examples shall be discussed in the next section. Illustration Example 1. Solve the equation: Following the finding in the previous section, we shall find such that and

. This results in and therefore, √ √ .

Replacing √ , the equation becomes:

√ ( ) √

Solving this elementary trigonometric equation, we get:

( )

Consequently, solutions for the given quintic equation are:

√ (

* ( * +)

Particularly, the above solutions could be expressed as follows:

√ (

* √ ( )

4 √ √ √ 5

√ (

* √ ( )

4 √ √ √ 5

√ .

/ √ ( )

4 √ √ √ 5

√ (

* √ ( )

4 √ √ √ 5

√ (

* √ ( )



Example 2. Solve the equation: Regarding this problem, gives us , of which . Replacing , the equation becomes:

( )

Finally, solutions for the given quintic equation are:

(

* ( * +)

The above expression contains an inversed trigonometric function of complex numbers. Interestingly, according to Bézout’s theorem for odd-ordered polynomials, there exists * + such that . Illusion In the equation as discussed in Example 1, it is observable that is a comfortable solution. This section re-solves Example 1 in a purely algebraic method and then compares the two methodologies. Interestingly, the trigonometricalizable quintic polynomial seems more convenient. Factoring the quintic polynomial as mentioned in Example 1, we get:

( )( ) Considering the generalized quartic equation:

By replacing , we get the reduced form quartic equation:

Where , , and is respectively determined as follows:

{

Next, we shall find two quadratic polynomials that their product is exactly the above reduced form. Accordingly, it is supposed to find the triplet ( ) such that:

( )( ) As coefficients from both sides are homogeneous, we get:

{

( )

{

.

/

.

/

The resolvent cubic equation is obtained through the following process:

.

/ .

/ ( )

( ) Based on the above resolvent cubic equation, solutions for the reduced quartic form are determined as follows:

√

46 √ .

/75

4 √

5

√

46 √ .

/75

4 √

5



And finally, . This method deals with the quartic equation by solving resolvent cubic and quadratic polynomials, respectively. Therefore, this approach is somehow like a Russian doll, namely Matryoshka. Considering the quartic equation as obtained in Example 1:

The Matryoshka approach for ( ) ( ⁄ ) gives us:

{

{

Solving the resolvent cubic equation:

8

√

√

9

Choosing ⁄ , which implies that √

⁄ , solutions for the quartic equation are:

4 √

5

. √ √ √ /

4 √

5

. √ √ √ /

In Example 1, solving a trigonometricalizable quintic equation is more convenient than solving an order of quartic, cubic, and quadratic equations. Accordingly, we may name this fact “Matryoshka paradox” and further discuss when this paradox happens. Particularly, we shall find (which is not a solution for the generalized quartic polynomial) such that:

( )( ) As both sides are homogeneous quintic polynomials, we get:

( ) ( )

Thus, the Matryoshka paradox happens for a specific class of quartic equations as follow: ( ) ( ) , ( ) -

Exempli gratia, for and , we get:

{( )( )

{

And the solutions are:

{ (

* (

*}

Imagination Considering the following equation:

( )

By choosing one value of √ and replacing √ , we get:

√ (

√ *

(

√ *

( )

Finally, solutions for the given quintic equation are:



√ [

(

√ *

] ( * +)

Furthermore, the so-called trigonometricalizing method as discussed in this note could be applied to a specific class of odd-ordered algebraic equations.

ABOUT NAGEL’S AND GERGONNE’S CEVIANS (II)

By Bogdan Fuștei-Romania

Note by Editor: The article is written as a story of discovery triangle inequalities. The author give us a detailed mind process of these discoveries. I consider it an innovative and outstanding method to show results to readers. Let be any triangle. We’ve proved that:

(and the

analogs), so we have:

(and the analogs);

But from

(Tereshin’s inequality) we will obtain:

(and the

analogs); We will prove the identity: ( )

( )

(and the analogs); ∑

( )

;

Using the inequality for real numbers, we have: and we will obtain: ∑( )

We know that: ( )√ ( )

( )√ ( ). Taking into account all the above:

( ) ( )√ ( ) ∑(

)

( ) ( )√ ( );

( )

( )

(and the analogs); ( ) (and the analogs);



(and the analogs) because

(and the analogs);

(and the analogs);

( )

.

/ (and the analogs);

(and the analogs) ( ) (and the analogs);

( )

(and the analogs);

( )

( )

( )

( )( ) (and the analogs);

So we will obtain:

( )( )

( ).

Finally we will remember that: ( ) .

/ (and the analogs)

∑

In any acute-angled triangle we have:

(and the analogs) because

if the triangle is acute-angled and

(and the analogs)

So, we will have: ( )( ) if triangle is acute-angled.

if triangle is acute-angled. ∑

for any acute-

angled triangle. ∏

∏

for any acute-angled triangle.

∏

So, if is acute – angled triangle we have: ∏

∏

(Panaitopol inequality); ∏

∏

(and the analogs). Summing we have the

following: ∏

∏

∑

for any acute-triangle.

√

(and the analogs);

(and the analogs) we will obtain the

following inequality: ∏

∏

∑

for any acute – angled triangle;

We’ve proved that (and the analogs). Using the inequality between

squared means and arithmetic means we will have: √ √ (and the analogs)

√ ∑ ∑√

(and the analogs);

, so we will remember that

(and the

analogs);

√

√

(and the analogs); (and the analogs);

∑

∑

∑

;

So

∑

( ); after calculating we will

have:

∑

; The triangle’s fundamental inequality:



( )√ ( )

( )√ ( )

Hence: .

/

( )√ ( )

∑

.

/

( )√ ( )

(and the analogs);

(and the analogs);

√ √

(and the analogs); √ √

(and the analogs); so we will

obtain: ( √

*

√ (and the analogs)

∏: √

;

√

√ (Lessel – Pelling inequality) (and the analogs);

√ √ (and the analogs). Summing we will obtain:

√ √ √

(and the analogs);

So (and the analogs) √ √ √

(and the analogs);

But (and the analogs) √ √ √

(and the analogs);

(Tereshin’s inequality) summing we will have the following:

( )

∑

;

.

/ ∑

;

√ √ (and the analogs); ∑

∑

;

√

√

(and the analogs); Summing we will have the following:

√ ∑

∑√

√

√

(and the analogs). Summing we will have the following:

√ ∑

∑√

(and the analogs); We apply the inequality between squared means and

arithmetic means and we will obtain:

√ (

√

* (and the analogs);

∏

√ ∏:

√

;

( √ ) (Blundon – Klamkin’s inequality). So we will obtain:

√

√ ∑

∑√



√

√ ∑

∑√

We’ve proved that: (and the analogs);

( ) (and the analogs); ( )

( );

( )

;

After calculating we will obtain the following:

( ) ( )

;

We will prove that:

√ √

; We know that: .

Squaring we will obtain: .

√ /

( )

( ) (( ) )

(Gerretsen’s Inequality); √ ; So the inequality is proved. (and the analogs). After some simplifications from

the proved inequality we will obtain: √

( ) √ ;

√ (Lessel Pelling inequality) (and the analogs)

From the above we will obtain: √

(and the analogs);

We’ve proved that:

√ ∑

∑√

. Using the inequality between arithmetic means and

geometric means we will have: ∑√

. Using the inequality between arithmetic means and

geometric means we will have: ∑√

√√∏

√

; taking into account the above

we have a new inequality, namely:

√ ∑

√

;

√ ∑

∑ √

√

(and the analogs); (and the analogs);

(and the analogs);

(and the analogs)

√ (and the analogs);

√ (and the analogs);

(and the analogs);

(and the analogs);

(and the analogs);



√ (and the analogs); But (and the analogs), so we can write

the following:

√

(and the analogs);

√

(and the

analogs);

(and the analogs); Replacing the above we have the following:

√

( ) (and the analogs);

√ ( ) (and the analogs); From the above we will obtain the following:

∑

∑√

( ); ∑

∑

√ ( );

From

√

(and the analogs). Using the inequality between the squared

means and the arithmetic means we will have:

√

√

√

√

√ .

