romanian mathematical magazine solutions · 2019. 7. 31. · romanian mathematical magazine...
Post on 28-Mar-2021
67 Views
Preview:
TRANSCRIPT
ROMANIAN MATHEMATICAL MAGAZINE
Founding EditorDANIEL SITARU
Available onlinewww.ssmrmh.ro
ISSN-L 2501-0099
Number 17 SUMMER 2020
SOLUTIONS
www.ssmrmh.ro
SOLUTIONS
www.ssmrmh.ro
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
www.ssmrmh.ro
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
www.ssmrmh.ro
JP.241 Let , , be the circumpedal extensions of cevians of centroid in
. Prove that:
+ + ≥ √
Proposed by Daniel Sitaru-Romania
Solution by proposer
= ; = ; = ; ∩ ∩ = { }
= =
( ) = ⋅ = ⋅ (the power in circumcircle)
⋅ = ⋅ ⇒ =
= = + = +
Analogous: = + ; = +
+ + = + + + + + ≥
≥ ⋅ ⋅ ⋅ = √
www.ssmrmh.ro
JP.242 Let , , be the circumpedal extensions of cevians of centroid in
. Prove that:
≥
Proposed by Daniel Sitaru-Romania
Solution by proposer
= ; = ; = ; ∩ ∩ = { }
= =
( ) = ⋅ = ⋅ (the power of in circumcircle)
⋅ = ⋅ ⇒ =
= = + = + ≥ ⋅ = ⋅ ⋅ =
Analogous: ≥ ; ≥ . By multiplying: ≥
JP.243. If , , > 1; = 8 then:
+ + ≥
Proposed by Daniel Sitaru-Romania
Solution 1 by Florentin Vișescu-Romania
Let be : ( , ) → ℝ, ( ) = = ( )
( ) = ( − + ) ( )
www.ssmrmh.ro
( ) = ( − + ) + ( ) > 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: { , , } = , { , , } =
www.ssmrmh.ro
( − )( − ) ≤ → − − + ≤
( − ) ≥ ( − ) + ( − )( − )
( − ) ≥ − + + − − +
( − ) ≥ + + − − −
− ≥ + + − − −
{ , , }− { , , } ≥ + + − − −
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
www.ssmrmh.ro
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
Solution 1 by Florentin Vișescu-Romania
+ + + = ↔ + + + =
, , , =+ + +
= =
, , , = ∙ ∙ ∙ =
, , , = , , , → = = = = → =
Solution 2 by proposer
If > 0 ⇒ ( − ) ( + ) ≥
( − + )( + ) ≥
www.ssmrmh.ro
+ − − + + ≥
− − + ≥ ⇒ ≥ + −
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:
√ + √ + √ + + + ≥ ( + + )
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution 1 by Marian Ursărescu-Romania
Let √ = ,√ = ,√ = , , , ≥ , + + = .We must prove:
( + + ) + + + ≥ ( + + ), ( )
Let = , = , =
www.ssmrmh.ro
( ) ⇔( + + )
+ + ++ +
( + + ) ≥( + + )
( + + )
( + + ) ( + + ) + + + − ( + + ) ≥ ( )
− , ( , , ) ≥ ⇔ ( , , ) ≥
( , , ) = ( + + ) ( + + ) + + + − ( + + ),∀ , , ≥
(Cîrtoaje’s theorem)
( , , ) = ( + + ) ( + + ) + + + − ( + + )
( , , ) = ( + ) ( + ) + + − ( + ) =
= + − + = ( − ) ( + ) ≥
Solution 2 by proposer
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:
( + )− ( + + ) = ( − ) ≥
www.ssmrmh.ro
JP.248. If , , > 0 then:
+ + + + + + + + + + + ≥ ( + + )
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution 1 by Marian Ursărescu-Romania
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.
www.ssmrmh.ro
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:
+ + + + + + + + + + + ≥ ( + + )
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
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:
www.ssmrmh.ro
+ + + + + + + + + + + ≥
≥ + + + + + ≥ ( + + )
+ + + + + + + + + + + ≥ ( + + )
Equality holds for = = .
Remark:
3) If , , > 0 and ≥ then:
+ + + ( + )+ +
++ +
++ +
≥ ( + + )
Marin Chirciu
Solution:
By Bergström’s inequality: + + ≥ ( )∑( )
=( )( )
(1)
By (1) and Schur’s lemma: + + + ( + ) + + ≥
≥ + + + + + ≥ ( + + )
+ + + ( + )+ +
++ +
++ +
≥ ( + + )
Equality holds for = = .
4) If , , > 0 then:
+ + + + + + + + ≥ ( + + )
Marin Chirciu
Solution
By Bergström’s inequality: + + ≥ ( )∑( )
=( )
(1)
By (1) and Schur’s lemma:
+ + + + + + + + ≥ + + + + + ≥
≥ ( + + )
+ + + + + + + + ≥ ( + + )
www.ssmrmh.ro
Equality holds for = = .
