romanian mathematical magazine solutions · 2019. 7. 31. · romanian mathematical magazine...

ROMANIAN MATHEMATICAL MAGAZINE

Founding EditorDANIEL SITARU

Available onlinewww.ssmrmh.ro

ISSN-L 2501-0099

Number 17 SUMMER 2020

SOLUTIONS

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SOLUTIONS

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Proposed by Daniel Sitaru-Romania

Nguyen Viet Hung-Hanoi-Vietnam

Marin Chirciu-Romania

Florentin Vișescu – Romania

D.M. Bătinețu – Giurgiu – Romania

Neculai Stanciu – Romania

Pedro H. O. Pantoja – Natal/RN – Brazil

Marian Ursărescu – Romania

Vasile Mircea Popa – Romania

Marian Voinea-Romania

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Solutions by Daniel Sitaru-Romania,Florentin Vișescu-Romania

Ravi Prakash-New Delhi-India,Michael Sterghiou-Greece

Marian Ursărescu-Romania,Nguyen Viet Hung – Hanoi – Vietnam, Do Chinh-

Vietnam,Michael Sterghiou-Greece,Marin Chirciu-Romania

Orlando Irahola Ortega-La Paz-Bolivia,D.M. Bătinețu – Giurgiu – Romania

Avishek Mitra-West Bengal-India,Soumava Chakraborty-Kolkata-India

Khaled Abd Imouti-Damascus-Syria,Pedro H. O. Pantoja – Natal/RN – Brazil

Adrian Popa – Romania, Rohan Shinde-India,Srinivasa Raghava-AIRMC-India

Naren Bhandari-Bajura-Nepal, Remus Florin Stanca-Romania

Mokhtar Khassani-Mostaganem-Algerie, Ali Jaffal-Lebanon, Sanong

Huayrerai-Nakon Pathom-Thailand,Marian Voinea-Romania,

Neculai Stanciu – Romania,Soumitra Mandal-Chandar Nagore- India

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JP.241 Let , , be the circumpedal extensions of cevians of centroid in

. Prove that:

+ + ≥ √


Solution by proposer

= ; = ; = ; ∩ ∩ = { }

= =

( ) = ⋅ = ⋅ (the power in circumcircle)

⋅ = ⋅ ⇒ =

= = + = +

Analogous: = + ; = +

+ + = + + + + + ≥

≥ ⋅ ⋅ ⋅ = √

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JP.242 Let , , be the circumpedal extensions of cevians of centroid in

. Prove that:

≥



= ; = ; = ; ∩ ∩ = { }

= =

( ) = ⋅ = ⋅ (the power of in circumcircle)

⋅ = ⋅ ⇒ =

= = + = + ≥ ⋅ = ⋅ ⋅ =

Analogous: ≥ ; ≥ . By multiplying: ≥

JP.243. If , , > 1; = 8 then:

+ + ≥


Solution 1 by Florentin Vișescu-Romania

Let be : ( , ) → ℝ, ( ) = = ( )

( ) = ( − + ) ( )

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( ) = ( − + ) + ( ) > 0

−convexe. By Jensen’s inequality: ≤ ( ( ) + ( ) + ( ))

+ +≤ + +

+ + ≥+ +

≥⏞

≥ =√

= ∙ ( ) =

Equality holds for = = = .

Solution 2 by proposer

= + − ≥ + − ≥ −

⇔ + − ≥ −

− + ≥ ⇔ − + ≥ ⇔ ( − ) ≥

≥ − ; ≥ − ; ≥ −

By adding: + + ≥ + + − ≥

≥ − = √ − = ⋅ − =

Equality holds for = = = .

JP.244 If , , ∈ ℝ then:

{ , , } ≥ { , , } + + + − − −

When equality holds?

Proposed by Nguyen Viet Hung-Hanoi-Vietnam

Solution by Ravi Prakash-New Delhi-India

WLOG: ≥ ≥ . Hence: { , , } = , { , , } =

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( − )( − ) ≤ → − − + ≤

( − ) ≥ ( − ) + ( − )( − )

( − ) ≥ − + + − − +

( − ) ≥ + + − − −

− ≥ + + − − −

{ , , }− { , , } ≥ + + − − −

JP.245. Let , , be non-negative real numbers such that + + = .

Find the minimum and maximum value of .

= √ + + + + √ +

Proposed by Nguyen Viet Hung – Hanoi – Vietnam

Solution by Michael Sterghiou-Greece

= ∑ √ + (1)

Let ( , , ) = ∑ ,∑ , . The function ( ) = √ is convex therefore by

Jensen ≤ + (2) But = ∑ ≥ → ≤ √ so √ becomes ≤ + √

which is the required maximum. Equality for = = = √ . We will show that

≥ + √ (3)

(2) becomes: ∑ √ ≥ + √ ↔

↔ + + ( + )( + ) ≥ + √

or ∑ ( + )( + ) ≥ √ ( ) = ( ) → ∑ ( + )( + ) +

+ ∑ √ + ∏ √ + ≥ ( ) → ( + + ) + + + + ≥ ( )

which becomes ≥ ( ) = ( ) ( ) with equivalence through as all squared

amounts are positive. As ( ) is a decreasing function of according to V. Cîrtoaje

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theorem with , fixed becomes maximal when = assuming WLOG ≤ ≤ .

It is enough therefore to show (3) for = in which case + = or ≤ .

(3) → + + + − ≥ + √ with ∈ ,√

. Let

( ) = + + + − − + √ . ( ) = − where

= + − − ⋅ ( − ) + and > 0. It is easy to show that

( ) > 0 [sum of positive factors] so ( ) has only one root equal to √ . This root is

unique for ( ) and the minimum value of ( ) on ,√

is ( ) = , hence

( ) ≥ . Equality for = = , = and permutations + √ ≤ ≤ + √

JP.246 If , , , > 0; + + + = then:

− + − + − + − =

Proposed by Daniel Sitaru – Romania


+ + + = ↔ + + + =

, , , =+ + +

= =

, , , = ∙ ∙ ∙ =

, , , = , , , → = = = = → =


If > 0 ⇒ ( − ) ( + ) ≥

( − + )( + ) ≥

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+ − − + + ≥

− − + ≥ ⇒ ≥ + −

For = and successively = ; = ; = then:

≥ + − (1)

≥ + − (2)

≥ + − (3)

≥ + − (4)

By adding (1); (2); (3); (4):

≥ + − = + + + + − =

= ++ + + −

=

= +−

=

For = = = = , = which results by condition from enunciation.

JP.247. Let , , be non-negative real numbers such that + + = .

Prove that:

√ + √ + √ + + + ≥ ( + + )


Solution 1 by Marian Ursărescu-Romania

Let √ = ,√ = ,√ = , , , ≥ , + + = .We must prove:

( + + ) + + + ≥ ( + + ), ( )

Let = , = , =

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( ) ⇔( + + )

+ + ++ +

( + + ) ≥( + + )

( + + )

( + + ) ( + + ) + + + − ( + + ) ≥ ( )

− , ( , , ) ≥ ⇔ ( , , ) ≥

( , , ) = ( + + ) ( + + ) + + + − ( + + ),∀ , , ≥

(Cîrtoaje’s theorem)

( , , ) = ( + + ) ( + + ) + + + − ( + + )

( , , ) = ( + ) ( + ) + + − ( + ) =

= + − + = ( − ) ( + ) ≥


We rewrite the inequality in homogenous form as follows:

√ + √ + √ ( + + ) + + + ≥ ( + + )

This is equivalent to:

√ + √ + √ ( + + ) + + + ≥ ( + + )

After replacing ( , , ) by ( , , ) the inequality becomes:

+ + + ( + + )( + + ) ≥ ( + + )

Or equivalently

+ + + ( + + ) + ( + ) ≥ ( + + )

According to Schur’s inequality of fourth degree we have:

+ + + ( + + ) ≥ ( + )

Hence it suffices to prove:

( + ) ≥ ( + + )

But this is true because:

( + )− ( + + ) = ( − ) ≥

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JP.248. If , , > 0 then:

+ + + + + + + + + + + ≥ ( + + )



By Bergstrom’s inequality: + + ≥( )

Remains to prove:

( + + )( + + ) + − ( + + )( + + ) ≥ , ( )

Reminder Cîrtoaje’s theorem:

If ( , , ) is a symmetric and homogenous polynomial og degree then:

( , , ) ≥ ⇔ ( , , ) ≥ , ( , , ) ≥ ,∀ ≥

In our case ( , , ) = + − =

( , , ) = ( + )( + ) − ( + ) = ( − ) ( + ) ≥ ⇒ ( )

Solution 2 by Do Chinh-Vietnam

+ + ≥⏞ ∙ ( + + ) = + +

Remains to prove:

( + + )( + + ) + − ( + + )( + + ) ≥

+ ≥ ( + )

( + )− − ≤

( + − )( + − )( + − ) ≤

( + − )( + − )( + − ) ≤ | + − | ∙ | + − | ∙ | + − | =

= | + − | ∙ | + − | ≤⏞( )

| | =

Observation: ( ) is true if , , are sides in triangle.