Analogous, we have:

√

√

√ ;

√

√

√ and the analogous;

√

( ) ( )

( )

(and the analogous);

(and the analogous) sine theorem

(and the analogous);

Summing we will obtain: ∑

( ) (and the analogs);

( ) (and the analogs); Summing we will prove a new inequality, namely: ( )

( )

( )

(and the analogs);

(and the analogs);

From the above we will obtain a new inequality: ∑

(

)∑

But

, so, we will have a new inequality: ∑

.

/;

( √ ) (Blundon-Klamkin’s inequality) ∑

.

/ .

√

/

We know that ( )( ) (and the analogs). Using the inequality between arithmetic means and geometric means we will obtain:

√( )( )

√

(and the analogs)

Summing we will obtain: ∑

. Taking into account the above inequality we have:

∑

;

( )

( )

(and the analogs);

Summing we have a new inequality, namely: ∑ .

/

∑

.

/. We know that ∑

;

( )

( )

;



( )

( )

.

/

√

(Euler’s inequality refinement)

(and the analogs) ∑

;

Summing we will obtain the following ( )∑

∑

(and the analogs). Summing we will obtain a new inequality:

∑

; (and the analogs)

∑

So finally we have a new inequality: .∑

/

We’ve proved that

√

√

√

and the analogs

(and the analogs)

We will prove a new identity: .

/

(

)(

)(

)

;

Taking into account the above we have the following:

(

*

(

)(

)(

)

(

*

(

)(

)(

)

(and the analogs);

√

( )( )( )

;

(and the analogs);

(and the analogs);

∑

so we will obtain the following:

( )

( )

Taking into account the above we have the following:

( )( )( )

But

we will have the following inequality:

( )( )( )

But and the analogs we will obtain a weaker inequality:

( )( )( )

( )( )( )

(and the analogs) we will have the following:



( )( )( )

( )( )( )

( )( )( )

( )( )( )

We’ve prove that

√

( )( )( )

√ ∑

√

We know that

√

(and the analogs). But

(and the analogs) so we

will have: √

√

(and the analogs);

(Panaitopol inequality)

So we will have the following:

1)

√ ∑

√

√

√

2)

√ ∑

√

√

√

3)

√ ∑

√

( )( )( )

√

√

(and the analogs)

4)

√ ∑

√

√

√

(and the analogs)

5)

√ ∑

√

( )( )( )

√

√

(and the analogs)

We’ve proved that √

( ) √ so we have √

√

√

We will have the following:

6)

√ ∑

√

( )( )( )

√

√

√

(and the analogs)

7)

√ ∑

√

( )( )( )

√

√

√

(and the analogs)

√

(and the analogs);

(and the analogs);

(and the

analogous); ( ) (and the analogs);

(and the analogs). Taking into

account the above we can write that: ( ) (and the analogs)

(Panaitopol). So we will have

(and the analogs); √

(and the analogs)



√

(and the analogs)

√

(and the analogs)

So ( )√

∑

. Finally: √

;

So we can write:

8)

√ ∑

√

9)

√ ∑

√

√

√

10)

√ ∑

√

√

√

( )( )( )

√

We’ve showed that √

and √

( ) √

√

( ) √

so finally we will have the following inequality:

√

√

11)

√ ∑

√

√

√

( )( )( )

√

(and the

analogs)

√

(and the analogs). Summing we will have the following:

√

( ) ( )

√

( )

We’ve proved that

√ ∑

∑√

and

√ ∑

∑√

and summing we will prove

a new inequality:

√ ∑.

/ ∑(√

√

*

√

( ) √ √

( )

√ √ √

(and the analogs)

(√ √ ) √ (and the analogs)

We will obtain that .

√

√ /√

(√ √ )

( ) .

√

√ /√



√

(and the analogs)

√

Summing the two inequalities we will obtain a new result.

√ (

√

√ )

( )( √ )

( )( ) (and the analogs)

√

(and the analogs) √

(and the analogs)

12)

√ ∑

√

√

√

√

∑

√

∑

13)

√ ∑

√

∑

14)

√ ∑

√

√

√

(and the analogs)

Practically any expression smaller than

can be used in the following inequality:

√ ∑

√

in order to obtain a new inequality.

There are limitless possibilities, but

∑

, replacing in the above obtained inequalities

it will follow a new series of inequalities.

(and the analogs); √( )( ) (and the analogs)

√

(and the analogs);

√

(and the analogs)

From the above we have the following:

√

(and the analogs);

∑√

, replacing in this expression in the above inequities we will obtain a series of

equivalent inequalities.

√

√

( )( )

( )( )

(and the analogs)

So we will obtain a new identity and namely: ∑

(This identity can be found as a

proposed problem by Prof. Mehmet Șahin)

We will remember that ∑

∑√

∑

We’ve proved that

(and the analogs);

√

√

(the inequalities between arithmetic means and geometric means)

√

(and the

analogs). We’ve proved that

√ √

√ √

(and the analogs)



( ) √ ∑ √

(and the analogs), so we will have

√

∑ √

.

/ (and the analogs)

(

*

√

∑ √

But

(and the analogs) so we will obtain a new inequality:

∑

√

∑ √

;

(and the analogs)

So we will obtain a new inequality, namely: ∑

√

∑ √

;

But ∑

∑

∑

√

∑ √

;

(Tereshin’s inequality)

.

/ (and the analogs)

∑

∑

∑

∑

so we can write that:

∑

√

∑ √

; We know that

√

√ ( ) (and the analogs);

√ ( ) √ (and the analogs);

√

√

(and the analogs); squaring we will obtain the following:

.

/ (and the analogs) ∑

∑

; we can write the following:

∑

√

∑ √

.

/

(and and analogs);

(and the analogs);

√

√

(and the analogs);

√

√

.


We will obtain:

√

√

(and the analogs); taking into account the above we

obtain the following: ∑

√

√

(

∑ √

√ *;

We proved that:

(and the analogs);

(and the analogs); (and the analogs); We will obtain the following identities, namely: ( ) (

) (and the analogs); ( )

(and the analogs);

Using the inequality between the squared means and arithmetic means:



( √ √ ) (and the analogs);

We also know that: √ √ (and the analogs);

Summing we obtain: √ √ √ (and the analogs);

Taking into account the above we will have the inequality:

∑( √ √ )

( ) ( ) (and the analogs); (sine theorem)

( )

( √ √ ) ( )

( √ √ ) (and the analogs); summing we will obtain the

following:

∑ ( √ √ )

√ √

(and the analogs);

Summing we have the following: ∑

∑

√ √

(and the analogs)

Summing we have the following: ∑

∑

√ √

;

We know that ( ) ( )( ) (and the analogs)

( )( )

( √ √ )

√ √

; ∑

∑

√ √

;

∑

(and the analogs)

√ √

(and the

analogs) ∑

∑

√ √

(and the analogs);

; so we will have:

(and the analogs)

√ √

(and the analogs);

( ).

/;

( ) .

/

√

∑ √

;

√ √

∑

∑

√ √

∑

∑

∑

∑

∑

√ √

;

.


√

(and the analogs);

√

(and the analogs). From the above we will write:

√

.

/;

√

.


Summing we have: ∑

√

.

/ ∑

√

∑

√ √



We’ve proved that:

√ √

√

√

We will obtain the following: √

( √ √ )

√

(and the analogs)

√

( )( )

√

(and the analogs);

( √ √ )

(and the analogs);

(and the analogs);

( √ √ )

√ √

(and the analogs);

∑

∑( √ √ )

We know that (and the analogs) ∑

∑( √ √ );

But √ ( ) (and the analogs)

∑√

√ ∑( √ √ );

We easily prove that:

(and the analogs);

( √ √ ) (and the analogs);

√

(and the analogs)

√ ( );

√ ( )

√ √

(and the analogs);

∑√

√ ∑

√ √

It is known that ( )( ) (and the analogs); applying the inequality between

the arithmetic means and geometric means we obtain:

(and the analogs);

√ √

(and the analogs)

We will remind that:

√ √

(and the analogs);

Summing we will obtain two new inequalities:

∑

∑( √ √ )

∑

√ √

We’ve prove that: (and the analogs); ( ) (and the

analogs);

(and the analogs)

( )

( ) (and the

analogs); Summing we have:

( ) ∑

( ) ( )( )

;

Summing we have:

( ) ∑

. But (and the analogs) we will obtain:



( ) ∑

( ) ∑

;

∑

we will obtain:

∑

Using the inequality between squared means and arithmetic means:

√

√ ( )( )

√

( )

√( )( )

( ) (and the analogs), summing we will obtain a new

inequality: √ ∑√

( )

We will finish with the following:

( )√ ( ) ( )√ ( )

( ) (and the analogs). We will obtain:

( ) ( )√ ( )

( ) ( )√ ( ) (and as always, the analogs).

