RMM - Triangle Marathon 601 - 700 · RMM - Triangle Marathon 601 - 700. RMM TRIANGLE MARATHON 601...

ROMANIAN MATHEMATICAL MAGAZINE

Founding EditorDANIEL SITARU

Available onlinewww.ssmrmh.ro

ISSN-L 2501-0099

RMM - Triangle Marathon 601 - 700

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RMM

TRIANGLE

MARATHON

601 – 700

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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


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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


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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

= + + −

− → = → + + − = →


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→ = + +

=− ( + )

=+ + − − −

=

603. In ∆ the following relationship holds:

−= + +

Proposed by Abdilkadir Altintas-Afyonkarashisar-Turkey,

Designed by Miguel Ochoa Sanchez-Peru


=⏞ ∙∑∑ −

− =− ∙ ∑

∑

= ↔

↔ −∑∑ = ↔

( )∙∑( ) −

∑∑ = ↔

− = ; − =

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

= − ( − − − ) = =


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604. If in ∆ the orthocentre lies on the incircle then:

+ + =


Design by Miguel Ochoa Sanchez-Peru


= + + − = → = + +

= − ( + ) = + + − − − = =

605. In the following relationship holds:

− =−


Solution 1 by Mehmet Sahin-Ankara-Turkey

−= − = ⋅ −


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=( − )( − )

⋅( − )

−( − )( − )

⋅( − )

=−

⋅( − )

−−

⋅( − )

=( − )

⋅−

−−

=( − )

⋅−

=−

= ⋅ (New equality). Let = ⋅ ⋅

=( − )

⋅−

⋅( − )

= ⋅( − )

⋅−

= ⋅ ( − )( − )

= ⋅ ( − )( − ) = ⋅ ( − )( − )

= ( )( ) = ( )( ). Let = −

= − ⇒ = − − − =− − ( − )

( − )( − )

= ⋅( − )

( − )( − )( − )( − ) = ⋅

( − ) ⋅ ⋅ ( − )( − )( − )( − )

= ( )⋅ ( ) = ( )⋅ ( ) , = ( )( ). = as desired ∴

Solution 2 by Soumava Chakraborty-Kolkata-India

− = −

= − =−

= ⋅−

= ⋅−


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= ∑ ⋅−

= + + ⋅−

=⋅

⋅−

= (proved)

606. If , , − sides in∆ ,√ ,√ ,√ − sides in ∆ ′ ′ ′ then :

√′

+√

′+

√′

= ( + + )

Proposed by Mehmet Sahin-Ankara-Turkey


√′ =

√+ − =

√( − ) =

=√ ( − )

= √ ∙∑( − )( − )

( − )( − )( − ) =

=√

∙ ( + ) =√ ( + ) = ( + + )

607. In ∆ , , , - medians in Gergonne’s triangle. Prove that:

+ + = ( + + )



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Let ∆ be the Gergonne’s triangle:

+ + = = ( ( − ) − ( − ) ) =

= ( − ) ( − ) = ( − ) =( − )( − )( − )

=

= ( − ) = ( − + + ) =

= ∙ ( + ) = ∙ ( + ) = ( + + )

608. Let ∆ ,∆ be the contact respectively the excentral triangle of

∆ . If = [ ], = [ ], = [ ], = [ ],

= [ ] then: = , =


Design from http://cut-the-knot.org/


= ∙ ∙ =( − )

=( − )

, =

= ∙( − ) ( − ) ( − )

=


http://cut-the-knot.org/

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= ( ) ∙ ∙∙

= ∙ = ; = ∙ =

609. If in ∆ , , , − altitudes of ∆ , where

, , -Gergonne’s cevians then: + + = −



=

[ ]

=[ ]

∙ = ∙ =

= ( − ) ∙ =∙

( − ) ∙ = ( − )

= ( − ) = − = ∙( − )

=−

= −

610. Let , , , , , be the altitudes respectively the medians of

intouch triangle in∆ . Prove that:


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+ +

+ += ∙

+ ++ +



Let ∆ be the intouch triangle = [ ] = .

+ +

+ +=∑

∑= ∙

∑∑ =

= ∙∑

∑= ∙

+ ++ +

= ∙+ ++ +

611.