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Solution 3 by Michael Sterghiou-Greece

+ + + + + + + + + + + ≥ ( + + )( )

Inequality is homogenous so WLOG we can assume:

+ + = , ( , , ) = ( + + , + + , ), =

of ( ) ≥⏞ − + ∙ ≥ (to prove)

This becomes the stronger inequality: − + ≥ ⇔ − + ≥

Which is -rd degree Schur’s inequality-usually written: ≤

Solution 4 and generalizations by Marin Chirciu-Romania

1) If , , > 0 then:

+ + + + + + + + + + + ≥ ( + + )


Solution

Schur’s lemma:

2) If , , > 0 then:

+ + + + + ≥ ( + + )

Proof.

Inequality can be written:

+ + + ≥ ( + ) + ( + ) + ( + ), which follows by Schur’s

inequality:

( − )( − ) + ( − )( − ) + ( − )( − ) ≥ , , , > 0, > 0, for

= . Equality holds for = = .

Back to the main problem: By Bergström’s inequality:

+ + ≥ ( )∑( )

=( )

(1)

By (1) and Schur’s lemma:

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+ + + + + + + + + + + ≥

≥ + + + + + ≥ ( + + )

+ + + + + + + + + + + ≥ ( + + )

Equality holds for = = .

Remark:

3) If , , > 0 and ≥ then:

+ + + ( + )+ +

++ +

++ +

≥ ( + + )

Marin Chirciu

Solution:

By Bergström’s inequality: + + ≥ ( )∑( )

=( )( )

(1)

By (1) and Schur’s lemma: + + + ( + ) + + ≥

≥ + + + + + ≥ ( + + )

+ + + ( + )+ +

++ +

++ +

≥ ( + + )


4) If , , > 0 then:

+ + + + + + + + ≥ ( + + )

Marin Chirciu

Solution


=( )

(1)


+ + + + + + + + ≥ + + + + + ≥

≥ ( + + )

+ + + + + + + + ≥ ( + + )

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5) In the following relationship holds:

+ + + + − + + − + + − ≥ ( + + )

Marin Chirciu

Solution


= (1)


+ + + + − + + − + + − ≥ + + + + + ≥

≥ ( + + )

+ + + + − + + − + + − ≥ ( + + )


6) In , ≤ the following relationship holds:

+ + + ( − )+ − + + − + + − ≥

≥ ( + + )

Marin Chirciu

Solution

By Begström’s inequality: + + ≥ ( )∑( )

=( )( )

(1)


+ + + ( − )+ − + + − + + − ≥

≥ + + + + + ≥ ( + + )

+ + + ( − )+ − + + − + + − ≥

≥ ( + + )


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++

≥

Marin Chirciu

Solution

By inequality 4) above, we replace ( , , ) with , , and we use the

known relationships: ∏ = ,∑ = . Equality holds for = = .

Solution (alternative):

By the known relationships:

=+

− ,+

= ++

, =

The inequality can be written:

− + ⋅ + ≥ ⇔ + ≥ − , which follows by

Blundon-Gerretsen’s inequality ≤ ( )( )

. It remains to prove:

( )( )( )

+ ≥ − ⇔ ≥ (Euler). Equality holds for = = .


+ ≥+

Marin Chirciu

Solution

Taking = − in the above inequality. Equality holds for = = .

Solution (alternative):


=+

− ,+

= ++

, =

Inequality can be written:

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+ − ≥ ⋅ + ⇔ + − ≥ , obviously by

Euler’s inequality ≥ . Equality holds for = = .

9) If in ,− ≤ ≤ the following relationship holds:

++

≥ +

Marin Chirciu

Solution


=+

− ,+

= ++

, =

The inequality can be written:

− + ⋅ + ≥ + ⇔ + ≥ + − , which

follows by Blundon – Gerretsen’s inequality ≤ ( )( )

. It remains to prove:

( + )( + )( − )

+ ≥ + − ⇔ ( − ) + ( − ) − ≥ ⇔

⇔ ( − )[( − ) + ] ≥ , obviously by Euler’s inequality ≥ and

− ≤ ≤ , which assure us [( − ) + ] ≥


JP.249. ABOUT PROBLEM JP.249

17 RMM SUMMER EDITION 2020

By Marin Chirciu – Romania

1) If , , > 0 then:

++

++

+≥

++

++

++


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Solution

By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

+ +≥

+=

+=

=( + )( + )( + )

≥ √ = ⋅ =


Remark.

We propose and solve a few problems of the same kind:


− + − + − ≥ − + − + − +

Marin Chirciu

Solution

By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

− − ≥ − = − = ∏( − ) =

= = ≥ √ = ⋅ =

Equality holds for an equilateral triangle.


−+

−+

−≥

−+

−+

−+

Marin Chirciu

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Solution

By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

− −≥

−=

−=

∏( − )=

= = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥



+ + ≥ + + +

Marin Chirciu

Solution

By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

≥ = = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥



+ + ≥ + + +

Marin Chirciu

Solution

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By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

≥ = = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≤ ⇔ ≤√



+ + ≥ + + +

Marin Chirciu

Solution

By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

≥ = = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ √ .



+ + ≥ + + +

Marin Chirciu

Solution

+ ≥ + ⇔ ≥

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≥ = = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ √



+ + ≥ + + +

Marin Chirciu

Solution

+ ≥ + ⇔ ≥

≥ = = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥

⇔ ≥ √



+ + ≥ + + +

Marin Chirciu

Solution

By squaring:

+ ≥ + ⇔ ≥

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≥ = = ≥ √ = ⋅ =



+ + ≥ + + +

Marin Chirciu

Solution

By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

≥ = = ≥ ⋅ =


11) In the following relationship hold:

+ + ≥ + + +

Marin Chirciu

Solution

By squaring:

∑ + ∑ ≥ ∑ + ⇔ ∑ ≥

≥ = = ≥

≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔

⇔ ≥ ⇔ ≥ √ .


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JP.250. ABOUT PROBLEM JP.250

17 RMM Summer Edition 2020

By Marin Chirciu – Romania

1) If , , > 0 are such that + + + ( ) + ( ) + ( ) ≤ , then:

+ + ≥


Solution

By Bergström’s inequality + ≥ , equality holds if = .

By summing: + + ≥ ∑ + (1)

≥ + + + ( + ) + ( + ) + ( + ) ≥( )

+ + ( + ) =

= ∑ ( ) +( )

≥ ∑ ( ) ⋅( )

= ∑ , hence ≥ ∑

So, ∑ ≤ (2)

Finally, ∑ ≥∑

≥( )

= . Equality holds if = = .

EXTENSION FOR 4 VARIABLES

2) If , , , > 0 are such that:

+ + + +( )

+( )

+( )

+( )

≤ , then:

+ + + ≥

Solution

By Bergström’s inequality + ≥ . Equality holds if = .

By summing: + + + ≥ ∑ + (1)

≥ + + + + ( + ) + ( + ) + ( + ) + ( + ) ≥( )

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≥ + + ( + ) =

= ∑ ( ) +( )

≥ ∑ ( ) ⋅( )

= ∑ , so ≥ ∑

Hence ∑ ≤ (2)

Finally, ∑ ≥∑

≥( )

=

Equality holds if and only if = = = .