References: 1. Themistocles M. RASSIAS, Hari M. SRIVASTAVA – Analytic and Geometric Inequalities and Applications. 2. Bogdan FUȘTEI, Romania – About Nagel and Gergonne’s cevians, Romanian Mathematical Magazine, www.ssmrmh.ro

RMM 2019, NUMBER 18 – AUTUMN 2020,PROBLEM SP.263



∑(

*

(

*

Proposed by George Apostolopoulos – Greece

Solution: We prove the following lemma: Lemma:


∑(

*

( ) ( )

Solution: We have

∑(

*

∑ ( )

( ) ( )

which follows from the known identity in triangle

∑ ( ) , ( ) ( )-

http://www.ssmrmh.ro/



Let’s get back to the main problem.

The left hand inequality. Using the Lemma the inequality can be written:

( ) ( )

, which follows from Gerretsen’s inequality

It remains to prove that:

( )( ) ( )

( )( ) , obviously from Euler’s inequality

.


The right hand inequality. Using the Lemma the inequality can be written:

( ) ( )

.

/

, which follows from Gerretsen’s inequality:


( )( ) ( )

(

*

( )( )

Obviously from Euler’s inequality . Equality holds if and only if the triangle is

equilateral.

Remark: Inequality 2) can be strengthened.


∑(

*


Solution:

The left hand inequality. Using the Lemma the inequality can be written:

( ) ( )

, which follows from Gerretsen’s inequality


( )( ) ( )

true with equality. Equality holds if and only if the triangle is equilateral.



The right hand inequality. Using the Lemma the inequality can be written:

( ) ( )

which follows from Gerrentsen’s inequality . It remains to prove that:

( )( ) ( )

true with equality. Equality holds if and only if the triangle is equilateral.

Remark:Inequality 3) is stronger than 2)


∑(

*

(

*

Solution: See 3) and the inequalities:

i)

( )( )

obviously from Euler’s inequality .

ii)

.

/


Refferences:


SOME APPLICATIONS OF CHEBYSHEV INEQUALITY IN GEOMETRY

By Florică Anastase-Romania

Theorem(Chebyshev):

Suppose ( ) and ( ) are two increasing sequence of real numbers,

then:

( )( )

Proof: By direct expending, we have

( ) ( )( )



∑ ( )( )

Comment. By the same proof , we also conclude that if the sequence ( ) is

increasing but the sequence ( ) is decreasing, then

( )( )

For symmetric problems, we can rearrange the order of variables so that the condition of

Chebyshev inequality is satisfied.Generally ,solutions by Chebyshev inequality are more

concise those that by order basic inequalities. Let’s consider the following apllications.

Application: In acute ∆ABC the following relationship holds:

( )

By Florică Anastase – Romania

Solution:

If {

⏞

( )

( )

( )

( )


( ) ( )( )


Solution:

Let 8

⏞

( ) (

*

and from sine theorem it follows:

.

/ (

*



{

( ) (

*

( )

( )

( ) ( )( )


, ( )-


Solution:

If 8

⏞

( ) (

*

( ) (

* ( )

With the relationships:

{

⏞( )

( )

( )

.

/

( )

, ( )-


( )( ) , ( )( )-


Solution:



, ( )( )-

{

( )

⏞

( )

( )( ) , ( )( )-


( ) ( )


Solution:

If 8

⏞

(

*(

*

(

*(

* ⏞

(

*( )

(

*

4

54

5.

/

(

* ( )

( ) ( )

Application:

1) In any ∆ABC the following relationship holds:

)

)

)

G.M.3/1972




)

(

*

)

Marin Chirciu ‘Inegalități Geometrice de la inițiere la performanță’ Ed.Paralela 45


) ∑

√

) ∑

√

( )

Marin Chirciu ‘Inegalități Geometrice de la inițiere la performanță’ Ed.Paralela 45

STRUCTURI ALGEBRICE (V)

By Vasile Buruiană – Romania

ASPECTE METODICE PRIVIND INELELE FACTORIALE ȘI APLICAȚIILE LOR

Cred că multe din problemele prezentate mai inainte au, în învățământul matematic, un caracter urgent și important.În adevăr, pe de o parte elevii fac de timpuriu cunoștință cu divizibilitatea, numerele prime, cmmmdc și cmmmc, scrierea zecimală a fracțiilor ordinare, unele numere iraționale și calcul cu fracții algebrice. O defecțiune în înșiruirea acestor cunoștințe are efecte iiiiiiiii ulterior și cu greu se mai poate îndrepta ceva sau avansa. Pe de altă parte importanța acestor probleme decurge din: a) caracterul fundamental al noțiunilor ce intervin. Nu pot fi deprinse calculele în dacă nu se cunoaște descompunerea în factori primi și calculul cmmmc. Nu pot fi mai târziu înțelese noțiunile de limită și de aproximare dacă nu a fost înțeleasă scrierea zecimală a numerelor și distincția între numerele raționale și iraționale. b) varietatea locurilor în care se aplică noțiunile și teoremele privind factorialitatea: De la

demonstrarea că un număr de forma √ este irațional pentru prime și până

la descompunerea în fracții simple utilizată în calculul integral, pentru fracțiile din ( ), sunt încă multe locuri unde întâlnim noțiuni de factorialitate. De pildă: reducerea gradului unei ecuații dacă îi cunoaștem rădăcini (cu ajutorul teoremei lui Bezout), găsirea rădăcinilor comune pentru ecuațiile algebrice ca fiind rădăcinile cmmdc al polinoamelor ce dau ecuațiile respective; adunarea și scăderea fracțiilor algebrice, etc. c) faptul că exemplele, contraexemplele și metodele utilizate sunt pur și simplu modele pe care se învată matematică.



Astfel teoremele ce apar sunt exemple de teoreme de existență și unicitate; metodele utilizate cuprind inducția, reducerea la absurd și calcule de găsire a unui obiect în mod algoritmic, ceea ce oferă metode de efectuat programe pe un computer; Rezolvarea unor probleme formează experiența, fortifică gândirea logică și stimulează imaginația elevilor, făcându-I să înțeleagă mai bine ce înseamnă o analiză, o generalizare, o trecere de la euristic la riguros. Astfel, rezolvarea unor ecuații în numere întregi, cunoașterea normei ca generalizare a modului de definirea a inelelor factoriale (ca inele Krull în care divizorii sunt principali ca o “măsură” a factorialității); criteriile de stabilit ca un element este ireductibil sau nu; sunt situații care nu numai că îl conving pe un elev, dar îl și lămuresc asupra naturaleții fenomenelor ce apar. Exemplele altor inele factoriale permit o paletă mai largă de exemple utile când face conștință cu obiecte mai abstracte. Astfel se pot obține noi corpuri finite când se iau clase de resturi modulo în , - pentru prim, în afară de inelul al claselor de resturi modulo prim. Exemplele de inele nefactoriale, formează situații

care să-i atragă atenția asupra importanței unor ipoteze, definiții și interpretări geometrice. d) perspectiva deschisă asupra matematicii: divizibilitatea în este o relație de ordine; inele euclidiene, principale sau Krull sunt noi clase de inele ce pot constitui ele însele interes; e) în există subinele interpretabile geometric și care nu mai au proprietăți “estetice” asemănătoare lui ; dacă se trece la inele de polinoame, problemele formulate în cazul inelelor de numere, capătă noi aspecte și apar noi dificultăți (de exemplu în privința găsirii elementelor inductibile). Pentru a ameliora însușirea temeinică a cunoștințelor, o serie de probleme și aplicații pot fi luate fie în cadrul lecțiilor, fie în cadrul cercului de matematică și fructificarea rezultatelor nu întârzie să apară. Am făcut o astfel de experiență la două clase paralele pe care le-am condus între 1995 și 1999, și rezultatele la examenul de bacalaureat s-au resimțit în mod favorabil. Voi prezenta în continuare câteva probleme de care m-am folosit în sensul celor spuse mai sus, indicând după numărul problemei clasa și perioada aproximativă în care a fost propusă și precizarea dacă a fost luată la lecția curentă sau la cercul de matematică.