5) In the following relationship holds:
+ + + + − + + − + + − ≥ ( + + )
Marin Chirciu
Solution
By Bergström’s inequality: + + ≥ ( )∑( )
= (1)
By (1) and Schur’s lemma:
+ + + + − + + − + + − ≥ + + + + + ≥
≥ ( + + )
+ + + + − + + − + + − ≥ ( + + )
Equality holds for = = .
6) In , ≤ the following relationship holds:
+ + + ( − )+ − + + − + + − ≥
≥ ( + + )
Marin Chirciu
Solution
By Begström’s inequality: + + ≥ ( )∑( )
=( )( )
(1)
By (1) and Schur’s lemma:
+ + + ( − )+ − + + − + + − ≥
≥ + + + + + ≥ ( + + )
+ + + ( − )+ − + + − + + − ≥
≥ ( + + )
Equality holds for = = .
www.ssmrmh.ro
7) In the following relationship holds:
++
≥
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 = = .
8) In the following relationship holds:
+ ≥+
Marin Chirciu
Solution
Taking = − in the above inequality. Equality holds for = = .
Solution (alternative):
By the known relationships:
=+
− ,+
= ++
, =
Inequality can be written:
www.ssmrmh.ro
+ − ≥ ⋅ + ⇔ + − ≥ , obviously by
Euler’s inequality ≥ . Equality holds for = = .
9) If in ,− ≤ ≤ the following relationship holds:
++
≥ +
Marin Chirciu
Solution
By the known relationships:
=+
− ,+
= ++
, =
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 [( − ) + ] ≥
Equality holds for = = .
JP.249. ABOUT PROBLEM JP.249
17 RMM SUMMER EDITION 2020
By Marin Chirciu – Romania
1) If , , > 0 then:
++
++
+≥
++
++
++
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
www.ssmrmh.ro
Solution
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
+ +≥
+=
+=
=( + )( + )( + )
≥ √ = ⋅ =
Equality holds for = = .
Remark.
We propose and solve a few problems of the same kind:
2) In the following relationship holds:
− + − + − ≥ − + − + − +
Marin Chirciu
Solution
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
− − ≥ − = − = ∏( − ) =
= = ≥ √ = ⋅ =
Equality holds for an equilateral triangle.
3) In the following relationship holds:
−+
−+
−≥
−+
−+
−+
Marin Chirciu
www.ssmrmh.ro
Solution
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
− −≥
−=
−=
∏( − )=
= = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥
Equality holds for an equilateral triangle.
4) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
≥ = = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥
Equality holds for an equilateral triangle.
5) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
www.ssmrmh.ro
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
≥ = = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≤ ⇔ ≤√
Equality holds for an equilateral triangle.
6) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
≥ = = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ √ .
Equality holds for an equilateral triangle.
7) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
+ ≥ + ⇔ ≥
www.ssmrmh.ro
≥ = = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ √
Equality holds for an equilateral triangle.
8) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
+ ≥ + ⇔ ≥
≥ = = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔ ≥
⇔ ≥ √
Equality holds for an equilateral triangle.
9) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
By squaring:
+ ≥ + ⇔ ≥
www.ssmrmh.ro
≥ = = ≥ √ = ⋅ =
Equality holds for an equilateral triangle.
10) In the following relationship holds:
+ + ≥ + + +
Marin Chirciu
Solution
By squaring: ∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
≥ = = ≥ ⋅ =
Equality holds for an equilateral triangle.
11) In the following relationship hold:
+ + ≥ + + +
Marin Chirciu
Solution
By squaring:
∑ + ∑ ≥ ∑ + ⇔ ∑ ≥
≥ = = ≥
≥ ⇔ ≥ ⇔ ≥ ⇔ ≥ ⇔
⇔ ≥ ⇔ ≥ √ .
Equality holds for an equilateral triangle.
www.ssmrmh.ro
JP.250. ABOUT PROBLEM JP.250
17 RMM Summer Edition 2020
By Marin Chirciu – Romania
1) If , , > 0 are such that + + + ( ) + ( ) + ( ) ≤ , then:
+ + ≥
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
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)
≥ + + + + ( + ) + ( + ) + ( + ) + ( + ) ≥( )
www.ssmrmh.ro
≥ + + ( + ) =
= ∑ ( ) +( )
≥ ∑ ( ) ⋅( )
= ∑ , 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.
