rmm - triangle marathon 601 - 700 · rmm - triangle marathon 601 - 700. rmm triangle marathon 601...
Post on 24-Jun-2020
69 Views
Preview:
TRANSCRIPT
ROMANIAN MATHEMATICAL MAGAZINE
Founding EditorDANIEL SITARU
Available onlinewww.ssmrmh.ro
ISSN-L 2501-0099
RMM - Triangle Marathon 601 - 700
www.ssmrmh.ro
Proposed by Daniel Sitaru – Romania
Aatman Supkar-India, Adil Abdullayev-Baku-Azerbaidian
Abdilkadir Altintas-Afyonkarashisar-Turkey,Mehmet Sahin-Ankara-Turkey
Theodoros Sampas-Greece, Mihály Bencze – Romania
Do Huu Duc Thinh-Ho Chi Minh-Vietnam, George Apostolopoulos-
Messolonghi-Greece,Vadim Mitrofanov-Kiev-Ukraine
Nicolae Papacu-Romania, D.M. Bătinețu-Giurgiu – Romania
Neculai Stanciu – Romania, Lelia Nicula-Romania
Hung Nguyen Viet-Hanoi-Vietnam, Athanasios Mplegiannis-Greece
Rovsen Pirguliyev-Sumgait-Azerbaidian, Marian Ursarescu-Romania
Do Quoc Chinh-Vietnam, Nguyen Van Nho-Nghe An-Vietnam
Marin Chirciu-Romania
www.ssmrmh.ro
Solutions by Daniel Sitaru – Romania
Ravi Prakash-New Delhi-India, Mehmet Sahin-Ankara-Turkey
Soumava Chakraborty-Kolkata-India, Do Huu Duc Thinh-Ho Chi Minh-Vietnam
Myagmarsuren Yadamsuren-Darkhan-Mongolia, Soumitra Mandal-Chandar Nagore-India,
Pham Quoc Sang-Ho Chi Minh-Vietnam
Vaggelis Stamatiadis-Greece, Boris Colakovic-Belgrade-Serbia
Rozeta Atanasova-Skopje, Rovsen Pirguliyev-Sumgait-Azerbaidian
Seyran Ibrahimov-Maasilli-Azerbaidian, Geanina Tudose-Romania
Serban George Florin-Romania, Rajeev Rastogi-India
Abdul Aziz-Semarang-Indonesia, Abdelhak Maoukouf-Casablanca-Morocco
Dimitris Kastriotis-Athens-Greece, Kunihiko Chikaya-Tokyo-Japan
Hoang Le Nhat Tung-Hanoi-Vietnam, Do Quoc Chinh-Vietnam, Mihalcea Andrei Stefan-
Romania
www.ssmrmh.ro
601. If in , (∢) = °, ∈ ( ), ∥ , ∥ ,
= = , then:
+ + + =
Proposed by Aatman Supkar-India
Solution by Ravi Prakash-New Delhi-India
= , = , = ⇒ =
= + = + = ( + )
= + = + = ( + )
= + = ( + ); =++ =
∴ + + + = ( + ) + ( + ) =
= ( + ) + ( + ) = ( + ) = ( + ) =
602. In ∆ , – orthocenter
– lies on the incircle ↔ =
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution by Daniel Sitaru-Romania
= + + −
− → = → + + − = →
www.ssmrmh.ro
→ = + +
=− ( + )
=+ + − − −
=
603. In ∆ the following relationship holds:
−= + +
Proposed by Abdilkadir Altintas-Afyonkarashisar-Turkey,
Designed by Miguel Ochoa Sanchez-Peru
Solution by Daniel Sitaru-Romania
=⏞ ∙∑∑ −
− =− ∙ ∑
∑
= ↔
↔ −∑∑ = ↔
( )∙∑( ) −
∑∑ = ↔
− = ; − =
= ( − ( − ) + ( + ) ) − ( − ( + ) + ( + ) )
= − ( − − − ) = =
www.ssmrmh.ro
604. If in ∆ the orthocentre lies on the incircle then:
+ + =
Proposed by Adil Abdullayev-Baku-Azerbaidian
Design by Miguel Ochoa Sanchez-Peru
Solution by Daniel Sitaru-Romania
= + + − = → = + +
= − ( + ) = + + − − − = =
605. In the following relationship holds:
− =−
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution 1 by Mehmet Sahin-Ankara-Turkey
−= − = ⋅ −
www.ssmrmh.ro
=( − )( − )
⋅( − )
−( − )( − )
⋅( − )
=−
⋅( − )
−−
⋅( − )
=( − )
⋅−
−−
=( − )
⋅−
=−
= ⋅ (New equality). Let = ⋅ ⋅
=( − )
⋅−
⋅( − )
= ⋅( − )
⋅−
= ⋅ ( − )( − )
= ⋅ ( − )( − ) = ⋅ ( − )( − )
= ( )( ) = ( )( ). Let = −
= − ⇒ = − − − =− − ( − )
( − )( − )
= ⋅( − )
( − )( − )( − )( − ) = ⋅
( − ) ⋅ ⋅ ( − )( − )( − )( − )
= ( )⋅ ( ) = ( )⋅ ( ) , = ( )( ). = as desired ∴
Solution 2 by Soumava Chakraborty-Kolkata-India
− = −
= − =−
= ⋅−
= ⋅−
www.ssmrmh.ro
= ∑ ⋅−
= + + ⋅−
=⋅
⋅−
= (proved)
606. If , , − sides in∆ ,√ ,√ ,√ − sides in ∆ ′ ′ ′ then :
√′
+√
′+
√′
= ( + + )
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
√′ =
√+ − =
√( − ) =
=√ ( − )
= √ ∙∑( − )( − )
( − )( − )( − ) =
=√
∙ ( + ) =√ ( + ) = ( + + )
607. In ∆ , , , - medians in Gergonne’s triangle. Prove that:
+ + = ( + + )
Proposed by Mehmet Sahin-Ankara-Turkey
www.ssmrmh.ro
Solution by Daniel Sitaru-Romania
Let ∆ be the Gergonne’s triangle:
+ + = = ( ( − ) − ( − ) ) =
= ( − ) ( − ) = ( − ) =( − )( − )( − )
=
= ( − ) = ( − + + ) =
= ∙ ( + ) = ∙ ( + ) = ( + + )
608. Let ∆ ,∆ be the contact respectively the excentral triangle of
∆ . If = [ ], = [ ], = [ ], = [ ],
= [ ] then: = , =
Proposed by Mehmet Sahin-Ankara-Turkey
Design from http://cut-the-knot.org/
Solution by Daniel Sitaru-Romania
= ∙ ∙ =( − )
=( − )
, =
= ∙( − ) ( − ) ( − )
=
www.ssmrmh.ro
= ( ) ∙ ∙∙
= ∙ = ; = ∙ =
609. If in ∆ , , , − altitudes of ∆ , where
, , -Gergonne’s cevians then: + + = −
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
=
[ ]
=[ ]
∙ = ∙ =
= ( − ) ∙ =∙
( − ) ∙ = ( − )
= ( − ) = − = ∙( − )
=−
= −
610. Let , , , , , be the altitudes respectively the medians of
intouch triangle in∆ . Prove that:
www.ssmrmh.ro
+ +
+ += ∙
+ ++ +
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
Let ∆ be the intouch triangle = [ ] = .
+ +
+ +=∑
∑= ∙
∑∑ =
= ∙∑
∑= ∙
+ ++ +
= ∙+ ++ +
611.
If in ∆ , − incentre, ⊥ , ⊥ , ⊥ then:
www.ssmrmh.ro
[ ]+
[ ]+
[ ]≥
( − )
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
[ ] = = = ∙+ +
≥
≥⏞ ∙− + +
=−
=( − )
612. Prove that:
+ + = + + + +
Proposed by Mehmet Sahin-Ankara-Turkey
Solution 1 by Do Huu Duc Thinh-Ho Chi Minh-Vietnam
In homogeneous barycentric coordinates: ( : : ), ( : : ), ( : : ). The
Brocard point has coordinates , , , so barycentric coordinates are:
www.ssmrmh.ro
∑ ,∑ ,∑
If ⃗ = ( , , ) in normalized barycentric coordinates then:
= − − − . Using that formula we have:
= + + ⇒ =+ +
Similarly: = ; =
⇒ + + =+ +
⋅ =
= + + + + (Q.E.D.)
Solution 2 by Soumava Chakraborty-Kolkata-India
∠ = ° − ( − + ) = ° −
In , =( ° )
= ⇒ =( )
∵ = ,∴ =( )
∴ = ⋅∑
=∑
(using (1), (2))⇒ =∑
⇒ = ∑ (a)
Similarly, = ∑ (b) and, = ∑ (c)
(a)+(b)+(c) ⇒ = ∑ + + = ∑ + +
www.ssmrmh.ro
613.
In ∆ the following relationship holds:
+ + + + + =
Proposed by Theodoros Sampas-Greece
Solution by Daniel Sitaru-Romania
+ =( + )( + − )
=
=+ + − −
=
= + ∙ + − − =
614. In ∆ the following relationship holds:
+ + ≥
Proposed by Daniel Sitaru-Romania
Solution 1 by Soumava Chakraborty-Kolkata-India
≥ ≥ → ≥ ≥
+ + ≥⏞ ∙ = ( + ) ≥
≥⏞ ∙ ≥⏞ ∙ = ≥⏞ ∙ =
www.ssmrmh.ro
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
+ + ≥⏞ =
=( ) ∙
( − )( − )( − ) =( )
= ≥
≥⏞ ∙ √ = ∙ √ ∙ = √ ≥⏞ ∙ =
Solution 3 by Soumitra Mandal-Chandar Nagore-India
= ≥⏞(∑ )
∑=∑ − =
=∙
= ≥⏞ ∙ =
615. If in , , , - cevians, ∩ ∩ = { } then:
+ + ≥ , + + ≥
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution by Mehmet Sahin-Ankara-Turkey
(Proof of first inequality)
Euler-Gergonne Theorem: In triangle let = , = , = , then
www.ssmrmh.ro
+ + = .
In that case: + + = . Using Arithmetic Harmonic Mean Inequality, then
+ + ≥ ⇒ + + ≥ ∴
(Proof of second inequality)
If = ⇒ = ⋅ and = + ⇒ = + ,
Similarly, = + and = +
+ + = + + + = + + + (1)
Let = + + , ≥ (?)
+ =+
++
++
Using Arithmetic-Harmonic Mean Inequality, we get
+ + ≥ ≥
+ ≥ − =
≥ (2)
From (1) and (2): + + ≥ + = (proof is completed)
616. If in ∆ , -Lemoine’s point then the following relationship holds:
√ ∙ ≥√ √ + √ + √
√ + √ + √
Proposed by Daniel Sitaru-Romania
Solution by Soumava Chakraborty-Kolkata-India
If , , > then:
+ + ≥⏞ (1)
www.ssmrmh.ro
For = √ , = √ , = √ in (1):
√ ∙ ≥√ √ + √ + √
√ + √ + √
617. In ∆ the following relationship holds:
≥ +
Proposed by Daniel Sitaru-Romania
Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia
≥ ( − ) → ≥ ( − ) = = ( )
= + ≥⏞ + ≥⏞( )
+
618. In each triangle prove that:
( + − )( + − )( + − ) ≦
When is equality true?