1. (clasa a IX a, început, lecție). Scrieți sub formă zecimală numerele

. Dați un

exemplu de număr rațional care scris sub formă zecimală să aibă perioada cu lungimea perioadei de maximum cifre. Elevii au rezolvat problema cu mici sugestii si au reușit să înțeleagă prin demonstrație: teorema fundamentală a aritmetici, submulțimile lui care se scrie zecimal sub formele ( ) și ( ) și și-au fixat clar noțiunea de număr prim (ireductibil).

Reformulând teorema împărțirii cu rest în și , - au înțeles de ce

are cel mult cifre

în perioadă, dacă în cercurile s-a reamintit construcția lui ( ) ( ) și a divizorilor și cum din condițiile ( ) ( ), ceea ce duce la soluția problemei. S-a definit de asemenea relația de ordine (totală, parțială) și s-a arătat ca în ordinea naturală este o relație de ordine totală, pe când ordinea dată de divizibilitate este parțială. 5. (Clasa a IX a, început, lecție) a) Să se arate că pentru avem și

b) Fie . Atunci √ √ Pentru a) s-au reamintit noțiunile de fracție ireductibilă, număr prim, s-a explicat semnul de „ ” și ce fel de demonstrație trebuie făcută (prin absurd), cum la implicația mai dificilă



intervine torema fundamentală a artmeticii și faptul că un număr sau este

prim. Apoi prim și și deci se contrazice că

era ireductibilă.

S-a atras atenția semnificațiilor distincte, utilizate, ale iiiiii ireductibil (număr; fracție) Pentru b) în lumina celor spuse la a) s-a folosit iarăși metoda reducerii la absurd: dacă

√ √ √ și √

deci din a) .

Luând descompunerile unice în factori primi

este par.

Analiza când și sunt ambii pari sau ambii impari conduce la situația sunt pari deci sunt pătratele unor numere: . Cu ocazia ridicării la pătrat s-au reamintit teremele de echivalență pentru egalități. 6) Clasa a X a(la capitolul logaritmi, lecție) Să se arate că există și iraționale astfel că să fie număr natural. s-au sugerat: radicalii și logaritmii produc în general numere iraționale (s-a reamintit

deosebirea numerelor raționale cu cele iraționale): √ și √ sunt iraționale (s-a

folosit metoda reducerii la absurd și definiția numărului prim, plus teorema fundamentală a

aritmeticii; de exemplu se ajunge, când presupunem că √

ireductibilă, ca

, ceea ce arată că obținem descompuneri distincte. Pentru a fi mai convingător, am folosit și calculul ultimei cifre pentru a arăta ca egalitatea este imposibilă).

5-CLASS-STANDARD

V.1. Let be If

. Find . Proposed by Petre Stângescu-Romania

V.2. Let be . If then

can not be a perfect square. Proposed by Petre Stângescu-Romania

V.3. If then can

not be a perfect square. Proposed by Petre Stângescu-Romania

V.4. Let be sequence: . Find the rank term of perfect squares from

the sequence. Proposed by Petre Stângescu-Romania

V.5. If then can not be a perfect square.

Proposed by Petre Stângescu-Romania



V.6. Find prime number such that and prime numbers.

Proposed by Ștefan Marica-Romania

V.7. Find a perfect square such that


V.8. In a sequence of natural consecutive numbers, the difference between the biggest and

the smallest number is equal to , and the sum of the smallest four numbers from the

sequence is equal to . Find how many terms the sequence has and how many numbers

from the sequence are divisible to . Proposed by Constantin Ionică – Romania

V.9. On chess table are written the following numbers: . Both Aurel

and Barbu are deleting four numbers, and they notice that the sum of the deleted numbers

by Barbu is with smaller than a third from the sum of the numbers deleted by Aurel.

a) What number remained on the table?

b) What number did Aurel deleted? Proposed by Constantin Ionică – Romania

V.10. Prove that in a class of pupils there are pupils that are not friends or there is a

pupil with at least friends. Proposed by Daniel Stretcu – Romania

V.11. A pupil cuts the first digit of the number and adds it to the remained number.

Prove that the pupil repeat this operation until he remains with a digits number, this

number has at least two equal digits. Proposed by Daniel Stretcu – Romania

V.12. Find three natural numbers, knowing that the difference between the first and the

third is , dividing the second to the third we obtain the quotient and the reminder

5, and dividing the first to the difference between the second and the third we obtain the

quotient and the reminder . Proposed by Daniel Stretcu – Romania

V.13. Let be the set: 2

3.

Find , for which doesn’t have elements that don’t belong to the set .

Proposed by Daniel Stretcu – 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. Prove that for can not be written as a sum of two perfect squares.


VI.2. Find , - prime number such that be a perfect square.


VI.3. Find such that be a perfect square.


VI.4. Find prime numbers; such that exist exactly nine pairs ( ) such

that:


VI.5. Find prime numbers such that


VI.6. Find such that: ( ) ( ) ( )


VI.7. Find prime numbers such that: (( ) ( ) )


VI.8. Let be coliniar points such that ( ) ( )

and the points and are the middle of the segments , -, respectively , - and

. Find the lengths of the segment , -.

Proposed by Constantin Ionică – Romania

VI.9. Let and be adjacent supplementary angles, and the angles and

adjacent and complementary angles such that ( )

a) Find ( ) ( ) and ( )



b) If in the interior of we build semilines distinct with the origin in , such that the

angles formed have the measures expressed in grades by natural numbers, prove that at

least two of the angles are congruent. Proposed by Constantin Ionică – Romania

VI.10. Find the natural numbers that satisfy the condition

. Proposed by Daniel Stretcu – Romania

VI.11. Find from

. Proposed by Daniel Stretcu – Romania

VI.12. Find the natural numbers , knowing that are directly proportional

with and


VI.13. Let be a triangle with cm, cm, and ( ) . On

bisector we take a point such that cm. Find the angles measures of

triangle. Proposed by Daniel Stretcu – Romania

VI.14. The natural numbers are written in a random order than are put

together in a single number. Prove that the obtained number it is not a perfect cube.


VI.15. Solve in real numbers the following equation:


VI.16. Find all such that: ( )

Proposed by Seyran Ibrahimov-Azerbaijan

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. In parallelogram the following relationship holds:

( ) , -


VII.2. Let be the contact triangle of . If

4( ) then .


VII.3. Find such that

. Proposed by Gheorghe Calafeteanu-Romania

VII.4 In ( ) ( ) internal

bisector, ( ) * + * + * + * +

Prove that:

( ) ( )


VII.5. Find the set * + all natural numbers such that:


VII.6. Find the area of triangle, knowing that and that:

√ √ √


VII.7. Find the last two digits of the number , knowing that is a

natural number bigger than and its last digits is .


VII.8. Find the natural numbers , nonzero and distinct, knowing that and

are directly proportional with and .


VII.9. Prove that √ is an integer, for none of a natural nonzero number .


VII.10. Solve for real numbers:



{

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

Proposed by Daniel Sitaru,Carina Maria Viespescu – Romania

VII.11 Solve for integers: {

( ) ( ) ( )

Proposed by Daniel Sitaru,Virginia Grigorescu – Romania



8-CLASS-STANDARD

VIII.1. If thhen:


VIII.2. If then:


VIII.3. If then find :




VIII.4. If then:


VIII.5. Find such that:

{


VIII.6. Find such that:

{


VIII.7. If then:


VIII.8. Find all integers such that: .

Proposed by Gheorghe Calafeteanu-Romania

VIII.9. Solve the following equation:

|

| |

| {

}


VIII.10. Find the nonzero natural numbers knowing that

.


VIII.11. Prove that , for any natural number .


VIII.12. Prove that is a composed number.