www.ssmrmh.ro
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
Solution by proposer
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:
++ =
+− + − + + + +
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Florentin Vișescu-Romania
+ + + + − + + − =
= − ( + ) + + ( + ) − + ( + )
+ + − − ( + ) + + − −
− + ( + ) + + − ( + ) =
+ + − ( + )+ + ( + ) +
+ + + ( + )+ − ( + ) − +
+ + − ( + )+ + ( + ) =
+ ( + ) ( ) + − + ( + ) ( ) +
+ ( + ) ( ) =
− ( + )− − ( + )− ( + ) = ( + ) = → = -not solution because >
= → =
Solution 2 by Orlando Irahola Ortega-La Paz-Bolivia
+ + = + − + [ − ( + )] + [ + ( + )]
+ > 0 ∧ + > 0 ⇒ ∈ ] ,∞[
= + − ⇒ ( + + ) = + +
= − ( + ) ⇒ ( + + )( + + ) =
= + ( + ) ⇒ ( + )( + )( + ) = ⇒+ = ( )+ = ( )+ = ( )
www.ssmrmh.ro
+ + = + +
(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:
⋅ <
Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania
www.ssmrmh.ro
Solution by proposers
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
⇔ = +( )
= + + − ≥
≥ + + ( + ) − ( + )
⇒ ≥ + − − + ( + )
⇒ ≥ ( + ) ≥ ( + ) ( )
⇒ ≥ ( + ) ( ) ⋅
⇒ ≥ ( + ) ⋅ ⋅ ⋅ √ ⋅ ∵ ≥ , ≥ √
www.ssmrmh.ro
⇒ ≥ ( + ) ⋅ ( ) ⋅ ( ) ⋅ ( ) ⋅√
∵ ≥ √
⇒ ≥ ( + ) ⋅ ⋅ ⋅ √
⇔ + ∑( )
≥ ( + )√ (proved)
Solution 2 by Soumava Chakraborty-Kolkata-India
= + + −
≥ + + ( + ) − (∵ + ≥ )
= + + ( + ) − −
= − + ( + ) = ( + )
=( − ) ( )
=∑ −∑
( + )
=( − − ) − ( − − )
( + )
= ( + )( + ) ≥ ⋅ ( + ) ≥ √ ⋅ ( + )
= √ ( + ) (Proved)
SP.241 Let , , be the circumpedal extensions of cevian of incentre in
. Prove that:
≥ ( + )( + )( + )
Proposed by Daniel Sitaru-Romania
www.ssmrmh.ro
Solution by proposer
= ; = ; = ; ∩ ∩ = { }
= ⇒ + = +
= + ⇒ = + ; = +
( ) = ⋅ = ⋅
+ ⋅ + = ⋅
= ( + )
= = + = + ( + ) ≥ ⋅ ( + ) = + √
Analogous: = √ ; = √
By multiplying: ≥ √ ⋅ √ ⋅ √ =
=⋅
( + )( + )( + ) ≥ ( + )( + )( + )
SP. 242 Let , , be the circumpedal extensions of cevians of incentre in
. Prove that:
+ + ≥ ( + )( + )( + )
Proposed by Daniel Sitaru-Romania
www.ssmrmh.ro
Solution by proposer
= , = , = , ∩ ∩ = { }
= ⇒ + = +
= + ⇒ = + ; = +
( ) = ⋅ = ⋅
+ ⋅ + = ⋅
= ( + )
= = + = + ( + )
Analogous: = +( )
; = +( )
By summing: + + = +( )
+ +( )
+ +( )
≥
≥ ⋅⋅ ⋅
( + ) ( + ) ( + ) =
= ( + ) ( + ) ( + ) = ( + )( + )( + )
www.ssmrmh.ro
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
Solution 2 by proposer
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 √
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Soumava Chakraborty-Kolkata-India
( + ) − <( )
√
(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
Solution 2 by Khaled Abd Imouti-Damascus-Syria
( + ) − <?
√ , ⋅ + − <?
√
+ − <?
√ , + − <?
√
+ − <?
√ . Suppose: =
( + )√ − <?
√ , < < 2 ⇒ < < 2
< < 2, ∈ ] , [. Now, let us prove:
( + )√ − <?
√ , ∈ ] , [. Let be the function:
www.ssmrmh.ro
( ) = ( + ) ⋅ √ − , ∈ ] , [, →
( ) = √ ,→
( ) =
( ) = ( + )√ − + ⋅ ( + ), ( ) = ( )
( ) = , ( ) = ; ( ) = ⇒ − − + + =
Let be the function: ( ) = − − + + , = ] , [
→( ) = ,
→( ) = −
( ) = − − + , ( ) = ⇒ − − + =
= − (− )( ) = + ( ); = ( + ) = ⋅ ⇒ √ =
=( )
= = + > 0, =( )
= = − < 0
√ = − √ − + √ + = −√ − < 0
= − − + + =−
− +
=− −
+ = > 0; √ = √ − < 0, = > 0
, √
( ) + + + + + + + + − − − − −−−
( )
,
√ +√
( )
( ) √ √ , ( ) , √ √ +
, ,
Not: Local max of ( ) on the interval [ , ] is nearly .
www.ssmrmh.ro
So: ∀ ∈ ] , [: ( ) ≤ , ( + )√ − < 5, 5 < 3√ , ( ) < 5√
If < < < 2 then: ( + ) − <?