Proposed by Theodoros Sampas-Greece
Solution 1 by Pham Quoc Sang-Ho Chi Minh-Vietnam
We have ( + − )( + − ) = − ( − ) ≤
⇒ [( + − )( + − )( + − )] ≤
⇒ ( + − ) ≤
Solution 2 by Vaggelis Stamatiadis-Greece
In :
( + − )( + − )( + − ) ≤ (1)
( + − )( + − )( + − ) ≤ (1)
www.ssmrmh.ro
+ − = >+ − = >+ − = >
⇒
=+
=+
=+
⎭⎪⎬
⎪⎫
(1): ≤ ⋅ ⋅ (2)
(2) ⇐ ( + )( + )( + ) ≥ ⇐
+ + + + + − ≥ ⇐
( + − ) + ( + − ) + ( + − ) ≥ ⇐
( − ) + ( − ) + ( − ) ≥ ; " = " ↔ = =
Solution 3 by Daniel Sitaru-Romania
If (∢ ) > ° → = < → < <
If (∢ ) = ° → = → + − = → = <
If its acute , , can be the sides of a triangle because:
+ = + > and analogous
By Padoa’s inequality for the triangle with sides , , :
( + − )( + − )( + − ) ≤ . For = , = , = :
( + − )( + − )( + − ) ≤ ;“=” ↔ = =
619. Let be a triangle, denote , , the distances from the centroid
to the sides , , respectively. Prove that:
≤ + + ≤+ +
Proposed by Mihály Bencze – Romania
www.ssmrmh.ro
Solution by Soumitra Mandal-Chandar Nagore-India
In , we have and similar triangles. So,
= ⇒ = = ⇒ ( ) = ,
similarly = , = . Now, ∑ ≤ ∑ = ∑
= ∵ = = = =
=( + + )
=( + + )
⋅ =+ +
≥ =∏
= =
620. In acute the following relationship holds:
√ + + + ≥ + +
Proposed by Daniel Sitaru – Romania
Solution 1 by Mehmet Sahin-Ankara-Turkey
Define ( ) = − function.
www.ssmrmh.ro
This function is convexe in ( , ). Using the Jensen’s inequality we get,
≤ [ ( ) + ( ) + ( )] ⇒ ( ) + ( ) + ( ) ≥ ⋅
≥ − ≥ − ≥ − = −√
⇒ √ + + + ≥ + +
Solution 2 by Soumava Chakraborty-Kolkata-India
In any acute – angled ,√ + ∑ ≥ ∑
Let ( ) = √ + − ,∀ ∈ ,
( ) = ( + − − ), where = and
= . Now, = − ∴ = = = ( = )
∴ ( ) = + − − = ( + )( − ) → (1)
Now, ∵ < < ∴ < < ⇒ < < ⇒ < <
⇒ < < = (say) → (2)
= − = = (say) → (3)
Now, = = ⇒ = √ − ⇒ = √ + → (4)
(3), (4) ⇒ = √ + → (5)
Now, = √ + (from (2), (4))⇒ = √√
<√→ (6)
(2), (6) ⇒ < <√⇒ < ⇒ − > → (7)
(1), (7) ⇒ ( ) > ∀ ∈ , ⇒ ( ) is convex on ,
∴√
+−
−+
≥√
+−
−+
www.ssmrmh.ro
(Jensen) =
√+
√−
√= ⇒ √ + ∑ ≥ ∑
621. Let be a triangle, the incircle with center intersect the segments
, , , in points , , . Prove that:
+ + ≤√
Proposed by Mihály Bencze – Romania
Solution by Ravi Prakash-New Delhi-India
= + ; = + ; = + = + − ( )( )
= + + = +
∴ =+
= ⋅( )
+
Thus, + + = . Where = ∑ ( ) . Let ( ) = ( ) , < <
( ) =( + ) − ( )
( + )
= ( + ) [ ( + ) − ( − )]
= ( + ) [( )( + ) − ]
www.ssmrmh.ro
= ( + ) [( − )( + ) − ]
= ( + ) [ + − − − ]
= ( + ) [ − − ] = +( − − )
= − + +
( ) = − −( )
= − +( )
< , for < <
∴ is concave function. Thus, ∑ ( ) = ∑ ≤ = ⇒
⇒ ∑ ≤ = √ . Hence + + ≤ √
622. In the following relationship holds:
( + − ) + ≥ √
Proposed by Daniel Sitaru – Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India
∑ ( + − ) = ∑ ( − ) , where + + =
let = − , = − , = − then ∑ ( − )( − ) = ( + ),
( − ) =
= − + = − ⋅ ( + ) +
= − , so, ∑ ( + + ) + = ∑ ( − ) +
= ( − ) + = ≥ √ ∵ ≥ √ ≥ √ (proved)
www.ssmrmh.ro
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
1. ∑( + − ) ≥ ⋅ (∑( + − )) = ⋅ ( + + ) = ⋅
2. ∑( + − ) + ≥ + = + ≥⏞
√
≥ ⋅ √ ⋅ ⋅ + = ⋅ √ = √
Solution 3 by Soumava Chakraborty-Kolkata-India
In any ,∑( + − ) + ≥ √
= ( − ) + ( ) ≥ ( − ) + = ( + )
≥ √ ( ) (∵ ≥ √ and ≥ ) = √ (proved)
623. If ∈ ( ) then:
+ + ≥+ +
Proposed by Daniel Sitaru – Romania
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
+ + ≥ ( + + ) (1)
= + + = + + = ⇔ = ≥ > >> > Chebysev
www.ssmrmh.ro
≥( + + )( + + )
⇔ ≥ ( + + )
⋅≥
( )⇔ ( + + ) ≥ (2)
(1); (2) ⇒ + + ≥ = (*)
2) RHS: = …
= = ; ∑ ⋅
=∑ ( )
=
= ∑ − ∑ + + + ≤ − − ( + ) + =
=( )
=( )
=( )
= ≤ = = (LHS)
≥ ≥∑ ⋅
Solution 2 by Soumava Chakraborty-Kolkata-India
For any point in the plane of a , ∑ ≥
For any point in the plane of , we have: ∑ ≥∑
∑ → (a)
We shall now prove: ∑
√ ≥ → (b)
⇔ ≥ ⇔ − ( ) ≥ ( − − )
⇔ ( + + ) ≥ ( − − ) +
⇔ + ( + )( + ) ≥ ( + ) + ( + ) → (1)
Now, LHS of (1) ≥ ( − ) + ( + )( + )
≥ ( + ) + ( + )
⇔ ( − ) + ( + )( + ) ≥ + ( + ) → (2)
Again, LHS of (2) ≥ ( − )( − ) + ( + )( + )
(Gerretsen) ≥ + ( + )
⇔ ( + ) ≥ + ( + ) → (3)
www.ssmrmh.ro
Now, RHS of (3) ≤ ( + + ) + ( + )
≤ ( + ) ⇔ ( − + ) ≥ ( + ) → (4)
Now, LHS of (4) ≥ ( − )( − + ) ≥ ( + )
⇔ − + − ≥ (where = )
⇔ ( − ){( − )( + ) + } ≥ → true ∵ ≥ (Euler)
⇒ (b) is true and from (a), we get: ∑ ≥ → (i)
Now, ∑ = ∑ ⋅ ( )( )( )( )
= ∑ ( − )
= ( − + ) = − ( + + ) +
≥ [ − ( + )] (∵ ∑ ≥ by AM ≥ GM)
= ( − ) = ( ) → (ii)
Now, given inequality ⇔ (∑ ) ∑ ≥ → (*)
(i), (ii) ⇒ of (*) ≥ ⋅ ( ) ≥ ⇔ ( − ) ≥ ⇔ ≥
→ true (Euler) (Proved)
624. In ∆ the following relationship holds:
°+
° °+
°>
Proposed by Daniel Sitaru-Romania
Solution by Soumitra Mandal-Chandar Nagore-India
= ° = ( ° + °) =° + °
− ° ° →
→ = ° + ° ° + °
° + ° ° + ° >⏞( + + )
° + ° ° + ° =
= ( + + ) ≥⏞ √ =
www.ssmrmh.ro
625. In ∆ the following relationshiop holds:
+ + −+ +
≤√
( + + )
Proposed by Daniel Sitaru-Romania
Solution by Boris Colakovic-Belgrade-Serbia
+ + − + + =+ +
−+ +
+ + =
=( + + ) − ( + + )
( + + ) ≥( + + )− ( + +
( + + ) =
=( + + )
( + + ) ≥( + + )
( + + ) = + + ≤√
( + + )↔
( + + ) ≤ ( + + ) ; ≤ + +
≤ + + = + + ↔ ≥ − ( )
≥⏞ − ≥ − ; ≥ ↔ ≥
626. If in acute , , , – altitudes, - orthocentre, -
circumcentre then: + + ≥
Proposed by Mehmet Șahin – Ankara – Turkey
Solution by proposer
www.ssmrmh.ro
| | = ⋅ ; = − ⋅ +
= − + ⋅ ; = − +
By adding: + + =⋅
+⋅
+⋅
−
−( ⋅ + ⋅ + ⋅ ) + + +
= + ( − ) + ( − ) + ( − )
= + (− ⋅ ) + ⋅ (− ) + (− )
= − ( + + )
= − ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
= − + +
( + − ) + ( + − ) ≥ ( + − )( + − )
≥ ( + − )( + − ) ⇒( + − )( + − )
≤
In that case ≥ − and + + ≤
≥ − = as desired.
627. In acute ∆ the following relationship holds:
+ + ≥ + +
Proposed by Do Huu Duc Thinh-Ho Chi Minh-Vietnam
Solution by Daniel Sitaru-Romania
=+
, =+ −
∑ = ∑ ≤( )
,(1) (C. Mateescu – 2016)
www.ssmrmh.ro
= ≥⏞( ) (∑ )
∑ =+
+ − =
=( + )
+ − ≥ ( + ) ≥ ( ) ↔
↔ ( + − ) ≤ ( + ) ∙ ( + )( )
( + − ) ≤⏞ ( + + + − ) =
= ∙ ( + ) ≤ ( + ) ∙ ( + ) ↔ ≤ ( + )( + )
≤⏞ + + ≤ + + ↔ ≥
628. In ∆ the following relationship holds:
+ + + + + ≥ √
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Daniel Sitaru-Romania
+ = + ≥√
√ ≥⏞
=√
∙ √ =√
∙ √ ≥⏞√
∙ ≥⏞
≥√
√ =√ ∙ √
=√
≥ √ ∙ ↔
≥ ↔ ≥
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
+ ≥⏞√
( + ) = √ =
www.ssmrmh.ro
= √ = √ ∙ ≥⏞ √ ∙√
= √ ∙ ∙ ≥⏞
≥ √ ∙ ∙ = √
629. In acute the following relationship holds:
≤
Proposed by Daniel Sitaru – Romania
Solution 1 by Mehmet Sahin-Ankara-Turkey
= and = . It is known that ∑ = +
( ) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ≤
⇔ ( ) ⋅( )
⋅ ≤ ( )⋅
⇔ ≥ (?) + + = +
+ + = + ⇒ + + = +
⇒ + + ≤ + =
≤ + + ≤ ⇔ ≥ ⇔ ≥
Solution 2 by Rozeta Atanasova-Skopje
If < , , < then ≤ ≤ = ,
www.ssmrmh.ro
then = ( )
≤ = ⋅ ⋅ =
Solution 3 by Soumava Chakraborty-Kolkata-India
∵ in any acute-angled , , , > ,∴ ≥ ⇒
≤ ( + + ) = +
= ∵ ≥ ⇒ ≤ ⇒ ≤ (1)
Also, ∵ is acute, ∴ < , , <
and hence, >
= ( )( )( )( )
= ( ) ≤ (using (1)) = = (proved)
630. In ∆ the following relationship holds:
+ ≥ + + ≥( − )
Proposed by Vadim Mitrofanov-Kiev-Ukraine
Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia
≤⏞ =( − )
= ∙∑ − ∑
=
=( − + + )
= ( + ) = +
≥⏞( − )( − )( − )
( ) = =
= ≥ ( − ) ( ) ↔ ( − ) ≥
www.ssmrmh.ro
( − ) ≥⏞ − = ∙ =
= ∙ ≥⏞ ∙ =
631. If in ∆ , − ninepoint center then:
a. =
b. + + =
c. + + ≤
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution by Daniel Sitaru-Romania
a. ∆ , − → = ( + ) − =
= ( + ) − ( − − − ) = − − + =
= − ++ +
= − ++ +
=+ + −
b. + + = + + =
c. + + = =
= ≤⏞ =
632. PAPACU’S INEQUALITY
In nonacute :
−,−
,−
≤
Proposed by Nicolae Papacu-Romania
www.ssmrmh.ro
Solution by Soumava Chakraborty-Kolkata-India
WLOG, we may assume , , =
Then, ≤ ⇔ ≥ , and ≤ ⇔ ≥ ∴ > , >
So, is the largest side, and ∵ is non-acute, ∴ ∢ ≥ °
We are to prove: ≤ ⇔ ( ) ≤ ⇔ + ≤ +
⇔+
≤ + + ⇔( + )
≤ +
⇔+ −
≤ +
⇔−
≤ +
⇔− − +
≤
⇔ ≤
⇔ ≥ ⇔ ≥√⇔ ≥ ° ⇔ ≥ ° → true (proved)
633. If in , = then:
≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India
= ⇒( − )( − )
⋅( − )( − )
=
www.ssmrmh.ro
⇒−
= ⇒ = = ⇒ =( − )
⇒ = ⋅ ⋅ ⇒ = . We need to prove,
≥ ⇔ ≥ √ . We know, = + + ⇒ = + ⇒ ≥ √
∴ ≥ (proved)
Solution 2 by Mehmet Sahin-Ankara-Turkey
⋅ ⋅ = and = ⇒ ⋅ ⋅ = (1)
⋅ = ⇒ − ⋅ − = ⇒ = ( − )( − )
⇒ ( − ) ⋅ = ( − )( − )( − )
⇒ ( − ) ⋅ = = ⇒ − = ⇒ = (2)
⋅ = ⇒ = ⋅ ⇒ = ⋅ = ⋅( − )
From (2) = ⋅ = ⋅ =
⋅ =⋅
= =
⋅ = ⋅ ⋅ = ⋅
From (2) ⋅ = ⋅ = ⋅ (3)
= ⇒ = ⇒ =( )
(4)
+ ≥ √ ⇒ ≥ √ ⇒ ≤ ⇒ ≤
and form (4) ≤( )
= =
from (3) ⋅ = ≤ ⋅ = ⋅
≤ ⋅ = as desired.
www.ssmrmh.ro
Solution 3 by Soumava Chakraborty-Kolkata-India
, = ⇒ ≥
+ =+
−= +
= + =+
=+
=−
=
∴+
= ⇒ − =
⇒ = ⇒ =
⇒ = ⇒ =
⇒ ∑ = ⇒ =( )
⇒ = (1)
(1) ⇒ it suffices to prove: ≥
⇔ ≥ ⇔ ≥
⇔ ≥ √ ⇔ ≤ ° ⇔ ≤ ° (i)
(a) ⇒ ≤ √ ⇒ ≤ √ ⇒ ≤ ° ⇒ (i) is true (proved)
www.ssmrmh.ro
634. In the following relationship holds:
+ + ≥
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Rovsen Pirguliyev-Sumgait-Azerbaidian
+ + ≥ (1)
=+ −
, =+ −
, =+ −
⇒⊕
⇒ + + = , we prove that ≥ ⇒
⇒ + + ≥ , (where = , =( )
)
we have = ⋅⋅
= ( + + )
hence + + ≥ ( + + ) and this is true.
