RMM INEQUALITIES MARATHON 1 – 100 · Shivam Sharma - New Delhi – India . Solutions by Daniel...
Transcript of RMM INEQUALITIES MARATHON 1 – 100 · Shivam Sharma - New Delhi – India . Solutions by Daniel...
www.ssmrmh.ro
RMM
INEQUALITIES
MARATHON
1 – 100
www.ssmrmh.ro
Proposed by Daniel Sitaru – Romania
D.M.Batinetu-Giurgiu-Romania
Neculai Stanciu-Romania
Rustem Zeynalov – Baku – Azerbaidian
Nguyen Ngoc Tu – HaGiang – Vietnam
Sladjan Stankovic – Skopje
Nguyen Viet Hung – Hanoi – Vietnam
Ibrahim Abdulazeez – Zaria – Nigeria
Ilkin Guliyev-Azerbaidian
Mihály Bencze-Romania
Adil Abdullayev – Baku – Azerbaidian
Hoang Le Nhat Tung – Hanoi – Vietnam
Madan Beniwal – Varanasi – India
Shivam Sharma - New Delhi – India
www.ssmrmh.ro
Solutions by Daniel Sitaru – Romania
Mihalcea Andrei Ștefan – Romania, Soumitra Mandal-Chandar Nagore-India
Uche Eliezer Okeke-Anambra-Nigerie, Soumava Chakraborty-Kolkata-India
Anas Adlany - El Zemamra-Morocco, Antonis Anastasiadis – Katerini – Greece
Kevin Soto Palacios – Huarmey – Peru, Șerban George Florin – Romania
Fotini Kaldi – Greece, Ravi Prakash - New Delhi – India
Myagmarsuren Yadamsuren – Darkhan – Mongolia,
Redwane El Mellas – Casablanca – Morocco, Richdad Phuc – Hanoi – Vietnam
Nirapada Pal – Jhargram – India, Chris Kyriazis – Greece
Rustem Zeynalov – Baku – Azerbaidian, Leonard Giugiuc – Romania
Dat Vo-Quynh Luu – Vietnam, Khanh Hung Vu – Ho Chi Minh – Vietnam
Seyran Ibrahimov-Maasilli-Azerbaidian, Nguyen Thanh Nho-Tra Vinh-Vietnam
Nguyen Ngoc Tu-Ha Giang – Vietnam, Abdallah El Farissi – Bechar – Algerie
SK Rejuan-West Bengal – India, Geanina Tudose – Romania
Tuk Zaya-Ulaanbaatar-Mongolia, Abdul Aziz-Semarang-Indonesia
Abdelhak Maoukouf-Casablanca-Morocco, Abdallah Almalih-Damascus-Syria
Pham Quoc Sang – Ho Chi Minh – Vietnam, Le Khanh Sy - Long An – Vietnam
Marian Cucoanes – Romania, Ngo Minh Ngoc Bao – Vietnam
Marian Dincă – Romania, Sanong Hauyrerai-Nakonpathom – Thailand,
Sladjan Stankovic – Skopje,Rovshan Pirguliyev – Sumgait – Azerbaidian,
Rozeta Atanasova – Skopje,Boris Colakovic-Belgrade – Serbia
www.ssmrmh.ro
Bedri Hajrizi – Mitrovica – Kosovo,Nikola Djurici – Serbia
Togrul Ehmedov – Baku – Azerbaidian, Ngoc Minh Ngoc Bao - Gia Lang –
Vietnam, Le Minh Cuong - Ho Chi Minh – Vietnam
Hoang Le Nhat Tung – Hanoi – Vietnam, Khalef Ruhemi – Iordania
Mohammed Hijazi – Iordania, Imad Zak – Saida – Lebanon
Subhajit Chattopadhyay - Visva Bharati – India
www.ssmrmh.ro
1. From the book: “Math Phenomenon”
If < ≤ ≤ then:
( + )( + )( + ) ≤ ∏ ≤ ( + )( + )( + )
Proposed by Daniel Sitaru – Romania
Solution by Mihalcea Andrei Ștefan – Romania
+ ≤ ⇔ ≤ ⇔ ≤ true
≤ + ⇔ ≤ ⇔ ≤ true
Multiplying the analogs ⇒ q.e.d.
2. From the book: “Math Phenomenon”
If ≥ ≥ > 0, + + = 10 then:
+ + ≤+ +
+≤ + +
Proposed by Daniel Sitaru – Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India ,
Solution 2 by Uche Eliezer Okeke-Anambra-Nigerie
Solution 1 by Soumitra Mandal-Chandar Nagore-India
+ ++
≥ ( + )
⎣⎢⎢⎢⎢⎡ ∵ + + ≥ ( + ) ,
+ + ≥ ( + )
+ + ≥ ( + ) ⎦⎥⎥⎥⎥⎤
= ( + + ). We need to prove,
( + + ) ≥ + + ( + + ) ⇔ ≥ , which is true. So,
www.ssmrmh.ro
∑ ≥ + + We need to prove,
+ + ≥+ +
+⇔ + + ≥
+ ++
⇔ + −( + + )
++ + −
( + + )+
+
+ + −( + + )
+≥
⇔ ( ) + ( ) + ( ) ≥ , which is true ∵ ≥ ≥
∴ + + ≥+ +
+≥ + +
Solution 2 by Uche Eliezer Okeke-Anambra-Nigerie
Left 1) + + ≤ ∑
⇒ + + ≤( + ) −
+
⇒ + + + + ≤ ( + )
⇒ + + + + ≤ + + +
To show ⇒ ∑ ≤ + - Goal. Now,
∑ ≤ ∑ ( ) = ∑( + ) = + + ≤ + (True)
Right 2) ∑ ≤ + +
⇒( + )( + ) ≤
++ + +
www.ssmrmh.ro
+ + + ≤+
+ + +
To show + ≤ ∑ - Goal. Now,
∑ = + + (*)
Using: + ≤ ; + ≤ ; + ≤
(*) ≥ + + = + ⇒ + ≥ ⋅ ∑ (True)
3. From the book “Math Phenomenon”
If , , , ∈ ( ,∞) then:
+ + ≤ ( + )( + )
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
To prove
⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ≤ + + +
⇔ + + + ≤ + + +
⇔ + ≤ + (1)
Let = , = , = , = → , , , > 0 ∵ , , , > 0
∴ + − −
= + − −
= − − − = ( ) − ( ) ( − )
= −( − )( − ) ( ) + ( ) ( ) + ( ) ( ) + ( )( ) + ( )
= −( − ) ( ) ≤ (∵ > 0, as , , , > 0)
www.ssmrmh.ro
equality at = ⇒ = ⇒ =
∴ of (1) – RHS of (1) ≤
⇒ (1) is true. Hence proved. (equality at = )
4. If , > 0, + + = 0 then:
+ + ≥ ( + ) √ +
Proposed by Daniel Sitaru – Romania
Solution 1 by Anas Adlany-El Zemamra-Morocco, Solution 2 by Antonis
Anastasiadis-Katerini-Greece
Solution 1 by Anas Adlany-El Zemamra-Morocco
First, note that: ( + + ) = + + + ( + )( + )( + )( + + + + + )
Then when + + = , we will have the following
+ + = ( + + + + + );
Now, let’s go back to the main problem and check out what we are really
dealing with the problem asks us to show that:
+ + ≥ ( + ) √ + whenever , > 0
We have + + ≥ ( + ) √ +
⇔ ( + + + + + ) ≥ ( + ) √ +
⇔ ( + + )( + + + + + ) ≥ ( + ) √ +
and the last step can be explained as follows:
+ + = ⇒ ( + ) + =
⇒ + + + ( + ) = ⇒ + + = ( + = − )
Now, since , > 0; by the AM-GM inequality we get:
+ + ≥ √ + − √ + (1)
www.ssmrmh.ro
Also, ∑( + ) − − = + + + +
= ( + ) + + ( + )( + = − )
= (( + ) − ) − = − − = − = ( − ) ≥
Which prove that ∑( + ) ≥ + (2)
Finaly, from results (1) & (2) the proof is completed.
Solution 2 by Antonis Anastasiadis-Katerini-Greece
+ + = ⇔ = − −
+ + ≥ ( + ) √ +
= + − ( + ) ≥ ( + ( + ) ) √ − ( + ) =
≤+
+ + ++
− ( + ) ⇒
≤ ( + + )( + )( − ( + ) ) It’s enough to prove:
+ − ( + ) ≥ ( + + )( + )( − ( + ) ) (1)
It is ( + ) − − = ( + )( + + ) (Cauchy)
(1) ⇔ − ( + )( + + ) ≥ ( + + )( + )( − ( + ) )
− ≥ ( − ( + ) ) ⇔ ( + ) ≥ which stands
5. If , , , ∈ ℝ then:
( − ) + ( + ) ≥ ( + ) ( + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru , Solution 2 by Serban George Florin-
Romania , Solution 3 by Fotini Kaldi-Greece , Solution 4 by Ravi Prakash-New Delhi-India ,
Solution 5 by Myagmarsuren Yadamsuren-Darkhan-Mongolia , Solution 6 by Soumava
Chakraborty-Kolkata-India
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Si , , , ∈ . Probar que
( − ) + ( + ) ≥ ( + ) ( + ) Ahora bien
www.ssmrmh.ro
( + )( + ) = ( + ) + ( − )
La desigualdad es equivalente
( − ) + ( + ) ≥ (( + ) + ( − ) )
⇔ (( + ) − ( − ) ) ≥
Solution 2 by Serban George Florin-Romania
( + )( + ) = ( + ) + ( − )
+ = , + = , − = , + =
⇒ = + , + ≥
= ( + ) ≤ ( + ) ⋅ ( + ) ⇒ ≤ +
Solution 3 by Fotini Kaldi-Greece
( + ) ≥ ( + ) , “ = “ if and only if =
( − ) + ( + ) ≥ (( − ) + ( + ) ) =
= ( + )( + )
Solution 4 by Ravi Prakash-New Delhi-India
Let = + , = + , then = ( − ) + ( + )
Now, ( + ) ( + ) = | | | | = (| |)
= [( − ) + ( + ) ] ≤ [( − ) + ( + ) ]
∵ ( + ) ≤ ( + )
Solution 5 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
+ + + = + + +
: ( + − ) + ( + + ) =
= ( − ) + ( + ) = + + + =
www.ssmrmh.ro
= ( + )( + ) =
LHS: ( − ) + ( + ) = ( + )( + ): RHS ( ) = ( )
( − ) + ( − ) ⋅ ( + ) + ( + ) = = ( + ) ⋅ ( + ) ; ( − ) ≥ ; ( + ) > 0
( + ) ( + ) = ( − ) + ( − ) ( + ) + ( + ) ≤⏞ ≤ ( − ) + ( + )
Solution 6 by Soumava Chakraborty-Kolkata-India
( + ) + ( − ) = ( + )( + )
Given inequality ⇔ + ≥
⇔ ( + ) ≥
But, ( + ) ≥ ( + ) =
∵ + ≥ ( + )
6. From the book: “Math Phenomenon”
If , ∈ [ , ] then:
++
++ ( − ) ≤
Proposed by Daniel Sitaru – Romania
Solution 1 by Redwane El Mellas-Casablanca-Morocco
Solution 2 by Richdad Phuc-Hanoi-Vietnam
www.ssmrmh.ro
Solution 1 by Redwane El Mellas-Casablanca-Morocco
Let ( , ) = + + ( − ) .
∴ ( , ), ( > 0,0), ( , > 0) = , , + ≤
Suppose that , > 0:
If ≤ : ( , ) ≤ + + ( − ) = + ≤ ≤
If > 1: ∴ ( , ) − ( , ) = + + ( − ) − +
= + − − ( + )
=+
− − ( + ) ≤ − − ( + ) < 0
⇒ ( , ) < ( , ) = + ≤
Finally ( , ) ≤ with equality if and only if = , = .
Solution 2 by Richdad Phuc-Hanoi-Vietnam
We have + + ( − ) ≤ ⇔ − + − + − ≤
⇔( − )( + + ) −
+ + ( − )( + ) + + − ≤
⇔( − )( + + )
+ + ( − )( + ) + + − − + ≤
which is true, because − − ≤ − − ≤ (∀ , ∈ [ , ])
Equality holds if and only if = ; =
www.ssmrmh.ro
7. If < ≤ ; , , > 0
= , = , = then:
+ + ≥+ ++ +
Proposed by Daniel Sitaru – Romania
Solution by Nirapada Pal-Jhargram-India
=∑
≥⏞∑
= ∑ =
− =∑
−∑∑ =
(∑ ) − ∑∑ =
∑ − ∑∑ ≥
− =∑∑ − ∑ =
(∑ ) − ∑∑ ∑ =
∑( ) − ∑( )( )∑ ∑ ≥
since + + ≥ + + , so ≥ ≥
And < ≤ ⇒ ≤ ≤
So, ≤⏞( )
= + +
8. If < ≤ ≤ , , ≥ then:
( + + )( + ) + ( + ) ≥
≥ ( + + )( + + )
Proposed by Daniel Sitaru – Romania
Solution by Kevin Soto Palacios – Huarmey – Peru
Si < ≤ ≤ , , ∈ ℝ. Probar que
( + + )( + ) + ( + ) ≥
≥ ( + + )( + + )
www.ssmrmh.ro
Condición ≥ ≥ ⇔ − − = ( − ) + ( − ) ≥
Desarrollan do la desigualdad
+ + + + + + + ≥
≥ + + + + + +
+ + + ⇔ (− + − ) + (− + − ) + ( − + ) ≥
⇔ − ( − ) − ( − ) + ( − ) = ( − ) ( − − ) ≥ (LQQD)
9. If ≤ , ≤ then:
++
++ ( − ) ≤ +
Proposed by Daniel Sitaru – Romania
Solution by Chris Kyriazis-Greece
The function ( , ) = + + ( − ) − −
≤ ≤ , ≤ ≤ is convex due to or
(it’s easy to check it with positive derivatives)
So the function achieves its maximum to one of the vertices of the square
[ , ] × [ , ]; ( , ) = + + − − =
( , ) = + + − − = − − < 0
( , ) = + + − − = − − < 0
( , ) = + + − − = − < 0
So the maximum is zero when = and = .
www.ssmrmh.ro
10. If , , > 0 then:
( + + )( + + + + − ) + ≥ + +
Proposed by Rustem Zeynalov-Baku-Azerbaidian
Solution by proposer
( + ) + ( + ) (( + ) − ( + )( + ) + ( + ) ) + ≥
≥ + + + ; + = ; + =
( + )( − + ) + ≥ + ; + + ≥ +
+ + ≥ =+ + ≥ =
+
+ + ≥ +
11. If , , > 0, 6 = then:
( + )( + )( + )≥ ( + + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Leonard Giugiuc – Romania, Solution 2 by Kevin Soto Palacios –
Huarmey – Peru, Solution 3 by Myagmarsuren Yadamsuren-Darkhan-
Mongolia
Solution 1 by Leonard Giugiuc – Romania
= , = , = ⇒ ( + + ) = ⇒
⇒ = , = , = ⇒ + + = and
=√
, =√
, =√
. After usual algebraic computations, we
get the equivalent + ∑ + ( + + ) ≥ ( )
www.ssmrmh.ro
By AM – GM, ≤ . So that
≥ ( ) , by AM – GM ∑ ≥ ( ) ; but
( ) ≥ ( ) ⇒ ≥ ( )
and by AM – QM + + ≥ ⇒ ( + + ) ≥ ( ) .