√
+ − <?
√ , + − <?
√
+ − <?
√ ; + − <?
√ . Suppose: =
⋅ ( + )√ − <?
√ , < < 2 ⇒ < < 2
< < 2, ∈ ] , [. Now, let us prove: ( + )√ − <?
√ , √ − <? √
( )
Let be the function: ( ) = √ − , ∈ ] , [
→[ ( )] = √ ,
→[ ( )] = ; ( ) =
−√ −
; ( ) = ⇒ = ∉ ] , [
( ) −−− − − −− −− −− −− −− −−−
( ) √
Let be the function: ( ) = √ , ] , [, →
[ ( )] = √ ,→
[ ( )] = √
( ) = √ ( ), ( ) = ⇒ = − ∉ ] , [
( ) −−− − − −− −− −− −− −− −−−
( ) √ √
∀ ∈ ] , [: ( ) < ( )
www.ssmrmh.ro
Solution 3 by proposer
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:
< ⇒ < ⇒ < )
www.ssmrmh.ro
SP.245. If < < < 2 then:
( + ) > 3( − ) ( − )
Proposed by Daniel Sitaru – Romania
Solution 1 by Khaled Abd Imouti-Damascus-Syria
If < < < 2 then: ( + ) > 3( − ) ( − )
+ > 3 − −
+ > 3 − −
+ > 3 − −
Suppose: = > 0
( + ) > 3( − ) ( − )
< < < 2 ⇒ < < 2 ⇒ 1 < < 2
So: let us prove ∀ ∈ ] , [:
( + ) >?
( − ) ( − )
+− >
?√ ⋅ −
Let be the function: ( ) = , ] , [
→[ ( )] = +∞,
→[ ( )] =
( ) =( + )( − ) − ( )( + )
( − ) =− − − −
( − )
( ) =− −
( − ) , ( ) = ⇒ − − =
= − ( )(− ) = + = > 0,√ = √
www.ssmrmh.ro
=− √
= − √ ∉ ] , [
=+ √
= + √ ∉ ] , [
( ) − −− −− −− −− −− − − −−− −− −
( ) +∞
Let be the function: ( ) = √ ⋅ √ − , ∈ ] , [
→( ) = √ ,
→[ ( )] =
( ) = √ ⋅ −√ −
; ( ) =− √√ −
( ) = ⇒ = ∉ ] , [
( ) − −− −− −− −− −− − −− −− −− −
( ) √
Note: ∀ ∈ ] , [; ( ) > ( )
+− > 3√ ⋅ −
Solution 2 by Soumava Chakraborty-Kolkata-India
( + ) >( )
( − ) ( − )
www.ssmrmh.ro
∵ , + , − > 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)
Solution 3 by proposer
Let’s consider ; = = ; = = ; = = − .
< + ⇔ < − + ⇔ < (true)
< + ⇔ < + − ⇔ ( + ) > 2
But: ( + ) > ( + ) =
< + ⇔ − < + ⇔ < ( + )
⇔ + − > 0 ⇔ 2 − − > 0; =
= ; =+
= ; =−
= −
( − − ) = −
www.ssmrmh.ro
> 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:
( + )( + ) ≥ ⋅ ⋅ ⋅
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Soumava Chakraborty-Kolkata-India
=( )
+ , =( )
+ , =( )
+ , =( )
+
(1), (2), (3), (4) ⇒ ( + )( + ) =
= ( + + )( + + )
= ⋅ ( + ) + ⋅ ( + ) + ( + )( + ) +
≥( )
( + ) + ( + ) + ( + ) ( + ) +
Now, =( )
⇒ = ⇒ = √ (China, IMO TST, 2003)
= (∵ is bicentric)⇒ + =( ) ⋅
Similarly, using (a), + =( ) ⋅
(m), (n), (i)⇒ ( + )( + ) ≥
≥ ⋅⋅
+ ⋅⋅
+ ( ) + ≥
≥ + + ∙ ∙ ∙ = ( ⋅ + ⋅ )
= ⋅ ⋅ ⋅ (China IMO TST, 2003)
www.ssmrmh.ro
Solution 2 by proposer
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
Solution 1 by proposer
www.ssmrmh.ro
( ) =+
, = −+
⇒
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 ∑ = ∑( + ) = − ∑
≤ + = +
Solution 2 by Soumava Chakraborty-Kolkata-India
www.ssmrmh.ro
From ,∠ = ° − ° − + ° − = = ° = ° −
From , = = ⇒ = ⇒ = =( )
Again, using , = = ⇒ = ⇒ =
⇒ =( )
. Similarly, =( )
∴ = − = =( ),( )
− +
= −( + )
= −( + )
= −+
= ∴ =( )
Applying sine rule on , °
=°
⇒ = (by (4))⇒ =( )
. Similarly, =( )
(1)+(5)+(6)⇒
=+
+
=⋅
( − ) ( − ) ⋅( − )
+ ⋅( − )
+
= ( − )( − ) ( − ) + ( − ) + −
= { ( + )− } + − =( + )−
+ −
=( − )( + )− ( − ) +
( − )
=( + )(∑ )( + − ) − ( + − ) + ( + − )
( − )
= ⋅{( + + ) − }
( − ) = ⋅{( + ) − }
( − )
= ⋅( ) ( − )
( − ) = ⋅ ⇒ = ⇒ =( )
www.ssmrmh.ro
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
.