EXTENSION FOR n VARIABLES:

3) If , , … , > 0 are such that:

+ + ⋯+ + ( + ) + ( + ) + ⋯+ ( + ) ≤

then: + + ⋯+ ≥ .

Marin Chirciu

Solution

By Bergstöm’s inequality + ≥ . Equality holds if = .

By summing − inequalities: + + ⋯+ ≥ ∑ + (1)

≥ + + ⋯+ + ( + ) + ( + ) + ⋯+ ( + ) ≥( )

≥ + + ( + ) =

=( + )

+ ( + ) ≥( + )

⋅ ( + ) =

So, ≥ ∑ , hence ∑ ≤ (2)

Finally, ∑ ≥∑

≥( )

= .

Equality holds if and only if = = ⋯ = .

Remark.

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For = we obtain problem JP.250 from 17-RMM SUMMER 2020, proposed by

Nguyen Viet Hung – Hanoi – Vietnam

JP.251. Let be , , , ∈ ( ,∞) with < < . Solve the following

equation:

( + ) + ( + ) + ( + ) = ( + ) + ( + ) + ( + )

Proposed by Florentin Vișescu – Romania


Obviously = is a solution. For ≠ . Considering the functions , : ( ,∞) → ℝ;

( ) = and ( ) = ( + ) . Applying Cauchy’s theorem on the intervals [ , ] and

[ , ]. Then ∃ ∈ ( , ) and ∈ ( , ) ( )( )

= ( ) ( )( ) ( )

and ( )( )

= ( ) ( )( ) ( )

or ( )

=( ) ( )

and ( )

=( ) ( )

we will prove that ( ) ( )

=( ) ( )

⇔ ( + ) − ( + ) − ( + ) + ( + ) = ( + ) − ( + ) −

− ( + ) + ( + ) ⇔

( + ) + ( + ) + ( + ) = ( + ) + ( + ) + ( + ) (true)

hypothesis. So, = ⇔ =

(as + ≠ + ; ≠ ) ⇒ − = ; = . Then, = { , }

JP.252. Solve for real numbers:

++ =

+− + − + + + +


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+ + + + − + + − =

= − ( + ) + + ( + ) − + ( + )

+ + − − ( + ) + + − −

− + ( + ) + + − ( + ) =

+ + − ( + )+ + ( + ) +

+ + + ( + )+ − ( + ) − +

+ + − ( + )+ + ( + ) =

+ ( + ) ( ) + − + ( + ) ( ) +

+ ( + ) ( ) =

− ( + )− − ( + )− ( + ) = ( + ) = → = -not solution because >

= → =

Solution 2 by Orlando Irahola Ortega-La Paz-Bolivia

+ + = + − + [ − ( + )] + [ + ( + )]

+ > 0 ∧ + > 0 ⇒ ∈ ] ,∞[

= + − ⇒ ( + + ) = + +

= − ( + ) ⇒ ( + + )( + + ) =

= + ( + ) ⇒ ( + )( + )( + ) = ⇒+ = ( )+ = ( )+ = ( )

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+ + = + +

(1) + = ⇒ = ⇒ = ( + )( + ) ⇒

⇒ = ∧ ( + + ) = ⇒ = ; + + ∈ ℂ

(2) + = ⇒ = ⇒ ( + ) = + ⇒

⇒ ( − )( + + ) = ⇒ =

(3) + = ⇒ = ⇒ = ; = { }

JP.253. If ∈ ℕ, ≥ and ∈ , ,∀ = , , then:

⋅ <

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

Solution by proposers

Because ∈ , it follows that or < , > ,∀ = , ⇔ or <

= < and then:

⋅ < ⋅ ≤

≤ ⋅ ⋅+

∙ ⋅<⏞_

∙∙ ⋅

∙ ⋅=

JP.254. Let be ∈ ℕ, ≥ and ∈ , ,∀ = , . Prove that:

⋅ <


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Because ∈ , it follows that or < and > ,∀ = , ⇔ or

< and = < ;∀ = , . So:

⋅ < <

<⏞ ⋅+

⋅ ⋅=

⋅

⋅= ⋅ =

JP.255. If ∈ ℕ then in the following relationship holds:

+ + + ≥ ( + )√

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

Solution 1 by Avishek Mitra-West Bengal-India

⇔ = +( )

= + + − ≥

≥ + + ( + ) − ( + )

⇒ ≥ + − − + ( + )

⇒ ≥ ( + ) ≥ ( + ) ( )

⇒ ≥ ( + ) ( ) ⋅

⇒ ≥ ( + ) ⋅ ⋅ ⋅ √ ⋅ ∵ ≥ , ≥ √

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⇒ ≥ ( + ) ⋅ ( ) ⋅ ( ) ⋅ ( ) ⋅√

∵ ≥ √

⇒ ≥ ( + ) ⋅ ⋅ ⋅ √

⇔ + ∑( )

≥ ( + )√ (proved)

Solution 2 by Soumava Chakraborty-Kolkata-India

= + + −

≥ + + ( + ) − (∵ + ≥ )

= + + ( + ) − −

= − + ( + ) = ( + )

=( − ) ( )

=∑ −∑

( + )

=( − − ) − ( − − )

( + )

= ( + )( + ) ≥ ⋅ ( + ) ≥ √ ⋅ ( + )

= √ ( + ) (Proved)

SP.241 Let , , be the circumpedal extensions of cevian of incentre in

. Prove that:

≥ ( + )( + )( + )


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= ; = ; = ; ∩ ∩ = { }

= ⇒ + = +

= + ⇒ = + ; = +

( ) = ⋅ = ⋅

+ ⋅ + = ⋅

= ( + )

= = + = + ( + ) ≥ ⋅ ( + ) = + √

Analogous: = √ ; = √

By multiplying: ≥ √ ⋅ √ ⋅ √ =

=⋅

( + )( + )( + ) ≥ ( + )( + )( + )

SP. 242 Let , , be the circumpedal extensions of cevians of incentre in

. Prove that:

+ + ≥ ( + )( + )( + )


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= , = , = , ∩ ∩ = { }

= ⇒ + = +

= + ⇒ = + ; = +

( ) = ⋅ = ⋅

+ ⋅ + = ⋅

= ( + )

= = + = + ( + )

Analogous: = +( )

; = +( )

By summing: + + = +( )

+ +( )

+ +( )

≥

≥ ⋅⋅ ⋅

( + ) ( + ) ( + ) =

= ( + ) ( + ) ( + ) = ( + )( + )( + )

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SP.243. Let be a polynomial such that ( ) + = ( − ), for all

real numbers. Prove that is constant.

Proposed by Pedro H. O. Pantoja – Natal/RN – Brazil

Solution 1 by Khaled Abd Imouti-Damascus-Syria

Suppose ( ) =

( )− + √ ( )− − √ = ( ) + − ( ),∀ ∈ ℝ

( ) − + √ ( ) − − √ = ( − ) − ( )

( ) − + √ ( )− − √ = [ ( − )− ( )] ,∀ ∈ ℝ

So: it must = = ⋯ = =

So: ( )− + √ ⋅ ( ) − − √ =

( ) = − √ = − √( ) = − √

( ) = − + √

or ( ) = + √ = + √( ) = + √

( ) = − − √

So: ( ) is constant


Suppose that is not a constant. Fixing ( ) = and comparing coefficients of

both sides we deduce that the coefficients of polynomial must be rational. On the

other hand, setting = with = − , that is, = ±√ , we obtain ( ) = ,

where − + = , i.e. = ± ± √ = ± ± √ . However, this is

impossible because ( ) must be of the form + √ for some rational , for the

coefficients of are rational. It follows that ( ) is constant.