Solution 2 by Kevin Soto Palacios – Huarmey – Peru
Siendo , , > 0, 6 = ( ). Probar que
( + )( + )( + )≥ ( + + ) =
Realizando los siguientes cambios de variables
= > 0, 2 = > 0, 3 = > 0, con , , > 0
La condición es equivalente
( + + ) = ⇔ + + =
⇔ + = + + + = ( + )( + )
⇔ + = ( + )( + ), + = ( + )( + )
La desigualdad propuesta es equivalente
( + )( + )( + ) ≥
( + )( + )( + )( + )( + )( + ) ≥
⇔ ( + ) ( + ) ( + ) ≥ ( )( )( ) =
(Válido por MA ≥ MG)
Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
= ⋅ ( + + )
LHS: a) + ⋅ ( + + ) ⋅ + ⋅ ( + + ) ⋅
www.ssmrmh.ro
⋅ + ⋅ ( + + ) =
= ⋅ + ( + + ) ⋅ + ( + + ) ⋅
⋅ ( + + + ) =
= ⋅ ( + + + ) ⋅ ( + + + ) ⋅
⋅ ( + + + ) ≥⏞
≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ = ⋅ ⋅ =
= ⋅ ⋅ = a)
:≥⋅
= = ( + + )
12. If < < < then:
√ + √ + + √ − √ + < √ + √ +
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
Let ( ) = + , ≥ , ( ) = − = − > 0,∀ > 2
Thus, increases on [ ,∞). For < < < , then
( ) < ( ) < ( ) < ( ). We have
√ + √ + + √ − √ +
= √ + √ ( ) + √ − √ ( ) < √ + √ ( ) + √ − √ ( )
www.ssmrmh.ro
= √ + √ ( ) < √ + √ ( ) = √ + √ +
13. If ≤ < < < then:
√ + √ + √ + √ > √ + √ + √
Proposed by Daniel Sitaru – Romania
Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia
= √ + √ + √ + √ > √ + √ + √
:
√ > √
√ > √
√ > √
√ > √ ⎭⎪⎪⎬
⎪⎪⎫
(*)
> √ + √ + √ + √ (**)
√ = + + √√ = + + ⋅ √√ = + ⋅ √
√ = √ ⎭⎪⎪⎬
⎪⎪⎫
(***)
(**), (***) ⇒ > + + ⋅ √ + + + ⋅ √ +
+ + ⋅ √ + ⋅ √ =√
+√
+√
+
+√
+ ⋅ √ + √ + √ + √ + √ + √
www.ssmrmh.ro
≥ ⋅√ ⋅ √ ⋅ √
+ ⋅√
⋅√
⋅√
⋅√
⋅√
⋅√
+
+√
⋅√
⋅√
⋅√
⋅√
⋅√
⋅√
⋅√
⋅√
⋅√
=
= √ + √ + √ ⋅ ⋅ ⋅
14. If , , > 0, + + = then:
− + <
Proposed by Daniel Sitaru – Romania
Solution 1 by Dat Vo-Quynh Luu-Vietnam,
Solution 2 by Khanh Hung Vu-Ho Chi Minh-Vietnam
Solution 1 by Dat Vo-Quynh Luu-Vietnam
− + =+ −
→ inequality ↔ + − < 3( + + )
↔ ( − ) − ( − ) + > 0 ↔ ( − + ) + ( − − ) > 0
Solution 2 by Khanh Hung Vu-Ho Chi Minh-Vietnam
If , , > 0, + + = . Prove that − + < (1) We have
(1)⇒ < ⇒ + + > 6 − +
⇒ + − + + − > 0 ⇒
⇒ + ( − ) + + − > 0 (2)
www.ssmrmh.ro
( ) = + ( − ) + + −
We have = ( − ) − ⋅ ( + − ) = − ( − ) < 0
⇒ (2) true ⇒ QED
15. If ≤ < 3, 0 ≤ < 5, 0 ≤ < 7 then:
√ + + √ + + √ + < 6
Proposed by Daniel Sitaru – Romania
Solution 1 by Nirapada Pal-Jhargram-India, Solution 2 by Seyran Ibrahimov-
Maasilli-Azerbaidian, Solution 3 by Nguyen Thanh Nho-Tra Vinh-Vietnam,
Solution 4 by Nguyen Ngoc Tu-Ha Giang-Vietnam
Solution 1 by Nirapada Pal-Jhargram-India
√ + + √ + + √ + < 6
≤⏞ + + = + +
< + + . As ≤ < 3, 0 ≤ < 5, 0 ≤ < 7
= + + =
Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian
1) √ + < 2 ⇒ < 7 ⇒ 0 ≤ < 3 True
2) √ + < 2 ⇒ < 31 ⇒ 0 ≤ < 5 True
3) √ + < 2 ⇒ < 127 ⇒ 0 ≤ < 7 True
(1)+(2)+(3) < 6 (proved)
Solution 3 by Nguyen Thanh Nho-Tra Vinh-Vietnam
< 3 ⇒ + 1 < 4 < 8 ⇒ √ + < √ =
www.ssmrmh.ro
< 5 ⇒ + 1 < 6 < 32 ⇒ √ + < √ =
< 7 ⇒ + 1 < 8 < 128 ⇒ √ + < √ =
√ + + √ + + √ + < 6
Solution 4 by Nguyen Ngoc Tu-Ha Giang-Vietnam
We have √ + < √ + < 2√ + < √ + < 2√ + < √ + < 2
⇒ √ + + √ + + √ + < 6,
∀ ≤ < 3, 0 ≤ < 5, 0 ≤ < 7
16. If < ≤ then:
≤++
+++
+++
≤
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Ravi
Prakash-New Delhi-India, Solution 3 by Soumitra Mandal-Chandar Nagore-
India, Solution 4 by Abdallah El Farissi-Bechar-Algerie, Solution 5 by SK
Rejuan-West Bengal-India, Solution 6 by Antonis Anastasiadis-Katerini-Greece
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Siendo < ≤ . Probar que ≤ + + ≤
Como ≥ > 0
⇒ = ⋅ ≥ , = ⋅ ≥ , = ⋅ ≥
⇒ = ⋅ ≤ , = ⋅ ≤ , = ⋅ ≤
Por lo tanto
⇒ + + ≥ + + = + + =
www.ssmrmh.ro
⇒ + + ≤ + + = + + =
Solution 2 by Ravi Prakash-New Delhi-India
For < ≤ , ∈ ℕ, + ≤ +
≤ + ≤ + = + ⇒ ≤++ ≤
Taking = , , and adding we get the desired inequality.
Solution 3 by Soumitra Mandal-Chandar Nagore-India
++
+++
+++
≥
⇔++ − +
++ − +
++ − ≥
⇔ ( ) + ( ) + ( ) ≥ , which is true ∵ ≥
≥++
+++
+++
⇔ ( ) + ( ) + ( ) ≥ , which is true ∵ ≥
∴ ≤++
+++
+++
≤
Solution 4 by Abdallah El Farissi-Bechar-Algerie
Let , ∈ ℝ∗, ≤ for all ∈ [ , ], ≤ + ( − ) ≤
([ , ] is a covex set)
We have = + = + −
Then for ≥ , ≤ ≤ it follow that ≤ ∑ ≤
Solution 5 by SK Rejuan-West Bengal-India
www.ssmrmh.ro
Applying Cauchy inequality we get,
≥ ≥⇑
[ ]
(1)
Again applying Cauchy inequality we get,
≥ ≥⏞ ≥ (i2)
[from (i)]. Again applying Cauchy inequality we get,
≥ ≥⏞ ≥ (iii)
[from (ii)]. Adding (1), (ii) & (iii) we get,
+ + ≥ ( + ) ≥ [Proved]
[as ≥ ⇒ + ≥ ]. Applying Cauchy inequality we get,
≤ ≤⏞ (1)
≤⏞ ≤⏞ (2)
From (1) & (2) we get, ≤ ; ≤
∴++
+++
+++
≤++
[ ≤ ] ⇒≤+ ⋅+
= ⋅++
=
⇒ + + ≤ [proved]
Solution 6 by Antonis Anastasiadis-Katerini-Greece
www.ssmrmh.ro
17. If , , ≥ then:
+ + ≥ √ + √ √
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Seyran
Ibrahimov-Maasilli-Azerbaidian, Solution 3 by Abdallah El Farissi-Bechar-
Algerie
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Siendo , , ≥ . Probar que + + ≥ √ + √ √
Como , , ≥ . Aplicando ≥
+ + + ≥ √ (A)
+ + + ≥ √ (B)
+ + + ≥ √ (C)
Sumando (A) + (B) + (C) ( + + ) ≥ √ + √ + √ √ ⇔
⇔ + + ≥ √ + √ + √ √ ≥ √ + √ √
Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian
= , = , = , + + ≥ ( + )
+ + ≥ ( + + )( + + ) ≥ ( + + )
www.ssmrmh.ro
( + + ) ≥ ( + ) ⇒ ≥
Solution 3 by Abdallah El Farissi-Bechar-Algerie
√ + √ √ = √ √ + √ √ ≤+ √ + + √
≤
≤+ + + + +
=+ +
≤ + +
18. If ≤ < < then:
−+
−+
−<
−+
−+
−
Proposed by Daniel Sitaru – Romania
Solution 1 by SK Abdul Halim-India, Solution 2 by Ravi Prakash-New Delhi-
India
Solution 1 by SK Abdul Halim-India
If , be a positive real number but is not equal to and , are rational
numbers then: > if >
Proof: > 0 and ≠ . > 0 and ≠
If = then both sides are equal with zero.
Let two unequal numbers and with associated positive rational
weights and − respctively, then we know. ⋅ ( )⋅ > [ − ] or, + − >
or, ( − ) > ( − ) or, > , since > 0, > 0
Now, for = , = ; > (i) ≤ < <
www.ssmrmh.ro
Similarly, > (ii)
From (i) – (ii) we get, − > −
i.R. >
Rest are in similar member. Hence the proof.
Solution 2 by Ravi Prakash-New Delhi-India
Let ∈ ℕ and > 1. Consider ( ) = − , ∈ [ , ],
( ) = − , ∈ [ , ]
By the Cauchy’s mean value theorem ∃ ∈ ( , ) such that
( ) − ( )( ) − ( ) =
( )( ) ⇒
−−
=( + )
=+
>+
; [∴ > 1] ⇒−
+>
−
If > ≥ 1, then
−+
>−
⇒−+
>−
≥−
Thus, > ,∀ > ≥ 1, ∈ ∴ if ≤ < < , then
−+
−+
−<
−+
−+
−
19. If < ≤ ≤ then:
( + )√ + ( + + )√ +( + + ) + √ + √
≥
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Ravi Prakash-New Delhi-India, Solution 2 by Geanina Tudose-
Romania
Solution 1 by Ravi Prakash-New Delhi-India
Let < ≤ ≤ 1. Consider ( + ) + ( + + )( ) + −
−( + + ) + + ( )
= ( + ) + ( + + )( ) +
− − − − ( + ) − − ( + + )( ) − ( )
= ( + − ) + ( − − ) + ( + − )( )
= ( − − ) − − ( ) ≥
as < , ≤ , < ; ( ) ≤ ; ( ) ( )( )
( ) ( )≥
Put = , = , we get ( )√ ( )( )
( ) √ ( )≥
Solution 2 by Geanina Tudose-Romania
( + )√ + ( + + )√ + ≥
≥ ( + + ) + √ + √
⇔ ( + )√ + ( + + )√ + ≥
≥ ( + + ) + ( + )√
+ √ + ( + + )√ + + √ + √
⇔ ( + )√ + ( + + )√ + ≥ ( + + ) +
+ √ + √
www.ssmrmh.ro
⇔ ( + − )√ + ( + − )√ + ( − − ) ≥
⇔ ( − − ) − √ − √ ≥ (1)
AM ≥ GM we have √ ≤ ⋅ = ( + ) ≤
√ ≤ ≤ = . Hence − √ − √ ≥
and − − ≥ . Therefore (1) is true
20. If , , > 0, = 1 then:
+ + ( + )( + ) + ( + )( + )( + ) ≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Nguyen Minh Tri-Ho Chi Minh-Vietnam, Solution 2 by Abdallah
El Farissi-Bechar-Algerie, Solution 3 by Ravi Prakash-New Delhi-India, Solution
4 by Tuk Zaya-Ulaanbaatar-Mongolia, Solution 5 by Nguyen Thanh Nho-Tra
Vinh-Vietnam, Solution 6 by Soumitra Mandal-Chandar Nagore-India
Solution 1 by Nguyen Minh Tri-Ho Chi Minh-Vietnam
( + )( + ) + ( + ) + ≥ ( + )( + )( + )
⇔ ( + + + ) + + + ≥ ( + )( + + + )
⇔ + + + + + + ≥ ( + + + + + + + )
⇔ ( + + ) + ( + + ) + ≥ ( + + ) + ( + + ) +
⇔ ( + + ) + ( + + ) ≥ (1)
We have: + + ≥ √ = ;
+ + ≥ = ⇒ (1) true⇒ Q.E.D
Solution 2 by Abdallah El Farissi-Bechar-Algerie
If = , then + ( )( ) + ( )( )( ) ≥
www.ssmrmh.ro
We have ( + )( + )( + ) ≥ √ = then − ( )( )( ) ≥
and + ( )( ) + ( )( )( ) = − ( )( )( )
generalization if ≥ ( = , , … , )−∏ = , then
∏ ( + ) ≥−
Solution 3 by Ravi Prakash-New Delhi-India
As = , = , = + ( )( ) + ( )( )( )
= + + ( + )( + ) + ( + )( + )( + )
=( + )( + ) + ( + ) +
( + )( + )( + )
=+ + + ( ) + + +
+ + + + + + + ( )
= − ( )( )( ). But + ≥ √ , + ≥ √ , + ≥ √
⇒ ( + )( + )( + ) ≥ ⇒ ≥ ( + )( + )( + )
Thus, ≥ − =
Solution 4 by Tuk Zaya-Ulaanbaatar-Mongolia
, , > 0 = 1; +
+ ( + )( + ) + ( + )( + )( + ) ≥
+ = ⇒ = − ; + = ⇒ = − ; + = ⇒ = −
−+
−+
−≥ ⇔ − + − + − ≥
www.ssmrmh.ro
≥ ⇔ ≥ ⇔ ( + )( + )( + ) ≥
+ ≥ √+ ≥ √+ ≥ √
⇔ ( + )( + )( + ) ≥ √ = ; = = =
Solution 5 by Nguyen Thanh Nho-Tra Vinh-Vietnam
, , > 0, = 1; ( + )( + )( + ) ≥ + √ =
⇔ ( + )( + )( + ) ≤ ⇔ − ( + )( + )( + ) ≥ −
⇔ − + − ( )( ) + ( )( ) − ( )( )( ) ≥ − +
⇔+
+ ( + )( + ) + ( + )( + )( + ) ≥
Solution 6 by Soumitra Mandal-Chandar Nagore-India
++ ( + )( + ) + ( + )( + )( + )
= − + + ( + )( + ) + ( + )( + )( + )
= − ( )( ) + ( )( )( ) = − ( )( )( )
≥⏞ − = − =
21. If , , , ≥ then:
( + ) ( + ) ( + )( + )( + )( + ) ≤ ( + )( + )( + )
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Abdul Aziz-Semarang-Indonesia, Solution 2 by Geanina Tudose-
Romania
Solution 1 by Abdul Aziz-Semarang-Indonesia
Since , , ≥ , we have
)( − )( − ) ≥ ⇒ ( + )( + ) ≤ ( + ) (1)
)( − )( − ) ≥ ⇒ ( + )( + )( + ) ≤ ( + ) (2)
)( − )( − ) ≥ ⇒
( + )( + )( + )( + ) ≤ ( + )(3)
Multiplying (1), (2), (3)
( + ) ( + ) ( + ) ( + ) ≤ ( + )( + )( + )
⇔( + ) ( + ) ( + )
( + )( + )( + ) ≤ ( + )( + )( + )
Solution 2 by Geanina Tudose-Romania
Rewritting the inequality
( + ) ⋅ ( + ) ⋅ ( + ) ⋅ ( + ) ≤ ( + )( + )( + )
We show ( + )( + ) ≤ ( + ) (1)
( + )( + )( + ) ≤ ( + ) (2)
and ( + )( + )( + )( + ) ≤ ( + ) (3)
In fact, we have more generally
( + )( + ) … ( + ) ≤ ( + … )
for (∀) ≥ , ≥ , > we show it by induction:
( ): ( + )( + ) ≤ ( + )
⇔ + + ≤ + ⇔ ( − )( − ) ≥ true
( ) → ( + )
www.ssmrmh.ro
( + ) … ( + )( + ) ≤⏞( )
( + … )( + ) ≤⏞( )
≤ ⋅ ⋅ ( + … )
Thus multiplying (1), (2), (3) we have the desired inequality
22. If < ≤ ≤ < 1 then:
( − ) + ( − ) + ( − ) < −
Proposed by Daniel Sitaru – Romania
Solution by Abdelhak Maoukouf-Casablanca-Morocco
< ≤ ≤ < 1
= ( − ) + ( − ) + ( − )
= ( − ) + ( − + − ) + ( − )
= ( − )( + ) + ( − )( + )
= ( − )+− + ( − )
+−
“ ≤ < 1”; “ < 1; < 1”
< ( − ) + ( − ) = ( − ) < < −
23. If , , , ≥ then:
+ + + + + + + >+ √
++ √
++ √
Proposed by Daniel Sitaru – Romania
Solution by Kevin Soto Palacios – Huarmey – Peru
Siendo , , , ≥ . Probar que
www.ssmrmh.ro
+ + + + + + + >+ √
++ √
++ √
Es suficiente probar + ≥√
(A)
⇔+
− ++
≥+ √
− ⇔+
−+
≥− √+ √
⇔
⇔−
( + )( + ) ≥− √+ √
⇔ √ √( )( ) ≥ √
√⇔ −√ ( √
( )( ) − √=
= − √+ √ − ( + )( + )
( + )( + ) + √=
= √ √ √( )( ) √
≥ , lo cual es cierto ya que → ≥
Analogamente para los siguientes términos
+ + +√
≥√
+⋅ √
≥√
⇔
⇔ + + ≥√
(B)
⇒ + + + ≥√
+√
≥√
(C)
Sumando (A)+(B)+(C)
⇒+
++
++
++
≥+ √
++ √
++ √
Por lo tanto
+ + + = + + + + + ≥
www.ssmrmh.ro
≥+ √
++ √
++ √
+ + + + >
>+ √
++ √
++ √
24. If < ≤ ≤ then:
( − ) ⋅ ( ) + ( − ) ⋅ ≤ ( − ) ⋅ ( ) + ( − ) ⋅
Proposed by Daniel Sitaru – Romania
Solution by Abdelhak Maoukouf-Casablanca-Morocco
∀ < ≤ ′ ( ) = ( − ) − ( − )
⇒ ( ) =−
−( − )
=− ( − ) ( − )
( − ) > 0
≤ ⇔ ( ) ≤ ( ) ⇔ ( ) ≤
< ≤ ≤ ⇒ ( ) ≤ & ( ) ≤ & ( ) ≤
⇒ ( ) + ( ) + ( ) ≤
⇔ [ ( − ) − ( − ) ] + [ ( − ) − ( − ) ] +
+[ ( − ) − ( − ) ] ≤
⇔ ( − ) ( ) + ( − ) ≤ ( − ) ( ) + ( − )
25. If , > 0, + + = 0 then:
+ + ≥ ( + ) √ +
Proposed by Daniel Sitaru – Romania
Solution by Anas Adlany-El Zemamra-Morocco
First, note that: ( + + ) = + + + ( + )( + )( + )( + + + + + )
www.ssmrmh.ro
Then when + + = , we will have the following
+ + = ( + + + + + );
Now, let’s go back to the main problem and check out what we are really
dealing with the problem asks us to show that:
+ + ≥ ( + ) √ + whenever , > 0
We have + + ≥ ( + ) √ +
⇔ ( + + + + + ) ≥ ( + ) √ +
⇔ ( + + )( + + + + + ) ≥ ( + ) √ +
and the last step can be explained as follows:
+ + = ⇒ ( + ) + = ⇒ + + + ( + ) =
⇒ + + = ( + = − )
Now, since , > 0; by the AM-GM inequality we get:
+ + ≥ √ + − √ + (1)
Also, ∑( + ) − − = + + + +
= ( + ) + + ( + )( + = − )
= (( + ) − ) − = − − = − = ( − ) ≥
Which prove that ∑( + ) ≥ + (2)
Finaly, from results (1) & (2) the proof is completed.