VIII.13. If such that and , prove that:



Proposed by Marin Chirciu Romania

VIII.14. If then:

√

√

√ √

Proposed by Marian Ursărescu – Romania

VIII.15. If then:

√ ( ) ( )

Proposed by Daniel Sitaru,Sorin Pîrlea – Romania

VIII.16. In the following relationship holds:

∏4

5

( )( )( )

( )( )( )

Proposed by Daniel Sitaru,Doina Cristina Călina – Romania

VIII.17. Find at least three solutions for such that:

( )

( )

( )

( )

( ) is a perfect square number.

Proposed by Naren Bhandari-Nepal

VIII.18. If then:

(

√

*

( ) ( )

√

Proposed by Daniel Sitaru,Dan Grigorie – Romania

VIII.19. If , - - great integer function, then:

[

] [

] [

] [

]

Proposed by Daniel Sitaru,Dana Cotfasă – Romania

VIII.20. Solve for :



{

( )

(

* ( )

Proposed by Daniel Sitaru,Elena Alexie – Romania

VIII.21. Solve for :

{

(

* ( )

Proposed by Daniel Sitaru,Elena Iacob Meda – Romania

VIII.22. Solve for integers: ( ) ( ) ( )( )

Proposed by Rovsen Pirguliyev-Azerbaijan

VIII.23. Solve for real numbers: 2

3 2

3 , - , -

* + , - , - - great integer function.




9-CLASS-STANDARD

IX.1. If such that and , prove that:

Proposed by Marin Chirciu, Octavian Stroe – Romania







IX.4. Let and . Prove that:

4

5

4

5

4

5

( )


IX.5. Let and . Prove that:

( )( )

( )( )

( )( )


IX.6. Let such that and . Prove that:

∑4

5


IX.7. Let such that and

. Prove that:


IX.8. Let and be positive real numbers such that . Prove that:

4

5

4

5

4

5


IX.9. Let be such that and . Prove that:



√

√

√

√


IX.10. If such that and , find the minimum of the expression:


IX.11. Let be such that . Prove that:

( )( )( )


IX.12. In the following relationship holds:

( ) , where



( ) ( ) ( ) √

Proposed by D.M. Bătinețu – Giurgiu,Cristina Spiridon – Romania

IX.14. If then in the following relationship holds:

( ) ( )

Proposed by D.M. Bătinețu – Giurgiu,Gheorghe Boroica – Romania


∑

( )

√

Proposed by D.M. Bătinețu – Giurgiu,Nicolae Mușuroia – Romania


∑

( )

√

Proposed by D.M. Bătinețu – Giurgiu,Gheorghe Stoica – Romania




∑

( )

√

Proposed by D.M. Bătinețu – Giurgiu,Cătălin Pană – Romania


∑

( )

(√ )

Proposed by D.M. Bătinețu – Giurgiu,Claudiu Ciulcu – Romania


∑

( )

√

Proposed by D.M. Bătinețu – Giurgiu,Elena Nicu – Romania

IX.20. If in - Gergonne’s cevians then:

√


IX.21. If then:

√ ( ) ( )

Proposed by Daniel Sitaru,Roxana Vasile – Romania


√ √

√

√


IX.23. In scalene the following relationship holds:

∑√





∑ √

( )√



:∑

∑

∑

;



∑

√( )( )

∑

∑

√( )( )

∑


IX.27. In – incenter the following relationship holds:


IX.28. In the following relationships holds:

∑

√

∑

√

∑√

√ ∑


IX.29. In – Bevan’s point, – excentral triangle, – incenter,

– circumradii of . Prove that:



√

√

√

√ ( )



(√ √ √ ) ( )√



(

* (

* (

* √ (

*



√

Proposed by Daniel Sitaru,Cătălin Spiridon – Romania

IX.33. If then:

( )( )( )

(

*

Proposed by Daniel Sitaru,Simona Radu – Romania


∑(

*

∑(

*

Proposed by Daniel Sitaru,Nicolae Radu – Romania

IX.35. Prove that in any triangle:

Proposed by Adil Abdullayev-Azerbaijan






( )

( )

( )






IX.40. If then: .


IX.41. If then .



:∑

;:∑

; ∑

Proposed by Mustafa Tarek – Egypt



IX.43. In - incenter the following relationship holds:

.

/

.

/

.

/


IX.44. In – Bevan’s point, – incenter, – Nagel’s cevian, – Gergonne’s cevian

the following relationship holds:

√ .

/

( ) ( )


IX.45. In – Bevan’s point, the following relationship holds:


IX.46. Let and are three real numbers when no pair of , - , - and , - are

simultaneously zero, prove:

[

] [

] [

]

, - , -

, - , -

, - , -

, - is the greatest integer part of . Proposed by Jalil Hajimir-Canada


(( )

*

(( )

*

(( )

*

Proposed by Mustafa Tarek-Egypt



IX.48. In – Nagel’s cevians, the following relationship holds:


IX.49. In – Bevan’s point the following relationship holds:

(

*


IX.50. In – Bevan’s point, – incenter, -excenters, the following

relationship holds:


IX.51. In acute – orthocenter the following relationship holds:

∑:

√

;

√


IX.52. Let be the sides of extouch triangle of then:


IX.53. In – Geronne’s cevian, – Nagel’s cevian, – Gergonne’s point, –

Nagel’s point. Prove that:

∑






10-CLASS-STANDARD

X.1. Solve in the following equation:

√ √ Proposed by Daniel Stretcu – Romania

X.2. Solve the equation: ( )

Proposed by Daniel Stretcu – Romania X.3. Sole the equation:

Proposed by Daniel Stretcu – Romania X.4. Solve the equation:

Proposed by Daniel Stretcu – Romania X.5. In the following relationship holds:

.

/

(

*

.

/

Proposed by Marin Chirciu – Romania X.6. In the following relationship holds:

, where


, where



( )

Proposed by Marin Chirciu – Romania X.10. Let be such that and . Prove that:




X.11. In the following relationship holds:

( ) , where .



( )

( )

( ) (

*


X.13 Let and . Prove that:

√ √

√


X.14. Let be such that . Prove that:

.

/

.

/

.

/

, where .



, where



.

/

.

/

.

/

, where



∑( ) , where



(

*




X.19. Solve the following equation:

( ) , where is given.


X.20. In having the perimeter equal with the following relationship holds:

( )( )( )



∑ ( )

√

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

X.22. If then in the following relationship holds:

∑

( ) √ ( )

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


Proposed by D.M. Bătinețu – Giurgiu,Gabriel Tică – Romania


Proposed by D.M. Bătinețu – Giurgiu,Gabriela Vasile – Romania


√

Proposed by D.M. Bătinețu – Giurgiu,Iulia Sanda – Romania

X.26. In acute the following relationship holds:

( )

( )

( )





.

/

.

/

.

/

.

/



( ) √ (( ) ( ) ( ) ) , -

Proposed by Daniel Sitaru,Paula Țuinea – Romania

X.29. In – Bevan’s point, – excenters, ( ),

( ) ( ), the following relationship holds:

.

/

Proposed by Mehmet Șahin-Turkey

X.30. In – incenter, – circumradii of . Prove that:

(

*

(

*

(

*

.

/



(

*

(

*

(

*


X.32. – excenters, – Bevan’s point. Prove that are sides of a

triangle. Proposed by Mehmet Șahin-Turkey

X.33. If in – excenters, then:

√ √

∑ √ .

/ .

/

.

/





, -


X.35. – bicentric quadrilateral, – circumradii, inradii. Prove

that: . Proposed by Mustafa Tarek-Egypt

X.36. In – bisectors of – excentral triangle. Prove that:

∑

( )

( )




X.38 In – incentre, - Bevan’s point, – area, the following relationship holds:

∏(

*


X.39 In the following relationship holds:

∑( ) ( √ )

√

√( )( )



∑√





( )( )( ) √∏(

*


X.42. Find at least three solutions for such that:

( )

( )

( )

( )

( )

is a perfect square number. Proposed by Naren Bhandari-Nepal


√


X.44. In – Nagel’s cevians the following relationship holds:



∑:

√

√

√

;

√

∑(

*


X.46. In – incenter the following relationship holds:


X.47. In - incenter, – circumradii of . Prove that:



∑(

*

∑


X.48. If in – Nagel’s cevian, – Gergonne’s cevian then:

∑√



√

√

√

√

∑

∑



∑

√


X.51. Find such that in any scalene the following relationship holds:

∑

( )

∑

- Nagel’s cevians.