SP.244. If < < < 2 then:

( + ) − < 3 √


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( + ) − <( )

√

(1)⇔ ( + ) ( − ) < 27

⇔ + − − − + > 0 =

⇔ ( − + ) + − − + > 0

⇔ ( − ) + − − + >( )

∵ ( − ) ≥ , ∴ it suffices to prove: − − + >( )

∵ < < < 2 ∴ < 1 < < 2 ⇒ 1 < < 2

Let − = of course, > 0

Then, (3)⇔ ( + ) − ( + ) − ( + ) + > 0

⇔ + + − − + >( )

Now, + ⋅ + ⋅ ≥ × ( ⋅ ) > 11 ⇒ − + >( )

Also, discriminant of − + = − ( ) < 0 ⇒ 7 − + >( )

and of course, ∵ > 0, 2 + >( )

(i)+(ii)+(iii)⇒(4)⇒(3)⇒(2)⇒(1) is true


( + ) − <?

√ , ⋅ + − <?

√

+ − <?

√ , + − <?

√

+ − <?

√ . Suppose: =

( + )√ − <?

√ , < < 2 ⇒ < < 2

< < 2, ∈ ] , [. Now, let us prove:

( + )√ − <?

√ , ∈ ] , [. Let be the function:

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( ) = ( + ) ⋅ √ − , ∈ ] , [, →

( ) = √ ,→

( ) =

( ) = ( + )√ − + ⋅ ( + ), ( ) = ( )

( ) = , ( ) = ; ( ) = ⇒ − − + + =

Let be the function: ( ) = − − + + , = ] , [

→( ) = ,

→( ) = −

( ) = − − + , ( ) = ⇒ − − + =

= − (− )( ) = + ( ); = ( + ) = ⋅ ⇒ √ =

=( )

= = + > 0, =( )

= = − < 0

√ = − √ − + √ + = −√ − < 0

= − − + + =−

− +

=− −

+ = > 0; √ = √ − < 0, = > 0

, √

( ) + + + + + + + + − − − − −−−

( )

,

√ +√

( )

( ) √ √ , ( ) , √ √ +

, ,

Not: Local max of ( ) on the interval [ , ] is nearly .

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So: ∀ ∈ ] , [: ( ) ≤ , ( + )√ − < 5, 5 < 3√ , ( ) < 5√

If < < < 2 then: ( + ) − <?

√

+ − <?

√ , + − <?

√

+ − <?

√ ; + − <?

√ . Suppose: =

⋅ ( + )√ − <?

√ , < < 2 ⇒ < < 2

< < 2, ∈ ] , [. Now, let us prove: ( + )√ − <?

√ , √ − <? √

( )

Let be the function: ( ) = √ − , ∈ ] , [

→[ ( )] = √ ,

→[ ( )] = ; ( ) =

−√ −

; ( ) = ⇒ = ∉ ] , [

( ) −−− − − −− −− −− −− −− −−−

( ) √

Let be the function: ( ) = √ , ] , [, →

[ ( )] = √ ,→

[ ( )] = √

( ) = √ ( ), ( ) = ⇒ = − ∉ ] , [

( ) −−− − − −− −− −− −− −− −−−

( ) √ √

∀ ∈ ] , [: ( ) < ( )

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Let’s consider ; = = ; = = ; = = −

< + ⇔ < − + ⇔ < (true)

< + ⇔ < + − ⇔ ( + ) > 2

But: ( + ) > ( + ) =

< + ⇔ − < + ⇔ < ( + )

⇔ + − > 0 ⇔ 2 − − > 0; =

= ; = ; = − ; ( − − ) = −

> 1 ⇒ 2 − − > 2 ⋅ − − = (true)

Denote – semiperimeter; – circumradii

= = = (1)

= = = = = < 1 (true)

= − = − =−

= =⋅ −

=−

By Mitrinovic’s inequality: ≤ √

By (1); (2): ( ) < √ ⋅

( + ) − < 3 √

(Equality doesn’t hold because can’t be an equilateral one:

< ⇒ < ⇒ < )

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SP.245. If < < < 2 then:

( + ) > 3( − ) ( − )



If < < < 2 then: ( + ) > 3( − ) ( − )

+ > 3 − −

+ > 3 − −

+ > 3 − −

Suppose: = > 0

( + ) > 3( − ) ( − )

< < < 2 ⇒ < < 2 ⇒ 1 < < 2

So: let us prove ∀ ∈ ] , [:

( + ) >?

( − ) ( − )

+− >

?√ ⋅ −

Let be the function: ( ) = , ] , [

→[ ( )] = +∞,

→[ ( )] =

( ) =( + )( − ) − ( )( + )

( − ) =− − − −

( − )

( ) =− −

( − ) , ( ) = ⇒ − − =

= − ( )(− ) = + = > 0,√ = √

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=− √

= − √ ∉ ] , [

=+ √

= + √ ∉ ] , [

( ) − −− −− −− −− −− − − −−− −− −

( ) +∞

Let be the function: ( ) = √ ⋅ √ − , ∈ ] , [

→( ) = √ ,

→[ ( )] =

( ) = √ ⋅ −√ −

; ( ) =− √√ −

( ) = ⇒ = ∉ ] , [

( ) − −− −− −− −− −− − −− −− −− −

( ) √

Note: ∀ ∈ ] , [; ( ) > ( )

+− > 3√ ⋅ −


( + ) >( )

( − ) ( − )

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∵ , + , − > 0 ∴ (1) ⇔

( − )( − ) − ( + ) > 0

⇔ ( − − + − ) > 0 =

⇔ − − + − >( )

∵ < < < 2 ,∴ < 1 < < 2 ⇒ > 1

Let − = of course, > 0

Then (2) ⇔ ( + ) − ( + ) − ( + ) + ( + ) − > 0

⇔ + + + >( )

Now, + ≥ √ >( )

=

∵ > 25,∴ 7 > ⇒ √ >

Also, + ≥ √ >( )

=

∵ > 169,∴ 45 > ⇒ √ >

(i)+(ii)⇒ + + + > 18 > 17

⇒ (3)⇒(2) ⇒(1) is true (Proved)


Let’s consider ; = = ; = = ; = = − .

< + ⇔ < − + ⇔ < (true)

< + ⇔ < + − ⇔ ( + ) > 2

But: ( + ) > ( + ) =

< + ⇔ − < + ⇔ < ( + )

⇔ + − > 0 ⇔ 2 − − > 0; =

= ; =+

= ; =−

= −

( − − ) = −

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> 1 ⇒ 2 − − > 2 ⋅ − − = (true)

Denote – semiperimeter; – area; – inradii

= = = (1)

=+ −

=+ ( − ) −

( − ) =

= = = < 1 (true)

= √ − = − = (2)

= = ⋅ ⋅ ( − ) ⋅−

= (2)

By Mitrinovic’s inequality:

> 3√ (Equality doesn’t hold because can’t be an equilateral one

< ⇒ < ⇒ < )

> 3√ ⋅ ⇒ > 3√

By (1); (2): ( ) > 3√ ⋅ ( )( )

( + ) > 3√ ( − ) − ( + )( − ) > ( − )

( + ) > 3( − ) ( − )

SP.246. If bicentric quadrilateral; – incircle then:

( + )( + ) ≥ ⋅ ⋅ ⋅


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=( )

+ , =( )

+ , =( )

+ , =( )

+

(1), (2), (3), (4) ⇒ ( + )( + ) =

= ( + + )( + + )

= ⋅ ( + ) + ⋅ ( + ) + ( + )( + ) +

≥( )

( + ) + ( + ) + ( + ) ( + ) +

Now, =( )

⇒ = ⇒ = √ (China, IMO TST, 2003)

= (∵ is bicentric)⇒ + =( ) ⋅

Similarly, using (a), + =( ) ⋅

(m), (n), (i)⇒ ( + )( + ) ≥

≥ ⋅⋅

+ ⋅⋅

+ ( ) + ≥

≥ + + ∙ ∙ ∙ = ( ⋅ + ⋅ )

= ⋅ ⋅ ⋅ (China IMO TST, 2003)

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Let , , , be sides in quadrilateral; = – semiperimeter; –

tangential quadrilateral ⇒ − = ; − = ; − = ; − = .