Proposed by Marian Ursărescu – Romania
Solution by proposer
The matrix being orthogonal one ⇒ | | = | | = | | = | | = | | =
( ) = − + − + − = (1)
= = (2)
www.ssmrmh.ro
= ∗ = (( ) − ) = (3)
= = + + + +
= ⋅ + + + + = ⋅ ( + + + + )
= ⋅ = (A)
= = = ⋅
= ⋅ = (5)
From (1)+(2)+(3)+(4)+(5)⇒ ( ) = − ⇒
=( ) = ⇒ ( ) = ⇒ = ±
⇒
⇒ = ± ⇒ = ⇒ =
SP.249. Let be the circumcevian triangle of centroid in . Prove
that:
[ ][ ] ≤
Proposed by Marian Ursărescu – Romania
Solution by proposer
www.ssmrmh.ro
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 = = .
Proposed by Marian Ursărescu – Romania
Solution 1 by Khaled Abd Imouti-Damascus-Syria
| | = | | = | | = , ( ), ( ), ( )
| − − | + | − − | + | − − | =
www.ssmrmh.ro
| − ( + )| + | − ( + )| + | − ( + )| =
| − ( + + − )| + | − ( + + − )| + | − ( + + − )| =
| − | + | − | + | − | =
| − | + | − | + | − | = | | + | | + | |
(| − | − | |) + (| − | − | |) + (| − | − | |) = (*)
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.
Solution 2 by proposer
( ), ( ), ( ) ; ⊂ ( , )
| − − | = | − − − | = −+ +
=
= − = | − | = , where is Euler’s ninepointcenter. The
relationship from hypothesis can be written: + + = (1)
Applying median theorem in ⇒
www.ssmrmh.ro
= ⇒ = ( + ) − , 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
1) In the following relationship holds:
+ + ≥ √
Proposed by D.M. Bătinețu – Giurgiu and Neculai Stanciu – Romania
Solution
We can give a refinement:
2) In the following relationship holds:
+ + ≥ ( − )
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):
www.ssmrmh.ro
≥ ⋅ ( − ) = ( − )
Equality holds for an equilateral triangle. Inequality 2) it’s stronger than 1)
3) In the following relationship holds:
+ + ≥ ( − ) ≥ √
Solution
By 2) and ( − ) ≥ √ ⇔ ( − ) ≥ √ , (Doucet + ≥ √ ). It
remains: ( − ) ≥ ( + ) ⇔ ≥ ⇔ ≥ , (Euler). Equality holds for
an equilateral triangle.
Guba’s inequality:
4) In the following relationship holds:
≥ ( − )
S.G. Guba, 1963
Proof.
By ∑ = ( − − ), the inequality can be written:
( − − ) ≥ ( − ) ⇔ ≥ − , (Gerretsen ≥ − )
It remains: − ≥ − ⇔ ≥ , (Euler)
Equality holds for an equilateral triangle.
Remark.
The same kind of problems can be proposed:
5) In the following relationship holds:
≥
Marin Chirciu
Solution
Lemma:
6) In the following holds:
=+ ( − ) + ( + − ) − ( + )
Proof.