√



∑

∑( )


X.53. Let be a triangle, its centroid. Prove that the following inequality holds:

:∑

; :∑

; ( ) (

*

with the usual notations in triangle. Proposed by Radu Diaconu, Emil Popa – Romania

X.54. Let be a triangle, the incenter and the circumcenter. If intersects

the circumcenter in prove that the following inequality holds:

:∑

; :∑

;

with the usual notations in triangle.

Proposed by Radu Diaconu, Emil Popa – Romania

X.55. Let be an acute-angled triangle, its orthocenter, and the circumcenter. The

perpendiculars from on the triangle’s sides cut in . Prove that the

following inequality holds:

:∑

; :∑

; √ ( ) ( ) ( )


X.56. Let be a triangle with the perimeter . Prove that the following inequality holds:

:∑√

;:∑

;

√




X.57. Prove that in any triangle the following inequality holds:

( ) ∑:

;

( )

Proposed by Radu Diaconu – Romania

X.58. Prove that in any triangle where , the following inequality holds:

∑( √

*

√∑

√

with the usual notations in triangle. Proposed by Radu Diaconu – Romania

X.59. Let the distances from gravity center of triangle to the sides

and

. Prove that the following inequality holds:

:∑( )

;

(

∑

√

)

√

with the usual notations in triangle. Proposed by Radu Diaconu – Romania

X.60. Let be a convexe quadrilateral having the length’s sides , the perimeter

and . Prove that the following inequality holds:

:∑

( )

; :∑

;


X.61. Prove that in any triangle where , the following inequality holds:

:∑

√

;:∑

; √




X.62. In tringle let be the interior bisectors and their intersection point.

Prove that the following inequality holds:

6(

*

(

*

(

*

7 (

)

with the usual notations in triangle, and .


X.63. We denote with and the middle of , - , - and , - sides of an any

triangle. The sides and intersect the circumcenter, in and . Prove the

following inequality:

6(

*

(

*

(

*

7 6(

*

(

*

(

*

7

Proposed by Radu Diaconu, Emil Popa – Romania

X.64. If are different in pairs then:

( √

)

( √

)

( √

)

( √ )

Proposed by Daniel Sitaru,Sorin Dumitrescu – Romania

X.65. Solve for real numbers:

( ) ( )

Proposed by Daniel Sitaru,Patricia Anicuța Bețiu – Romania


√4

54

54

5

Proposed by Daniel Sitaru,Cristian Moanță – Romania




∑. √(( ) )(( ) )/

√

Proposed by Daniel Sitaru,Delia Popescu – Romania


.

/

.

( )

( )

( ) /

.

( )

( )

( ) /

Proposed by Daniel Sitaru,Delia Schneider – Romania


∑

:∑

; :∑

;

Proposed by Daniel Sitaru,Gilena Dobrică – Romania

X.70. If – sides in a bicentric quadrilateral with – inradii then:

Proposed by Daniel Sitaru,Nineta Oprescu – Romania

X.71. If * + then:

| √

|

| √

|

( )

Proposed by Daniel Sitaru,Dorina Goiceanu – Romania

X.72. If in then the following relationship holds:

(

√ *

Proposed by Daniel Sitaru,Mioara Mihaela Mirea – Romania

X.73. Solve for real numbers:




X.74. In – internal bisectors, – lies on circumcircle,

( ),( ), ( ) – collinear points. Prove that:



( )

4

5


X.76. In – internal bisectors, ( ) ( ) .

Prove that:

.

/


X.77. In – incenter, – centroid the following relationship holds:



(

* √


X.79. If then:


X.80. If then:



( ) √ 4

5



√ √

√ ( )


X.82. In – Bevan’s point, – Lemoine’s point, – excenters,

– circumradii of then the following relationship holds:

∑

√ ∏

√


X.83. In acute – Bevan’s point, – incenter the following relationship holds:

( )

( )

( )


X.84. In - incenter, – circumradii in the following

relationship holds find where:

.



∑

∑





∑

∑

∑




X.88. If in – Nagel’s cevian then:

( ) ( )


X.89. In – Bevan’s point the following relationship holds:

( )


X.90. Let and be non – negative real numbers and ∑ , - . Prove

that:

∑, -

∏, -

, - is the greatest integer part of .

Proposed by Jalil Hajimir-Canada

X.91. In – Bevan’s point the following relationship holds:



∑

∑ ( )

∑

.

/


X.92. In – incenter, – Gergonne’s cevians the following relationship holds:

∑

∑

∑


X.93. In – incenter, – Bevan’s point the following relationship holds:



√ √ √ √

√


X.95. If in acute – circumcevians, – circumcenter then:



∑:

√

;




X.97. In - sides of intouch triangle. Prove that:

√ √

√

√

√




X.99. In acute – circumcevians. Prove that:



∑

∑(

*

∑


X.101. In – Bevan’s point, – incenter, – sides of extouch triangle the

following relationship holds:

∑

∑





∑

∑:

.

/;


X.103. In – incenter, – Bevan’s point. Prove that:



√

√ ( )



√

√

√


X.106. In – Nagel’s cevians the following relationship holds:


X.107. In – Bevan’s point, the following relationship holds:





( )

: ∑

( )

;


X.109. In – Gergonne’s cevian, – Nagel’s cevian. Prove that:

(

)



∑(

*

( )


X.111. In – area the following relationship holds:

( ) ( ) ( ) √







∑(

*

∑( )



∑

∑( )( )

( ) ( )( )

∑

∑ ( )

( ) ( )( )


X.115. Let be the circumcevians, – incenter in acute . Prove that:


X.117. TAREK’S LEMMAS: In – Nagels’ cevians. Prove that:

√ ( )


X.118. If in – sides of intouch triangle, – sides of extouch triangle,

– are then the following relationship holds:



( ) (

*





( ) (

*



√

√

√

√


X.122. Let be the extouch triangle of . Prove that:

( )

( )

( )




11-CLASS-STANDARD

XI.1. Let such that . Prove that:

√

√

√

√


XI.2. ( ) ( ) . Find: ( )


XI.3. If ( ) – symmetric, invertible then:



( ) Proposed by Marian Ursărescu – Romania

XI.4 If ( ) then: , ( ) ( ) -


XI.5.

√

√

√ . Find:

√ ∏4

√ 5


XI.6. If ( ) ( ) then: ( ) ( ) Proposed by Marian Ursărescu – Romania

XI.7. If ( ) ( ) then find:

( ) Proposed by Marian Ursărescu – Romania

XI.8. ( ) (( ) ) ( ) . Find: ,( ) -


XI.9. √ . Find:

√

Proposed by Marian Ursărescu – Romania XI.10. Find ( ) such that:

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

XI.11. If ( ) then:

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

XI.12. If then:

√

√

√

Proposed by Daniel Sitaru,Nicolae Oprescu – Romania

XI.13. If * + then:

4√

√( )

√

√( )

√

√( ) 5

( )( )( )

( )( )( )

Proposed by Daniel Sitaru,Daniela Beldea – Romania

XI.14. Solve for real numbers:



|

|

Proposed by Daniel Sitaru,Marian Voinea – Romania

XI.15. If (( )) ( ) ( ) ( ) ( ) then:

(

* (

* ( ) ( )

Proposed by Daniel Sitaru,Amelia Curcă Năstăselu – Romania XI.16. Find:

4√( ) √( )

√( ) √

5

Proposed by Daniel Sitaru,Alecu Orlando – Romania XI.17. . Find:

( )

(

∑(

)

+

Proposed by Daniel Sitaru,Mihaela Stăncele – Romania

XI.18. If

then:

( √ )

(√ )

Proposed by Daniel Sitaru,Eugenia Turcu – Romania

XI.19. Let be . Solve in the following equation: ( )


XI.20. Let be the real fixed number and such that . Find the

maximum and the minimum of the expression: √

√


XI.21 Generalization from Kvant: If * + then:

∏(

)

:∏( )

;

Proposed by Daniel Sitaru,Laura Zaharia – Romania

XI.22. If then:

∑

.