26. If < ≤ ≤ then:
( − )( − )( − ) √ √ √ ≤ − ( − ) −
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Chris Kyriazis-Greece, Solution 2 by Abdelhak Maoukouf-
Casablanca-Morocco, Solution 3 by Ravi Prakash-New Delhi-India, Solution 4
by Soumitra Mandal-Chandar Nagore-India
Solution 1 by Chris Kyriazis-Greece
If = or = or = , the inequality is obvious
Let’s suppose that < <
The function ( ) = is strictly convex, so using the Hermite-
Hadamard inequality we have that: ∫ ( )
≥ ( ) ⇔ ≥
⇒⏞√
− ≥ ( − ) √ . Doing exactly the same:
− ≥ ( − ) √ ; − ≥ ( − ) √
Multiplying those three inequalities (everything is positive) we have
that − ( − ) − ≥ ( − )( − )( − ) √ √ √
Solution 2 by Abdelhak Maoukouf-Casablanca-Morocco
∵ ∀ ≥ ( ) ≥ So ∀ ≥ : ≥
⇔ ≤ ⇔ ( − ) ≤ − (1)
∵ ∀ , > 0: ≤ ⇔ ≤ (2)
(1)×(2) ⇒ ∀ ≥ > 0( − ) ≤ −
⇔ ( − ) ≤ ( − )
www.ssmrmh.ro
< ≤ ≤ ⇒( − ) √ ≤ −( − ) √ ≤ −( − ) √ ≤ ( − )
⇒ ( − )( − )( − ) √ √ √ ≤ − ( − ) −
Solution 3 by Ravi Prakash-New Delhi-India
If = , − = ( − ) √ . Suppose > > 0
−−
=−
( − ) +!
( − ) +!
( − ) + ⋯
= + !( + ) + ! ⋅
( + + )+ ! ⋅
( + + + ) + ⋯
> 1 + √ + !( ) + !
( ) + ⋯
= + √ + !√ + !
√ + ⋯ = √
⇒ − > √ ( − )∀ ≥
Thus, ( − ) √ ( − ) √ ( − ) √ ≤ − ( − ) −
⇒ ( − )( − )( − ) √ √ √ ≤ − ( − ) −
Solution 4 by Soumitra Mandal-Chandar Nagore-India
is convex function. Hence, applying Hermite – Hadamard – Inequality
∫ ≥ , ∫ ≥ and ∫ ≥
⇒ − − ( − ) ≥ ( − )( − )( − )
≥ ( − )( − )( − ) √ √ √
27. If < < < < then:
www.ssmrmh.ro
>
Proposed by Daniel Sitaru – Romania
Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco, Solution 2 by
Abdallah Almalih-Damascus-Syria
Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco
∴ let ( ; ) = ; ( ; ) ∈ (ℝ∗ )
⇒ ( , ) =( − ) − ( − )
( − ) =− +
( − ) ≤
∵ + ≤ :∀ > 0 ∗ ( ; ) = ( ; ) ⇒ ( ; ) ≤
∵ < << < ⇒ ( ; ) > ( ; ) ⇔
−− >
−−
⇔ > ⇔ >
Solution 2 by Abdallah Almalih-Damascus-Syria
If < < < < then
> this inequality equivalent to ( is increased function)
> ; > which it’s right.
because ( ) = is a concave function as ( ) = − < 0 So it satisfy
( ) ( ) ≤ ( ) ( ) ≤ ( ) ( ) for ∈] , [ take = ∈] , [
so ( ) ( ) < ( ) ( ) < ( ) ( ) (1)
www.ssmrmh.ro
take ∈ ∈] , [ so ( ) ( ) < ( ) ( ) < ( ) ( ) (2)
hence ( ) ( ) <( ) ( ) ( ) <
( ) ( ) ( ) so ( ) ( ) < ( ) ( )we have
( ) = so ( ) ( ) ≤ ; ≤
28. If ≤ , , < 1 then:
( + )( + )( + )( − )( − )( − ) ≥
+−
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
For ≤ , , < 1, consider = ( − ) ( + )( + )( + ) −
−( + ) ( − )( − )( − )
=( − ) ( + )
( − )( − )( − ) ( + )( + )( + )
Use → − to obtain
=( − ) +
( − )( − )( − ) ( + + ) +
Use → +
= + ++ + + + + +
Use → −
= + ++ + − + + −
Note that + − ( + ) = ( − ) > 0
⇒ + > 3 + (1)
www.ssmrmh.ro
Also, + + ≥ + + and ( − ) + ( − ) + ( − ) ≥
≥ ( − ) + ( − ) + ( − ) [∵ ≤ , , < 1]
⇒ ( + + )[( − ) + ( − ) + ( − ) ]
≥ ( + + )[ ( − ) + ( − ) + ( − ) ]
⇒ + + − ≥ + + − (2)
From (1), (2), we get ( + )( + + − ) ≥
≥ [ + + − ] ⇒ ≥
⇒ ( − ) ( + )( + )( + ) ≥ ( + ) ( − )( − )( − )
Put = , = , = to obtain
( + )( + )( + )( − )( − )( − ) ≥
( + )( − )
29. If < < < then:
−−
+>
−−
+
Proposed by Daniel Sitaru – Romania
Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco, Solution 2 by
Soumitra Mandal-Chandar Nagore-India
Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco
∴ let ( ; ) = ; ( ; ) ∈ (ℝ∗ )
⇒ ( , ) = + ( − ) − ( − )
( − )
www.ssmrmh.ro
= −( )
( ) ≤ ∵ ≤ ≤ :∀ ≥
∗ ( ; ) = ( ; ) ⇒ ( ; ) ≤
∵ < << < ⇒ ( ; ) > ( ; ) ⇔ >
∵
⎝
⎛
+
=+−
∀ < 1 ⎠
⎞ ⇔−
+− >
−+
−
Solution 2 by Soumitra Mandal-Chandar Nagore-India
LEMMA: For any concave function : [ , ] → ℝ and ∈ ( , ) we have ( ) − ( )
− ≤( ) − ( )
− ≤( )− ( )
−
Let ( ) = for any ∈ [ , ], ( ) = , ( ) = − ( ) ≤
so, is concave, hence for < < we have
< < (1)
for < < , < < (2)
combining (1) and (2) we have, <
(Proved)
30. If < ≤ ≤ , + + = 3 then:
+ + ≥ + + + ( − )( − )( − )
Proposed by Daniel Sitaru – Romania
Solution 1 by Abdul Aziz-Semarang-Indonesia, Solution 2 by Pham Quoc Sang-
Ho Chi Minh-Vietnam
Solution 1 by Abdul Aziz-Semarang-Indonesia
www.ssmrmh.ro
+ + + ( − )( − )( − ) = + ( )( )( )
+ + = ⇒ ≤⇓
≤+ + + − + − + −
=+ +
= + +
Solution 2 by Pham Quoc Sang-Ho Chi Minh-Vietnam
≥ + ( − )( − )( − )
Because + + = ⇒ ≤ =
We have: ∑ − ∑ = − + − + −
= [ ( − ) + ( − ) + ( − )]
= [ ( − ) + ( − + − ) + ( − )]
= [( − )( − ) + ( − )( − )] = ( − )( − )( − )
≥ ( − )( − )( − )
(because ≤ and − ≥ , − ≥ , − ≥ )
= ( − )( − )( − ); “=” = = =
31. Let , , be positive real numbers. Prove that
+ + ≥ ( + + ) + ( )( )( )( )( )
Solution by Nguyen Ngoc Tu – HaGiang – Vietnam
www.ssmrmh.ro
Solution by proposer
We have ∑ = ∑ ( )( ) ( )
=( − )( − )
++
( − )( − )+
+ ( + + )+
=+
( + )( + )( − ) + ( + + )
( )( )( − ) = ( )
( )( )( )( − ) ≥ ( )( )( )
( − )
= ( + )( + )( + ) ⋅( − )
Similarly ( )( )( − ) ≥ ( )( )( ) ⋅
( ) ,
+( + )( + ) ( − ) ≥ ( + )( + )( + ) ⋅
( − ).
Hence ∑ ( )( )( − ) ≥ ( )( )( ) ⋅
( )
=( − )
( + + )( + )( + )( + )
⇒ + + ≥ ( + + ) + ( )( )( )( )( )
32. If , , ≥ , + + = then:
( + + ) + ≥ ( + + )
Proposed by Sladjan Stankovic-Macedonia
Solution 1 by Soumitra Mandal-Chandar Nagore-India, Solution 2 by Le Khanh
Sy-Long An-Vietnam, Solution 3 by Marian Cucoanes – Romania, Solution 4
www.ssmrmh.ro
by Leonard Giugiuc – Romania, Solution 5 by Ngo Minh Ngoc Bao-Vietnam ,
Solution 6 by Marian Dincă – Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India
Let , , ≥ , + + = then ∑ + ≥ ∑
We will prove that for any , , ≥ with + + = the inequality
( ) ≥+ +
… . ( )
holds true where ( ) = − + for all ≥
Now, ( ) = − > 0 for all ≥ = = .
So, the function is right – convex for ≥ . By RCF theorem it suffices
to prove (1) for = ≥ ≥ with + + = . Hence we will show
( ) ≥ ( ) for ≥ ≥ ≥ and + =
where ( ) = ( ) ( ) = + − − . So, ( ) − ( )
= ( − ){ ( + + ) + ( + )− } = ( − )( − + )
[∵ + = ] = ( − )( − )( − ) ≥ [∵ + = ⇒ ≥ − > 0]
∴ ( ) ≥ ( ) for ≥ ≥ ≥ and hence the proof is complete. So,
+ ≥
(Proved) Equality at = = = and = = , = or any other
permutations.
Solution 2 by Le Khanh Sy-Long An-Vietnam
Let , , be nonnegative real numbers such that + + = and
www.ssmrmh.ro
= { , , }. Prove that
( + + ) + ≥ ( + + ) +( − )
Lemma: If , , ≥ and = { , , } → + ≥ , then
( + ) − ( + ) ≥( + )
−( + )
+( − )
Or ( )
≥ ( − )
We have =( ) ( )
≥
Thus, the proof is completed. Therefore, it suffices to ( ) + + ≥ + ( ) , Which is ( − ) ( + ) ≥
The equality occurs for = = = , and for = and = = .
Solution 3 by Marian Cucoanes – Romania
+ + = (1); ( + + ) + ≥ ( + + ) (2)
(2) ⇔( )
( + + ) + ( + + ) ≥
≥ ( + + ) ( + + ) and because:
( + + ) + ( + + ) − ( + + ) ( + + ) =
= ( + + + + + )− ( + +
+ ) − ( + + + + + ) ⇔
⇔ ( + + + + + ) ≥ ( +
+ + ) + ( + + + + + ) (3)
Using Schur inequality for degree = ⇒
( − )( − ) + ( − )( − ) + ( − )( − ) ≥ ⇔
⇔ + + + + + ≥ +
www.ssmrmh.ro
+ + + + + ⇒
( + + + + + ) ≥ ( +
+ + + + + ) (4)
Using AM – GM inequality ⇒ ( + + + + + ) ≥
≥ ( + + ) (5)
Then (4) + (5) ⇒ (3) ⇒ (2) ⇒ q.e.d.
Solution 4 by Leonard Giugiuc – Romania
It’s well known that,
( + + ) + ( + + ) ≥ ( + + )( + + ) ⇒
( + + ) ≥ ( + + ) − . Suffice it to show that
( + + ) − + ≥ ( + + ). But
+ + = ( + ) ⇒ + + = ( + ), where = .
The latter is + − ≥ . But if ≤ ≤ , then
≥ − − + , hence suffice it to show − ≥ , which is true.
If ≤ ≤ , then + − ≥ − ≥ .
Done! Equality at ( , , ) or , , and permutations.
Solution 5 by Ngo Minh Ngoc Bao-Vietnam
( + + ) + ≥ ( + + ) (*). We have lemma:
( , , ) = + + + + .