By Brahmagupta’s formula ( – cyclic):

[ ] = ( − )( − )( − )( − ) = √ (1)

√ ⋅ ⋅ ⋅ = √ =( )

[ ] = [ ] + [ ] + [ ] + [ ] =

=⋅ ⋅

+⋅ ⋅

+⋅ ⋅

+⋅ ⋅

≤

≤⋅

+⋅

+⋅

+⋅

=( + ) + ( + )

=

=( + )( + )

= ⋅+

⋅+

≤

≤ ⋅+

⋅+

= ( + )( + )

( + )( + ) ≥ √ ⋅ ⋅ ⋅

( + )( + ) ≥ ⋅ ⋅ ⋅

SP.247. In ; – incenter; , , – circumcenters of

, , . Prove that: + ≤ + + ≤ +

Proposed by Marian Ursărescu – Romania


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( ) =+

, = −+

⇒

cyclic ⇒ = ( ) = +

⇒ + = ⇒ cyclic ⇒ , , , , , are concyclic

We have: = = = ⇒

= = = ⇒ + ≤ ∑ ≤ +

For the left-hand side, we apply Popoviciu’s inequality for the concave function:

( ) = ⇒ = + ⇒

⇒+ +

++ +

≤+

= ⇒

+ ≤ ⇔ ≥ + + ⇒ ≥ +

For the right-hand side, we have: ∑ ≤ ∑

and ∑ = ∑( + ) = − ∑

≤ + = +


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From ,∠ = ° − ° − + ° − = = ° = ° −

From , = = ⇒ = ⇒ = =( )

Again, using , = = ⇒ = ⇒ =

⇒ =( )

. Similarly, =( )

∴ = − = =( ),( )

− +

= −( + )

= −( + )

= −+

= ∴ =( )

Applying sine rule on , °

=°

⇒ = (by (4))⇒ =( )

. Similarly, =( )

(1)+(5)+(6)⇒

=+

+

=⋅

( − ) ( − ) ⋅( − )

+ ⋅( − )

+

= ( − )( − ) ( − ) + ( − ) + −

= { ( + )− } + − =( + )−

+ −

=( − )( + )− ( − ) +

( − )

=( + )(∑ )( + − ) − ( + − ) + ( + − )

( − )

= ⋅{( + + ) − }

( − ) = ⋅{( + ) − }

( − )

= ⋅( ) ( − )

( − ) = ⋅ ⇒ = ⇒ =( )

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Similarly, =( )

and =( )

(i)+(ii)+(iii)⇒ ∑ = ∑ = ∑ =( )

∑ = ∑ ( )( )

≤ ( − )( − ) = ∑( − ( + ) + )

= − + + +

=+

=+

=+

= +

∴ ∑ ≤ + . Again, (iv) ⇒ ∑ = ∑ ≥ +

⇔ ≥+

=( + ) + ( + )

=+

+ +

⇔ ≥ + = ( + ) + = +

⇔ ∑ + ≤( )

∑ . Now, ∵ is acute, ∴ ( ) = ∀ ∈ , is

concave, ∴ applying Popoviciu’s inequality: ∑ + ≤ ∑

⇒ ∑ + ≤ ∑ ⇒ ∑ + ≤ ∑ ⇒ (a) is true ∴ ∑ ≥ +

(Hence proved)

SP.248. Let be ∈ (ℝ) such that = and = = . Find

.



The matrix being orthogonal one ⇒ | | = | | = | | = | | = | | =

( ) = − + − + − = (1)

= = (2)

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= ∗ = (( ) − ) = (3)

= = + + + +

= ⋅ + + + + = ⋅ ( + + + + )

= ⋅ = (A)

= = = ⋅

= ⋅ = (5)

From (1)+(2)+(3)+(4)+(5)⇒ ( ) = − ⇒

=( ) = ⇒ ( ) = ⇒ = ±

⇒

⇒ = ± ⇒ = ⇒ =

SP.249. Let be the circumcevian triangle of centroid in . Prove

that:

[ ][ ] ≤



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We have the following relationship: = ( )⋅ ⋅

= =

= (1)

But + + ≤ (2)

≥ √ ⋅ ⇒ ≥ (3)

From (1)+(2)+(3) ⇒ ≤ ⋅ (4)

But = and = (5)

From (4)+(5)⇒ ≤⋅

(6)

From Mitrinovic’s inequality ≥ + (7)

From (6)+(7)⇒ ≤⋅ ⋅

=

SP.250. Let be , , ∈ ℂ ∖ { } different in pairs:

| | = | | = | | = ; ( ); ( ); ( ).

If | − − | + | − − | + | − − | = then = = .



| | = | | = | | = , ( ), ( ), ( )

| − − | + | − − | + | − − | =

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| − ( + )| + | − ( + )| + | − ( + )| =

| − ( + + − )| + | − ( + + − )| + | − ( + + − )| =

| − | + | − | + | − | =

| − | + | − | + | − | = | | + | | + | |

(| − | − | |) + (| − | − | |) + (| − | − | |) = (*)

Suppose | − | ≠ and | − | ≠ , | − | ≠ ( − )( − ) ≠

− − + ≠

( ⋅ + ⋅ ) ≠ ⋅ ⋅ (I)

Similarly, ( + ) ≠ ⋅ (II)

and ( + ⋅ ) ≠ ⋅ ⋅ (III)

By adding (I), (II), (III): ⋅ + ⋅ + ≠ ⋅

⋅ − ⋅ ≠

⋅ ≠ ⇒ | | ≠ ⇒ | | ≠

≠ , ≢ and hence: | − | − | | ≠ , | − | − | | ≠

and | − | − | | ≠ this is in contradiction with relation (*)

So: = ⇒ ≡ and the triangle is equilateral.


( ), ( ), ( ) ; ⊂ ( , )

| − − | = | − − − | = −+ +

=

= − = | − | = , where is Euler’s ninepointcenter. The

relationship from hypothesis can be written: + + = (1)

Applying median theorem in ⇒

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= ⇒ = ( + ) − , because the radius of Euler’s circle

= , and the analogs. = ( + ) − , = ( + ) − ⇒

+ + =+ + +

+ + =⇒ + + ≤ ⇒

( + + ) ≤ :( + + ) ≤ ( + + ) ⇒

( + + ) ≤ ⇒ + + ≤ , but in our case = ⇒

⇒ + + ≤ with equality when the triangle is equilateral.

SP.251. ABOUT PROBLEM SP.251

17 RMM SUMMER EDITION 2020

By Marin Chiriciu – Romania


+ + ≥ √

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

Solution

We can give a refinement:


+ + ≥ ( − )

Marin Chirciu

Proof.

By Chebyshev: ∑ ≥ ∑ ∑ (1)

Known: ∑ = ∑ and ∑ = ∑

(1) can be written: ∑ ≥ ⋅ ∑ ⋅ ∑ = ∑ ∑ (2)

By CBS: ∑ ∑ ≥ (∑ ) (3)

By ∑ ≥ ( − ), (S.G. Guba, 1963), from (1), (2), (3):

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≥ ⋅ ( − ) = ( − )

Equality holds for an equilateral triangle. Inequality 2) it’s stronger than 1)


+ + ≥ ( − ) ≥ √

Solution

By 2) and ( − ) ≥ √ ⇔ ( − ) ≥ √ , (Doucet + ≥ √ ). It

remains: ( − ) ≥ ( + ) ⇔ ≥ ⇔ ≥ , (Euler). Equality holds for

an equilateral triangle.

Guba’s inequality:


≥ ( − )

S.G. Guba, 1963

Proof.

By ∑ = ( − − ), the inequality can be written:

( − − ) ≥ ( − ) ⇔ ≥ − , (Gerretsen ≥ − )

It remains: − ≥ − ⇔ ≥ , (Euler)


Remark.

The same kind of problems can be proposed:


≥

Marin Chirciu

Solution

Lemma:

6) In the following holds:

=+ ( − ) + ( + − ) − ( + )

Proof.