www.ssmrmh.ro
=+ −
= + + − =
= ( ) , 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:
( + − )+ +
( + − )+ +
( + − )+ ≤
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
Solution by Florentin Vișescu
( + − ) ≤⏞+ + −
=+
( + − ) ≤⏞+ + −
=+
www.ssmrmh.ro
( + − ) ≤⏞+ + −
=+
( + − )+ ≤
( + − )+ ≤
( + − )+ ≤
( + − )+ +
( + − )+ +
( + − )+ ≤
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 (∑ )(∑ ) ≤
www.ssmrmh.ro
≤ ≤
But < =
= − =√ + √
⇒ =√ + √√ − √
=
= − =√ − √
=+ √
= + √
< + √ = + √ ⋅( + )( + )
<+ √ ( + )( + )
+ √ ( + )( + ) <( + )
+ √ ( + ) < 49( + )
+ √ + + √ < 49 +
− − √ > 8 + 4√ −
− √ > √ − (True)
SP.254. Let … ; ≥ be a convexe polygon with area
= [ … ] with sides = ; ∈ , ; = ;
, ∈ , ; ∈ , then:
+> ∑ ( + )
Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania
Solution by Marian Ursărescu – Romania
Because , ∈ , ⇒ < and < ⇒ we must show:
www.ssmrmh.ro
∑ ≥∑ ( )
(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
Solution 1 by Marian Ursărescu-Romania
From Hölder’s inequality we have:
+ + ≥+ +
⇒ we must show: + + ≥ ⋅( )
⋅ ( ) ⇔
⇔ + + ≥ ⋅ √ (1)
+ + ≥ ( ) ⋅ ⋅ (2)
But = and ⋅ ⋅ = (3)
From (2)+(3)⇒ + + ≥ ⋅ =
= ⋅ ⋅ ≥ ⋅ √ ⋅ ⋅ =
www.ssmrmh.ro
= ⋅ √ = ⋅ √ ⇒ (1) it is true.
Solution 2 by Avishek Mitra-West Bengal-India
≥ ⋅
⇒ ≥ ( ) ( ) ⋅ ⇒ ≥ ( ) ( ) ⋅( )
⇒ ≥ ⋅ ⋅ ⋅ √( )
⋅√
( )
⇒ ≥ ( ) ( ) ⋅ ( ) ( ) ⋅ ⋅ ( ) ⋅( )
⇒ ≥ ( )( )
⋅ ( )( )
⋅( )
⋅( )
⇒ ≥ ⋅ ⋅ ( ) ⋅ ⋅( )
⇔ ≥ ⋅( )
⋅ (Proved)
Solution 3 by Soumava Chakraborty-Kolkata-India
= ≥⏞ ∙∑
=
= = =( − )
=
=∑ −∑
=( − − ) − ( − − )
=
= ( + ) ≥⏞;
( + ) ( √ ) =
= ⋅( )
⋅
www.ssmrmh.ro
UP.241. Let be ∈ ℕ; ≥ ; ≥ ; … ; a convexe polygon inscribed
in a circle with radius . If = ; ∈ , ; = ;
– semiperimeter then:
≥
Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania
Solution by Adrian Popa – Romania
: =
= = =
=
>. ( + + ⋯+ )
( + + ⋯+ ) = ( ) = ⋅
=+ + ⋯+
=+ + ⋯+
=
= + + ⋯+
But + + ⋯+ = ⇒ ⋯ = =
Let be ( ) = ⇒ ( ) = ⇒ ( ) = − < 0 ⇒ → concave
We apply Jensen’s inequality to function :
www.ssmrmh.ro
+ + ⋯+≤
+ + ⋯+⇒
⇒ + + ⋯+ ≤
So, ∑ ≥⋅ ⋅ ⋅
=⋅ ⋅
UP.242. =⋅ ⋅
⋅ ⋅⋅ ⋅
=
If , , ∈ ℝ then find:
=
Proposed by Daniel Sitaru – Romania
Solution 1 by Ravi Prakash-New Delhi-India
Let = =
= = [ ]
where + + = + + = + + =
=⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅
Where ⋅ = + + , etc
Sum of the elements of
= ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅
www.ssmrmh.ro
= ( + + )( + + ) + ( + + )( + + ) +
+( + + )( + + ) =
= + + + + + + + +
[∵ + + = , ]
= 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
= sum of the elements of
= 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
www.ssmrmh.ro
= ⇒ = ⇒+ ++ ++ +
=
∴ =
UP.243. Find:
=→
+ + + ⋯+ −
Proposed by Daniel Sitaru – Romania
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:
→ + + =→
( − ) = −∞
www.ssmrmh.ro
Solution 3 by Naren Bhandari-Bajura-Nepal
We write:
=→
−( − )
=→
− +−
=→
− −−
=→
−+
+( + ) −
=→
− + − ( + ) +( + ) − ( + )
( + )
≤→
− + − ( + ) ≤→
( + + ( !) − )
=→
( + − − + ( ) − ) = + −→
( + ) = −∞
UP.244. If ∈ (ℚ), ( − ) + √ = then:
( + ) ≥ , ∈ ℝ
Proposed by Marian Ursărescu – Romania
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
www.ssmrmh.ro
− √ ∈ 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:
=→
( − )
Proposed by Marian Ursărescu – Romania
Solution 1 by Florentin Vișescu-Romania
= ; = + ; =→
( − )
= + = ; = + = ; = + = ; = + =
We suppose > −( > 2) and we prove that >
= + > obviously.