( , , ) ≥ ,∀ , , ≥ ⇔
⇔
+ + + + ≥( + ) < + +
+ + + ≥( ) = ( + + )( + ) + ( + − − ) + ( − + − ) ≥ ,∀ ≥
www.ssmrmh.ro
The inequality (*) ⇔
⇔ ( + + ) + ( + + ) ≥ ( + + ) ( + + )
⇔ + + + + + ≥
≥ + + + +
⇔ − − − + ≥ ⇔
⇔ − − − + ≥
Consider quadratic symmetry polynomial:
( , , ) = − + − −
Use lemma with = − , = , = = − ⇒
⇒
+ + + + =( + ) = < + + =
+ + + = > 0 (1)
Consider function: ( ) = − − + ,∀ ≥ .
We have: ( ) = ( − − ), ( ) = ⇔ − − = ⇔
⇔=
= −
www.ssmrmh.ro
⇒ ( ) ≥ ( ) = > 0,∀ ≥ 0 (2)
With (1) and (2) ⇒ ( , , ) ≥ .
Equality when = = = or ; ; or ; ; or ; ;
Solution 6 by Marian Dincă – Romania
= − , = + , = ; + = −
+ = ( + ) − ( ) = ( − ) −
( + + ) + ≥ ( + + ) ⇔
[( − ) − + ( − ) ] + ≥ ( − + ( − ) )
Let : , →
( ) = [( − ) − + ( − ) ] + − ( − + ( − ) )
( ) = (− )( ) ( − ) − + =
= − + − + = − ≤ ⋅ − =−
< 0
= decreasing ⇒
( ) ≥ = − − + ( − ) + − − + ( − ) =
= − + ( − ) + − + ( − ) =
= + ( − ) + − ⋅ − ( − ) = ( )
( ) = − ( − ) − + ( − ) =
= [ − ( − ) ] − [ − ( − )] = = [ − ( − ) ]− ( − ) = [ − ( − )][ + ( − ) + ( − ) ] −
− ( − ) = ( − )[ + ( − ) + ( − ) ]− ( − ) =
= ( − )[ + ( − ) + ( − ) − ] = ( − )( − + )
www.ssmrmh.ro
− + = ⇒ , =± √ − ⋅
=± √
=
= ±√
, for = { , , }, result ≥ , + ≤ ⇒ ≤ and < < , we
obtain: − + = ( − )( − ) > 0
and: ( ) = ( − )( − + ) ≤
= decreasing ( ) ≥ ( ) = + + − ⋅ − − =
= − = , we shall obtain: ( ) ≥ ( ) ≥
33. If ∈ ℕ, ≥ , , ≥ then:
+ + + + ⋯+ + ≤ ( − ) +
Proposed by Daniel Sitaru - Romania
Solution 1 by Soumava Chakraborty-Kolkata-India ,
Solution 2 by Ravi Prakash-New Delhi-India
Solution 3 by Uche Eliezer Okeke-Anambra-Nigeria
Solution 1 by Soumava Chakraborty-Kolkata-India
+ ≤ + ⇔ ( + ) ≤ ( + ) (∵ , ≥ )
⇔ ≤ ( + ) ⇔ { ( + ) − } ≥
⇔ { ( − ) + ( + ) } ≥ → true
∴ + ≤ + (1)
Again, + ≤ + ⇔ + ≤ ( + )
⇔ ≥ → true ∵ + ≤ + (2)
(1), (2) ⇒ + ≤ + is true, for = , =
Let us assume + ≤ + holds true
for = (some integer ≥ ); we shall prove.
www.ssmrmh.ro
Then show that + ≤ + will hold true for = + as
well + ≤ +
⇔ + ≤ ( + ) (a)
By our assumption, + ≤ +
⇒ + ≤ ( + ) (3)
Now, ( + ) = ( + )( + )
≥ ( + ) + (by our assumption and by using (3))
= ( + ) + +
= + + + + + ≥?
+
⇔ + + ( + − ) ≥
⇔ + + ( + ) + ( − ) ≥ → true,
∵ , ≥ ⇒ a is true
So, whenever + ≤ + is true for = ( ≥ , ∈ ℕ), then,
+ ≤ + is true for = + as well.
Hence, by the principle of mathematical induction,
+ ≤ + (b) (∀) ≥ , ∈ ℕ
(b), (1) ⇒ + ≤ + ∀ ≥ , ∈ ℕ
∴ + ≤ ( − ) +
Solution 2 by Ravi Prakash-New Delhi-India
Let ≥ , ≥ . Suppose = , ≥
www.ssmrmh.ro
= ≥ ⇒ = ≥ =
⇒ ( + ) = + + ≥ + + = ( + )
Suppose = + , ≥ ( ) ≥ ≥ , [∵ ≥ ]. Now,
( + ) = + + ≥ + +
∴ ( + ) ≥ ( + ) ∀ ≥ , ≥ Suppose ≥ > 0 then
+ ≥ + ⇒ ( + ) ≥ ( + )
⇒ ( + ) ≤ + ⇒ ( + ) ≤ ( − ) +
Solution 3 by Eliezer Okeke-Port Anambra-Nigeria
If ∈ ℕ, ≥ , , ≥ . The ∑ + ≤ ( − ) +
Equality case follows if and only if ⇒ = = ⇒ ≠ ≠ , , ∈ ℝ
< . It suffices to prove
+ < + (1)
WLOG: Assume <
(*) + = + ⇒ + = + +
(**) + = + +
Clearly since < ∀ ≥ ⇒ <
www.ssmrmh.ro
Conclusion + < + (True)
+ ≤ + = ( − ) +
34. From the book: “Math Phenomenon”
If , ∈ ( ,∞) then:
++
++
+≥
Proposed by Daniel Sitaru – Romania
Solutions 1,2 by Fotini Kaldi-Greece
Solutions 3,4 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
Solution 1 by Fotini Kaldi-Greece
> 0, > 0
+ +≥
+ +⇒ + + ≥
≥ ( + + ) , , , > 0
≥+
++
+⋅ +
⋅=
≥⋅
+ + +⋅
≥
≥⋅
( + ) =
Solution 2 by Fotini Kaldi-Greece
Weighted Means > 0, > 0
www.ssmrmh.ro
≥+
⋅+
⋅⋅ +
⋅=
= ( + )( + ) +⋅
≥
≥ √ √ ⋅= = ⇒ ≥
Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
ln ⋅ ln + 12 ln ⋅ ln
=+ ⋅
( ) = ⇒ ( ) ≥ CONCAVE
++
+ ⋅ ++
≥
≥⋅
≥∙ √ ⋅ ⋅ =
Solution 4 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
==
= ∈ ℕ⇒
++
++
+≥
www.ssmrmh.ro
≥√
+
⎝
⎜⎛
⋅
⎠
⎟⎞
+ = √ + + ≥
( − )y ≥ ⋅ √ ⋅ ⋅ ( ) = .
35. If , , ∈ ( ,∞) then:
+ + ≥( + ) ( + ) ( + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Nirapada Pal-Jhargram-India , Solution 2 by Fotini Kaldi-Greece ,
Solution 3 by Eliezer Okeke-Anambra-Nigeria
Solution 1 by Nirapada Pal-India
= =
≥⏞+ + +
=( + ) ( + ) ( + )
Solution 2 by Fotini Kaldi-Greece
+ + =
= + + + + + + + + ≥⏞
≥+ + +
⇒
www.ssmrmh.ro
( + + ) ≥+ + +
⇒
⇒ ( + + ) ≥ ( ) ( ) ( ) ; “ =” ⇔ = =
Solution 3 by Eliezer Okeke-Port Anambra-Nigeria
( + + ) = ( ) + ( ) + ( ) = ( ) + ( ) + ( )
⇒ ( + + ) + ( + + ) =+ +
+ ( + )
=∑
+ ( + )
= + + ++
++
++
≥⏞
⋅ = ( ) ( ) ( )
36. If , , ∈ , then:
+ + + + + >
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
Solution 1 by Nirapada Pal-Jhargram-India , Solution 2 by Seyran Ibrahimov-
Maasilli-Azerbaidian , Solution 3 by Soumava Chakraborty-Kolkata-India
Solution 1 by Nirapada Pal-Jhargram-India
For , , ∈ , , < 1, < 1, < 1
So, + +
www.ssmrmh.ro
= + ⋅ + + ⋅ + + ⋅
>+
++
++
> 1, .
≥ by Nesbbit. ∴ + + >
Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian
≥( + + )
( + + ) >
( + + ) > 3( + + )
Jensen: + + ≥
> 3( + + ); + + < 9
Solution 3 by Soumava Chakraborty-Kolkata-India
∑ > ∑ ≥ (Nesbitt)
∵ > , < < 0 < , , <
37. Let ( , , ) is continuous and symmetric function with
, , > 0. If ( , , ) ≥ √ ,√ ,
Prove that: ( , , ) ≥ √ , √ , √
Proposed by Sladjan Stankovic-Skopje
Solution by Marian Dincă – Romania
Let: = , = , = ; ( , , ) = ( , , ) ≥ , ,
Let: ( , , ) = ( , , ) − continuous and symmetric equivalent to:
www.ssmrmh.ro
If: : ( , , ) ≥ , , . Prove that:
: ( , , ) ≥+ +
,+ +
,+ +
Let ≥ , let ∈ ,
→( − , + , ) =
+,
+, ≤ ( , , )
result to ( , ) such that: ∈ ( , )
( − , + , ) ≤ ( , , ), for ∈ ( , )
let: − = ⇒ = −
and: ( , , ) ≥ , + − ,
and: , + − , = + − , ,
if: + − ≥ , result:
+ − , , ≥ + − − , + ,
Let: + − − = ⇒ = + − ( + + )
We obtain: + − , , ≥ , , , or
( , , ) ≥+ +
,+ +
,+ +
⇔
( , , ) ≥ √ , √ , √
38. If , , > 0, + + = √ then:
++ ≤
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Ravi Prakash-New Delhi-India
Solution 2 by Soumitra Mandal-Chandar Nagore-India
Solution 3 by Sanong Hauyrerai-Nakon Pathom-Thailand
Solution 1 by Ravi Prakash-New Delhi-India
For ≤ ≤ √ , let ( ) = √ +
Note is continuous on ,√ and
( ) = −+ √ −
++
, < < √
=+ √ − − −
+ √ − ( + )=
√ − − √ − +
+ √ − ( + )
( ) > 0 if < <√
= if =√
< 0 if √
< < √
∴ attains its maximum value at =√
and
√=
√+
√=
Now, + ( ) = √ + ( ) ≤
Similarly, for other two expressions.
Solution 2 by Soumitra Mandal-Chandar Nagore-India
We have, ( ) = − ( ) ≤ for all ≥
So, is concave. Applying Jensen +
≤+ +
=
www.ssmrmh.ro
similarly, ∑ ≤ so, ∑ + ≤
Solution 3 by Sanong Haueyrerai-Nakon Pathom-Thailand
We will show that
+ + + + + ≤
Because = ,− < < , ∈ ℝ
If = ,− < < , ∈ ℝ and since + + = √ , , , > 0.
Hence = , < < , ∈ ℝ . If = , < < , ∈ ℝ
Give = , = , =
If = , = , =
and + + = + + = √
Hence + + = + + ≤ −
If = , = , =
and + + = + + = + + = √
Hence + + = + + ≤ =
Hence + + ≤ −
Therefore + + + +
+ ≤ + = . That is, it is to be true.
39. Prove that for any real numbers , ,
( + + )( + − )( + − )( + − ) ≤
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
www.ssmrmh.ro
Solution 1 by Sladjan Stankovic-Skopje , Solution 2 by Marian Dincă –
Romania , Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
Solution 4 by Soumava Chakraborty-Kolkata-India
Solution 1 by Sladjan Stankovic-Skopje
− = −( − − ) ≤
Solution 2 by Marian Dincă – Romania
( + + )( + − )( − + )(− + + ) =
= ( + + ) − − − =
= ( + − ) ( )
Let: + − = , + − = , + − =
Rezult: = + , = + , = +
( + − ) = ( + + )
= ( + )( + ) = ( + + + ) ≥
≥ ( + + ) ⇔ ≥
Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
(( + ) − ) ⋅ ( − ( − ) ) =
= − + (( + ) + ( − ) ) ⋅ − ( − ) ≤ ⇔
ASSURE
⇔ − (( + ) + ( − ) ) ⋅ + + ( − ) =
= − ( + ) + ( + ) = − ( + ) ≥
Solution 4 by Soumava Chakraborty-Kolkata-India
− ≤ ⇔ − − − + + − ≤
www.ssmrmh.ro
⇔ + + + − ( + ) ≥
⇔ − ( + ) + ( + ) ≥ ⇔ ( − − ) ≥ ,
which is true (Proved)
40. Prove that if , , , , , ∈ ( ,∞) and + + = ; + + =
then: ⋅ ⋅ ≤
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Solution 2 by Nirapada Pal-Jhargram-India
Solutions 3,4 by Uche Eliezer Okeke-Anambra-Nigeria
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Probar para todo , , , , , ∈< 0,∞ >, de tal manera que
+ + = , + + = lo siguiente ≤
Por la desigualdad ponderada ≥
⋅ + ⋅ + ⋅+ + ≥ =
= ≥ (LQQD)
Solution 2 by Nirapada Pal-Jhargram-India
≤⏞
= = =
Solution 3 by Uche Eliezer Okeke-Anambra-Nigeria
If , , , , , ∈ ( ,∞); + + = ; + + =
www.ssmrmh.ro
⋅ ⋅ ≤ ; = ⋅ ⋅
≤ = = = (Proved)
Solution 4 by Uche Eliezer Okeke-Anambra-Nigeria
= ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅
≤+ ++ +
⋅+ +
( + + )
= ( + + ) ⋅ ⋅ ≤( + + ) ( + + )
⋅ ⋅
= ⋅ ⋅ ⋅ = (Proved)
41. Prove that:
+ + + < 4
Proposed by Daniel Sitaru – Romania
Solution 1 by Rovshan Pirguliyev-Sumgait-Azerbaidian , Solution 2 by Uche
Eliezer Okeke-Anambra-Nigeria , Solution 3 by Soumava Chakraborty-
Kolkata-India , Solution 4 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
Solution 1 by Rovshan Pirguliyev-Sumgait-Azerbaidian
= , = , = . Ineq ⇔ + < 4. It is known that
< , < we have
www.ssmrmh.ro
+ < + = = ( ) =
=( + ) +
=+
<
<+
=⋅ + ⋅
=
Solution 2 by Uche Eliezer Okeke-Anambra-Nigeria
Let = ; = ; = . = +
= + ≤⏞ + = (*)
Now < ; 3 < ; < . So < ; < 2; < 2
⇔ (*) < ( ) ( ) ( ) = < 4. (Proved)
Solution 3 by Soumava Chakraborty-Kolkata-India
Let = , = , = . We have: < <
∴ = + . Now, <⏞ √ <⏞( )
√ = (∵ < )
Again, ∑ >⏞ ( ) ⇒ ∑ <( )
⇒ ∑ <( )
= √ <⏞( )
√ = (∵ < , < )
(1), (2) ⇒ < + = + = <
= = < 4 (∵ < ) = (Proved)
Solution 4 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
===
+ + = <+ = <
(*)
www.ssmrmh.ro
++
+ +=
++ ⋅
+ +≤⏞
≤ + + + + =
= ( + ) + ( + + ) <⏞(∗)
⋅ + =
42. From the book: “Math Accent”
If , , ∈ ℕ − { , }, + + = then:
+ + + + + > 200
Proposed by Daniel Sitaru – Romania
Solution 1 by Redwane El Mellas-Casablanca-Morocco
Solution 2 by Ravi Prakash-New Delhi-India
Solution 1 by Redwane El Mellas-Casablanca-Morocco
∴ + =( + )!
! ! =( + ) … ( + )
!
=[( + ) … ( − + )]
[ . … ]( − + )( + )
Also, ≥ ⇒ (∀ = , … , ): − + ≥ ⇒ [( )…( )][ . … ] ≥ .
So + ≥ ( )( ) (*)
Then ∑ + ≥ ∑ ( )( ) > ∑ ( ) = ∑ =
A Generalization of Daniel Sitaru’s binomial inequality
Let , … , ≥ ∈ ℕ such that + ⋯+ = + for ≥ .
www.ssmrmh.ro
So, +
++
+ ⋯++
> 66 + .