Solution 2 by Soumava Chakraborty-Kolkata-India
= =+ −
⋅
= ( + − ) = =− −
≥
⇔( − − )
≥ ⇔ ( − − ) ≥
LHS of (1) ≥ ( − − )( − ) ≥?
⇔ ( − − )( − ) ≥?
⇔ ( − ) ≥?
( − )( + ) (2)
LHS of (2) ≥ ( − )( − ) ≥?
( − )( + )
⇔ − + ≥?
⇔ ( − )( − ) ≥?
true ∵ ≥ (Euler)
www.ssmrmh.ro
635. In acute the following relationship holds:
( + )( + ) ≥
Proposed by Daniel Sitaru – Romania
Solution by Do Huu Duc Thinh-Ho Chi Minh-Vietnam
Let = , = , = ; since is acute then , , >
The inequality can be written as:
( + )( + ) ≥ ⇔ ( + ) ≥ ( + )( + )( + )
⇔ ( + ) ≥ ( + ) + ⇔ ( + ) ≥
It’s true by Cauchy since:
∑ ( + ) ≥ ⋅ + ⋅ + ⋅ = ⇒ Q.E.D.
636. If , , > then in :
( + )+
( + )+
( + )≥ √
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
Solution 1 by Ravi Prakash-New Delhi-India
++
++
+=
= + + + + + ≥ ( + + ) (1)
= + + . As ( ) = is convex,
( ) + ( ) + ( ) ≥+ +
= = √
∴ from (1), we get + + ≥ √
www.ssmrmh.ro
Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian
( + )≥
( + )( + )( + )≥ √
√ ≥ √ ; ⋅ ≥ √ ∴ ≤ √
Solution 3 by Soumava Chakraborty-Kolkata-India
In any , for , , > , we have the following:
( ) + ( ) + ( ) ≥ √ ; ≥ ( )( )( )( ) (A-G)
≥ ( ⋅ ) (∵ ( + )( + )( + ) ≥ − )
≥ ⋅ ( )√
(∵ ≥ by Euler and ≥√
by Mitrinovic)
=√
= ⋅√
( ) = √ (proved)
637. In the following relationship holds:
+ √ + √≤
Proposed by Daniel Sitaru – Romania
Solution 1 by Boris Colakovic-Belgrade-Serbia
+ √ + √ ≥ √ + √ = √ =
+ √ + √≤
+ √ + √≤ + + =
+ +=
= ⋅ =
Solution 2 by Geanina Tudose-Romania
+ √ ⋅ + √ = + √ ⋅ + √ ⋅ +
www.ssmrmh.ro
= + √ ⋅ ( + ) ≥ + √ ⋅ √ =
Hence ∑√ √
≤ ∑ = = =⋅
⋅ ⋅=
Solution 3 by Ravi Prakash-New Delhi-India
√ + √ + = + √ ( + ) +
≥ + √ √ + =
∴+ √ + √
≤ ( + + ) = = ( ) =
Solution 4 by Serban George Florin-Romania
+ √ + √≤
+ √ + √ = √ ⋅ √ + √ + √ ⋅ √ ⋅ √ + √ + √
≥ √ ⋅ √ ⋅ √ ⋅ √ ⋅ √ ⋅ √ ⋅ √ = √ ⋅ √ ⋅ =
= √ ⋅ ( ) = √ ⋅ √ = ≥
∑√ √
≤ ∑ = ⋅ = ⋅ = = (A)
638. In the following relationship holds:
+ + ≥ ( − − )
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania
Solution 1 by Mehmet Sahin-Ankara-Turkey
= + + ≥+ ++ + =
( + + )+ +
www.ssmrmh.ro
≥ ( + + ) ⋅ ( + + ) ⋅ ⋅
+ + = ⇒ ( + + ) ≤ ( + + )
≤ ( + + ) ⇒ + + ≥
Also, ≥ ( + + ) ⋅ ⋅ ⋅ ⋅ ; ≥ ( + + ) ⋅ ⋅ ⋅
= ⋅ , ≥ , + + = ( − − )
⇒ ≥ ⋅ ⋅ ( − − ), ⋅ ⋅ ⇒ ≥ ( − − )
Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian
∑ ≥ ∑ ≥ ∑∑
≥ ∑ = ∑ = ( − − )
+ + = ; + + = + +
= − − − = − −
Solution 3 by Soumava Chakraborty-Kolkata-India
In any , let ( ) = , ( ) = and ( ) = and also,
( ) = , ( ) = and ( ) = . Then + + ≥ ( − − )
( ) = = , ( ) = = , ( ) = =
∴ = ( + + ) (1)
WLOG, we may asuume ≤ ≤ ∴ ( ) ≥ ( ) ≥ ( )
⇒ ≥ ≥ and also, ≥ ≥ Hence, applying Chebyshev and using (1),
≥ ∑ ⋅ ∑ ≥ ∑ (∑ ) (Chebyshev)
= ∵ =
≥ √ ⋅ √ = = ( − − ) =
Hadwiger-Finsler inequality ⇒ ∑ ≥ √ + ∑ ≥ √ by
www.ssmrmh.ro
Weitzenbock’s inequality ⇒ ∑ ≥ √
∴ ∑ ≥ √ and of course ∑ ≥ √
639.
Proposed by Lelia Nicula-Romania
Design by Miguel Ochoa Sanchez-Peru
Solution by Daniel Sitaru-Romania
∙ ≥⏞ ≥⏞ ∙ ≥
≥ ≥⏞ ∙ √ =√
≥
≥⏞ ≥ − ↔ ≥ − + ↔
≥ + ↔ − + − ≥ , = ≤⏞ , − ≤
↔ − + − ≤ ↔ ( − )( − ) ≤
640.
Proposed by Hung Nguyen Viet-Hanoi-Vietnam
www.ssmrmh.ro
Design by Miguel Ochoa Sanchez-Peru
Solution by Daniel Sitaru-Romania
+ − + + − ≤⏞
+ − ≤ + + = ( )
√ = ( + − ) = ( + − ) ≤⏞( )
≤ ∙ = = = ; √ ≤
641. In the following relationship holds:
( + + ) + + ≥ √
Proposed by Daniel Sitaru – Romania
Solution 1 by Athanasios Mplegiannis-Greece
From cosines law:
= + −= + −= + −
⇒( )
+ + = ( + + ) ⇒
⇒ + + ≥ ⋅ √ ⋅ ⋅ ⇒
⇒ + + ≥ ⇒
⇒ ( + + ) ≥ ⇒
⇒ + + ≥ √ √ ⇒
⇒ ( + + ) + + ≥ √
Equality holds for = = ⇒ equilateral.
Solution 2 by Soumava Chakraborty-Kolkata-India
Given inequality ⇔
www.ssmrmh.ro
⇔ ( + + ) ≥ ⋅ (1)
Now, =( )
⋅ ⋅
Numerator = (∑ − )(∑ − )(∑ − )
= − + −
= − + − ( ) −
= − − −
= {( + + ) − ( − − ) − } −
= { ( + ) − }−
= ( − − ) − =( )
( − − − )
(a), (b) ⇒ = =( )
Ionescu – Weitzenbock. Again, (∑ ) ≥ (2)
(1), (c), (2) ⇒ it suffices to prove: ( − − ) ≥ ⋅ ⋅
⇔ − − ≥ ( − − − )
⇔ ≤ + + ⇔ ≤ + + (3)
Now, LHS of (3) ≤ + + ≤?