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=+ −

= + + − =

= ( ) , from

= , = + ( − ) + + ( + )

( + ) = [ + ( − ) + ( + − ) − ( + ) ]

Back to the main problem:

By Lemma:

+ ( − ) + ( + − )− ( + )≥

≥ [ + ( − ) + + ( + ) ] ⇔

⇔ − − ( − ) − ( + ) ≥ ⇔

⇔ [ ( − ) − ( − ) ] ≥ ( + )

(Gerretsen: ≥ − ≥ ( ) ). It remains:

( + )+

[( − )( − − )− ( − ) ] ≥ ( + ) ⇔

⇔ − + ≥ ⇔ ( − )( − ) ≥

Obviously by Euler ≥ . Equality holds for an equilateral triangle.

SP.252. If , ≥ then in ∆ the following relationship holds:

( + − )+ +

( + − )+ +

( + − )+ ≤


Solution by Florentin Vișescu

( + − ) ≤⏞+ + −

=+

( + − ) ≤⏞+ + −

=+

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( + − ) ≤⏞+ + −

=+

( + − )+ ≤

( + − )+ ≤

( + − )+ ≤

( + − )+ +

( + − )+ +

( + − )+ ≤

Equality holds for: + − =+ − =+ − =

→ ( + + ) + ( + + ) = ( + + )

, ≥ , + = → = =

+ =+ =+ =

→ = =

SP.253. If ∈ ℕ; ≥ ; ∈ , ; ∈ , then:

<( + )

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

Solution by Florentin Vișescu – Romania

≥ , ∈ ; ,∀ = ,

= <( + )

Let be { , , … , } = { , , … , } with ≤ ≤ ≤ ⋯ ≤

Then (∑ )(∑ ) ≤

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≤ ≤

But < =

= − =√ + √

⇒ =√ + √√ − √

=

= − =√ − √

=+ √

= + √

< + √ = + √ ⋅( + )( + )

<+ √ ( + )( + )

+ √ ( + )( + ) <( + )

+ √ ( + ) < 49( + )

+ √ + + √ < 49 +

− − √ > 8 + 4√ −

− √ > √ − (True)

SP.254. Let … ; ≥ be a convexe polygon with area

= [ … ] with sides = ; ∈ , ; = ;

, ∈ , ; ∈ , then:

+> ∑ ( + )


Solution by Marian Ursărescu – Romania

Because , ∈ , ⇒ < and < ⇒ we must show:

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∑ ≥∑ ( )

(1)

From Bergström’s inequality we have: ∑ ≥ ∑∑ ( )

(2)

From (1)+(2) we must show:

≥ ⇔ ≥

Relation which is true by Schanmberger’s inequality.

SP.255. If ∈ ℕ then in the following relationship holds:

+ + ≥

≥ ⋅( )

⋅

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


From Hölder’s inequality we have:

+ + ≥+ +

⇒ we must show: + + ≥ ⋅( )

⋅ ( ) ⇔

⇔ + + ≥ ⋅ √ (1)

+ + ≥ ( ) ⋅ ⋅ (2)

But = and ⋅ ⋅ = (3)

From (2)+(3)⇒ + + ≥ ⋅ =

= ⋅ ⋅ ≥ ⋅ √ ⋅ ⋅ =

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= ⋅ √ = ⋅ √ ⇒ (1) it is true.

Solution 2 by Avishek Mitra-West Bengal-India

≥ ⋅

⇒ ≥ ( ) ( ) ⋅ ⇒ ≥ ( ) ( ) ⋅( )

⇒ ≥ ⋅ ⋅ ⋅ √( )

⋅√

( )

⇒ ≥ ( ) ( ) ⋅ ( ) ( ) ⋅ ⋅ ( ) ⋅( )

⇒ ≥ ( )( )

⋅ ( )( )

⋅( )

⋅( )

⇒ ≥ ⋅ ⋅ ( ) ⋅ ⋅( )

⇔ ≥ ⋅( )

⋅ (Proved)


= ≥⏞ ∙∑

=

= = =( − )

=

=∑ −∑

=( − − ) − ( − − )

=

= ( + ) ≥⏞;

( + ) ( √ ) =

= ⋅( )

⋅

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UP.241. Let be ∈ ℕ; ≥ ; ≥ ; … ; a convexe polygon inscribed

in a circle with radius . If = ; ∈ , ; = ;

– semiperimeter then:

≥


Solution by Adrian Popa – Romania

: =

= = =

=

>. ( + + ⋯+ )

( + + ⋯+ ) = ( ) = ⋅

=+ + ⋯+

=+ + ⋯+

=

= + + ⋯+

But + + ⋯+ = ⇒ ⋯ = =

Let be ( ) = ⇒ ( ) = ⇒ ( ) = − < 0 ⇒ → concave

We apply Jensen’s inequality to function :

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+ + ⋯+≤

+ + ⋯+⇒

⇒ + + ⋯+ ≤

So, ∑ ≥⋅ ⋅ ⋅

=⋅ ⋅

UP.242. =⋅ ⋅

⋅ ⋅⋅ ⋅

=

If , , ∈ ℝ then find:

=


Solution 1 by Ravi Prakash-New Delhi-India

Let = =

= = [ ]

where + + = + + = + + =

=⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

Where ⋅ = + + , etc

Sum of the elements of

= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅

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= ( + + )( + + ) + ( + + )( + + ) +

+( + + )( + + ) =

= + + + + + + + +

[∵ + + = , ]

= sum of the elements of .

Thus, if sum of the elements of each row of B is 1, sum of the elements of ( )

= sum of the elements of

We have: =

Note that the sum of the elements of each row is .

Thus, the sum of the elements of

= the sum of the elements ( )

= then sum of the elements of

=

Assume the sum of the elements of is for the same ∈ ℕ

≥

The sum of the elements of



=

∴ the sum of the elements of is ;∀ ≥ ⇒ ∑ =

Solution 2 by Ravi Prakash-New Delhi-India

Let =

=+ +

+ ++ +

= =

Thus ( ) = ( ) = = . Continuing in this way, we get

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= ⇒ = ⇒+ ++ ++ +

=

∴ =

UP.243. Find:

=→

+ + + ⋯+ −


Solution 1 by Rohan Shinde-India

=→

+ − =→

− ( − )

∀ ≥ we know that ( − ) > 1

⇒ ( − ) > = −

⇒ =→

− ( − ) <→

( − + )

As → ∞; ∼ ( ) + ⇒ < → ( ( ) − + + )

But as → ∞, the linear function grows more rapidly than the logarithmic function.

⇒ as → ∞; ( ) − → −∞ ⇒ ≤ −∞ ⇒ = −∞

Solution 2 by Srinivasa Raghava-AIRMC-India

We have: = − ( )( )

= ( ) − ( )( )

It is easy to see that: ∑ ( )( )

= ( ). Then, we have:

→ + + =→

( − ) = −∞

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Solution 3 by Naren Bhandari-Bajura-Nepal

We write:

=→

−( − )

=→

− +−

=→

− −−

=→

−+

+( + ) −

=→

− + − ( + ) +( + ) − ( + )

( + )

≤→

− + − ( + ) ≤→

( + + ( !) − )

=→

( + − − + ( ) − ) = + −→

( + ) = −∞

UP.244. If ∈ (ℚ), ( − ) + √ = then:

( + ) ≥ , ∈ ℝ


Solution by Florentin Vișescu – Romania

In these conditions: ( − ) + √ = ⇔ + √ √ =

We denote = √ √ ⇒ root for

( + ). But ( + ) has rational coefficients ⇒ root of it

As ( + ) has the grade IV, we obtain:

( + ) = ( − )( − )( + + ) = − √ + ( + + ) =

= + + − √ − √ − √ + + +

= + − √ + − √ + + − √ + ∈ ℚ[ ] ⇒ ∈ ℚ

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− √ ∈ ℚ ⇒ − √ = ∈ ℚ ⇒ = + √

− √ + ∈ ℚ ⇒ − √ ∈ ℚ ⇒ − √ ∈ ℚ ⇒

⇒ − + √ √ ∈ ℚ ⇒ − √ − ∈ ℚ

⇒ − √ ∈ ℚ ⇒ − √ = ∈ ℚ

or − √ = − or √ = − ∈ ℚ

If ≠ ⇒ √ = (False) so, = ⇒ = √ ⇒ √ − √ ∈ ℚ

√ ( − ) ∈ ℚ ⇒ − = ⇒ =

Then, ( + ) = − √ + + √ +

( + ) = ( + ) − = + + − = +

+ ≥

− + ≥ ( − ) ≥ (True)

UP.245. If = , = + , ∈ ℕ then find:

=→

( − )



= ; = + ; =→

( − )

= + = ; = + = ; = + = ; = + =

We suppose > −( > 2) and we prove that >

= + > obviously.