For a proof, see my proof in the case = .
Solution 2 by Ravi Prakash-New Delhi-India
For ≤ ≤ − , ≥ . As , , ≥ , + ≥ +
+ ≥ + , + ≥ +
∴ + + + + + > 2( + + ) =
43. If ∀ ∈ ℝ,+ ≥ +
+ + ≥ + ++ + + ≥ + + +
, , , , > 0 then:
⋅ ⋅ ⋅ =
Proposed by Daniel Sitaru – Romania
Solution 1 by Leonard Giugiuc – Romania
Solution 2 by Șerban George Florin – Romania
Solution 1 by Leonard Giugiuc – Romania
Let the function : → , ( ) = ∑ . Observe that ( ) = ∑ .
From the Fermat’s mean value theorem, since ( ) ≥ ( )∀ ∈ , we
get ( ) = ⇒ ∏ = .
So the first condition gives us = , the second gives = and
the third = . Multiplying, get .
Solution 2 by Șerban George Florin – Romania
( ) = + , + ≥ + (∀) ∈ ℝ
⇒ ( ) ≥ ( ) ⇒ = minimum point. ( ) = +
www.ssmrmh.ro
T. Fermat ⇒ ( ) = = +
( ) = + + , + + ≥ + + (∀) ∈ ℝ
( ) ≥ ( ) ⇒ = minimum point T. Fermat ( ) = ( ) = + + =
( ) = + + + , + + + ≥ + + + (∀) ∈ ℝ
( ) ≥ ( )|(∀) ∈ ℝ ⇒ = minimum point
T. Fermat ⇒ ( ) = = + + + = ⇒
= ; ≠ ⇒ = ⇒ = ⇒ = , ≠ ⇒ = ⇒ =
+ = ⇒ ⋅ = ⇒ ⋅ = ⇒ ⋅ =
⇒ ⋅ ⋅ ⋅ = ⋅ ⋅ =
44. In :
+ + ≥
+ + ≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Solution 2 by Leonard Giugiuc – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Probar en un triàngulo
1) + + ≥
2) + + ≥
Saberes previos
www.ssmrmh.ro
+ + = , + + =
Por la desigualdad de Cauchy
+ + ≥+ +
+ +=
+ + ≥+ +
+ +=
Solution 2 by Leonard Giugiuc – Romania
Let = , = and = .
Obviously , , > 0 and + + = .
By Cauchy, + + ( + + ) ≥ ( + + ) ⇒
+ + ≥ ⋅ ( + + ) =
The second is identical, with the specification that + + = .
45. If , ∈ ℝ , < then:
√ + − √ +−
<+ √ ++ √ +
Proposed by Daniel Sitaru – Romania
Solution by Rozeta Atanasova-Skopje
Let = and = .
( ) = and ( ) = and > ⇒
− < − … (1)
www.ssmrmh.ro
= and ( ) = < 1,∀ ∈ ℝ ⇒
⇒ < … (2)
=√ + − √ +
− =( − )
− <( ) ( − )
+
=⋅
= <( )
− = −
= + + − + + =+ √ ++ √ +
=
46. If , > 0; ∈ ℝ then:
( + )( + )( + ) + + ≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Ravi
Prakash-New Delhi-India, Solution 3 by SK Rejuan-West Bengal-India, Solution
4 by Seyran Ibrahimov-Maasilli-Azerbaidian
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Siendo , > 0, ∈ . Probar que
( + )( + )( + ) + + ≥
De las condiciones se puede afirmar lo siguiente
= + > 0; = + > 0
⇒ + = ( + ) + ( + ) = +
La desigualadad es equivalente
www.ssmrmh.ro
⇒ ( ) + + ≥ (Lo cual es cierto por ≥ )
Solution 2 by Ravi Prakash-New Delhi-India
Put = + > 0
= + > 0 ⇒ + = +
∴( + )
( + )( + ) + +
=( + )
+ + = + + + + ≥
∵ + ≥ ∀ , > 0
Solution 3 by SK Rejuan-West Bengal-India
Given that, , > 0, ∈ ℝ
Now, ( + ) = ( + ) + ( + )
= ( + ) + ( + )
Hence by AM ≥ GM we get,
+=
( + ) + ( + )
≥ ( + )( + )
⇒ ( )( )( ) ≥ (1)
Solution 4 by Seyran Ibrahimov-Maasilli-Azerbaidian
= =
++
+≤ +
+≤
+≤
www.ssmrmh.ro
+ ≥ +
≥ ⇒ + ≥ ( + ) (Proved)
47. Let be a tetrahedron with ∠ = ∠ = ∠ = ° and let
be any point inside the triangle . Denote respectively by , , the
distances from to faces ( ), ( ), ( ). Prove that
(a) + + = .
(b) ≤ ⋅ ⋅
(c) ⋅ + ⋅ + ⋅ ≥ .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution by Ravi Prakash-New Delhi-India
Let’s take ⃗ = ̂, ⃗ = ̂, ⃗ = , where , , are positive.
Equation of plane is + + = .
Let ( , , ) be any point inside , then
+ + + = ,< < ,< < ,< <
Note that = , = , =
Now, = , = , = .
(a) + + = + + =
(b) = + + ≥ ⋅ ⋅ ⇒ ≤
(c) ( )( ) + ( ) + ( )( ) =
= + + + +
www.ssmrmh.ro
= + + + + + + + +
≥ + + + ⋅ + ⋅ + ⋅
= + + =
48. If , , , > 0 then:
( + ) ( + ) ( + ) ≤ ⋅ ⋅ ( + + + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Soumitra
Mandal-Chandar Nagore-India
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Siendo , , , > 0. Probar
( + ) ( + ) ( + ) ≤ ⋅ ⋅ ( + + + )
⇔+ + +
+≥
+ +
⇔+ + +
+≥ + +
Por la desigualdad ponderada MA ≥ MG
≥ + + ⇔ ≥ + +
⇔+ + +
+≥ + +
Solution 2 by Soumitra Mandal-Chandar Nagore-India
Applying Weighted AM ≥ ;
www.ssmrmh.ro
⋅ + + ⋅ + +
≥+
⋅+
⇒+ + +
+≥
+⋅
+
⇒ ( + + + ) ⋅ ⋅ ≥ ( + ) ⋅ ( + ) ⋅ ( + )
49. If ∈ ( , ); > 0 then:
+ + + ≤√
+
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Seyran
Ibrahimov-Maasilli-Azerbaidian, Solution 3 by Soumitra Mandal-Chandar
Nagore-India
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Si ∈< 0, >, > 0. Probar que
+ + + ≤√
+
Dado las condiciones se puede afirmar que > 0, > 0
Aplicando MA ≥ MG
+ + + ≤ +√
= +√
Además por MA ≥ MG
√+
√+ + ≥ = (A)
www.ssmrmh.ro
+ +√
≥√
=√
(B)
Sumando (A) + (B) √
+ ≥ +√
⇔
⇔√
+ ≥ +√
≥+
++
Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian
= = ; + ≤ + ; ≤
≤ ; + ≥ + ; ≥ ⇒ + ≥ ( + )
Solution 3 by Soumitra Mandal-Chandar Nagore-India
++
+≤⏞
√+
We need to prove, +√
≥√
+
⇔ +√
−√
≥ , which is true
∵ > 0, > > 0, > 0 Hence proved
50. Prove that if , ∈ ( ,∞) then:
( ) ≤+ √ ++ √ +
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
Let ( ) = + √ + − + , ≥
( ) =+ √ +
+√ +
− + , > 0
www.ssmrmh.ro
=√ +
+ ( + ) − > ( + ) ( + ) − = ∀ > 0
∴ ( ) increases on [ ,∞) ⇒ ( ) > ( ) ∀ > 0
⇒ ( ) > 0∀ > 0 ⇒ 6 + + > 6 − ∀ > 0
⇒ + √ + > ∀ > 0 (1)
Next, let ( ) = − √ + + + , ≥
( ) = −√ +
+ + , > 0
⇒ ( ) = −√ +
+ > 0∀ > 0
∴ ( ) increases on [ ,∞) ⇒ ( ) > ( )∀ > 0
⇒ − + > −6 − ∀ > 0
⇒ −√ + > ∀ > 0 (2)
Putting = in (1), = in (2), (with , > 0) we get
− + + + > ⋅
⇒+ √ +
+ √ +> ( ) ,∀ , > 0
51. If , , , ∈ ℝ then:
+ + + ( + + )( + + ) ≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Ravi Prakash-New Delhi-India, Solution 2 by Boris Colakovic-
Belgrade-Serbia, Solution 3 by Bedri Hajrizi-Mitrovica-Kosovo
www.ssmrmh.ro
Solution 1 by Ravi Prakash-New Delhi-India
= ( + + ) + ( + + )( + + ) ≥
If + + ≥ , then write
= ( + + − − − ) + ( + + )( + )( + )
= ( − ) + ( − ) + ( − ) + ( + + )( + )( + ) ≥
If + + < 0 and + + < 0, still ≥ .
If + + < 0 and + + > 0,
then + + ≤ √ + < 2 Now, write
= ( + + ) + ( + + )
{( + + ) − − − }
= ( + + )( − − − ) +
+( + + ) ( + + ) ≥
⇒ ≥ in this case. Thus, ≥ ∀ , , , ∈ ℝ
Solution 2 by Boris Colakovic-Belgrade-Serbia
+ = , =−
⇒ + + =
= + − =( + )
− ; + + ≤ + + ⇔
( + )− ( + + ) ≤
( + )− ( + + ) ⇔
⇔ ≤ + + + ( ) − ( + + ) ≤ ( ) ( + + )
Solution 3 by Bedri Hajrizi-Mitrovica-Kosovo
+ + + ( + + ) − ( + + ) + ( + + )( + + ) ≥
( + + ) + ( + + )∈ ,√
( + + ) ≥
www.ssmrmh.ro
It’s enough to prove that: ( + + ) − ( + + ) ≥
+ + ≥ + + which is true for all , , ∈ ℝ
52. :ℝ → ℝ, ( ) = , , sides in scalene
Prove that:
( )( ) >
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
( ) =
As ( ) = when = , = , = , = , = , =
( − ), ( − ), ( − ), ( − ), ( − ), ( − ) are factors of ( )
Also, ( ) is a homogenous expression of degree in , , ,
∴ ( ) ≡ ( − )( − )( − )( − )( − )( − )( + + + )
where is a constant. When = ,
( ) = − = −− −− −− −
= −( − )( − ) + ++ + + + + +
→ − , → −
www.ssmrmh.ro
( ) = −( − )( − )+ +
= − ( − )( − )+ +
= − ( − )( − )+ +
= − ( − )( − )( − )( + + )
Also, ( ) = ( − )( − )( − )(− ) ( + + )
= ( − )( − )( − )( + + ) ∴ = − . Thus,
( ) = ( − )( − )( − )( − )( − )( − )( + + + )
| ( )| = |( − )( − )( − )| + | − | + | − | +
+ | − | + | + + + |
⇒( )( ) = − + − + − + + + +
( )( ) = − + − + − + + + +
≥ ( − )( − ) + ( − ) + = + = >
53. If , , ∈ ℝ∗; , , ∈ ℝ
= + +
= + +
then: + ≥
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
www.ssmrmh.ro
Now, + = + + + + +
( + + )
+ + + + ( + + ) + + +
= + + + + + +
+ + + + + + +
+ + + + + +
= + + + + + + + + ≥
54. If , ∈ , then:
++
++
+<
√
Proposed by Daniel Sitaru – Romania
Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam
If , ∈ ; then + +⋅
≤ √
Put = , =⋅
, =⋅
( , , > 0). We have:
+ ( ⋅ ) + ( ⋅ ) = + = ⇒
⇒ + + = . We have + ⋅ + ⋅⋅
≤ √
www.ssmrmh.ro
⇒+
++
++
≤√
⇒+
++
++
≤√
We have ≤ √ ⋅ + √ ⇒ ≤ √ √ ⋅ ⇒
⇒ ≤ √ + √ ( + )
⇒ √ ⋅ − + √ ⋅ + √ ⋅ + √ ≥ ⇒
⇒ − √ √ − − √ − − √ − ≥ ⇒
⇒ −√ √ ⋅ + ⋅ + √ ⋅ + + √ ≥
Similarly, we have ≤ √ ⋅ + √ and ≤ √ ⋅ + √
⇒ + + + + + ≤√
+ + +√
=√
=√
The equality occurs when
= = = √ ⇒ = ⋅ = ⋅ =√
We have = √ ⇒ = − √ = √ and = √
We have ⋅ = ⋅ = √ ⇒ = = √ ⇒ =
55. Prove that if , , ∈ ( ,∞); = then:
+ + ≥
Proposed by Daniel Sitaru – Romania
Solution by Boris Colakovic-Belgrade-Serbia
+ + ≥ =
www.ssmrmh.ro
+ + ≥ + = + ≥ +
+ + ≥ = ⋅ = >
56. If , , ≥ then:
− ≥ ( − )( − )( − )
Proposed by Daniel Sitaru – Romania
Solution by Soumitra Mandal-Chandar Nagore-India
( − )( − )( − ) + + + − − − = −
− ≥ ( − )
⇔ ( − )( − )( − ) + ( − ) + ( − ) + ( − ) ≥
( − )( − )( − )
⇔ ( − ) {( + )( + + ) − } + ( − ) +
+ ( − ) + ( − ) ≥ , which is true since , , ≥ and
( + )( + + ) ≥ ∴ − ≥ ( − )
(Proved)
57. If , , ≥ , + + = then:
( ) + ( ) + ( ) ≥( + + )
Proposed by Daniel Sitaru – Romania
Solution by Sanong Hauyrerai-Nakon Pathom-Thailand
www.ssmrmh.ro
+ + ≥( + + )
, , ≥ and + + =
Definition = , ∈ ℝ and − < <
Iff = , ∈ ℝ and – < <
Proof give = Iff =
= Iff = ; = Iff =
consider ≤ + + = + + ≥ ( )
Iff ( + + ) ≥ ( ) and since + + =
Hence, , , ≥ , , , ≥
and ≥ , ≥ , ≥
Hence, + + ≥ ( + + ) ≥ ( )
Therefore + + ≥ ( )
58. If , > 0, ∈ ℕ, ≥ then:
+√+
+ −√+
< 2
Proposed by Daniel Sitaru – Romania
Solution 1 by Ravi Prakash-New Delhi-India, Solution 2 by Geanina Tudose –
Romania, Solution 3 by Sanong Hauerai-Nakon Pathom-Thailand, Solution 4
by SK Rejuan-West Bengal-India, Solution 5 by Nikola Djurici-Serbia, Solution
6 by Ngo Minh Ngoc Bao-Vietnam
www.ssmrmh.ro
Solution 1 by Ravi Prakash-New Delhi-India
We assume ≥ ; + √ = √ √ = √ √√
and − √ = √ √ = √ √√
. Note that
√ √ + √ √ = ( ) = . Put √ √√
= √ ,
and √ √√
= √ , ≤ < . Now,
= +√+
+ −√+
=√ + √√ +
+√ − √√ +
= √ + √ = ( ) where ( ) = ( ) + ( )
≤ ≤ ; ( ) = ( ) (− ) + ( ) ( )
= ( ) ( ) − ( ) ( ) ; ( ) = ⇒ =
( ) = ( ), , = ,( )
, =( )
∴ ≤
Solution 2 by Geanina Tudose –Romania
By GM ≤ AM we have ⋅ ⋅ … ⋅ ⋅ + √ ≤... ⋯ √
⋅ … ⋅ ⋅ −√+ ≤
+. . . + + + √+
www.ssmrmh.ro
Summing up, we have + √ + − √ ≤√ √
=
The inequality is strict, since ± √ ≠ , , > 0
Solution 3 by Sanong Hauyrerai-Nakon Pathom-Thailand
, > 0, is integer ≥
+ ≤ ( + )
+ ≤ ( + )
+ ≤ ( + )
…
…
+ ≤ ( )( + )
hence + ( )( ) + −
≤ ( ) + + + = ( − )( ) = ( ) =
Solution 4 by SK Rejuan-West Bengal-India
If , > 0, ∈ ℕ, ≥ , then, by mth power theorem we get,
+ √+ + − √
+<
+ √+ + − √
+
www.ssmrmh.ro
⇒ +√+
+ −√ +
< 2 =
⇒ +√+ + −
√+ < 2
Solution 5 by Nikola Djurici-Serbia
≤ weighted mean inequality. Put = + ( )( ) ,
= −( )+ ,
+= ; ⋅ = ⋅ =
Equality holds only if = , but that would mean that = or = , so
equlity isn’t reached, so it’s strict <.