/

.

/

Proposed by Daniel Sitaru;Ileana Stanciu – Romania



XI.23. Find 0

/ such that:

|

|




12-CLASS-STANDARD

XII.1. Let , - a continuous function that admits the symmetry center of point

.

/ graphic. Let be a natural odd number. Find: ∫ ( )

.


XII.2. Find such that divide ( ) ( ) .


XII.3. ( ) , -

If then: .

/ ( )

( )


XII.4.

∫( ) ( )

Find such that .


XII.5. Find:



:

∑ (( ) )

;

Proposed by Daniel Sitaru,Ionuț Ivănescu – Romania

XII.6. If

then:

:∫

√

;

(

∫

)

:∫

√

;

(

∫

)

Proposed by Daniel Sitaru,Ileana Duma – Romania

XII.7. Find:

( ) ∫4( )( )

( )( )5

Proposed by Daniel Sitaru – Romania

XII.8. Prove without computer:

∫

∫

∫

∫

∫

Proposed by Daniel Sitaru,Mădălina Giurgescu – Romania

XII.9. Find:

(

∑(, - [ , -])

+

– golden ration, , - - great integer function

Proposed by Daniel Sitaru,Lucian Tuțescu – Romania

XII.10. Solve for : ∫ . ( )/

.

Proposed by Daniel Sitaru,Lucian Lazăr – Romania

XII.11. , - ( ) – continuous. If: ∫ ( )

then:



∫

( ) ∫ ( )

Proposed by Daniel Sitaru,Alina Tigae – Romania

XII.12. Solve for real numbers:

∫(

*

(

*

Proposed by Daniel Sitaru,Aurel Chriță – Romania

XII.13. If

then:

∫( )




UNDERGRADUATE PROBLEMS

U.1. Find:

∫( ( ) ( ( )) )

Proposed by Ajao Yinka-Nigeria

U.2. Find:



:∫ ( )

√

;


U.3. Prove that:

( ) ∫

( )

∑ 8∏ ( )

[

( ) ]9

Hence evaluate ( ) where ( )

* +


U.4. If then:

∫ ∫∫.

/

( )

Proposed by Daniel Sitaru,Lavinia Trincu – Romania

U.5. If then:

∫:∫ :∫(√( )( )( )

( )( )( )+

;

;

(

*

Proposed by Daniel Sitaru,Camelia Dană – Romania

U.6. If

then:

∫∫ 4 ( )

( )5

√ ( )

Proposed by Daniel Sitaru,Luiza Cremeneanu – Romania

U.7. Let are acute angles,

( ) ( ) ( ) ( ) ( ( ) ( ) ( ) ( )) √

then find the value of ( ( ) ( ) ( ) ( ))

Proposed by K. Srinivasa Raghava-AIRMC-India

U.8. Let are the sides of a triangle and for then show that:

( )( )( )

( )( )( ) ( ) where

( )


U.9. Let for



( ) ∫ ( )

( )

then evaluate the sum ∑

( )


U.10. Let be the Ramanujan Matrix which is given by Pr. Srinivasa Raghava:

(

, prove that

(

,

(

)

Where are the Eigen Values of , and

FETI SINANI


U.11. If we define the function ( ) for any real number

( ) ∫ 4 ( )

5

then prove the limits

∫ ( )

∫ ( )


U.12. Prove this sharp inequality:

∫ (

*

( )




U.13. If

( ∫

( )

( )

+ ( )

then for any integer , show that: √

4

(√ )5

√

( )( ) is Fourier Transform with variable .


U.14. Evaluate the integral in a closed – form:

∫( ( ))

( )

.

/


U.15. Evaluate the following limit in a closed form:

(∑( ) ( )

( )( ) ( )

+


U.16. If we define:

( ) ∫( ) ( )

then prove that:

( ) ( ) ( )

( )

( )


U.17. If , - denotes the greatest integer not exceeding , then prove that:

∑, -

√


U.18. If we define the function ( ) for any with

, -



( ) ∫ 4 ( )

5

√

Then show that:

∑ (

( )+

(

*4 (

* (

*5

where ( ) – Hurwitz Zeta Function.


U.19. If, for

( ) ∫ ( )

( )

then evaluate the integral in a closed – form:

∫ ( )

( ) ( )


U.20. If √

. Evaluate the product in a closed form:

∏

and establish this inequality

∏

( )


U.21. Let ( ) √

( )

( ) √

Prove that:

( ) √ √ √ √

Proposed by Mohammed Bouras – Maroc

U.22. Prove that:



4√ √

√ √

5

√

Proposed by Mohammed Bouras – Maroc

U.23. Find:

∫∫(

. .

//

*

Proposed by Abdul Mukhtar-Nigeria

U.24. Find:

∑:∫∫4 ( )

5

;


U.25. Find:

∫ 4 ( )

√ √

5


U.26.

∫( )


U.27. Find:

∫( ∑( )

+


U.28.



∫( ) ( )


U.29.

∫ ( )

( ) .√ /


U.30.

∫( ( ) ) ( )

( )


U.31. Prove that:

∫ ( )

( )

∫ ( )

( )

: ( ) .

/

;

Proposed by Abdul Hafeez Ayinde-Nigeria

U.32. Find:

∫( )


U.33.

∫∫ ( )

√

.

/ 4

5


U.34. Prove that:



∫ ( )

.

( )

( )/ 4

( )

.

/5, for


U.35. Integral form for Prof. Dan Sitaru’s sum

Let be an even integer then generalize for

( ) ∫

and prove that:

( )

√ (

*


U.36. If:

( ) (∑( ( ))

+ ( )

then:

∫ (( ( )) 4 (

*5

+

.√ √ √ √ √ /

√ √

( √ √ )

√ √


U.37.

∫

then show that for



∫ (

√ *

(

√ * (

√ *

and evaluate in closed form.


U.38. For all , if

( ) {∑

}

then show that:

∑ 4

( ) ( )

( ( ) ) ( )

( )( ( ) )5

√ 4

√

5

Note: * + does not represent fractional part.


U.39. Find:

∑ :∑( )

( ( ));

Proposed by Daniel Sitaru,Constantina Prunaru – Romania

U.40. Find:

∑ (∑ ( (

* (

( ) **

+

Proposed by Daniel Sitaru,Alina Georgiana Ghiță – Romania

U.41. If – continuous, ( ) ( ) ( ) then:

∫∫∫ ( )

∫ ( )

Proposed by Daniel Sitaru,Mihaela Stăncele – Romania



U.42. If ( ) then:

∫ ∫4( ( ) ( ) 4 ( ) ( )

55

( ) ∫( ( ) ( ( )))

Proposed by Daniel Sitaru,Mihaela Dăianu – Romania

U.43. Inspired by Narendra Bhandari. Find:

:∫4

5

;


U.44. Find:

∑( )

( ) (

*


U.45. Prove that:

∑( )

(

*

( )

where is zeta function.


U.46. Evaluate ∫ .

/


U.47.

( ) ∫( ( ))

√( )



Find:

( )




ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-SPRING 2021

PROBLEMS FOR JUNIORS

JP.286. If then:

( )( )

( )( )

( )( )

Proposed by Daniel Sitaru-Romania

JP.287. If .