So, > − 1,∀ > 2, ∈ ℕ ⇒ → = ∞
As = + ⇒ = − + ⇒
⇒ − = ⇒ ( − ) = −
www.ssmrmh.ro
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
( + ) = ( ) +
www.ssmrmh.ro
+ √= +
+ √=
+ √+
+ √
− √= +
− √=
− √+
− √
=√ +
−− √
++ √
−− √
= √ ++ √
−− √
=
= √ + −√
= √ +√
Solution 2 by Remus Florin Stanca-Romania
+ √−
− √= +
+ √− +
− √=
=√ +
+ −+ √
−− √
− −− √
=
=√ +
−− √
+ −− √
−− √
=
= ∙√
√ + √ −+ −
√= √ +
√
Solution 3 by Mokhtar Khassani-Mostaganem-Algerie
+ √−
− √=
= ++ √
− −+ √
− +− √
−− √
=
=√ +
+ −√
−− √
= √ +√
www.ssmrmh.ro
UP.247. If < ≤ then:
≥√ √ − √
+√ − √
( )+
( − )
Proposed by Daniel Sitaru – Romania
Solution 1 by Ali Jaffal-Lebanon
=√
+√
+
≥ √ − ++
− √ + −+
≥ √ √ − √ + √ − √ +−
≥√ √ − √
+√ − √
( )+
−
Solution 2 by proposer
www.ssmrmh.ro
≥ [ ] + [ ] + [ ]
( , ); √ , ;+
, ; ( , )
, ( ) ; √ , √ ;+
,+
;
≥ √ − ++
− √( )
+ −+
=
=√ √ − √
+√ − √
( )+
−
UP.248. If , , … , ≥ , … = , ≥ then:
+ + ⋯+ − − −⋯− ≥
Proposed by Marin Chirciu – Romania
www.ssmrmh.ro
Solution 1 by Florentin Vișescu-Romania
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)
www.ssmrmh.ro
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 ( … ) + ( … ) + ( … ) + ⋯+ ( … )
≥ − + … + − + … + … +
+ … + ⋯+ … + …
⇒ + + + ⋯+ ≥ + + + + + + + + + + ⋯+ +
www.ssmrmh.ro
= + + + ⋯+ +
Therefore, it is true.
UP.249 If ∝> 0 , ∈ ( ,∝) , , , … , ∈ such that | −∝| < , = ,
then:
+ + ⋯+ + + ⋯+ >(∝ − )∝
Proposed by Marian Voinea-Romania
Solution by proposer
Let be = + , with , ∈ , = , .
By | −∝| < , we obtain ( −∝) + <
∝ > + +∝ − ≥ ( + )(∝ − ) ⇒ > 0, = , .
∝ > ( + )(∝ − ) ,∝, > 0,
By squaring: > ∝∝
> ∝∝
(1)
+ + ⋯+ = + + ⋯+ + ( + + ⋯+ )
+ + ⋯+ =−+ +
−+ + ⋯+
−+
| + + ⋯+ | + + ⋯+ =
= ( + ⋯+ ) + ( + ⋯+ )+ +⋯+ + + + + ⋯+ +
≥ ( +⋯+ ) +⋯+ = ( +⋯+ ) + ⋯+ >( )
( + ⋯+ ) ∝∝
+ ⋯+ ∝∝
=∝∝
( + ⋯+ ) + ⋯+ ≥ ∝∝
( + ⋯+ ) + ⋯+ ≥
www.ssmrmh.ro
Hence:
| + + ⋯+ | + + ⋯+ > ∝∝
.
UP.250 If , , > 0 then:
+ > + +
Proposed by Daniel Sitaru-Romania
Solution by proposer
Let be : ( ,∞) → ℝ; ( ) = + −
( ) =−
+ + ( + ) =− ( + )
( + )( + ) =−
( + ) < 0
decreasing ⇒ ( ) = → ( ) =
⇒ ( ) > 0 ⇒ + − + > 0
+ − + > 0 ⇒ + > +
+ > + ⇒ + > +
+ >
+ > ≥ ⋅ =
UP.251. If , , , > 0 then in the following relationship holds:
+ ( + )≥ ( + )( + ) ( + )
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
www.ssmrmh.ro
Solution by Marian Ursărescu – Romania
∑≥
∑
( )⋅∑ (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:
⋅ + + ≥
Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania
Solution by proposers
According to AM-GM inequality:
= + + ≥
≥ ⋅ = ⋅ (1)
But = (2)
So, ≥ ⋅ ≥ ⋅√
= − ⋅ =
= ⋅ q.e.d with equality ⇔ the triangle is equilateral, = = .