+ +
⇔ ≥ → true by Euler (proved)
Solution 3 by Soumitra Mandal-Chandar Nagore-India
( + − ) + ( + − )≥ ( + − )( + − ) ⇒
≥ ( + − )( + − ) similarly,
≥ ( + − )( + − ) and ≥ ( + − )( + − )
( ) ≥ ( + − ) = ( ) ⇒
www.ssmrmh.ro
⇒+ +
≥ ( )
≥ √
642. In ∆ , − Lemoine’s point.Prove that:
++
++
+≤
+ ++ +
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution by Daniel Sitaru-Romania
= ∑ − (∑ ) =∑ −(∑ ) =
=( + ) −
(∑ ) = (∑ ) → = ∑
+ = + + + ≤⏞ + +
643. In the following relationship holds:
+ + ≤√ ( + )
√ + √+
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
= = ; = = ; = =
∴ given inequality ⇔ √√ √
( + ) + √√ √
+ √√ √
+ ∑ ≥ ∑ (1)
Now, √√ √
≤ √√ √
⇔ √ + ≤ + √
www.ssmrmh.ro
⇔ ( − ) + √ √ − √ ≤ ⇔ √ − √ √ + √ + √ ≤
⇔ √ ≤ √ ⇔ ≤ (i). Similarly, √√ √
≤ √√ √
⇔ ≤ (ii)
Also, + ≤ + ⇔ − − ( − ) ≤
⇔ ( − )( + − ) ≤ ⇔ − ≤ (∵ + > ) ⇔ ≤ (iii)
Similarly, + ≤ + ⇔ ≤ (iv). WLOG, let’s assume ≤ ≤
∴ LHS of (1) ≥ ∑ √√ √
(∑ + ∑ ) + ∑ (using (i), (ii), (iii), (iv))
≥ ⋅ + + = +
∴ it suffices to prove: ∑ + ∑ ≥ ∑
⇔ ∑ + ∑ ≥ ∑ ⇔ ∑ ≥ ∑ → true ∴ (1) is true (proved)
644. In the following relationship holds:
+ + + ( + ) ≥√
+ +
Proposed by Daniel Sitaru – Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India
Let = + + , = + + and =
+ + = , + + = + + and =
∴ + ( + ) = − + + − = −
= − = ( − − ), we will prove
( − − )( + + ) ≥ √
⇔ ( − − )( + + ) ≥ ∴ ≥ √
⇔ − ( + ) ≥ ⇔ ≥ ( + ) +
We know, ≥ − now we will prove,
( − ) ≥ ( + ) + ⇔ − + ≥
www.ssmrmh.ro
⇔ ( − )( − ) ≥ , which is true
∴ + ( + ) ≥√
+ +
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
1. + + ≥
∑ ( + )−
2. + + ≥ ; 3. ≥ √
⇒ + ( + ) ≥( )
( + ) − =
= − − = − =
= − ≥( )
( − ) =
= ⋅ = ≥( )
√ ⋅ ≥( ) √
+ + ≥√
+ +
Solution 3 by Soumava Chakraborty-Kolkata-India
= + − + ( − )
= + ∑ − ∑ + ∑ − = ∑ (1)
Now, ∑ ≥ √ + ∑ (Hadwiger-Finsler) ≥ √
(Ionescu – Weitzenbock) ⇒ ∑ ≥ √
⇒ √∑
≤ √√
= (2)
(1), (2) ⇒ it suffices to prove: ≤ ∑
⇔ ( − − ) ≥ (3)
Now, LHS of (3) ≥ ( − )( − ) ≥
⇔ − + ≥ ⇔ ( − )( − ) ≥
→ true ∵ ≥ (Euler) ⇒ (3) is true
www.ssmrmh.ro
645. In the following relationship holds:
+ + + ≤
Proposed by Vadim Mitrofanov-Kiev-Ukraine
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
=∑⋅ ⋅
∏≤∑
∏=
∑ ( − )( − )
( − )( − )( − )⋅
=
=
∑ ( − )( − )
= ⋅ ⋅ ( − )( − ) =
= − ( + ) + − + =
= ( − − − + ) = ( − ) =−
≤−
; + = − + =
Solution 2 by Soumava Chakraborty-Kolkata-India
,∑ ≤ ; = ∑( )( )
≤ ∑ ∑( )( )
(CBS)
=−
∏( − ) = = ≤−
⇔ ∑ ≤ − + ⇔ ≤ − + (1)
Now, Gerretsen ⇒ of (1) ≤ + + ≤ − +
⇔ ≥ ⇔ ≥ → true (Euler) ⇒ (1) is true (proved)
www.ssmrmh.ro
646. If in , - incentre, , , circumradii of , ,
with circumcentres , , then:
+ + ≥
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution 1 by Mehmet Sahin-Ankara-Turkey
= ° −
is a quadrilateral. Using Ptolemy theorem we get,
| | = ⋅ + ⋅ ⇒ | | = + ; | | = ⋅ + ⋅ ⇒ | | = +
| | = ⋅ + ⋅ ⇒| |
= ⇒ + + ≥ (Nesbitt)
Solution 2 by Soumava Chakraborty-Kolkata-India
< + (from )
www.ssmrmh.ro
For an equilateral , = + ∴ combining above 2 situations,
≤ + = + ⇒ ≥ (1)
Now, ⋅ ⋅ = ⋅ ⋅ ⇒ = ⋅ ⋅ = =( )
(1), (2) ⇒ ≥ = = =
= − + = +
∴ ≥ (a)Similarly, ≥( )
& ≥( )
(a)+(b)+(c)⇒ ≥ ∑ ≥
647. If in : = then:
√ ≤ + +
Proposed by Daniel Sitaru – Romania
Solution 1 by Mehmet Sahin-Ankara-Turkey
In : = ( ) , = ( )
www.ssmrmh.ro
=( − )( − )
, =( − )( − )
⋅ = ⋅ ⋅ ⇔
⇔( − )
⋅( − )
= ⋅( − )( − )
⋅( − )( − )
; =( − )
= ⇒ + + = ⇒ = + ≥ √ ⇒ ≥ √ (1)
Using the cosinus theorem, we get;
= + − ⋅ and + = ; + + =
+ = − ⇒ = − − ⋅
⇒ ( + ) = ⇒ + − = ⇒ ⋅ =
⇒ = ⇒ = √ ⋅√
(2)
From (1) and (2) ⇒ ≥ √ ⇒ ( ) ≤ °
√ ⋅ ⋅ ≤ + + = ⇒ ≤√
⇒ ( ) = °
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
⋅ = ⋅ ⋅ ⇔
⇔ = ⋅ = − ⋅ − = ( − )( − ) =
= ( − )( − ) =−
⇔ =−
⇔ = ⇔ = ⇒
⇒ = + (*)⇒ ≥ (1) ⇔ ≥ (Assure)
≥−
= ⋅ + ⋅ =
= ⋅ + ⋅ = ⋅ ⇔ ≥ (II)
www.ssmrmh.ro
(I), (II) ⇒ ≥( )
= ≥( )
⇒ = (3)
√ ⋅ ⋅ =( )
√ ⋅ ⋅ = ⋅ √ ⋅ ⋅√
= = + =(∗)
+ +
Solution 3 by Soumava Chakraborty-Kolkata-India
− = ⇒ = (1)
⇒ − =
⇒ = = ≤ ⇒ ≤ ⇒ ≤ ⇒ ≤ ⇒ ≤ (a)
Now, √ ≤ + +
⇔ √ ( ) ≤ ⇔ √ ≤
⇔ √ ≤⏞( )
=+
+−
(1) ⇒ = − ⇒ = =( )
(i), (2) ⇒ given inequality ⇔ √ ≤
⇔ ≤√⇔ ≤ ⇔ ≤ true by (a)
648. If in :
= , = , = then:
≤√
Proposed by Daniel Sitaru – Romania
Solution 1 by Rajeev Rastogi-India
Given = ⇒ =
= = ( ) ⇒ = . Similarly, =
www.ssmrmh.ro
= ⇒ ⋅ = ( )( )( ) (1)
Now, AM ≥ GM ≥ ( − )( − )( − ) ⇒ ( )( )( ) ≤√
= √
Solution 2 by Ravi Prakash-New Delhi-India
+ = ( + )
= + − =+
−
⇒ =−
+
⇒−
+=
−
+⇒ =
⇒ = ≤√
Solution 3 by Soumava Chakraborty-Kolkata-India
= ⇒ = ⇒ = ⇒ = (1)
Similarly, =( )
and =( )
; (1)×(2)×(3)⇒ =
=( − )( − )( − )
≤?
⇔ ≥ ( − )( − )( − )
⇔ ≥ ( − )( − )( − ) = ⇔ ≥ (4)
Now, Gerretsen ⇒ ≥ − ≥ − = ⇒ (4) is true
649. In the following relationship holds:
++
++
+≥ +
( − )( − )
Proposed by Adil Abdullayev-Baku-Azerbaidian
www.ssmrmh.ro
Solution by Myagmarsuren-Yadamsuren-Darkhan-Mongolia
In : ( )√
≤ + + ≤ ( )√
(*) ⇒ ≥ √( )
⋅ ∑ ≥ (**)
(**)⇒ ∑ = ∑( )
= ∑ ( − ) = [∑ ( − + )] =
= ⋅ − + =
= ⋅ [ − ( − − ) + ( − − )] =
= ⋅ [ − + ( + ) + − ( + )] =
= ⋅ ( + − − ) = ( − ) ⇒ ∑ = ⋅ ( − ) (***)
(**); (***) ⇒ ≥ √( )
⋅ ( − ) ≥ , ≥ √ = √ ≥ (****)
(****)⇒ : ≥ √ ; : ≥√
650. If in , , , - internal bisectors,
∈ ( ), ∈ ( ), ∈ ( ) then:
+ + ≤ ( + )
Proposed by Vadim Mitrofanov-Kiev-Ukraine
Solution by Soumava Chakraborty-Kolkata-India
Let = , = , = , = , = , =
Angle – bisector theorem ⇒ = → (1)
Stewart’s theorem ⇒ + = ( + ) (where = )→ (2)
Now, = √ ( − ) = ( )( )( )
→ (3)
(2) ⇔ + = ( )( )( )
+ (from (1), (3))
www.ssmrmh.ro
⇒ ( + ) − ( + ) + ( + + )( + − ) =
⇒ = ( ) ± ( ) ( ) ( )( )( )
→ (4)
The discriminant = ( + ) [( + ) − {( + ) − }]
= ( + ) {( + ) − } → (5)
(4), (5) ⇒ = ( ) ± ( ) ( )( )
If = ( ) ( ) ( )( )
= ( )( )
, then,
= ⋅ ( )( )
(from (1))= ( )( )
⇒ + = ( )( )
⋅ ( + ) = ( )
But ∵ + = , ( ) = ⇒ = ( )( )
= and = ( )( )
=
If, = ( ) ( ) ( )( )
, then also, = and,
= ⋅ = ∴ = → (i), = → (ii)
Similarly, = → (iii), = → (iv)
= → (v), = → (vi)
Now, = + −
=( )
+( )
− ⋅ ⋅ (from (vi), (ii))
=( ) ( )
⋅ (− − + + + + + + − − )
=( ) ( )
(say) → (6)
= − − + + + + + + − −
= −( − ) − ( + ) + ( + ) + ( + ) + +
≤ ( + )( + − + − ) + + (∵ ( − ) )
= ( + )( − ( − ) ) + + ≤ ( + ) + + (∵ ( − ) )
= ( + + + ) = ( + )( + )
⇒ ≤ ( + )( + ) ∴ ( + ) ( + ) ≤ ( + )( + )
www.ssmrmh.ro
⇒ ⋅
( )( ) (from (6)) → (a). Similarly, ≤
( )( )→ (b) and,
≤( )( )
→ (c); (a)+(b)+(c)⇒ ∑ ≤ ∑ ( )( )( )( )
=⋅ ( + + )
+ ∑ ( − ) =( + + )
( + + ) −
= ∴ ∑ ≤ → (7)
Now, + + ≤ √ ∑ (by CBS)
it suffices to prove: √ ∑ ≤ ( + )
⇔ ∑ ≤ ( + ) ⋅ ⇔ ∑ ≤ ( ) → (8)
(7), (8) ⇒ it suffices to prove: ≤ ( )
⇔ ( + ) ( + + ) ≥ ( + + ) → (9)
Gerretsen ⇒ LHS of (9) ≥ ( + ) ( − ) ≥ ( + + )
⇔ ( + ) ( − ) ≥ ( + + ) → (10)
Gerretsen ⇒ RHS of (10) ≤ ( + + ) ≤ ( + ) ( − )
⇔ − − ≥ = ⇔ ( − )( + ) ≥ → true ∵ ≥ (Euler) (proved)
651. In the following relationship holds:
( − ) ( − )+
≥ ( + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
[( + ) ⋅ ( + )]+ ≥
ö ∑( + ) ⋅ ( + )∑ + ∑ =
(∑ + ∑ )∑ + ∑
= ; = =( + )
+ =( + + )
+ ≥
www.ssmrmh.ro
≥( + ) ⋅
+ = = ⋅ ( + + ) =
= ⋅ ( + + ) ≥ ( + ) = ( + )
Solution 2 by Soumava Chakraborty-Kolkata-India
= ∑ ( ) ( ) ≥ ∑ ∑∑
(Bergström)
=(∑ + )∑ + ∑ =
( + + + )− − =
( + + )− − ≥
≥ ( + ) ⇔ + ( + ) + ( + ) ≥
≥ ( + ) − ( + )( + )
⇔ + ( + )( + ) ≥ ( + ) → (1)
Now, LHS of (1) ≥ ( − ) + ( + )( + ) (Gerretsen)
≥ ( + ) ⇔ ( − ) + ( + )( + ) ≥
⇔ ( − ) + ( + )( + ) ≥ → (2) Now, LHS of (2) ≥ ( − )( − ) + ( + )( + )(Gerretsen) ≥
⇔ ≤ ( − )( − ) + ( + )( + ) → (3)
Now, LHS of (3) ≤ ( + + ) (Gerretsen)
≤ ( − )( − ) + ( + )( + )
⇔ − + ≥ ⇔ ( − )( − ) ≥
→ true ∵ ≥ (Euler) (proved)
652. In the following relationship holds:
++
++
+≤
√⋅
( − )
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution by Soumava Chakraborty-Kolkata-India
In any ,∑ ≤ √ ⋅ ( ); ≥ ( ) ∴ it suffices to prove:
www.ssmrmh.ro
+ ≤( − )
⇔ + + − ≤( − )
⇔ (∑ ) ∑ ≤ ( ) ⇔ (∑ ) ⋅ ∑∏
≤ ( ) → (1)
Now, ∑ ( + )( + ) = ∑ (∑ + ) = ∑ + (∑ ) →(a)
Now, (∑ )(∑ ) = ∑ + ∑ ( − ) = ∑ + (∑ ) − (∑ )
⇒ ∑ = (∑ )(∑ ) + (∑ ) − (∑ ) → (b)
(a), (b) ⇒ ∑ ( + )( + ) = (∑ )(∑ ) + (∑ )
= + − + ( + + )
= + ( − − ) + ( + + )
= {( − − )( − − ) + ( + + )}
= { − ( + ) + ( + )( + )} → (c)
Again, ∏( + ) = + ∑ (∑ − )
= (∑ )(∑ )− = (∑ ) (∑ ) − ( ) −
= (( + + ) − ) −
= (∑ )( − ( − ) + ( + ) ) − → (d)
(c), (d) ⇒ (1) becomes: (∑ )[( + ){ − ( − ) + ( + ) }
− { − ( + ) + ( + )( + )}] ≥ ( + )
⇔ ( − ) − − − + ( + ) − − ≥
≥ ( + ) + (∑ ) → (2)
of (2) ≥ ∑ ( − ) − − − − +
( + ) − − ≥?
( + ) + ∑ → (3)
⇔ { ( − + ) + ( + )( − − )}
≥?