So, > − 1,∀ > 2, ∈ ℕ ⇒ → = ∞

As = + ⇒ = − + ⇒

⇒ − = ⇒ ( − ) = −

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As → = ∞ ⇒ → = ∞

⇒→

( − ) =→

− = ∞( − ) = −∞

Solution 2 by Remus Florin Stanca-Romania

We prove by using the Mathematical induction that > 0;∀ ∈ ℕ:

( ): > 0 is true ( = )

We suppose that ( ): > 0 is true and we prove that ( + ): > 0 is true by

using the fact that ( ) is true: > 0 and > 0 ⇒

+ > 0 ⇒ > 0 ⇒

⇒ proved ⇒ > 0;∀ ∈ ℕ.

≥ , = ≥ , = + for ≥ ⇒ ≥ ;∀ ≥ ⇒ ≥ ;∀ ∈ ℕ

≥ ⇒ ≤ ⇒ + ≤ + ⇒ ≤ + , − = > 0 ⇒

⇒ − > 0 ⇒ > ⇒ < ≤ + ⇒→

= ∞

= − + ⇒ − = − ⇒ ( − ) = − (1)

= + ⇒ + = + + ( + ) ⇒→

= + ∞ = ⇒( )

⇒→

( − ) = −∞ = −∞ ⇒ = −∞

UP.246.Prove that:

+ √−

− √= √ +

√

where ( ) is the digamma function.

Proposed by Vasile Mircea Popa – Romania

Solution 1 by Rohan Shinde-India

( + ) = ( ) +

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+ √= +

+ √=

+ √+

+ √

− √= +

− √=

− √+

− √

=√ +

−− √

++ √

−− √

= √ ++ √

−− √

=

= √ + −√

= √ +√


+ √−

− √= +

+ √− +

− √=

=√ +

+ −+ √

−− √

− −− √

=

=√ +

−− √

+ −− √

−− √

=

= ∙√

√ + √ −+ −

√= √ +

√

Solution 3 by Mokhtar Khassani-Mostaganem-Algerie

+ √−

− √=

= ++ √

− −+ √

− +− √

−− √

=

=√ +

+ −√

−− √

= √ +√

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UP.247. If < ≤ then:

≥√ √ − √

+√ − √

( )+

( − )


Solution 1 by Ali Jaffal-Lebanon

=√

+√

+

≥ √ − ++

− √ + −+

≥ √ √ − √ + √ − √ +−

≥√ √ − √

+√ − √

( )+

−


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≥ [ ] + [ ] + [ ]

( , ); √ , ;+

, ; ( , )

, ( ) ; √ , √ ;+

,+

;

≥ √ − ++

− √( )

+ −+

=

=√ √ − √

+√ − √

( )+

−

UP.248. If , , … , ≥ , … = , ≥ then:

+ + ⋯+ − − −⋯− ≥

Proposed by Marin Chirciu – Romania

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Let : ;∞ → ℝ; ( ) = − ; ( ) = + ; ( ) = − =

For − ≥ ; ≥ ; ≥ ; ≥ is convexe. Then:

+ + + ⋯+≤

( ) + ( ) + ⋯+ ( )

For , , … , ∈ ;∞ . Let be = , = , … , =

Then we obtain: ⋯ ≤ ( ) ⋯ ( )

…⋅ ≤ ( ) + ⋯+ ( )

⋅ ≤ − + ⋯+ −

( ) ≤ − + ⋯+ −

− ≤ + + ⋯+ − − −⋯−

− ≤ + + ⋯+ − + …

≤ + + ⋯+ − − …−

The inequality holds for ≥ √ ; ≥ √ ; = ,

Solution 2 by Michael Sterghiou-Greece

= , ≥

∑ −∑ ≥ (1)

∑ ≥ ⋅ ∏ = (2)

∑ ≥ ∏ = (3)

≤ → ∑ ≤ (4)

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The function ( ) = − is concave and (1)→ ∑ ≥

By generalized Jensen on ( ) = − with “weights” , = , we get:

−≥

⎣⎢⎢⎡ ∑

∑−

⎦⎥⎥⎤

≥( ),( )

⋅

⎣⎢⎢⎡

∑−

⎦⎥⎥⎤

≥∑

− which suffices to ≥ . This reduces to ∑

− ≥ → ∑ ≤

which is (4). Done!

Solution 3 by Sanong Huayrerai-Nakon Pathom-Thailand

For , , , … , ≥ and … = , where ≥ we have:

+ + + ⋯+ ≥

⇒ ( + ) + ( + ) + ⋯+ ( + ) ≥

and ( … ) = ( ) ⇒ + + + ⋯+ ≥

and since , , , … , ≥

Hence + + + ⋯+ ≥ ( + ) + ( + ) + ( + ) + ⋯+ ( + ) and

since … + … + … + ⋯+ …

= … + … + … + ⋯+ …

Hence + + ⋯ ( … + … + ⋯ … ) ≥

≥ ( + ) + ( + ) + ⋯ ( + ) ( … + + ⋯+ +⋯+ … )

By matching we have:

( … ) + ( … ) + ( … ) + ⋯+ ( … )

≥ ( + )( … ) + ( + )( … ) + ( + )( … ) + ⋯+

+( + )( … )

Hence ( … ) + ( … ) + ( … ) + ⋯+ ( … )

≥ − + … + − + … + … +

+ … + ⋯+ … + …

⇒ + + + ⋯+ ≥ + + + + + + + + + + ⋯+ +

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= + + + ⋯+ +

Therefore, it is true.

UP.249 If ∝> 0 , ∈ ( ,∝) , , , … , ∈ such that | −∝| < , = ,

then:

+ + ⋯+ + + ⋯+ >(∝ − )∝

Proposed by Marian Voinea-Romania


Let be = + , with , ∈ , = , .

By | −∝| < , we obtain ( −∝) + <

∝ > + +∝ − ≥ ( + )(∝ − ) ⇒ > 0, = , .

∝ > ( + )(∝ − ) ,∝, > 0,

By squaring: > ∝∝

> ∝∝

(1)

+ + ⋯+ = + + ⋯+ + ( + + ⋯+ )

+ + ⋯+ =−+ +

−+ + ⋯+

−+

| + + ⋯+ | + + ⋯+ =

= ( + ⋯+ ) + ( + ⋯+ )+ +⋯+ + + + + ⋯+ +

≥ ( +⋯+ ) +⋯+ = ( +⋯+ ) + ⋯+ >( )

( + ⋯+ ) ∝∝

+ ⋯+ ∝∝

=∝∝

( + ⋯+ ) + ⋯+ ≥ ∝∝

( + ⋯+ ) + ⋯+ ≥

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Hence:

| + + ⋯+ | + + ⋯+ > ∝∝

.

UP.250 If , , > 0 then:

+ > + +



Let be : ( ,∞) → ℝ; ( ) = + −

( ) =−

+ + ( + ) =− ( + )

( + )( + ) =−

( + ) < 0

decreasing ⇒ ( ) = → ( ) =

⇒ ( ) > 0 ⇒ + − + > 0

+ − + > 0 ⇒ + > +

+ > + ⇒ + > +

+ >

+ > ≥ ⋅ =

UP.251. If , , , > 0 then in the following relationship holds:

+ ( + )≥ ( + )( + ) ( + )


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∑≥

∑

( )⋅∑ (1)

But ∑ = (2)

From (1)+(2) we must show: ∑ ⋅

( )( )≥

( )( ) ( )⇔

∑ ≥( )

⇔ ∑ ≥ (3)

But ∑ = ∑ ≥ ( )( )( )

(4)

But ( + + ) ≥ ( + + ) (5)

From (4)+(5)⇒ ∑ ≥ ⇒ (3) it is true.