Solution 6 by Ngo Minh Ngoc Bao-Vietnam
Let = √ ≤ √√
= . Considering function
( ) = ( + ) + ( − ) ,∀ ∈ ( , ]
⇒ ( ) = ( + ) − ( − ) < 0,∀ ∈ ( ; ] ⇒ ( ) < ( ) =
59. If , , ≥ , + + = then:
√ + √ + √ ≤
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Ravi
Prakash-New Delhi-India, Solution 3 by Togrul Ehmedov-Baku-Azerbaidian
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Siendo , , ≥ , de tal manera que + + = . Probar que
www.ssmrmh.ro
√ + √ + √ ≤ . Para ello probaremos que
+ + = ⇔ + + =
Partimos de + = ⇔ = − ⇔ ( ) = ( − )
⇔ = − ⇔−
− = −−
− +
⇔ ( − )( − + ) =
= ( − )(− + )
⇔ − + − + − =
= − + −
⇔ − + − = , cuyas raíces son = ,
donde → = , , , , , ,
⇔ ( − + − ) =
⇒ − + − = , cuyas raíces son = ,
donde → = , , , , ,
Ahora bien llevando a “ ” , donde ≠
⇔ ( − + − ) =
⇔ − + − =
Haciendo =
⇔ − + − = , cuyas raíces son
= , ,
Aplicando Vietta ⇒ + + = − =
Aplicando Cauchy en la desigualdad propuesta
www.ssmrmh.ro
√ + √ + √ ≤
≤ ( + + ) + + ≤
≤ ( + + ) = √ =
Solution 2 by Ravi Prakash-New Delhi-India
Let = , = ⇒ = ( − )
⇒ − =
, , are roots of ( − ) = ( − )( − )
⇒ − + − =
+ + =∑
= =
Now, by C-S inequality √ + √ + √ ≤
≤ √ + + + +
≤ + + √ = √ √ =
Solution 3 by Togrul Ehmedov-Baku-Azerbaidian
√ + √ + √ ≤ ( + + ) + +
≤ ( + + ) + + ≤ + +
= + + = + +
www.ssmrmh.ro
=+ +
=− + − + −
=∑ − + +
=− + +
=
−
=+
=√
=
√ + √ + √ ≤
√ + √ + √ ≤
60. If , ∈ ℝ, ≠ then:
( + + ) − ( + + )−
> ⋅ √
Proposed by Daniel Sitaru – Romania
Solution 1 by Chris Kyriazis-Greece,
Solution 2 by Khanh Hung Vu-Ho Chi Minh-Vietnam
Solution 1 by Chris Kyriazis- Greece
The function ( ) = + +
www.ssmrmh.ro
∈ ℝ is strictly convex and positive for every ∈ ℝ (easy)
So, using the Hermite – Hadamard inequality we have: (if > )
∫ ( )
−>
+⇔
+ + − ( + + )−
> + + =
= + ⋅ +
> ⋅ ⋅
= ⋅ = ⋅ √
Solution 2 by Khanh Hung Vu-Ho Chi Minh-Vietnam
WLOG > . We have: ( ) ( ) = ( ) ( ) ( ) ( ) =
=( − )( + + + + + )
−=
=( − ) ⋅ ( + + + + + )
( − ) ⋅
⇒ ( ) ( ) = ( )⋅ ( )
( )⋅ (1)
We have
+ + + + + ≥ + √ ⋅ + √ ⋅ + =
= + ⋅ + ⋅ + = + ⋅ + ⋅
www.ssmrmh.ro
On the other hand by AM-GM we have
+ ⋅ + ⋅ = + + + + + ≥
≥ ⋅ ⋅ ⋅ ⋅ ⋅ = √ = ⋅
⇒ + + + + + ≥ ⋅ ⇒ ( + + + + + ) ≥
≥ ⋅ ⋅ = ⋅ ⋅( )
= ⋅( )
= ⋅ √ (2)
(1), (2) ⇒ ( ) ( ) ≥ ( )⋅ ⋅ √
( )⋅ (3)
We need to prove that ( )⋅ ⋅ √
( )⋅> √ (4)
We have (4) ⇒ ( )⋅ √
( )⋅> 6 ⋅ √ ⇒
( )⋅> (5)
Put = ( > 0). We have (5) ⇒⋅
> ⇒ − > ⋅ ⋅ ⇒
⇒ − ⋅ ⋅ − > 0 (6)
( ) = − ⋅ ⋅ − ⇒ ( ) = ⋅ − ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅
⇒ ( ) = ⋅ − ⋅ − ⋅ ⋅ = ⋅ ( − ⋅ − )
( ) = − ⋅ − ⇒ ( ) = ⋅ − > 0 ⇒ ( ) is a increasing function
⇒ ( ) > ( ) = ⇒ ( ) > 0 ⇒ ( ) is a increasing function ⇒
⇒ ( ) > ( ) = ⇒ (6) true ⇒(5) true ⇒ (4) true
(3), (4)⇒ ( ) ( ) > ⋅ √
Q.E.D.
61. If , ∈ , then:
( + ) < +
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
www.ssmrmh.ro
Let’s prove that ∀ ∈ , , >
( ) > ⇔ >
Let ( ) = − ( ) =
( ) = + ( )−
= + − = ( ) ( ) =
( ) = + − = ( ) ; ( ) =
( ) = ( ) + ( )( )( ) + ( − ) −
= ( + ) + ( + ) + ( − ) −
= ( + ) + ( + ) ++
− =( + ) + ( + ) + − ( + )
+
=+ + ( ) + + ( ) + + − ( )
+=
++
> 0,
( = > 0)
∴ ( ) > 0 and ( ) = ⇒ ( ) > ℎ( ) = ⇒ ( ) > 0 and ( ) =
⇒ ( ) > ( ) ⇒ ( ) > 0 and ( ) = ⇒ ( ) > ( ) =
⇒ > ⇒ > ∀ ∈ ,
∴ > and >
⇒ + > + = ( + )
62. If ∈ ℕ, ≥ then:
+ + + + > 2 +
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam
If ∈ , ≥ then
+ + + + > 2 +
We have
> = ⇒ + + + > 1
⇒ + + + + > + 1 ≥ 3
⇒ + + + + > 3 (1)
On the other hand, we have ≥ (Since ≥ )
⇒ ≤ ⇒ + ≤ (2)
(1) and (2) ⇒ + + + + > 2 +
63. If , , … , > 0, ∈ ℕ∗, :ℝ → ℝ, ( ) = + +⋯+ then:
+ + ⋯+ + ≥( + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Nirapada Pal-Jhargram-India,Solution 2 by Soumitra Mandal-
Chandar Nagore-India, Solution 3 by Ngoc Minh Ngoc Bao-Gia Lang-Vietnam
Solution 1 by Nirapada Pal-Jhargram-India
= ⋅∑
≥
⎝
⎛∑
⎠
⎞
www.ssmrmh.ro
=
⎝
⎜⎛∑ ∑
⎠
⎟⎞
As ( ) = ∑
=
⎝
⎜⎜⎛
⎝
⎜⎛∑
⎠
⎟⎞
⎠
⎟⎟⎞
≥⏞
= =( + )
Solution 2 by Soumitra Mandal-Chandar Nagore-India
≥
= + + + ⋯+
≥⏞ ( + + + ⋯+ ) = ( ) (proved)
Solution 3 by Ngoc Minh Ngoc Bao-Gia Lang-Vietnam
Use Cauchy – Schwarz inequality:
+ + ⋯+ + ≥ + + ⋯+ + (*)
We have:
+ + ⋯+ + = +⋯+ +
www.ssmrmh.ro
= + + ⋯+ + + + + ⋯+ + + ⋯
… + + + ⋯+ +
≥ ⋅ ⋅… ⋅ ⋅ + ⋅ ⋅… ⋅ ⋅ +⋯
+ ⋅ ⋅… ⋅ ⋅ =( + )
⇒ (*) ≥ RHS (*) ≥ ⋅ ( ) = ( )
Equality when = = ⋯ =
64. If = + , = + , = + ,
, , = , then: √ + √ + √ ≤ √
Proposed by Boris Colakovic-Belgrade-Serbia
Solution 1 by Nirapada Pal-Jhargram-India, Solution 2 by Abdul Aziz-
Semarang-Indonesia, Solution 3 by Myagmarsuren-Yadamsuren-Darkhan-
Mongolia
Solution 1 by Nirapada Pal-Jhargram-India
+ + = gives ∑ =
Now, ∑√ = ∑√ ≤ ∑ = √ ∑ + = √
Solution 2 by Abdul Aziz-Semarang-Indonesia
Since + + = then + + =
⇔ + + − = ⇔ + + =
By CS, √ + √ + √ ≤ ( + + )( + + )
www.ssmrmh.ro
⇔ √ + √ + √ ≤ √ ⋅ = √
Solution 3 by Myagmarsuren-Yadamsuren-Darkhan-Mongolia
If
= ⋅ += ⋅ += ⋅ +
; ; > 0+ + = ⎭
⎪⎬
⎪⎫
then √ + √ + √ ≤ √
( ) = √ ⇒ ( ) ≤
⇒ √ + √ + √ ≤ ⋅ ∑ ⋅ =
= ⋅+ ∑ ⋅
=⋅ √
= √
65. If in , ≥ ≥ then:
+ + <
Proposed by Daniel Sitaru – Romania
Solution by Rozeta Atanasova-Skopje
≥ ≥ ⇒ ≤ ≤ ⇒by Chebyshev’s sum inequality
≤ ( + + ) + + (1)
But + + < + + (2)
because ( ) = + − ( + )
= + + < ( + ) ⇒ < + , and similarly
+ < +
< +
www.ssmrmh.ro
− − − − −− −
( + + ) < 2( + + )
On the other hand
+ + = + + + (3)
Let’s consider the solutions of − = ⇒
= ⇒ ( ) = ⇒
= + + + + + +
= + + + + + = − (4)
From (1), (2), (3) and (4) ⇒
< ( + + ) − = ⋅ = =
66. If , , ∈ ( ,∞), = then:
+ ( + )+ >
Proposed by Daniel Sitaru – Romania
Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam
If , , ∈ ( , +∞) and = . Prove that
⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ >
Lemma: > if ∈ ;
Since ∈ , and ∈ , , applying the lemma, we have:
> ⋅ = and > ⋅ =
www.ssmrmh.ro
We have ⋅ + ( + ) ⋅ > ⋅ + ( + ) ⋅ =
= + ( + ) = + > 2√ ⋅
Similarly, we have ⋅ + ( + ) ⋅ > 2√ ⋅ and
⋅ + ( + ) ⋅ + > 2√ ⋅
So
⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ >
> 2√ ⋅ ⋅ √ ⋅ ⋅ √ ⋅ = √ ⋅ ⋅
⇒ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ >
> 8 ⋅ ⋅ =
67. If ∈ ℕ∗, ≥ , , , > 1, + + = then:
+ √ + − √ < 18
Proposed by Daniel Sitaru – Romania
Solution 1 by Khanh Hung Vu-Ho Chi Minh-Vietnam, Solution 2 by SK Rejuan-
West Bengal-India, Solution 3 by Eliezer Okeke-Anambra-Nigeria
Solution 1 by Khanh Hung Vu-Ho Chi Minh-Vietnam
If ∈ ℕ∗, ≥ , , , > 1, + + = then
+ √ + − √ + √ + − √ + √ + − √ < 18
By AM-GM, we have:
+ √ = √ √ +√
≤( )⋅ √ √
√ =
www.ssmrmh.ro
= √ +√
Similarly, we have − √ ≤ √ −√
⇒ + √ + − √ ≤ ⋅ √ (1)
On the other hand, by AM-GM we have ⋅ ( ) ≤ ( )⋅
⇒ √ ≤ ( )⋅⋅ ( )
= ( )⋅⋅
(2)
(1), (2) ⇒ + √ + − √ ≤ ( )⋅⋅
Similarly, we have + √ + − √ ≤ ( )⋅⋅
and
+ √ + − √ ≤+ ( − ) ⋅
⋅
⇒ + √ + − √ + + √ + − √ + + √ + − √ ≤
≤+ ( − ) ⋅
⋅ ++ ( − ) ⋅
⋅ ++ ( − ) ⋅
⋅
⇒ + √ + − √ + + + − + + √ + − √
≤
≤( + + ) + ( − ) ⋅
⋅ =⋅ + ( − ) ⋅
⋅ =( + ) ⋅
⋅ =( + )
Since < 2 ⇒ ( ) < 18
So + √ + − √ + + √ + − √ + + √ + − √ ≤
The equality doesn’t exist. Therefore,
+ √ + − √ + + √ + − √ + + √ + − √ < 18
www.ssmrmh.ro
Solution 2 by SK Rejuan-West Bengal-India
If ∈ ℕ∗, ≥ , , , > 1, + + =
By mth power theorem we get = as, √ < ⇒ − √ > 0
+ √ + − √<
+ √ + − √
⇒
⎩⎪⎨
⎪⎧
+ √∥
( )
+ − √
⎭⎪⎬
⎪⎫
< 2 =
⇒ < 2∑ (1)
Again by mth power theorem, ∑ < ∑ = =
[∵ ∑ = ] ⇒ ∑ < 3 ⇒ ∑ < 9 ⇒ 2∑ < 18 (2)
Combining (1) & (2) we get,
< 2 < 18 ⇒ + √ + − √ < 18
Solution 3 by Eliezer Okeke-Anambra-Nigeria
for , , > 1 + + = ≥ ∈ ℕ
prove ∑ + √ + − √ < 18. Consider the composite
functions ( ) = + √ and ( ) = − √ both functions are
concave. Hence ( ) = ( ) + ( ) is concave.
Lemma: √ < √ for > 1 (1)
www.ssmrmh.ro
Lemma: AM ≤ AM (2)
We apply Jensen
+ √ + − √ ≤⏞
⎣⎢⎢⎢⎡ ∑
+∑
+∑
−∑
⎦⎥⎥⎥⎤
=
⎣⎢⎢⎢⎡
+ + −
⎦⎥⎥⎥⎤
= ( + ) + ( − ) <⏞( )
( + ) + ( − )
<⏞( ) ( + ) + ( − )
= ( ) =
68. If , , , > 0, + + + = 1 then:
+ + + + ( + + + + + ) ≥
≥ + √ + √
Proposed by Daniel Sitaru – Romania
Solution by Kevin Soto Palacios – Huarmey – Peru
Siendo , , , > 0, de tal manera que + + + = . Probar que
+ + + + ( + + + + + ) ≥
≥ + √ + √
Aplicando la siguiente identidad conocida
( + ) = + + ( + ), donde = + , = + ⇒ ( + + + ) = ( + ) + ( + ) + ( + + + )( + )( + )
⇒ = + + ( + ) + + + ( + ) + ( + )( + )
⇒ + ( + ) + ( + ) = + + + + ( + + + ) +
www.ssmrmh.ro
+ ( + + + ) + ( + )( + )
⇒ + ( + ) + ( + ) = + + + + + + ( + + + )
⇒ + ( + ) + ( + ) = + + + + ( + + + + + )
Como , , , > 0. Aplicando ≥
⇒ + + + + ( + + + + + ) =
= + ( + ) + ( + ) ≥ + √ + √
69. If ∈ ℕ∗, > 0 then:
!
(− ) −+ < +
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
Consider the expression = ∏
We split this expression into partial fractions.