www.ssmrmh.ro
UP.253. If ∈ ℕ; ≥ ; , , > 0; ( ) ; , > 0;
= + , (∀) ∈ ℕ then:
+ ++ +
≥( + )
Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania
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 ≥ | + ⇒ ≥ ⇒( )
⇒( )
+ ≥ + − ⋅+
+ ⇒( )
⇒( )
+ + + ≥
≥+
+ − + +
www.ssmrmh.ro
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:
= + + = + + =
www.ssmrmh.ro
=
⎝
⎛ ++
⎠
⎞ ≥
≥ ( + ) ⋅ ⋅ ⋅ … ⋅ ⋅" " +
=
= ( + )+
≥ ( + )∑ +
=
= ( )
( )∑= ( )
⋅ (2)
So, relationships (1) and (2) we deduce that:
= ⋅ ≥ ( + ) ⋅( + )
⋅=
( + ),∀ ∈ ℕ∗
UP.254. If , > 0; ∈ ℕ; ≥
→= > 0;
→= > 0
then find:
=→
⎝
⎛ −
⎠
⎞
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
Solution 1 by Marian Ursărescu-Romania
=→
……
…− =
= →… ⋅
→…
… − (1)
www.ssmrmh.ro
→
…=
→
…=. .
=→
…( + ) ⋅ … =
→ + ⋅ +
=→
=→
⋅ =→
=
=. .
→ ( + ) ⋅ =→ ( + ) ⋅ ( + )
= → ⋅ ⋅ = (2)
=→
…… −
……
⋅……
=
→
( … )… =
→
…… ⋅
…
=→
⋅…
=→ + ⋅
+⋅ … ⋅
=→
⋅…
=→
⋅ ⋅…
=( )
= ⋅ ⋅ = = (3)
From (1)+(2)+(3)⇒ = ⋅ =
Solution 2 by Remus Florin Stanca-Romania
= → ∏ ⋅∏
∏− (1)
→⋅ =
→⋅
→=
→⋅
→
( + )⋅ =
www.ssmrmh.ro
= ⋅→
⋅→
=
= ⋅ → = ⇒ → = (a)
⇒( )
=→
∏⋅ ⋅
⎝
⎛∏
∏−
⎠
⎞ =
=→
∏⋅ ⋅
⎝
⎛∏
∏−
⎠
⎞ =
=→
∏∏ ⋅ ( + ) ⋅
→
⎝
⎛∏
∏−
⎠
⎞ =
= ⋅→ + ⋅
→
⎝
⎛∏
∏−
⎠
⎞ =( )
= ⋅ → ⋅ ⎝⎜⎜⎜⎛
⎝
⎛∏
∏⎠
⎞
⎠
⎟⎟⎟⎞
∏
∏
⋅∏
∏ (2)
→
∏=
→
∏=
→
∏( + ) ⋅ ∏ = ⋅
→ + =
=( )
⇒→
∏=
So, →∏
=
→ ∏=
www.ssmrmh.ro
------------------------------- “⋅”
⇒"⋅"
→
∏
∏= ⇒
( )= ⋅
→
⎝
⎛∏∏ ⋅
∏ ⎠
⎞ =
=→
⋅∏
=
=→
( + )∏ ⋅
∏= ⋅ ⋅
→
+=
= ⋅ ⋅ = ⇒ =
Solution 3 by Soumitra Mandal-Chandar Nagore- India
→=
→=
→ ( + ) ⋅
= → ⋅ ⋅ = . Now, →∏
= →∏
=→
∏( + ) ⋅ ∏ =
→ + ⋅→ +
=
= → → = . Let =∏
∏ for all ∈ ℕ. Then
→=
→
∏
+∏
⋅+
=
Hence, → , for all → ∞
→=
→
⎝
⎛∏∏ ⋅
∏ ⎠
⎞ =→
⎝
⎛ ⋅+
∏⋅
+⎠
⎞
www.ssmrmh.ro
= ⋅ ⋅ =
→
⎝
⎛ −
⎠
⎞ =→
⎝
⎛∏
⋅−
⋅
⎠
⎞
= ⋅ ⋅ = (Answer)
UP.255. If > 0; , > 0; ∈ ℕ; ≥ ;
→ = > 0; → ⋅= > 0 then find:
=→
( + )
⋅−
⋅
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
Solution by Marian Ursărescu – Romania
=→
( + )⋅
⋅− =
= → ⋅( )
⋅− (1)
→=
→
( )=. .
→
( + )( )( )
⋅ ⋅⋅
( ) =
= → ⋅ ⋅ ⋅ ⋅ ( ) = (2)
→
( ) ⋅
( + ) ⋅
( + )⋅
=→
+⋅ =
www.ssmrmh.ro
= → ( ⋅ )⋅ (3)
→ = (4)
→ ( )=
→⋅ =
→⋅ ⋅ =
( )⋅ ⋅
= ( ) = −( + ) (5)
From (1)+(2)+(3)+(4)+(5)⇒ = − ( + ) = ( )
www.ssmrmh.ro
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
top related