( + ) + (∑ ) → (3)
www.ssmrmh.ro
RHS of (3) ≤ ( + ) + (∑ )( + + )
≤?
{ ( − + ) + ( + )( − − )}
⇔ (∑ ){ ( − + ) + ( + )( − − )} ≥?
( + ) → (4)
LHS of (4) ≥ ( − − ){( − )( − + ) +
( + )( − − )} ≥?
( + )
⇔ ( − ) − + + ( + ) − − − ( + ) ≥?
≥ ( + ) ( − ) − + + ( + ) − − → (5)
LHS of (5) ≥ ( − ){( − )( − + ) + ( + )( −
− ) − ( + )}
≥?
( + ){( − )( − + )} + ( + )( − − )
⇔ − + − + ≥ =
⇔ ( − )[( − ){( − )( + ) + } + ] ≥ → true ∵ ≥ (proved)
653. In the following relationship holds:
( + + ) + + ≤ ( + + ) + +
Proposed by Daniel Sitaru – Romania
Solution 1 by Abdul Aziz-Semarang-Indonesia
1) ( + + ) ≤ ( + + ) + +
2) + + ≤ + + + +
Multiplying (1) and (2),
( + + ) + + ≤
≤ ( + + ) + +↙
+ +
≤
( + + )( + + ) + + + +
www.ssmrmh.ro
then we have:
( + + ) + + ≤ ( + + ) + +
Solution 2 by Ravi Prakash-New Delhi-India
Let = , = , = , and consider
( + + )( + + )− ( + + )( + + )
= + + + + + + + + −
− + + + + ++ + + +
= ( + − − ) + ( + − − ) + ( + − − )
= ( − )( − ) + ( − )( − ) + ( − )( − ) ≥
∴ ( + + )( + + ) ≥ ( + + )( + + )
Solution 3 by Soumava Chakraborty-Kolkata-India
In any , let ( ) = , ( ) = and ( ) = , then,
( + + )( + + ) ≤ ( + + )( + + ) (1)
We shall show that for any , , , , , > (1) is true
(1) ⇔ + + + + + +
+ + + ≤ + + + + +
+ + + +
⇔ + − ( + ) + + − ( + ) +
+ + − ( + ) ≥ (2)
Now, + ≥ ( + ) (Chebyshev) ≥ ( + )
(AM ≥ GM) ⇒ + − ( + ) ≥
⇒ + − ( + ) ≥ (i)
Similarly, + − ( + ) ≥ (ii)
and, + − ( + ) ≥ (iii)
(i)+(ii)+(iii)⇒ (2) is true (Proved)
www.ssmrmh.ro
654. In the following relationship holds:
( + )≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco
( + )≥
√ ( + )= ( + ) ≥
≥(∑( + ))
=+ +
=
Solution 2 by Dimitris Kastriotis-Athens-Greece
( + ) ≥ (*)
( + )=
( + ) ( + )≥(∗) ( + )
=( + )
≥ ( + ) + ( + ) + ( + )
= ( + + )
=( + + )
=
Solution 3 by Kunihiko Chikaya-Tokyo-Japan
( + )+
( + )+
( + )≥
( ) − ( + ) = ( ) ≥ , etc.
( + )+
( + )+
( + )≥ ( + ) + ( + ) + ( + )
= ( + + )( + + ) ≥ ( ⋅ + ⋅ + ⋅ )
= ( ) =
www.ssmrmh.ro
Solution 4 by Soumava Chakraborty-Kolkata-India
=( + )
+≥
{∑( + )}
∑ +
≥∑ + ∵ + ≤
+ ≤ = =
655.
Prove that:
( ′) + ( ′) + ( ) ≥
Proposed by Abdilkadir Altintas-Afyonkarashisar-Turkey
Solution by Daniel Sitaru-Romania
( ′) = ( (∢ )) =+
=
= ≥ ∙ =
www.ssmrmh.ro
656.
Proposed by Mehmet Sahin-Ankara-Turkey
Design by Miguel Ochoa Sanchez
Solution by Daniel Sitaru-Romania
+ + + = + ( − ) = + ∙( − ) + ( + )
=
=( − ) + ( + )
≥( − )
↔ ( − − + ) ≤ ( + )
( − ) ≤⏞ ( + + )( − ) ≤ ( + ) ↔
↔ − + ≥ ↔ ( − ) ≥
657. , , : excenters
Prove that:
( ) + ( ) + ( ) ≤ √
Proposed by Mehmet Sahin-Ankara-Turkey
www.ssmrmh.ro
Design by Miguel Ochoa Sanchez-Peru
Solution by Do Huu Duc Thinh-Ho Chi Minh-Vietnam
Prove that: ⋅ + ⋅ + ⋅ ≤ √
Lemma: = + ; = + ⋅ ; = + ⋅ ;
+ + = +
Applying Cauchy – Schwarz:
⋅ + ⋅ + ⋅ ≤ ( + + ) + +
= ( − − ) + ⋅ ≤
≤ ( + + − − ) + ( + ) (Gerretsen)
= ( + )[ + ( + )] ≤ ( + ) + + =
= ⋅ = √
658. If in , , , are Gergonne’s cevians, - inradius in
then: ≥( )
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Do Huu Duc Thinh-Ho Chi Minh-Vietnam
www.ssmrmh.ro
In , , , are Gergonne’s cevians. If is the inradius of then:
≥( )
Lemma: = ⋅ ( ) ; = + − ⋅ ⋅ ( ) =
= ( − ) ( − ) = ⋅ ( − )
Similarly, = ⋅ ( − ); = ( − )
We have: = = ⋅
∑ ( )≥ ⋅
∑ ∑( ) (Cauchy – Schwarz)
= ⋅
⋅ ( )=
( )
≥( )
=( )
( )( )⇒
⇒ ≥( )
(Q.E.D.)
659. In the following relationship holds:
( + )+
( + )+
( + )≤
√
Proposed by Rovsen Pirguliyev-Sumgait-Azerbaidian
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
( + )=
⋅ ( + )⋅ = ( + )
= ⋅ ( + + ) = ( + ) + =
= ([∑ ⋅ ∑ − ( + + )] + [ (∑ ) − ( + + )]) =
= + − ⋅
= ( − − + + + ) − =
= [ ⋅ ∑ − ] = ( − + ) = ⋅ ( − + ) (*)
www.ssmrmh.ro
(*) ⇒ ∑ ( ) = ⋅ ( − + ) ≤ ( − + ) ≤
√
≤√
⋅ ⋅ ( − + ) = √ ⋅
Solution 2 by Soumava Chakraborty-Kolkata-India
=( − )
=− +
= ∑ − ( ) + ∑ =( ) (∑ ) − . Also, ≥
( )
⋅ =
(1), (2) ⇒ it suffices to prove + ≥
⇔ ( + ) ≥ + + ⇔ ≤ + − (a)
Now, LHS of (a) ≤ + + ≤?
+ −
⇔ ≥?
⇔ ≥?
→ true (Euler)
660.
Proposed by Mehmet Sahin-Ankara-Turkey
Design by Miguel Ochoa Sanchez-Peru
Solution by Daniel Sitaru-Romania
[ ] = √ + √ + √ √ + √ − √ =
www.ssmrmh.ro
= + − + √ − − + √ = − ( + − ) =
= + + − − − ≤ ( − ) ↔
+ + − − − ≤ ( + − ) ↔ − − ≤
↔ ( − ) ≥
661. In the following relationship holds:
( + )( + )( + ) ≥ +( − )−
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution 1 by Mehmet Sahin-Ankara-Turkey
( + )( + )( + ) ≥ +( − )−
⇔ ( + )( + )( + )( − ) ≥ ( − ) + ( − )
we should prove that it is − ≥
= ( + + + )( + )( − )
= ( + + + + + + + )( − )
= [ ( + ) + ( + ) + ( + ) + ]( − )
= [ ( − ) + ( − ) + ( − ) + ]( − )
= [ ( + + )− ( + + ) + ]( − )
we know it is + + = − −
where , , , are area, circumradius, inradius, semiperimeter, respectively.
= [ ⋅ ( − − ) − ( − − ) + ⋅ ]( − )
= ( + + )( − )
= ( − ) + − + −
= ( − ) − − + ; = ( − ) + ( − )
= ⋅ ( − ) + ( − ) = − + −
= − −
www.ssmrmh.ro
− = ( − ) − − + − ( − − ) ≥?
⇒ ( − ) ≥?
− − ⇔ ( − ) ≥ − −
≥ . Let’s use Gerretsen Inequality:
Gerretsen Inequality − ≤ ≤ + +
− −− ≤ ≤ + +
⇔ − − ≤ − + − + −
⇔ + − − ≥ ⇔ ( − )( − + ) ≥ as desired ∴.
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
In : ( + )( + )( + ) ≥ + ( )
1) ( + )( + )( + ) = ∑( + ) + =
= ⋅ − + = −
2) ∑ ⋅ ∑ − ≥ + ( ; ; )
⋅ − ≥ ( ; ; )
⋅ [ + + − ] = ( − + ) =
3) ( )
≤ ⋅ ⋅ ( − ) = ( − ) =
≥ (Assure) ( − + ) ≥ ( − )
− + ≥ −
≥ − ⇒ ≥ − ≥ − ; ≥ (Euler)
Solution 3 by Soumava Chakraborty-Kolkata-India
In any ,∏( + ) ≥ + ( ). Given inequality ⇔
+ ( − ) ≥ +( − )− ⇔ ≥ +
( − )−
⇔ ( + + ) ≥ + ( ) ⇔ − + ≥ ( ) (1)
Now, > ⇔ ( − ) > ⇔ > → true as
www.ssmrmh.ro
≥ (by Euler) > ∴ > → (2)
∴ of (1) ≥ − = ( − ) = ( )
≥ ( ) (using (2)) ⇒ (1) is true (proved)
662. Let , , be the sides of extouch triangle for ∆ . Prove that:
+ + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
+ + =−
= − =
= − = − =
= − ∙ = − ≥⏞ − =
663. Let be the excentral triangle of ∆ with semiperimeter . If
is the inradii of ∆ then:
√ ≥
www.ssmrmh.ro
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
[ ] = = = ≤ √
=[ ][ ] ≥
[ ]√
= ∙√
= ∙√
√ ≥ = = =
664. Let be the contact triangle of .
If = [ ], = [ ], = [ ], = [ ] then:
+ + + ≥ ⋅−
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Soumava Chakraborty-Kolkata-India
Let be the contact triangle of . If = [ ], = [ ], = [ ],
= [ ], then, + ∑ ≥( )
⋅
= = = ∴ (1) ⇔ ∑ ≥ ⋅ −
= − = ⋅ ⇔ ∑ ≥ ⋅ → (2).
Now, ∑ ≥ =[ ] [ ]
=
www.ssmrmh.ro
= =
( )⇒ (2) is true (proved)
665. In the following relationship holds:
+ + + + + ≥
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Mehmet Sahin-Ankara-Turkey
Let = + + . Using Bergström inequality
≥ ( ) = (1)
with Cauchy – Scwarz inequality ( + + ) ≤ + +
⇒ + + ≤ + + = ( + + )
⇒ + + ≤ ⋅ = (2)
Also, + ≥ ⋅ + ⋅ = , proof is completed.
Solution 2 by Soumava Chakraborty-Kolkata-India
In any ,∑ +∑
≥
+ ∑ ≥ö
∑ + ∑ = ∑
≥ ∑( + ) = ( − − ) + ( + + )
= − − ≥?
⇔ − − ≤?
⇔ ≤?
+ + → (1)
Now, LHS of (1) ≤ + + ≤?
+ +
⇔ − − ⇔ ( + )( − ) ≥ → true (Euler) (proved)
www.ssmrmh.ro
666. In ( – incircle) the following relationship holds:
≤ + + ≤
Proposed by Marian Ursarescu-Romania
Solution 1 by Hoang Le Nhat Tung-Hanoi-Vietnam
= ( + − )+ +
=+ ++ − = ( + + )
+ − ≥
≥ (∑ ) ⋅(∑ )
= ⇒ ∑ ≥ (1)
=+ ++ − ≤ =
⋅ + +=
( + + )⋅
⇔+ ++ − ≤
≤( + + )
( + + )( + − )( + − )( + − ) ⇔+ ++ − ≤
≤ ( + − )( + − )( + − ) ⇔∑( + + )[( + − )( + − )]
( + − )( + − )( + − ) ≤
≤ ( + − )( + − )( + − ) ⇔ ( + + ) − ≤ ⇔
⇔ ∑ + ≥ ∑ ( + ) ⇔ ∑ ( − )( − ) ≥ (true) (Schur) (2)
(1), (2) ⇒ ≤ + + ≤ ⇒ Q.E.D.