= (− + )( + ) … (− )( ) … ( + − ) ⋅ +
=(− )
!⋅
!( − )! ( + − )!
⋅+
=!∑ (− ) − . Also, + + > 1
∴!
(− ) − += <
+≤
+
⇒!
(− ) − +<
+
www.ssmrmh.ro
70. If , ≥ , ∈ ℕ∗ then:
⋅ ⋅ ≥ ( − )
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
∑ ⋅ ⋅ ≥ ( − ) (1)
If = or = , there is nothing to prove. ∴ suppose , > 0
Now, (1) can be written as
≥ − ⇔ ≥ −
If is odd,
≥ + −
= + ≥ = = −
If is even, ∑
≥ + − +
= + +
www.ssmrmh.ro
≥ + = = −
71. If , , > 0, + + + = 0 then:
| + + + | ≥ +
Proposed by Daniel Sitaru – Romania
Solution 1 by Kevin Soto Palacios-Huarmey-Peru, Solution 2 by Ravi Prakash-
New Delhi-India
Solution 1 by Kevin Soto Palacios-Huarmey-Peru
Siendo , , > 0, de tal manera que + + + = . Probar que
| + + + | ≥ +
De la condición ⇒ ( + ) = −( + ) ⇔ ( + ) + ( + ) =
⇒ + + + + ( + ) + ( + ) =
⇒ + + + − ( + ) − ( + ) =
⇒ + + + = ( + ) + ( + )
Por lo tanto, Como , , > 0; Aplicando MA ≥ MG
| + + + | = + + + ≥ +
Solution 2 by Ravi Prakash-New Delhi-India
Let , , > 0, + + + = 0
− = ( + + ) ≥ ( ) ⇒ − ≥
⇒ − − ≥ ⇒ + ≤ − < 0
Also, + + + = (− − − ) + + +
= −( + ) + ( + ) < 0
www.ssmrmh.ro
∴ the given inequality becomes
− ( + + + ) ≥ − − . Now,
− = (− ) ≥ + + (− ) ⇒ − ≥ + −
Similarly − ≥ + − . And − ≥ + − . Thus,
− − − − ≥ ( + + − )− − (1)
But ≥ ⇒ + + ≥ −
⇒ ( + + − ) ≥ − − > 0 (2)
From (1), (2) we get − − − − ≥ − −
or | + + + | ≥ +
Equality when = =
72. If , , ≥ − then:
− ≥ ( − )
Proposed by Daniel Sitaru – Romania
Solution by Richdad Phuc-Hanoi-Vietnam
WLOG, assume ≤ ≤ or ≥ ≥ . We have
− = ( − ) ( + ) − ( + ) + ( − ) − − +
Let ( ) = ( + ) , ≥ − , ( ) = ( + ) > 0,∀ > −2
is increasing function on [− , +∞)
⇒ ( − ) ( + ) − ( + ) ≥ ,∀ , ≥ −
Let ( ) = , ≥ − , ( ) = ( + ) > 0, for all ≥ −
is increasing function on [− ; +∞) case ≤ ≤
www.ssmrmh.ro
≥≥
⇒ ( − ) − + − ≥
we get − ≥ case ≥ ≥ ≤≤
− ≥ ⇒ . . . Equality hold if = =
73. If < < then:
>
Proposed by Daniel Sitaru – Romania
Solution by Abdelhak Maoukouf-Casablanca-Morocco
∀ > 1: + ≤ ⇔ ( − ) − ≤ ⇔( )
( ) ≤ ⇔ ≤
< < ⇒ ∫ > ∫ ⇔ > ⇔ >
74. Prove that + ≥√√ ,∀ ∈ ℝ.
Proposed by Ibrahim Abdulazeez-Zaria-Nigeria
Solution by Daniel Sitaru-Romania
+ ≥⏞ = =
= √ ≥⋅√ ≥ √ =
√√
75. Prove that:
° + ° + ° + ⋯+ ° < 54 ⋅ °
Proposed by Ilkin Guliyev-Azerbaidian
www.ssmrmh.ro
Solution by Ravi Prakash-New Delhi-India
Note: ( ) ≤ ∀ ∈ ℕ, < < [ = , ( ) ≤ ]
Assume ( ) ≤ for some ∈ ℕ.
( + ) = ( + ) = ( ) + ( )
≤ ( ) + ≤ + = ( + ) ∴ ( °) ≤ °
⇒ ( °) ≤ ° = ( °)
76. If , , ∈ , then:
( + + ) ≤ + +
Proposed by Daniel Sitaru-Romania
Solution 1 by Anas Adlany-El Zemamra-Morocco, Solution 2 by Abdelhak
Maoukouf-Casablanca-Morocco
Solution 1 by Anas Adlany-El Zemamra-Morocco
∑ ( ) ≥⏞ (∑ )(∑ ( )) ≥⏞ ∏ (∑ ( )) =
= ( ) .
Solution 2 by Abdelhak Maoukouf-Casablanca-Morocco
→ is a descending function on ; , So by Chebyshev:
≥ ⇔ ( + ) ≥
⇔ + ≥ +
www.ssmrmh.ro
if ∑ ≤ ∑ similarly we’ll have ∑ ≤ ∑
⇒ ∑ + ∑ ≤ ∑ + ∑ ∵False supposition
So ∑ ≥ ∑ ⇔ ∑ ≥ ∑
77. If ≤ , , < 1 then:
( + )( + )( + )( − )( − )( − ) ≥
+−
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
For ≤ , , < 1, consider
= ( − ) ( + )( + )( + ) −
−( + ) ( − )( − )( − )
= ( − ) ( + )( − )( − )( − ) ( + )( + )( + )
Use → − to obtain
= ( − ) +( − )( − )( − ) ( + + ) +
Use → +
= + ++ + + + + +
Use → −
= + ++ + − + + −
Note that + − ( + ) = ( − ) > 0
⇒ + > 3 + (1)
www.ssmrmh.ro
Also, + + ≥ + + and
( − ) + ( − ) + ( − ) ≥ ( − ) + ( − ) + ( − )
[∵ ≤ , , < 1] ⇒ ( + + )[( − ) + ( − ) + ( − ) ]
≥ ( + + )[ ( − ) + ( − ) + ( − ) ]
⇒ + + − ≥ + + − (2)
From (1), (2) we get ( + )( + + − ) ≥
≥ [ + + − ] ⇒ ≥
⇒ ( − ) ( + )( + )( + )
≥ ( + ) ( − )( − )( − )
Put = , = , = to obtain ( )( )( ) ≥ ( )
78. If , , > 0 then:
√+ √
++ + √
( + + ) ≥
Proposed by Daniel Sitaru – Romania
Solution by Nguyen Thanh Nho-Tra Vinh-Vietnam
∑ √√ √
= ∑√ √
≥ √ √ √
√ √ √=
√ √ √+
⇒ =√ √ √
+ √ √ √( ) + ⇒ ≥⏞ + =
79. If , , , > 0 then:
≥ √ ≥
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution by Ravi Prakash-New Delhi-India
Consider ∑ − = ∑[( − ) + ( − ) + ( −
−ܖ܉ܜ
= ( − )( − )
As is increasing on ( ,∞), ≥ ⇒ ≥
∴ ( − )( − ) ≥ ,∀ > 0, > 0
Thus, ∑ ≥ ∑ ≥ ∑( ) . Next,
( ) ≥ ( )
= [ ]
80. If , , ∈ ℝ, > 0, | | ≤ , | | ≤ , | | ≤ then:
( − ) + ( − ) + ( − ) + ≤
Proposed by Daniel Sitaru – Romania
Solution by Le Minh Cuong-Ho Chi Minh-Vietnam
Apply AM-GM we get:
( − ) = ( − )( + ) ≤ ( ) = (1)
( − ) ≤ (2); ( − ) ≤ (3)
and: ≤ + + (4)
From (1), (2), (3), (4), we get:
( − ) + ( − ) + ( − ) + ≤
www.ssmrmh.ro
“=” ⇔ = = =
81. If , , , , , > 0, + + ≥ 3 then:
( + + )( + + )( + + ) ≥+ +
Proposed by Daniel Sitaru – Romania
Solution by Pham Quoc Sang-Ho Chi Minh-Vietnam
We have ( + + )( + + )( + + ) ≥⏞
≥ + + = ( + + ) ≥
On the other hand, we have ≥
so ( + + )( + + )( + + ) ≥
“=” = = = and = =
82. If , , , , , ≥ , + + = then
+ + + + + ≥ + + +
Proposed by Daniel Sitaru – Romania
Solution by SK Rejuan-West Bengal-India
Let us consider , , with the associated weights , , respectively by
AM≥GM we get, ≥
⇒ + + ≥ [∵ + + = ]
www.ssmrmh.ro
⇒ ≥ ( ) (1)
Similarly we get, ≥ ( ) (2)
and ≥ ( ) (3)
Adding (1), (2) & (3) we get,
+ + + + + ≥
≥ ( ) + ( ) + ( ) (4)
Applying AM≥HM on right side of (4) we get
++
++
+≥
+ ( + + )( + + )
= (5)
Combining (4) & (5) we get,
+ + + + + ≥ + + +
83. If , , , > 0 then:
( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ≥
≥ + √ + √ + √ + √
Proposed by Mihály Bencze-Romania
Solution 1 by Ravi Prakash-New Delhi-India
( + )( + )( + ) − + ( )
= + ( + + ) + ( + + ) +
www.ssmrmh.ro
− + ( ) + ( ) +
= + + − ( ) + + + − ( ) ≥
[∵ ≥ ] ⇒ ( + )( + )( + ) ≥ + ( ) (1)
Similarly, ( + )( + )( + ) ≥ + ( ) (2)
( + )( + )( + ) ≥ ( ) (3)
and ( + )( + )( + ) ≥ + ( ) (4)
Multiplying (1), (2), (3), (4) we get the required inequality.
Solution 2 by Nguyen Thanh Nho-Tra Vinh-Vietnam
( + )( + )( + ) = + + +
≥ + ⋅ ⋅ = + √
⇒ ( + )( + )( + ) ≥ + √ (1)
Similarly, ( + )( + )( + ) ≥ + √ (2)
( + )( + )( + ) ≥ + √ (3)
( + )( + )( + ) ≥ + √ (4)
(1).(2).(3).(4) ⇒ ≥
84. In convexe quadrilater:
www.ssmrmh.ro
+ + + ≥
Proposed by Daniel Sitaru – Romania
Solution by Geanina Tudose – Romania
For convenience denote
= ; = ; = ; = ; = ; = ; = ; =
In by Sine Theorem = ⇒ = ⋅
In , = ⋅ . Thus, = ⋅⋅
= ⋅⋅
Similarly, = ⋅⋅
; = ⋅⋅
; = ⋅ . The inequality becomes
⋅ + ⋅ + ⋅ + ⋅ ≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
85. In cyclic quadrilateral, , , , – sides, - circumradius:
√( + + + ) ≤
√
Proposed by Adil Abdullayev-Baku-Azerbaidian
www.ssmrmh.ro
Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by
Myagmarsuren Yadamsuren-Darkhan-Mongolia, Solution 3 by Soumitra
Mandal-Chandar Nagore-India
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Probar en un cuadrilátero cíclico con lados , , , y –
Circunradio
( ) ≤√
. Aplicando ≥ ⇔ ( ) ≤ ≤√
Es suficiente probar
⇔ √ √ ≥ ⇔ ≥ ≥ ≥
www.ssmrmh.ro
Nuevamente por ≥ ≤ ( ) = ⋅ ≤
Usando Ptolemy theorem
⋅ + ⋅ = ⋅ ⇔ + = ⋅
( ) + ( ) + ( ) + ( ) =
=( )
=( ) + ( )
,
=( )
=( ) + ( )
⋅ =( ) + ( ) ( ) + ( )
≤
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
ó ( ) (1)
⇒ = ; ⇒ =
+= + ≤ ⋅
+
+ ≤ ⋅
+ ≤ ⋅
+ ≤ ⋅
+ ≤ ⋅
⎭⎪⎪⎬
⎪⎪⎫
( + + + ) ≤ + + + =
≤ + + + ≤
≤+
++
= ⋅ ⋅ = √
≤ √ ⇒√≤ (*)
www.ssmrmh.ro
(1)⇒√≥
√⇒
√≥(∗)
≥√
⇒
⇒ ≥ √ ⇔ ≥ √ ⇒+ + +
≥ √
Solution 3 by Soumitra Mandal-Chandar Nagore-India
A cyclic quadrilateral with sides , , , and semi-perimeter and
circum – radius is given by
=( + )( + )( + )( − )( − )( − )( − )
∴ ≥⏞ √ ⋅ √ ⋅ √
− + − + − + −
= ⋅√
( + + + ) ⇒√
( + + + ) ≤√
86. In tetrahedron, , , , − altitudes, − circumradius,
− inradius: ( + + + ) ≥
Proposed by Daniel Sitaru – Romania
Solution by Kevin Soto Palacios – Huarmey – Peru
A Simple Proof of Euler’s Inequality in Space
Zhang Yun – Jinchang City - Gasu Province – China
Let be the radius of the circumscribed sphere of a tetrahedron and let
be the radius of the inscribed sphere of the tetrahedron. Then Euler’s
famous inequality in space state that
www.ssmrmh.ro
≥ (1)
We give here a simple proof of this inequality.
Let be the circumcenter of the tetrahedron .
Let ( = , , , ) denote the area of the face opposite the vertex ,
let denote the distance from to its opposite face, and let denote
the distance from the point to the face opposite . Then
+ ≥ , and so + ≥ . Thus, + ≥ .
Adding the four inequalities, we obtain that
( + + + ) + + + + ≥
≥ + + + .
Let denote the volume of the tetrahedron .
Then = = ( + + + ), so
( + + + ) + ≥ × , from which it follows that
( + + + ) ≥ . Since = ( + + + ),
this gives ( + + + ) ≥ × ( + + + ).
Thus, ≥ , so the inequality (1) is proved.
In two dimensions rather that three, if is now the radius of the
circumscribed circle of a triangle and the radius of the inscribed circle,
then, by a similar argument, ≥ .
Solution by Kevin Soto Palacios – Huarmey – Peru
Si es un tetraedro, donde , , , , son las alturas, y r el
circunradio e inradio.
Probar que ( + + + ) ≥ .
Es un tetraedro se cumple la siugiente identidad y desigualdad
www.ssmrmh.ro
+ + + = , ≥
Ahora bien, por la desigualdad de Cauchy
+ + + = + + + ≥( + + + )
+ + +=
Por lo tanto ( + + + ) ≥ ⋅ =
87. If in – tetrahedron, , , , - altitudes, − inradii then:
−+ +
−+ +
−+ +
−+ ≥
Proposed by D.M. Bătinețu – Giurgiu and Neculai Stanciu – Romania
Solution by Kevin Soto Palacios – Huarmey – Peru
Si es un tetraedro, donde , , , , son las alturas y el
inradio. Probar que + + + ≥
Recordar la siguiente identidad + + + =
La desigualdad propuesta es equivalente −+
+ +−+
+ +−+
+ +−+
+ ≥ +
+ + + ≥ . Aplicando la desigualdad de Cauchy
++
++
++
+≥
( + + + )
+ + + +=
88. If in - tetrahedron = = , = = , = = ,
www.ssmrmh.ro
- radii of circumsphere then:
( − )( − )( − ) ≤
Proposed by Daniel Sitaru – Romania
Solution by Rozeta Atanasova-Skopje
Equifacial tetrahedrons exist only when the faces are congruent acute
triangles, and then = ⇒ > 0, > 0, > 0 ⇒
= ( − )( − )( − )
= ⋅+ −
⋅+ −
⋅+ −
= ( + − )( + − )( + − ) =
= ≤ + +
≤ + +
= = =
89. Let , , > 0 such that: + + = . Find the maximum of
expression: = + +
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
Solution by Hoang Le Nhat Tung – Hanoi – Vietnam
We have: − − + + = ( − ) + ( − ) + ( − ) − ( − ) − ( − ) − ( − )
= ( − ) + + − − −
= ( − ) ( + + + + ) ≥ ( > 0 and ( − ) ≥ )
⇒ − − + + ≥ ⇔ − + + ≥ ⇔
www.ssmrmh.ro
⇔ − + + + ≥ +
⇔ ≤ ⇔ ≤ (1)
By inequality AM – GM for 4 positive real numbers:
+ = + + + ≥ ⋅ ⋅ ⋅ = ⋅ =
= √ ⇔+
≤√
=√
Therefore (1) and by AM-GM:
⇒− + + +
≤√
≤ ⋅ + + + +
Similar: ≤ + ; ≤ +
Therefore: ⇒ = + + ≤
≤ + + + + +
⇔ ≤ + + + + + (2)
I have + + = and inequality:
( + + ) ≥ ( + + ) with: = , = , = :
= + + ≥ ⋅ + ⋅ + ⋅ = + + ⇔
⇔ + + (3)
Other let (3) and inequality AM-GM. I have:
≥ + + = + + + + + − ≥ + + − ⇔
⇔ + + ≤ (4)
Let (2), (3), (4): ⇒ ≤ ⋅ + ⋅ = = ⇒ ≤ ⇒ =
www.ssmrmh.ro
Equality occurs if:
⎩⎪⎨
⎪⎧ , , > 0; + + =
− = − = − == ; = ; =
= = = ; = =
⇔ = = = .