UP.252. If , , > 0 then in the following relationship holds:

⋅ + + ≥



According to AM-GM inequality:

= + + ≥

≥ ⋅ = ⋅ (1)

But = (2)

So, ≥ ⋅ ≥ ⋅√

= − ⋅ =

= ⋅ q.e.d with equality ⇔ the triangle is equilateral, = = .

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UP.253. If ∈ ℕ; ≥ ; , , > 0; ( ) ; , > 0;

= + , (∀) ∈ ℕ then:

+ ++ +

≥( + )


Solution 1 by Remus Florin Stanca – Romania

The inequality can be written as:

+ + ⋯+ + + + ⋯+ + + ≥

≥ ( ) . We use Cauchy’s inequality:

+ ⋯+ + + ⋯+ + + ≥

≥ + ∑ (1)

We use Bernoulli’s inequality: ( + ) ≥ + for ≥ − and ≥ and in our

particular case, we have:

− + ≥ ⋅ + (2) because

− ≥ − and ≥ because ≥ | + ⇒ ≥ ⇒( )

⇒( )

+ ≥ + − ⋅+

+ ⇒( )

⇒( )

+ + + ≥

≥+

+ − + +

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We need to prove that: ∑ − + + ≥ ( ) ⇔

⇔+

+ − + + ≥( + )

⇔

⇔+

+ −( + )

+ + ≥( + )

⇔

⇔+

+ ≥( + )

−( + )

⇔

⇔ + ≥+

− ⇔

⇔ ∑ ≥ − (3)

We need to prove that

≥ − ⇔ ≥ − ⇔

⇔+

+ ≥+

Case I: ≤ ⇒ ≤ ⇒ ≤ ⇒ + ≤ + ⇒

⇒ ≥ (4)

Let = and = we prove that: + ≥√

⇔ ≥√

⇔

⇔√

≥ ⇔ √ + +√

≥ (true) ⇒ + ≥ ⇒( )

⇒( )

+ ≥ ⇒ ≥ − (*)

Case II: > ⇒ > 1 ⇒ 1 < ⇒ < ⇒ + <

< + ⇒ > 1( )

(5), let , ≥ and

, ≥ :

We prove by using the Mathematical induction that ≥ :

www.ssmrmh.ro

1. we suppose that ( ):” ≥ ” is true

2. we prove that ( + ): “ ≥ ” is true by using the fact that ( ) is true:

= + , ≥ ⇒ ≥ ≥ ⇒ ≥ , ≥ ⇒ ≥ ≥ ⇒

⇒ ≥

So: ≥

≥

−− −− −− “+”

⇒" "

+ ≥ ⇒ ≥ ⇒ ≥ ∀ ∈ ℕ

≥ , let = ⇒ ≤ , we prove that + > (6)

Let = and = ⇔ + >√

⇔ >√

⇔

⇔ √ + +√

> 2| ⋅ √ + ⇔ √ + − √ + + > 0, let √ + = ⇔

⇔ − + > 0, = − ⇒ √ = √ − ⇒ we prove that √ + > 1 +

√ − such that − + > 0 ⇔ > 1 + − , > 4 ⇒ > 2 >

> 1 + − ⇒ true ⇒ (6) is true ⇒ (5) is true ⇒> − (**)

⇒(∗);(∗∗)

the given inequality is true ⇒

⇒ + ∑ + ∑ ≥ ( ) (Q.E.D.)

Solution 2 by proposers

We have:

= + = + ≥ ( + ) ⋅ ⋅… ⋅" "

⋅ =

= ( + )∑ = ( + ) (1)

Also, we have:

= + + = + + =

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=

⎝

⎛ ++

⎠

⎞ ≥

≥ ( + ) ⋅ ⋅ ⋅ … ⋅ ⋅" " +

=

= ( + )+

≥ ( + )∑ +

=

= ( )

( )∑= ( )

⋅ (2)

So, relationships (1) and (2) we deduce that:

= ⋅ ≥ ( + ) ⋅( + )

⋅=

( + ),∀ ∈ ℕ∗

UP.254. If , > 0; ∈ ℕ; ≥

→= > 0;

→= > 0

then find:

=→

⎝

⎛ −

⎠

⎞



=→

……

…− =

= →… ⋅

→…

… − (1)

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→

…=

→

…=. .

=→

…( + ) ⋅ … =

→ + ⋅ +

=→

=→

⋅ =→

=

=. .

→ ( + ) ⋅ =→ ( + ) ⋅ ( + )

= → ⋅ ⋅ = (2)

=→

…… −

……

⋅……

=

→

( … )… =

→

…… ⋅

…

=→

⋅…

=→ + ⋅

+⋅ … ⋅

=→

⋅…

=→

⋅ ⋅…

=( )

= ⋅ ⋅ = = (3)

From (1)+(2)+(3)⇒ = ⋅ =


= → ∏ ⋅∏

∏− (1)

→⋅ =

→⋅

→=

→⋅

→

( + )⋅ =

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= ⋅→

⋅→

=

= ⋅ → = ⇒ → = (a)

⇒( )

=→

∏⋅ ⋅

⎝

⎛∏

∏−

⎠

⎞ =

=→

∏⋅ ⋅

⎝

⎛∏

∏−

⎠

⎞ =

=→

∏∏ ⋅ ( + ) ⋅

→

⎝

⎛∏

∏−

⎠

⎞ =

= ⋅→ + ⋅

→

⎝

⎛∏

∏−

⎠

⎞ =( )

= ⋅ → ⋅ ⎝⎜⎜⎜⎛

⎝

⎛∏

∏⎠

⎞

⎠

⎟⎟⎟⎞

∏

∏

⋅∏

∏ (2)

→

∏=

→

∏=

→

∏( + ) ⋅ ∏ = ⋅

→ + =

=( )

⇒→

∏=

So, →∏

=

→ ∏=

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------------------------------- “⋅”

⇒"⋅"

→

∏

∏= ⇒

( )= ⋅

→

⎝

⎛∏∏ ⋅

∏ ⎠

⎞ =

=→

⋅∏

=

=→

( + )∏ ⋅

∏= ⋅ ⋅

→

+=

= ⋅ ⋅ = ⇒ =

Solution 3 by Soumitra Mandal-Chandar Nagore- India

→=

→=

→ ( + ) ⋅

= → ⋅ ⋅ = . Now, →∏

= →∏

=→

∏( + ) ⋅ ∏ =

→ + ⋅→ +

=

= → → = . Let =∏

∏ for all ∈ ℕ. Then

→=

→

∏

+∏

⋅+

=

Hence, → , for all → ∞

→=

→

⎝

⎛∏∏ ⋅

∏ ⎠

⎞ =→

⎝

⎛ ⋅+

∏⋅

+⎠

⎞

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= ⋅ ⋅ =

→

⎝

⎛ −

⎠

⎞ =→

⎝

⎛∏

⋅−

⋅

⎠

⎞

= ⋅ ⋅ = (Answer)

UP.255. If > 0; , > 0; ∈ ℕ; ≥ ;

→ = > 0; → ⋅= > 0 then find:

=→

( + )

⋅−

⋅

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


=→

( + )⋅

⋅− =

= → ⋅( )

⋅− (1)

→=

→

( )=. .

→

( + )( )( )

⋅ ⋅⋅

( ) =

= → ⋅ ⋅ ⋅ ⋅ ( ) = (2)

→

( ) ⋅

( + ) ⋅

( + )⋅

=→

+⋅ =

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= → ( ⋅ )⋅ (3)

→ = (4)

→ ( )=

→⋅ =

→⋅ ⋅ =

( )⋅ ⋅

= ( ) = −( + ) (5)

From (1)+(2)+(3)+(4)+(5)⇒ = − ( + ) = ( )

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It’s nice to be important but more important it’s to be nice.

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru

romanian mathematical magazine solutions · 2019. 7. 31. · romanian mathematical magazine...

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