Solution 2 by Mehmet Sahin-Ankara-Turkey
www.ssmrmh.ro
Let is incenter of . =
| |⇒ | | =
| | = = ⋅ = ⋅( − )( − )
= ( − )( − )
= | | = [( − )( − ) + ( − )( − ) + ( − )( − )]
= ( − ⋅ + + + ) = ( − + + + ) =+
≥ ⇒ ≤ ⇒ ≤ ; ≥ ⇒ ≥ as desired.
Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
=−
=( − )
( − ) =( − )
= ( − ) = ⋅−
= ⋅∑( − )( − )
∏( − ) =
= ⋅ ∑∏( )
=⋅∏( )
= ⋅ ( ) = (*)
1) ∑ = ≤ =
2) ∑ = ≤ =
667. In the following relationship holds:
+ + ≥ ( + )
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution 1 by Mehmet Sahin-Ankara-Turkey
+ + ≥?
( ); = + + = + + ≥ ( ) (1)
+ + = + ; = ; + + = ⋅ + ⋅ + ⋅
www.ssmrmh.ro
= ⋅ − + − + −
= ⋅( − )( − ) + ( − )( − ) + ( − )( − )
( − )( − )( − )
= ⋅ (− ( + + ) + ( + + ) + )
= (− ⋅ + ⋅ ( − − ) + ⋅ ) = ( − )
= ( − ) = ( − ) (2)
From (1) and (2): ≥ ( )( )
≥( )
⇔ ( + ) ≥ ( − )
⇔ ( )( )
≥ ≥ ( − ) (Gerretsen’s inequality)
⇔ ( + ) ≥ ( − )( − )
⇔ + ⋅ + ⋅ + ≥ ( − − + )
⇔ + + + − + − ≥
⇔ − + − ≥ ⇔ − + − ≥
⇔ ( − )( − + ⋅ ) ≥ ⇔ ≥ (Euler) ∴
Solution 2 by Serban George Florin-Romania
= ⋅ ( − ) = ⋅ ⋅ ( − ) ≥ö
≥ ⋅( + + )∑ ( − ) = ∑ − ∑ = − ( − − )
≥ − + + = ( + )
∑ ≥⋅ ⋅( )
=⋅( )
q.e.d.
Solution 3 by Soumava Chakraborty-Kolkata-India
= ( − )
www.ssmrmh.ro
={∑ ( − )( − )}( − )( − )( − ) =
∑{ ( − ( + ) + )}⋅
= − ( − ) +
= − + + − ( )
= − − = ( + ) −
= {( + + )( + ) − }
= { + ( + ) } =( ) + ( + )
(1) ⇒ it suffices to prove:
+ ( + )≥ ( + ) ⇔ ( + ) + ( + ) ≥
⇔ ( − ) ≤ ( + ) (2)
Now, LHS of (2) ≤ ( − )( + + ) ≤?
( + )
⇔ − − + ≥
⇔ ( − )( + − ) ≥ → true, ∵ ≥ (Euler) (proved)
668. If in the ninepoint circle and the circumcenter are tangents then:
<√
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
Let be the nine – point center. We know that the radius of the nine-point circle =
and = −∑ (1)
www.ssmrmh.ro
If the nine – point circle and the circumcircle touch each other externally, then,
= + = = −∑ (by (1)) ⇒ ∑ = → impossible.
∴ the nine – point circle and the circumcircle must touch each other internally.
∴ = − = = − ∑ (by (1))
⇒ = = − − = ⇒ = + + ≥
≥ − ⇒ − + ≥ ⇒ − + ≥ → (2) =
Now, < √ ⇔ < √ ⇔ > √ ⇔ >
⇔ > ( + + ) ⇔ − − − >
⇔ ( + + )( − + ) + ( − ) + > → true
∵ − + ≥ (by (2)) and ≥ (Euler) (Hence proved)
669. Let ∆ be the intouch triangle of ∆ and , , medians in
∆ . Prove that:
+ + ≤
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
≤⏞ ∙ ≤
≤ ∙ + + ∙ + + ≤⏞ ∙ =
= ∙ = ∙∙ ∙[ ] =
= ∙∏( )
= ∙ ∙ ∙ = ≤⏞ ∙ =
www.ssmrmh.ro
670. In the following relationship holds:
⋅ ≤ + + ≤ −
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Soumava Chakraborty-Kolkata-India
In any , ≤ ∑ ≤ −
≥ ( − ) etc ∴ ∑ ≤ ∑ = ∑ ( )
= − − + ( − ) = − − + = − −
=∑( − )( − )
∏( − ) − = ( − ( + ) + )−
=− ( ) + + +
− =( − )
≤ −
Also, ∑ ≥∑
=∑
≥ (Done)
671. In the following relationship holds:
+ + ≤+
++
++
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution by Soumava Chakraborty-Kolkata-India
= × = ( ) etc ∴ given inequality becomes:
( + ) ≥ { ( − ) + ( − ) + ( − ) }
Let − = , − = , − = ⇒ = ∑
∴ = + , = + , = + ( , , > ). Then (1) becomes:
www.ssmrmh.ro
[{( + )( + )} {( + ) + ( + ) }] ≥
≥ ( + ) ( + ) ( + ) { ( + ) ( + ) }
⇔ + + + + +
+ + + + + +
+ + + + +
+ + + + +
+ + + +
+ + ≥( )
+
+ + + +
+ + +
= ( + ) ≥ ( + )( + )
≥ ( + ) = ( + ) ≥ ( + )( + )
≥ ∑ ( + ) = (∑ ) ⇒⇒ (∑ ) ≥ (∑ ) (a)
Again, ∑ = ∑( + ) ≥( )
∑ = (∑ )
⇒ ∑ ≥ (∑ ) (b). Also, ∑ + ∑ ≥ ∑
(c) ≥⏞( )
(∑ )
Also, (∑ + ∑ ) ≥ ∑ ≥( )
( ) (∑ )
www.ssmrmh.ro
Also, (∑ + ∑ ) ≥ ∑ ≥( )
( )(∑ )
Lastly, (∑ + ∑ ) ≥ ∑ ≥( )
( )(∑ )
Now, (∑ ) = ( ∑ )
≥( )
⋅ + ∵ ≥ +
= +
Also, ∑ + ∑ = ∑ ( + )
≥ =
= ≥( )
+
Last by, (∑ + ∑ ) = {∑ ( + )}
≥ { } { ( + )} = +
Now, (∑ + ∑ ) = {∑ ( + )} ≥ (∑ )
= ( + ) ≥ ( + )( + )
≥( )
( + ) = ∑ +
Also, (∑ + ∑ ) ≥ ∑ ( + ) +
≥ +
= + ≥( )
( )+
In the end, ∑ + (∑ + ∑ ) +
+ + +
+ + + +
www.ssmrmh.ro
≥( )
( + + + + + ) =
(a)+(b)+(c)+(d)+(e)+(f)+(g)+(h)+(i)+(k)+(l)+(m)⇒ (2) is true
672. In the following relationship holds:
−+ + ≥ −
Proposed by Vadim Mitrofanov-Kiev-Ukraine
Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia
RHS: ∑ ⋅ = ∑ ⋅ − =
= ⋅+
⋅ ≥ ⋅+
⋅−
=
= ⋅ ( + ) = ⋅ ⋅ + ⋅ = ⋅
∑ ≥ Assure ≥ ⋅ ; − ≥ ; ≥ (Euler) RHS
673. In the following relationship holds:
+ + ≤ + + ≤ + +
Proposed by Marian Ursarescu-Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India
= ⋅ =( + − )
=( + − )
=
= ( + + )( + − )( + − )( + − ) ≤
≤+ +
∵ ≥ ( + − ) = ; =( − )
=
= ∑ ≥ ( )( )( ) since, ∏ ( − ) = = =
www.ssmrmh.ro
= , we need to prove, ≥ ⇔ ≥ ⇔ ≥ , which is true
∴ ≥
Solution 2 by Soumava Chakraborty-Kolkata-India
= =⋅
= ≤ ⇔ ≤ ⇔
⇔ ≥ → true (Euler) ∴ ∑ ≤ ∑ . Now, ∑ ≥ ∏
=∑
= ≥ = ⇔ ≥ ⇔ ≥ ⇔
⇔ ≤ √ → true (Mitrinovic) ∴ ∑ ≤ ∑ (proved)
Solution 3 by Daniel Sitaru-Romania
≤ ↔ ≤ ↔ ≤
= = ⋅ = ≤∑
≤ ↔
↔ ≤ ↔ ≤√
674. If in , = then:
( − )( − )( − )+ √ + √ + √
≥√
Proposed by Daniel Sitaru – Romania
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
: = . Prove that: ∏( )∏ √
≥∏
; =∏ ⋅
= =
www.ssmrmh.ro
:∏( − )
∏ +=
∏( + )
∏ +√
=∏( + )
⋅ ∏ +√
=
= ⋅∏( )⋅∏ √
= ⋅ ⋅ ∏( )∏ √
= ⋅ ∏( )∏ √
≥ = (Assure)
( + ) ≥ + √
( + ) + ≥ √ + √ +
( + ) + ( + ) ≥ √ + √
Solution 2 by Soumava Chakraborty-Kolkata-India
= ⇒ √ =√
. Similarly, √ =√
and √ =√
∴ given inequality becomes: ( )( )( )√ √ √
≥
⇔ ( + )( + )( + ) ≥ ( + )( + )( + )
(where = √ , = √ , = √ )
⇔ ∑ + ∑ ≥ (∑ ) + ∑ → (a)
Now, ∑ ∑ ≥ ∑ = ∑ → (1)
Again, ∑ ∑ = ∑ ≥ ∑ = (∑ ) → (2)
(1)+(2)⇒ (a) is true (Proved)
675. In the following relationship holds:
√ ++√ +
+√ +
≤−
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
√ +≤ ⋅ +
www.ssmrmh.ro
≤ ⋅ = ⋅+ +
≤
≤ ⋅ = ⋅ =−
≤−
=−
Solution 2 by Soumava Chakraborty-Kolkata-India
∵ + ≥ ∴ ≤ √√
(1). Similarly, ≤( ) √
√& ≤
( )√
√
(1)+(2)+(3)⇒ ≤ ∑ √√
≤ ∑ √√
= ∑ ≤?
⇔ ( − ) ≥?
( − − ) ⇔ − ≥?
− −
⇔ ≤?
+ + (a)
LHS of (a) ≤ + + ≤?
+ +
⇔ − − ≥?
⇔ ( − )( + ) ≥?
→ true ∵ ≥ (Euler) ⇒ (a) is true (proved)
Proof 2
+ ≥( + )
⇒√ +
≤( )
√ +
Similarly, ≤( )
√ & ≤( )
√
(1)+(2)+(3)⇒ ≤ √ ∑ ≤?
⇔∑ ( + )( + )
( + )( + )( + ) ≤? −
⇔(∑ ) + ∑
+ ∑ ( − ) ≤? −
⇔(∑ ) + + (∑ −∑ )
( + + ) − ≤? −
⇔( − − ) +
( + + ) ≤? −
⇔( − − )
+ + ≤? −
www.ssmrmh.ro
⇔ ( − )( + + ) ≥?
( − − ) (a)
LHS of (a) ≥⏟( )
( − )( + + )( − )
RHS of (a) ≤( )
( − − )( + + )
(4), (5) ⇒ in order to prove (a), it suffices to prove:
( − )( − )( + + ) ≥ ( + + )( − − )
⇔ ( − − ) + ( + + + ) ≥
⇔ ( − − ) + ( + + + ) ≥ (b)
∵ − − = ( − )( + ) ≥
∴ ( − − ) ≥ ( − )( − − )
∴ ( − − ) + ( + + + ) ≥( )
≥ {( − )( − − ) + + + + }
Also, RHS of (b) ≤( )
( + + )
(6), (7) ⇒ in order to prove (b), it suffices to show:
( − )( − − ) + + + + ≥
≥ ( + + ) ⇔ − − + ≥ =
⇔ ( − ){( − )( + ) + } ≥ → true ∵ ≥ (Euler)
⇒ (b) is true (Hence proved)
676. In ∆ the following relationship holds:
−≤ + + ≤
−
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
=
( − )
=( − )
= ∙( + − )
=
www.ssmrmh.ro
=+ −
≤⏞+ + + −
=−
=+ −
≥⏞− + −
=
=−
=−
677. In ∆ the following relationship holds:
−+
−+
−≥
Proposed by Marian Ursarescu-Romania
Solution 1 by Daniel Sitaru-Romania
− = ( − ) ≥⏞(∑ )
∑ − ∑ = − ∑ ≥⏞
≥− ∑ ∑
=−
= =
Solution 2 by Soumava Chakraborty-Kolkata-India
WLOG, we may assume ≥ ≥ ∴ ≥ ≥ and
− ≥ − ≥ − ∴ − ≥ −
= ⋅∑( − )( − )
( − )( − )( − ) =− + + +
=+
≥ = (proved)
678. a. Let , , be the lengths of the sides of a triangle such that
+ + = . Prove that:
( + )( + )( + ) ≤( + )( + + + )
b. Let , , be positive real numbers such that + + = . Prove that:
www.ssmrmh.ro
+ ( + ) ≤ ( + ) ( + ) ( + )
Proposed by Do Quoc Chinh-Vietnam
Solution by proposer
a. We have:
+ + + = + + + + + + +
We will prove that + + + ≤ ; + + + ≤ .