Maximum of be: then = = = .
90. , , … , > 0, ∈ ℕ∗,∑ =
Find:
= ( )
Proposed by Madan Beniwal-Varanasi-India
Solution by Ravi Prakash-New Delhi-India
Let = ∏ ( ) ⇒ = ∑ ( )
Let = ∑ ( ) + ∑ − ⇒ = + ( = , , … , )
Set = ⇒ − = ( = , , … , ). Thus,
= = ⋯ = = − ⇒ = − , = − , … , = − . Now,
= = (− ) ( ) ⇒ − = ( + ) = +
Thus, = − =
=+
⋯
=+
( )
www.ssmrmh.ro
91. If , , are maximum positives values such that:
( ) = − −
find: + + +
Proposed by Shivam Sharma-New Delhi-India
Solution 1 by Khalef Ruhemi-Iordania, Solution 2 by Mohammed Hijazi-
Iordania
Solution 1 by Khalef Ruhemi-Iordania
Find ∑ ( ), use ∫ ⋅ = ( ) + ; : Euler’s constant
∴ ≔ ( ) = − +−−
= − +−− ⋅ = − + − −
use ∑ = ,∑ =
Then = − + ∫ − ( )( )
∴ = − + ∫ + = − + ∫( )
( )( )
∴ = − +( − + − )
( − )
→
−−
( − )
Since → = → = =
Then, = − + − + ⋅ ∫
www.ssmrmh.ro
= − + − +( )( ) + = − + − + + ( )
∴ = − + ( ), But ( ) = − +
= − + (− + ) = − − +
∴ ∑ ( ) = − − ( − ). But = −
∴ ( ) = − − − + = − − − +
= − − ∴ ∑ ( ) = − − . Take =
⇛ ( ) = ( ) − ( ) − ( ) = − −
∴ = = = = ∴ + + + = ( )( ) =
Solution 2 by Mohammed Hijazi-Iordania
( ) = ( − ) = − +
using summation by parts:
⋅ = −+
= ( + ) − ( + )
so the red sum = ( ) − ( )
so ∑ ( ) = + −
so by comparing the result in the problem we get:
= = and = = − so + − − = ( ) =
92. Let , , be positive real numbers. Find the minimum possible value of
www.ssmrmh.ro
++
++ +
+
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution 1 by Kevin Soto Palacios – Huarmey – Peru , Solution 2 Seyran
Ibrahimov-Maasilli-Azerbaidian , Solution 3 Imad Zak-Saida-Lebanon ,
Solution 4 Myagmarsuren Yadamsuren-Darkhan-Mongolia ,
Solution 5 Soumava Chakraborty-Kolkata-India
Solution 1 by Kevin Soto Palacios – Huarmey – Peru
Siendo , , números . Hallar el mínimo valor + + +
( )( ) + ( ) ( )( ) ≥
( ) ( )
( )( ) + ( ) ( )( )
( )( ) + ( ) ( )( ) ≥
( ) ( ) ( )( )( ) + ( ) ( )
( )
Luego, aplicando MA ≥ MG
= ( ) ( ) ( )( )( ) + ( ) ( )
( ) + ( ) ( )( ) ≥
≥( + ) ( + ) + ( + )
( + )( + ) ⋅( + ) + ( + )
( + )
≥( ) ( )
( )( ) ≥ . Por transitividad + + + ≥
La igualdad se alcanza cuando = = .
Solution 2 Seyran Ibrahimov-Maasilli-Azerbaidian
+ + ≥ Nesbit = = =
www.ssmrmh.ro
+ + + ≥ − + + ; ( ) = − + +
( ) = − + ; ( ) = ⇒ − = ⇒ + = ⇒ =
( ) > 0 ⇒ = − + + = ; =
Solution 3 Imad Zak-Saida-Lebanon
= + + + ≥? ? homogeneous ⇒ + + =
= + < 3 = . By AM-GM ≤
Replace by − ; + by − + by − to get!
≥ − +−
− ++
−= − + − +
−
Note that ≥ ( − ) ( ) ⇔ ( − )( − ) ≥ True
Use AM – GM again to get
≥ − + − + − + − +−
+−
≥ − + ( − ) ( ) ≥ ≥ − + =
∴ ≥ << = >> when = = ≤
⇒ = = ⇒ = . In general = = .
Solution 4 Myagmarsuren Yadamsuren-Darkhan-Mongolia
www.ssmrmh.ro
++
++ +
+≥
1) ⋅ ⋅ ( )( ) + = ( ) ( )
( )
2) + ≥ ( )( ) ( ) (ASSURE)
≥ ≥ ; ≥
+ + + ≥ ⋅ ( + ) ⋅ + + +
≥ ( + ) ⋅ ( + ) + ( + ) =( + )
( + ) + ( + )
3) → LHS ⇒ ≥ ⋅( )( ) ( ) + ⋅ ( ) ( )
( ) =
=( + )
( + ) + ( + ) +( + ) + ( + )
( + ) +( + ) + ( + )
( + ) ≥
Solution 5 Soumava Chakraborty-Kolkata-India
If , , > 0, then, + + + =?
+ + + =( + ) + ( + )
( + )( + ) =( + ) + +
( + )( + )
≥( + ) + ( + )
( + )( + ) =( + )( + + + )
( + )( + )
≥( + ) ( + )( + )
( + )( + ) =+
( + )( + )
www.ssmrmh.ro
≥( )
( ) ∴ ≥ ( ) + ( ) (using (1))
= ( ) + ( ) + ( ) ≥ ( )( )( )( ) =
∴ required min value =
93. , , ∈ ( ,∞)
( , , ) =+ ⋅ +
+
Find ( , , )
Proposed by Daniel Sitaru – Romania
Solution 1 by Subhajit Chattopadhyay-Visva Bharati-India , Solution 2 by
Kevin Soto Palacios – Huarmey – Peru , Solution 3 by Geanina Tudose –
Romania
Solution 1 by Subhajit Chattopadhyay-Visva Bharati-India
( , , ) =+ ⋅ +
+
take = , = ; = clearly =
Hence the LHS turns out to be
∑ = ∑ ( ) = ( + + ) − + +
Now HM ≤ AM ⇒ − ≥ − ⇒ − ≥ − ⇒ − ≥ −
∴ = ≥ ( + + ) − ( + + ) = ( + + ) ≥ =
Hence, ( ) = . Equality occurs at = = = or, = = =
Solution 2 by Kevin Soto Palacios – Huarmey – Peru
www.ssmrmh.ro
Si , , ∈< 1,∞ > de tal manera que ( , , ) = ∑ ⋅
Hallar ( , , ). Realizamos los siguientes cambios de variables
= > 0, = > 0, = > 0 ⇔ =
La desigualdad propuesta es equivalente
+ ++
≥( + )( + ) = ( + ) = ( + + ) ≥
≥ = (Válidor por ≥ )
La igualdad se alcanza cuando = =
Solution 3 by Geanina Tudose – Romania
We can denote = ; = ; =
( , , ) = ( , , ) =+ +
+
Where , , > 0 subject to =
We have ( , , ) = ∑ ( ) = ∑ ( + ) −
From ≤ we have – ≥ −
⇒ ( , , ) ≥ ( + )−+
=( + )
=
= ( + + ) ≥ ⋅ =
( , , ) = ( , , ) ≥ (min. value attained for
= = = i.e. = = )
94. Let , , be positive real numbers such that: + + = . Find the
minimum of expression:
www.ssmrmh.ro
=√ +
++
+√ +
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
Solution by Hoang Le Nhat Tung – Hanoi – Vietnam
By inequality AM-GM. We have:
√ +=
( + )( − + )≥
⋅ + + − + = ( − + )
Similar: ≥ ; ≥ . Therefore:
⇒ = + + ≥ + + (1)
Other, by inequality CBS:
+ + = + + ≥
≥ (2)
Then (1), (2): ⇒ ≥ (3)
We will prove that: ≥ (4)
(4) ⇔ ( + + ) ≥ ( − + ) + ( − + ) + ( − + )
⇔ ( + + ) + ( + + ) ≥ ( + + ) −
−( + + ) + ( + + )
⇔ ( + + ) + ( + + ) ≥ ( + + ) −
− ( + + ) + ( + + )
⇔ ( + + ) + ( + + ) ≥
≥ ( + + )− ( + + )( + + ) +
+ ( + + ) ( + + )
(Because = + + and = ( + + ) )
⇔ ( + + ) + ( + + ) +
www.ssmrmh.ro
+ + + + + + + ( + + ) ≥
≥ ( + + ) + ( + + + + + )( + + )
⇔ ( + + ) + ( + + ) +
+ + + + + + + ( + + ) ≥
≥ ( + + ) + ( + + ) + ( + + ) +
+ ( + + )
⇔ ( + + ) + ( + + ) ≥ ( + + ) +
+ ( + + ) + ( + + ) (5)
By AM-GM I have:
+ + =+ + +
++ + +
++ + +
≥
≥ + +
⇒ + + ≥ + + ⇔ ( + + ) ≥ ( + + ) (6)
+ + =+ + +
++ + +
++ + +
≥
≥ + +
⇒ + + ≥ + + ⇔ ( + + ) ≥ ( + + ) (7)
+ + =( + )
+( + )
+( + )
≥
≥⋅
+⋅
+⋅
⇒ + + ≥ ( + + ) ⇔ + + ≥ ( + + ) (8)
Then (6), (7), (8):
⇒ ( + + ) + ( + + ) ≥ ( + + ) +
+ ( + + ) + ( + + ) ⇒ Inequality (5) True ⇒ (4) True
Then (3), (4): ⇒ ≥ ⇒ = . Equality occurs if:
www.ssmrmh.ro
⇔
⎩⎪⎨
⎪⎧
, , > 0; + + = 3+ = − + ; + = − + ; + = − +
( − + ) = ( − + ) = ( − + )= = > 0
Therefore Minimum of is: then = = = .
95. Let , , be non-negative real numbers such that + + = . Find the
maximum and minimum possible values of
( + )√ + + ( + ) + + ( + )√ + .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Solution by Kevin Soto Palacios – Huarmey – Peru
Siendo , , números reales no negativos de tal manera que
+ + = . Hallar el máximo y mínimo valor de
= ( + ) ( + ) + ( + ) ( + ) + ( + )√ +
Para hallar el máximo valor. Aplicamos la desigualdad de Cauchy
= ( + ) ( + ) + ( + ) ( + ) + ( + )√ + ≤
≤ ( + + + + ) ( + )( + ) + ( + )( + ) + ( + )( + )
= ( + + ) ( + + ) + ( + + ) =
= + ( + + ) ≤ +( + + )
=
= + = =√
La igualdad se alcanza cuando = = =
www.ssmrmh.ro
Para hallar el mínimo valor
Como , , ≥ ⇔ √ + ≥ , + ≥ ,√ + ≥
⇒ = ( + ) ( + ) + ( + ) ( + ) + ( + )√ + ≥
≥ ( + ) + ( + ) + ( + ) = ( + + ) =
La igualdad se alcanza cuando = , = = y sus permutaciones.
96. If , ∈ (ℝ), = then:
( + + )( + + )( + + ) ≥
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
Let =×
where ∈ ℂ, =×
( ) = ( ) ( ) = ( ) ( ) = | ( )| ≥
Now, + + = ( + )( + ) = ( + )( + )
where = √ ∴ ( + + ) = | ( + )| ≥
Similarly ( + + ) ≥ and ( + + ) ≥
∴ {( + + )( + + )( + + )} ≥
97. If , , ≥ , + + = then:
+ + + + + ++ + + + + ++ + + + + +
≤ ( − ) ( − ) ( − )
Proposed by Daniel Sitaru – Romania
Solution 1 by Nirapada Pal-Jhargram-India
Solution 2 by Ravi Prakash-New Delhi-India
www.ssmrmh.ro
Solution 1 by Nirapada Pal-Jhargram-India
= +
[All the other vanishes after splitting]
= ( − )( − )( − )∑ − ( − )( − )( − ) ∑
= ( − )( − )( − ) ( − )
= ( − ) ( − ) ( − )
Now, ≤ = As + + =
∴ ≤ ( − ) ( − ) ( − )
Solution 2 by Ravi Prakash-New Delhi-India
=+ + + + + ++ + + + + ++ + + + + +
=
= ⋅ = ( ) {( − ) ( − ) ( − ) }
But + + = , , , ≥ ⇒ ( ) ≤ ( + + ) = ⇒ ( ) ≤
∴ ≤ ( − ) ( − ) ( − )
98. If , , ∈ (ℝ), = = = = = = then:
www.ssmrmh.ro
( + + + + + + ) ≥
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
Let =×
where ∈ ℂ and =×
, then
( ) = ( ) ( ) = ( ) ( ) = | ( )| ≥
Assuming , , commute so that = = = = = =
Now, le = + + ⇒ = ( + + ) = + + t
Let = + + + + + + = + +
= ( − )( − ), = − + √ = ( − )( − )
Now, ( ) = | ( − )| ≥
99. If , , ≥ then:
+ ++ ++ +
≤ ( + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Serban George Florin-Romania
Solution 2 by Nirapada Pal-Jhargram-India
Solution 1 by Serban George Florin-Romania
−−
−=
www.ssmrmh.ro
= −−
−
= −−
=
= ⋅ −−
= ≤ ( + ) ( + ) ( + )
⇒ ≤ ( + )( + )( + ); + ≥ √ ; + ≥ √ ;
+ ≥ √
⇒ ( + )( + )( + ) ≥
Solution 2 by Nirapada Pal-Jhargram-India
+ ++ ++ +
=−
−−
− → , = , ,
= −−
−
= −−
−
= −−
−
− → , = , ,
= −−
= = ( )( )( )
≤ ( + ) ( + ) ( + ) , ( + ) − ( − ) = ⇒ ( + ) ≥
= ( + )
100. If , , ∈ [ , ] then:
www.ssmrmh.ro
+ + +
≤
Proposed by Daniel Sitaru – Romania
Solution by Ravi Prakash-New Delhi-India
Put = , = , = ; ≤ , , ≤
The given determinant becomes
=
= ( )( − ) + ( − )
+ ( − )
As ( − ) + ( − ) + ( − ) =
Either two of them non-negative and one is non-positive or one of them
is non-negative and two are non-positive.
Case 1 − ≥ , − ≥ , − ≤
then ≤ ( − ) + ( − )
≤ ( − + − ) ∵ ≤ , ≤
⇒ ≤ ( − ) ≤ ∵ ≤ , ≤
Similarly for other such cases.
Case 2 − ≥ , − ≤ , − ≤
∴ ≤ ( − ) ≤
Similarly for other such cases.
www.ssmrmh.ro
Its nice to be important but more important its to be nice.
At this paper works a TEAM.
This is RMM TEAM.
To be continued!
Daniel Sitaru