/ then:

:∏ (

;:∏ (

);

Proposed by Daniel Sitaru-Romania JP.288. Solve for real numbers:

√

(√ √ )

Proposed by Hoang Le Nhat Tung -Vietnam



JP.289. If ( ) then:

( )( )( ) √ ( )


JP.290. In the following relationship holds:


JP.291. Solve for real numbers:

{

√

√

√

√


JP.292. If then:

Proposed by Marin Chirciu-Romania

JP.293. If

then:

( )( )( ) ( ) Proposed by Marin Chirciu-Romania

JP.294. If then:

( )

( )

( )

(√ √ √ )


JP.295. If then:

( )

( )

( )

(√ √ √ )


JP.296. If then:

( )

Proposed by Marin Chirciu-Romania



JP.297. If then find:

( ) ( )

( )

( )

( )


JP.298. If such that:

{

( )

( )

( )

Proposed by Daniel Sitaru-Romania JP.299. If then:

( )

( )

( )

( )

( )

( )


JP.300. In incenter, symedians in ( )

( ) ( ) Prove that: , -

, -

Proposed by Marian Ursărescu-Romania

PROBLEMS FOR SENIORS

SP.286. In the following relationship holds:

√

√( )

√

√

√

√

√( )

Proposed by George Apostolopoulos-Greece


( )

( )


SP.288. If then find:

:

√

√

√

;


SP.289. If then in the following relationship holds:

∑

∑




SP.290. If then:

√ √ √ ( )

When equality holds? Proposed by Hoang Le Nhat Tung -Vietnam

SP.291. In acute let be the altitudes of orthic triangle. Prove that:

√ :∑

;:∑

( )

;

Proposed by Radu Diaconu-Romania

SP.292. Let be a tangential quadrilateral circumbscribed to a circle with radii Prove that:

, - ∑ ( )√ ( ) ( ) ( )

√

:∑

;


SP.293. If then:

( ) (

*

( )(√ )

√ ( ) (

*


SP.294. If then:

√ ( )

√ ( )

√ ( )



( )( )( ) ( √ )

Proposed by Florentin Vişescu-Romania

SP.296. In acute orthocenter, circumcentre, incenter the following relationship holds:

∑( )

( )

( )




SP.297. If then:

√ √

√

∑

√ ( )


SP.298. Find without softs:

∫( ( ) (

* ( )*

Proposed by Pedro Pantoja -Brasil SP.299. Solve for real numbers:

8√ √( ) √ ( ) √( ) ( )


SP.300. If ( ) then find: ( )

Proposed by Marian Ursărescu-Romania

UNDERGRADUATE PROBLEMS

UP.286. If then:

∫( )

√ √( √ ) ( √ )

Proposed by Florică Anastase-Romania UP.287. If then:

(

* ∫

√

(√ )

Proposed by Daniel Sitaru-Romania UP.288. Solve for real numbers:

{

√ ( )

√ √




UP.289. If then:

(

*

∫( ( ))

∫( ( ))

( ( ))


UP.290. If (, -) convexe, then:

∫ ( )

∫ ( )

∫ ( )


UP.291. Prove that:

∫( ( ) (

**

:∫ ( )

;


UP.292. Prove that:

∫.

/

∫

( )

( )


UP.293. Solve for real numbers:

{

√

√


UP.294. In medians; centroid; ( )

Prove that: is a cyclic quadrilateral. Proposed by Marian Ursărescu-Romania

UP.295. Find such that:

{.

/ .

/ .

/

.

/

.

/

.

/




UP.296. If then:

√

( )

√

( )

√

( )

Proposed by Pedro Pantoja-Brazil

UP.297. If ( )

then find:

√ √ √ √

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

UP.298. Find:

. √( ) ( )

√( )

/


UP.299. If fixed values; ( ) ( )

( )

( ) then find:

4( )

√

√

5


UP.300. If ( ) ( ) ( )

( ) and exist

( ( ))

then find: (( ( ))

( ( ))

( ) + ( ( ))






INDEX OF AUTHORS RMM-26

Nr.crt. Numele și prenumele Nr.crt. Numele și prenumele 1 DANIEL SITARU-ROMANIA 45 RADU DIACONU-ROMANIA

2 D.M.BĂTINEȚU-GIURGIU-ROMANIA 46 EMIL POPA-ROMANIA

3 FLORICĂ ANASTASE-ROMANIA 47 MOHAMMED BOURAS – MAROC

4 NECULAI STANCIU-ROMANIA 48 PEDRO PANTOJA -BRASIL

5 MARIAN URSĂRESCU-ROMANIA 49 AJAO YINKA-NIGERIA

6 BOGDAN FUSTEI-ROMANIA 50 HOANG LE NHAT TUNG-VIETNAM

7 DAN NĂNUȚI-ROMANIA 51 GEORGE APOSTOLOPOULOS-GREECE 8 MARIN CHIRCIU-ROMANIA 52 MEHMET ȘAHIN-TURKEY

9 VASILE BURUIANĂ-ROMANIA 53 SEYRAN IBRAHIMOV-AZERBAIJAN

10 FLORENTIN VIȘESCU-ROMANIA 54 ABDUL MUKHTAR-NIGERIA

11 GHEORGHE CALAFETEANU-ROMANIA 55 SRINIVASA RAGHAVA-INDIA

12 PETRE STÂNGESCU-ROMANIA 56 NAREN BHANDARI-NEPAL

13 BENNY LÊ VĂN 57 ABDUL HAFEEZ AYINDE-NIGERIA

14 ȘTEFAN MARICA - ROMANIA 58 ADIL ABDULLAYEV-AZERBAIJAN

15 CONSTANTIN IONICĂ-ROMANIA 59 MUSTAFA TAREK-EGYPT

16 DANIEL STRETCU - ROMANIA 60 JALIL HAJIMIR-CANADA

17 OCTAVIAN STROE-ROMANIA 61 ROVSEN PIRGULIYEV-AZERBAIJAN

18 VIRGINIA GRIGORESCU-ROMANIA 62 MIHAELA STĂNCELE-ROMANIA

19 SORIN PÎRLEA-ROMANIA 63 MIHAELA DĂIANU-ROMANIA

20 SORIN DUMITRESCU-ROMANIA 64 LUIZA CREMENEANU-ROMANIA

21 SIMONA RADU-ROMANIA 65 LUCIAN TUȚESCU-ROMANIA

22 ROXANA VASILE-ROMANIA 66 LUCIAN LAZĂR-ROMANIA

23 PAULA ȚUINEA-ROMANIA 67 LAVINIA TRINCU-ROMANIA

24 PATRICIA ANICUȚA BEȚIU-ROMANIA 68 LAURA ZAHARIA-ROMANIA

25 NINETA OPRESCU-ROMANIA 69 IULIA SANDA-ROMANIA 26 NICOLAE RADU-ROMANIA 70 IONUȚ IVANESCU-ROMANIA

27 NICOLAE OPREA-ROMANIA 71 ILEANA STANCIU-ROMANIA

28 MIOARA MIHAELA MIREA-ROMANIA 72 ILEANA DUMA-ROMANIA

29 GILENA DOBRICĂ-ROMANIA 73 GABRIELA VASILE-ROMANIA

30 DORINA GOICEANU-ROMANIA 74 GABRIEL TICA-ROMANIA

31 DELIA SHNEIDER-ROMANIA 75 EUGENIA TURCU-ROMANIA

32 DELIA POPESCU-ROMANIA 76 ELENA NICU-ROMANIA

33 CATALIN SPIRIDON-ROMANIA 77 ELENA IACOB MEDA-ROMANIA

34 DANIELA BELDEA-ROMANIA 78 ELENA ALEXIE-ROMANIA

35 CRISTIAN MOANȚĂ-ROMANIA 79 DOINA CRISTINA CĂLINA-ROMANIA

36 CONSTANTINA PRUNARU-ROMANIA 80 DANA COTFASĂ-ROMANIA

37 CRISTINA SPIRIDON-ROMANIA 81 DAN GRIGORIE-ROMANIA

38 MARIAN VOINEA-ROMANIA 82 CLAUDIU CIULCU-ROMANIA

39 CAMELIA DANĂ-ROMANIA 83 CĂTĂLIN PANĂ-ROMANIA-ROMANIA

40 AMELIA CURCĂ NĂSTĂSELU-ROMANIA 84 AUREL CHIRIȚĂ-ROMANIA

41 ALINA GEORGIANA GHIȚĂ-ROMANIA 85 ALINA TIGAE-ROMANIA

42 ALECU ORLANDO-ROMANIA 86 MĂDĂLINA GIURGESCU-ROMANIA

43 NICOLAE MUȘUROIA-ROMANIA 87 GHEORGHE BOROICA-ROMANIA

44 GHEORGHE STOICA-ROMANIA 88 CARINA MARIA VIESPESCU-ROMANIA NOTĂ: Pentru a publica probleme propuse, articole și note matematice în RMM puteți trimite materialele pe mailul: [email protected]

mailto:[email protected]

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