Without loss of generality, we assume that is the number between and . Then,
we have: ( − )( − ) ≤ ⇔ + ≤ +
Applying the AM-GM inequality, we have:
+ + + ≤ + + + = ( + ) = ⋅ ( + )( + )
≤ ⋅ ( ) = . Similarly, we have: + + + ≤ . Thus, we have:
( + )( + )( + ) ≤ + − + ( − ) +
= + + ( − ) ≤ + . Therefore, we need to prove that:
+ ≤ ( )( ) ⇔ ( + + ) + ≥ ( + )
⇔ ( + + )( + + ) ≥ ( + + ) + ⇔
⇔ [ ( + ) + ( + ) + ( + )] ≥ + + +
Let = + − , = + − and = + − then
⎩⎪⎨
⎪⎧ =
=
=
( , , > )
Therefore, it suffices to show that: [∑( + )( + )( + + )] ≥
≥ ( + ) + ( + ) + ( + ) + ( + )( + )( + ) ⇔ + + + ≥
≥ ( + ) + ( + ) + ( + ). Always true because this is Schur level 3
inequality. The equality holds when = = = .
b. From part a., we have the following result:
( + )( + )( + ) ≤ + . Therefore, we need to prove that:
www.ssmrmh.ro
+ ( + ) ≤ ( + ) ⇔ ( + ) [ + ( + )] ≤
Applying the AM-GM inequality, we have:
( + )( + )[ + ( + )] ≤[ ( + ) + + ( + )]
= [ + ( + ) ] ≤
from part a., we have ( + ) ≤ . The equality holds for = = = .
679. Let , , be the lengths of the sides of a triangle. Prove that:
+ ++ +
+ ++ +
+ ++
≥ ++
++
++
Proposed by Do Quoc Chinh-Vietnam
Solution by proposer
The inequality needs to prove to be equivalent to:
( + ) + ( + )+ +
( + ) + ( + )+ +
( + ) + ( + )+ ≥
≥ + + + . We will prove that: ( ) + ( ) ≥ + . Applying the
Cauchy-Schwarz and AM-GM inequality, we have:
( + )+ +
( + )+ =
( + )√( + ) +
( + )√( + )
≥( + )√ + ( + )√
( + ) + ( + )
=( + )√ + ( + )√
( + ) + ( + ) =( + )√ + ( + )√
( + − ) + ( + )
≥( + )√ + ( + )√
( + ) ( + − ) + ( + )=
( + )√ + ( + )√( + ) ( + + )
www.ssmrmh.ro
Therefore, we need to prove that:
( + )√ + ( + )√( + ) ( + + ) ≥ + + ⇔ ( + )√ + ( + )√ ≥
≥ ( + + ) ( + ) which is equivalent to:
( + )√ + ( + )√ − ( + + ) √ + √ ≥ ( + + ) ( + )− √ −√
⇔ ( − )√ + ( − )√ ≥( + + ) ( + ) − √ + √
( + ) + √ + √
⇔ √ − √ √ + √ ≥( + + ) √ − √
( + ) + √ + √
Therefore, we need to prove that: √ + √ ( + ) + √ + √ ≥ + +
By the Cauchy-Schwaraz inequality, we have:
√ + √ ( + ) + √ + √ ≥ √ + √ = + + √ > ( + ) >
> + + . Similarly, we have: ( ) + ( ) ≥ +
( ) + ( ) ≥ + . Therefore, we have:
( + ) + ( + )+ +
( + ) + ( + )+ +
( + ) + ( + )+ ≥
≥ + + + . The equality holds for = = .
Remark.
From the proof the above inequality, we have the following inequality:
( + )+ +
( + )+ +
( + )+ ≥
680. In the following relationship holds:
+ +≤
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
= ≤ ( − ); =
+ + =+ +
≤ ≤ ( − )( − )( − ) = ⋅ =
= ; LHS ≤ ≤ ⋅ ; ≤ ⇔ ≤ ⇔ ≤ √
≤√
⋅ ⋅ =√
Solution 2 by Soumitra Mandal-Chandar Nagore-India
We know, √ ≤ ≤ √ and ≥
∑ ≤ = ∏ ( + ) ⋅ ( − )
= ∏ ( + ) ⋅ =⋅
( + + )( + + )−
= ( + + ) − = + + ≤⋅ ⋅+ +
= ( + ) = + ≤⋅
+ =
681. If in acute , - centroid, - incentre, - orthocenter then:
( + ) + ≥
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
For acute-angled , ≤( )
( + ) etc. Also, ∑ = ∑ =
www.ssmrmh.ro
= ∏ = ⋅ =( )
. Moreover, ∑ = + ( − − ) =( )
= − − ; ≤∑
≤( )
≤ ( + ) = + =( )
= + ( − ) = + −∑
=
= + −( − − )
=+ − ( − − )
=+ + −
=( + + − )
∴ ≤( ) ( + + − )
=− +
+ + + − +
=( − + + + + − + )
=( )
; (a), (b) ⇒ it suffices to prove: + + − ≥
≥ + + − ⇔ ≤ − + ⇔
≤ − + (i)
Now, LHS of (i) ≤ + + ≤? − + ⇔
− + ≥?
⇔ ( − )( − ) ≥?
→ true ∵ ≥ (Euler) (proved)
682. In the following relationship holds:
+ ( − )( − ) ≥ √
Proposed by Daniel Sitaru – Romania
Solution 1 by Mihalcea Andrei Stefan-Romania
LHS ≥√∑( − )( − ) + =
√∑ ( − − ) =
√( − ) =
www.ssmrmh.ro
= √ . We will prove: √ ≥ √ ⇔ ≥ . By AM-GM ⇒
⇒ ( − )( − )( − ) ≤ = ⇒ ≤√
(1)
≥ √ . We will prove: ⋅ √ ≥ ⇔ ≥ √ ⇔ ≥ √ ⋅ ⇔
⇔ ≥ √ ⇔ (1) q.e.d.
Solution 2 by Myagmagsuren Yadamsuren-Darkhan-Mongolia
⋅ + ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅
= ⋅ + ≥ ⋅√
⋅ ⋅ ( + ) = √ ⋅ ( + + )
= √ ⋅ ( − )( − ) = √ ⋅ ⋅( − )⋅ ∏( − ) = √ ⋅ ≥ √ ⋅
Solution 3 by Soumava Chakraborty-Kolkata-India
+ ( − )( − ) ≥√
( + ) ( − )( − )
=√ − + −
( − )( − ) =√
( − + − )
=√∑ = √ ≥
√√ (proved)
683.
In ∆ the following relationship holds:
−≤ + + ≤
−
Proposed by Mehmet Sahin-Ankara-Turkey
www.ssmrmh.ro
Solution by Daniel Sitaru-Romania
= − ∙ = − = ∙( − )
=−
+ + =−
≤−
↔ ≤ ( − ) ↔ ≤
+ + =−
≥−
↔ ≥
684.
In the following relationship holds:
≤ + + + + + ≤−
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Mehmet Sahin-Ankara-Turkey
+ ≥ ⇒ + ≤
= + + + + + ≤ + + =+ +
+ + = − − (well known)
≤+ +
=− −
⋅ =− −
=− −
Using Gerretsen Inequality we get
≤ = = proof is completed.
www.ssmrmh.ro
Solution 2 by Soumava Chakraborty-Kolkata-India
+ ≤−
⇔ + + ≤+
⇔ + ≥+
⇔ ( + ) ( + ) ≥ ( + )( + )
⇔ ( + ) ( + ) ≥ +
⇔ ( + ){∏( + )} ≥ (∑ ){(∑ ) + ∑ } (1)
LHS of (1) ≥ ( + ) ⋅ (∑ )(∑ ) ≥?
(∑ )((∑ ) + ∑ )
⇔ ( + ) ≥?
+
⇔ ( + ) ≥?
⇔ ( + ) ≥?
+ ( + )
⇔ ( + ){ + ( + ) + ( + ) }
≥?
{ + ( + ) − ( + )} + ( + )
⇔ ( − ) + ( + )( + ) + ( + ) ( − ) ≥?
≥ + ( + )
LHS of (2) ≥ ( − ) ( − ) +
+ ( + )( + ) + ( + ) ( − )
RHS of (2) ≤ ( + + ) + ( + )
∴ it suffices to prove: ( − )( − ) + ( + )( + ) −
( + + ) − ( + ) +
+ ( + ) ( − ) ≥
⇔ ( − + ) + ( + ) ( − ) ≥ (3)
www.ssmrmh.ro
∵ − + > 0
∴ LHS of (3) ≥ ( − )( − + ) +
+ ( + ) ( − ) ≥?
⇔ − + − ≥?
=
⇔ ( − )( − + ) ≥?
⇔ ( − ){( − )( − ) + } ≥?
→ true (Euler) ∵ ≥
∴ ∑ ≤
,∑ ≥ (proved)
685.
In acute the following relationship holds:
++
++
+<
− −
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
In any acute – angled ,∑ <
− −=
( − − )=∑
=(∑ )
=
∴ given inequality ⇔ ∑ < ∑ → (1)
Now, < ⇔ < + ⇔ ( − ) <
⇔ < ⇔ > ⇔ > → which is true
www.ssmrmh.ro
∀ ∈ , . Hence, ∵ is acute – angled, ∴ < → (a)
Similarly, < → (b) and < → (c)
(a)+(b)+(c)⇒ (1) is true (Hence proved)
686.
In the following relationship holds:
( + ) + ( + ) + ( + ) ≤
Proposed by Marian Ursarescu-Romania
Solution 1 by Soumava Chakraborty-Kolkata-India
LHS ≤ ∑( + ) ∑ = ∑ + ∑ ∑
= ( − − )( − − ) ≤?
⇔ ( − − )( − − ) ≤?
⇔ − ( + ) + ( + ) ≤?
⇔ + ( + ) ≤( )
?+ ( + )
Now, LHS of (1) ≤ ( + + ) + ( + )
≤?
+ ( + ) ⇔ ( − ) ≥( )
?( + )
Now, LHS of (2) ≥ ( − )( − ) ≥?
( + )
⇔ − + ≥?
⇔ ( − )( − ) ≥?
→ true
⇒ (2) is true (proved)
www.ssmrmh.ro
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
( + ) + ( + ) + ( + ) ≤
≤ ( + ) ⋅ , ≤ ⋯, ≤ ⋯
( + ) ⋅ ≤ ( + )( + ) = ( + ) + ( + ) ⋅ =
= + ( + ) ⋅ = + ⋅ ( + )+ −
=
= + ⋅ ( + ) ( + − ) =
= +[ − + − + + ] +
+[ − + − + + ] ++[ − + − + + ]
=
= + ⋅ ( + + ) =
687.
In the following relationship holds:
( + )( + )( + )≥
+ ++ +
Proposed by Adil Abdullayev-Baku-Azerbaidian
Solution 1 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
:∏ ( + )⋅ ⋅ ⋅ ≥
∑∑ ⋅
1) ∏( )⋅ ⋅ ⋅
= ∑ ⋅∑ = ( ) = (*)
www.ssmrmh.ro
2) ≤ ≤ ( − )
≤ ≤ ( − )
≤ ≤ ( − )
=
=
=
∑∑ ⋅
≤ ∑ ( )∑ ⋅
=⋅ ∑
= ⋅⋅
= ⋅ ⋅⋅ ⋅
= (*)
=( )
≥( )
Solution 2 by Soumava Chakraborty-Kolkata-India
∏( + )=
+ ∑ ( + − )=
( + ) −
=( )
∴ (1) ⇔( )
∑ ≥( )
∑
Now, ∑ ≥( )
= ∑ = ∑ ⋅ = ∑( ) = ∑ =
(b) ⇒ in order to prove (2), it suffices to prove: ≥ ∑ ⇔ ∑ ≤( )
Now, ≤ ( − ) etc ⇒ ∑ ≤ ( − ) = ⇒ (3) is true (Proved)
688.
If in , - Lemoine’s point then:
√ + + ≤+ +
Proposed by Daniel Sitaru – Romania
Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia
www.ssmrmh.ro
1) =( ) ⋅
∑; … , …
) ∑ ≥ √ ( ) true
√ ⋅ =( )
√ ⋅⋅
∑ ⋅ = √ ⋅∑∑ ≤
( ) √ ⋅ ∑√ ⋅
=∑
689.
In the following relationship holds:
√ ⋅ ≤ + + ≤√
⋅
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Daniel Sitaru-Romania
≤ + + ≤ ( . − − )
√ ≤⏞ = ≤⏞ ∙ = ∙ =
= ∙ ∙ = ∙ ≤⏞ ∙ =
= ≤⏞ ∙√
=√
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
www.ssmrmh.ro
∑ = ∑ ⋅ = = ; √ ≤ ≤ √ ⋅ ; √ ≤ ≤ √
) = ⋅ ⋅ ≥√
√
) = ⋅ ⋅ ≤ ⋅ ≤√
√⋅
⎭⎪⎬
⎪⎫
690.
In ∆ the following relationship holds:
≤√√
Proposed by Nguyen Van Nho-Nghe An-Vietnam
Solution by Daniel Sitaru-Romania
≤√√
↔ √ ≤ √ ↔ √ ≤ √ ↔ √ ≤
≥⏞ ∙ ∙ ≥⏞ ∙ √ = √
691. In the following relationship holds:
www.ssmrmh.ro
+ + ++ +
≥+ +
+ +
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution by Soumava Chakraborty-Kolkata-India
In any ,∑ +∑
≥∑
∑
(∑ )(∑ ) +(∑ ) ≥
∑(∑ )
⇔ + ≥
⇔ + +
≥
⇔ + ≥
We shall now prove: (∑ )(∑ ) + (∑ ) ≥ (∑ ) (2)
⇔ ⋅ ( ) ≥ −
⇔ ≥ ( + + ){ ( + + ) − ( − − )}
⇔ ≥ ( + + )( + ) ⇔ ≥ ( + + )( + )
⇔ ( − ) ≥ ( + ) (3)
of (3) ≥ ( − )( − ) ≥?
( + )
⇔ − + ≥?
⇔ ( − )( − ) ≥?
→ true ⇒ (2) is true
Let us consider a triangle with sides , , and apply (2) on it we shall
have:
⋅ + ⋅ ⋅ ≥ ⋅
⇔ (∑ )(∑ ) + (∑ ) ≥ (∑ ) ⇒ (2) is true (proved)
www.ssmrmh.ro
692.
In ∆ the following relationship holds:
+ +≥
Proposed by Nguyen Van Nho-Nghe An-Vietnam
Solution by Daniel Sitaru-Romania
≥+
↔ ≥+
↔
( + )−≥ ( + )
( + ) −
↔ ( − ) ≥ ( − ) ↔ ( − ) ( − + )( + − ) ≥
www.ssmrmh.ro
≥+
≥⏞ ∙ = ∙ =
↔∏
≥ ↔ + + ≥
693.
In the following relationship holds:
( + + ) ( + + ) ≥
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
Given inequality ⇔ ∑ + ∑ + ∑ + ∑ + ∑ ≥
Now, ∑ + ∑ + ∑ = ∑( + ) + ∑ ( + )
= ( + )( + ) ≥ √ √ =
www.ssmrmh.ro
⇒ ( ∑ + ∑ + ∑ ) + ∑ + ∑ ≥ ∑ + ∑ ≥ ∑
≥ ⋅ √ ≥√
(proved)
694.
In the following relationship holds:
≤ + + ≤− +
Proposed by Marin Chirciu-Romania
Solution 1 by Soumava Chakraborty-Kolkata-India
In any , ≤( )∑ ≤
( ); ∑ = ∑ = ∑
=( − )
( − ) ( − ) =∑( − )
{ ( − )( − )( − )} =∑( − )
=( ) ∑( − )
( − ) = ( − + ) = + + − − +
= + + − −
∴ ( − ) =( )
+ + − ( )−
Now, ∑ = (∑ ) − ∑
www.ssmrmh.ro
= − − ( ) =( )
− +
Also, − (∑ ) = − ( + ∑ − ∑ )
=( )− − +
(3), (4) ⇒ ∑( − ) = + (∑ ) + (∑ )
− + − − + −
= − − + − − +
= − + − + − −
= − + (− )( + ) + ( − − ) −
= − − − − + + + + − − −
=( )
+ + + −
(1), (5) ⇒ ∑ =( )
∴ (ii) ⇔ ≤ ( − + ) (from (6))
⇔ + + + ≤( )
( + + )
Now, LHS of (iia)
≤ ( + + ) + + + ≤(?)
( + + ) ⇔
( + ) ≥( )
( + + )
Now, LHS of (iib) ≥ ( − )( + ) ≥(?)
( + + )
⇔ ≥ ⇔ ≥ → true (Euler)⇒ (ii) is true
(6) ⇒ (i) ⇔ + + + − ≥( )
LHS of (ia) ≥ ( − ) + + + − ≥(?)
⇔ ≤( )
+ +
www.ssmrmh.ro
Now, LHS of (ib) ≤ ( + + ) ≤(?)
+ +
⇔ − − ≥ ⇔ ( + )( − ) ≥ → true ∵ ≥ (Euler)
⇒ (i) is true (Done)
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
In : ≤ ∑ ≤
I) LHS: ⇒ = ; = ⇒ ∑ = ∑ ( ) = ∑ ( ) =
= ⋅ ∑ ( ) = ⋅ ∑ ( ) ≥ ⋅⋅
= = ⋅ ≥ (LHS)
II) RHS: ∑ ≤
= (I) ; ⋅ = − (II)
⋅ ⋅ =( );( ) −
⋅ ( − ) =
= − ( − ) = + − + −
= + − + −∑
=
= + ( − − ) − +∑
−⋅ ( − − )
=
= − + − − − + + +∑
=
= − + + ∑ = − + + =
= − − + + +
= − + + − − ++ +
=
= − ++ +
≤ − +
www.ssmrmh.ro
+ + − +( + ⋯ )
≤ − +
+ +≤ + ⇔ ( + + ) ≤ ( + )
( + + ) ≤ ( − )( − )
( + + ) ≤ ( − )( + )
+ + ≤ + − ⇒ ≥ (Euler)
695.
In any triangle the following relationship holds:
−=
+ ( + )≥ { , ( + )√ }
Proposed by Lelia Nicula-Romania
Solution 1 by Daniel Sitaru-Romania
+ ( + )≥ ↔ + ( + ) ≥ ↔ ( + ) ≥ ( )
+ ( + )≥⏞
( + ) + ( + )≥ ( + )√ ↔
↔ + + ≤ ( + )√ ↔ ( + )√ ≤ + +
( + )√ ≤⏞ ( + ) ≤ + + ↔ − − ≥ ↔
↔ − − ≥ ↔ ( − )( + ) ≥
Solution 2 by Mehmet Sahin-Ankara-Turkey
Let = + + ; = ( )( ) ( )( ) ( )( )( )( )( )
www.ssmrmh.ro
=(− + + ) + (− + + ) + (− + + )
= ( ) ( ) ( ) ( ) (1)
+ + = + +
( ) + ( ) + ( ) = + + + + + − (2)
From (1) and (2): =
= .
=+ + +
=( + + + )
= ( ) (3). Let , ( + )√ = (4). From (3) and (4):
+ ( + )≥ ⇒ ≤ + +
⇒ ≤ . Using the Gerretsen Inequality. − ≤
− ≤ + + ⇔ − + ≥
⇔ − + ≥ ⇔ ( − )( − ) ≥ ⇔ ≥ (Euler inequality)
Let , ( + )√ = ( + )√ (5)
From (3) and (5): ( ) ≥ ( + )√ ⇔ + + + ≥ √ ( + )
Using the Gerretsen inequality
− + + + ≥ √ ( + ) ⇔+ −√ ( + )
≥ ≥ √
⇔ + − ≥ ( + ) ⇔ − − ≥ ⇔ ( − )( + ) ≥
⇔ ≥ (Euler inequality). The proof is completed.
696.
www.ssmrmh.ro
In the following relationship holds:
+ +
+ +≤
Proposed by Marian Ursarescu-Romania
Solution 1 by Soumava Chakraborty-Kolkata-India
In any ,∑∑
≥ . We shall first prove: ∑∑
≥( )
(∑ )
(1) ⇔ ⋅∑
≥(∑ )
⇔ (∑ ) ≥( )
(∑ )
Let us consider a with sides , , .
Medians of are , , and its area =
Semi-perimeter of = ∑ and its inradius =∑
Applying ≥ √ on this triangle, we get: ∑ ≥ √ ∑
⇒ ≥ √ ⇒ ≥( )
√
(3) ⇒ in order to prove (2), it suffices to prove:
√ ≥ ⇔ ≥ + ( ) = +
⇔ + ≤( )
Now, LHS of (4) ≤ + ≤?
⇔ ≥?
⇔ ≥?
→ true ⇒ (4) is true ⇒ (2) is true. Applying (2) on the , we get:
www.ssmrmh.ro
∑
∑≥
∑⇒ ⋅ ∑
∑≥ ⇒ ∑
∑≥ = (Proved)
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
≥ ; ∑ ⋅ ∑ ≥
∑ ∑
∑≥ ⇔
∑
∑≥
∑≥
∑= =
697.
In the following relationship holds:
+ + ≥
Proposed by Marian Ursarescu-Romania
Solution 1 by Soumava Chakraborty-Kolkata-India
In ,∑ ≥ ; ∑ =⋅
= ⋅ = ⋅
= ⋅ = ≥ ⋅∑
= (∑( + )) = + + = ( ) ≥ = (proved)
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
www.ssmrmh.ro
= + ; … …
= ⋅ + = ⋅ + ⇒
∑ = ⋅ ∑ ⋅ ∑ − (*);
) ∑ =
) ∑ =
) = ⎭⎪⎬
⎪⎫
(**)
(*), (**) ⇒ ∑ = ⋅ ⋅ − = − = − ≥
Solution 3 by Soumitra Mandal-Chandar Nagore-India
≥ − and ≥
= ≥⏞∑
∑=
∑ −
∑=
∑
we need to prove, ≥ ⋅ ⇔ ≥ ⋅
⇔ ≥ ( + + ) ⇔ ≥ ( + + ) since, ≥
⇔ ≥ ( + ) we will prove, − ≥ ( + )
⇔ ( − ) ≥ , which is true. Hence proved.
698.
If in : = then: + ≥
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
Solution by Soumava Chakraborty-Kolkata-India
∵ = , ∴ given inequality ⇔ + ≥ ( )
⇔ − + ≥ ⇔ ( − )( − ) ≥ → true ∵ ≥ (Euler)
699.
If in , ≤ , , ≤ then:
( + + ) + + < ( + )
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
= ⋅ ⋅ ⋅ = ⋅
= = ( − ) = − = − −
=−
=( )
( − )
=⋅ ⋅
= = ( + )
www.ssmrmh.ro
= + + = ⋅( + )
=( ) ( + )
(1) × (2) ⇒ = ( − )( + ) > ( + ) ⇔ ( − ) > ( + )
⇔ > → true ∵ ≥ >
700. In any acute the following relationship holds:
− , − , − ≥
Proposed by Marian Ursarescu-Romania
Solution 1 by Soumava Chakraborty-Kolkata-India
WLOG, we may assume , , =
Now, ≥ ⇔ ≥ ⇔ ≥ ⇔
⇔ + − − ≥ + − −
(∵ + , + > )
⇔ ( − ) − ( + )( − ) − ( − ) ≥ ⇔
⇔ ( − ) − ≥ ⇔ ( − ) − − ≥ ⇔
⇔ − ≤ ⇔ ≤ ⇔ ≥ (1) (∵ , are acute angles)
Similarly, ≥ ⇔ ≥ (2)
(1), (2) ⇒ ≥ ⇒ ≥ ⇒ ≥ (3)
www.ssmrmh.ro
∵ ≥ , ∴ it suffices to prove: ≥ (a)
∵ < , < ⇒ − < < ⇒ < ≤ (4)
(4) ⇒ ≥ ≥( )
⇒ (a) is true (Proved)
Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia
−= ⋅ + ⋅ = ⋅ +
−⋅ =
+⋅
Similary = ⋅
= ⋅
() = + ; + ; +
≥ ≥⇓
⇔ + ; + ; + = +
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
top related