If in ∆ , − incentre, ⊥ , ⊥ , ⊥ then:


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[ ]+

[ ]+

[ ]≥

( − )



[ ] = = = ∙+ +

≥

≥⏞ ∙− + +

=−

=( − )

612. Prove that:

+ + = + + + +


Solution 1 by Do Huu Duc Thinh-Ho Chi Minh-Vietnam

In homogeneous barycentric coordinates: ( : : ), ( : : ), ( : : ). The

Brocard point has coordinates , , , so barycentric coordinates are:


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∑ ,∑ ,∑

If ⃗ = ( , , ) in normalized barycentric coordinates then:

= − − − . Using that formula we have:

= + + ⇒ =+ +

Similarly: = ; =

⇒ + + =+ +

⋅ =

= + + + + (Q.E.D.)


∠ = ° − ( − + ) = ° −

In , =( ° )

= ⇒ =( )

∵ = ,∴ =( )

∴ = ⋅∑

=∑

(using (1), (2))⇒ =∑

⇒ = ∑ (a)

Similarly, = ∑ (b) and, = ∑ (c)

(a)+(b)+(c) ⇒ = ∑ + + = ∑ + +


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613.

In ∆ the following relationship holds:

+ + + + + =

Proposed by Theodoros Sampas-Greece


+ =( + )( + − )

=

=+ + − −

=

= + ∙ + − − =


+ + ≥

Proposed by Daniel Sitaru-Romania


≥ ≥ → ≥ ≥

+ + ≥⏞ ∙ = ( + ) ≥

≥⏞ ∙ ≥⏞ ∙ = ≥⏞ ∙ =


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Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia

+ + ≥⏞ =

=( ) ∙

( − )( − )( − ) =( )

= ≥

≥⏞ ∙ √ = ∙ √ ∙ = √ ≥⏞ ∙ =

Solution 3 by Soumitra Mandal-Chandar Nagore-India

= ≥⏞(∑ )

∑=∑ − =

=∙

= ≥⏞ ∙ =

615. If in , , , - cevians, ∩ ∩ = { } then:

+ + ≥ , + + ≥


Solution by Mehmet Sahin-Ankara-Turkey

(Proof of first inequality)

Euler-Gergonne Theorem: In triangle let = , = , = , then


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+ + = .

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:

√ ∙ ≥√ √ + √ + √

√ + √ + √


Solution by Soumava Chakraborty-Kolkata-India

If , , > then:

+ + ≥⏞ (1)


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For = √ , = √ , = √ in (1):

√ ∙ ≥√ √ + √ + √

√ + √ + √


≥ +


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)


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+ − = >+ − = >+ − = >

⇒

=+

=+

=+

⎭⎪⎬

⎪⎫

(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


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Solution by Soumitra Mandal-Chandar Nagore-India

In , we have and similar triangles. So,

= ⇒ = = ⇒ ( ) = ,

similarly = , = . Now, ∑ ≤ ∑ = ∑

= ∵ = = = =

=( + + )

=( + + )

⋅ =+ +

≥ =∏

= =

620. In acute the following relationship holds:

√ + + + ≥ + +



Define ( ) = − function.


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This function is convexe in ( , ). Using the Jensen’s inequality we get,

≤ [ ( ) + ( ) + ( )] ⇒ ( ) + ( ) + ( ) ≥ ⋅

≥ − ≥ − ≥ − = −√

⇒ √ + + + ≥ + +


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 ,

∴√

+−

−+

≥√

+−

−+


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(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 ( ) = ( ) , < <

( ) =( + ) − ( )

( + )

= ( + ) [ ( + ) − ( − )]

= ( + ) [( )( + ) − ]


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= ( + ) [( − )( + ) − ]

= ( + ) [ + − − − ]

= ( + ) [ − − ] = +( − − )

= − + +

( ) = − −( )

= − +( )

< , for < <

∴ is concave function. Thus, ∑ ( ) = ∑ ≤ = ⇒

⇒ ∑ ≤ = √ . Hence + + ≤ √


( + − ) + ≥ √



∑ ( + − ) = ∑ ( − ) , where + + =

let = − , = − , = − then ∑ ( − )( − ) = ( + ),

( − ) =

= − + = − ⋅ ( + ) +

= − , so, ∑ ( + + ) + = ∑ ( − ) +

= ( − ) + = ≥ √ ∵ ≥ √ ≥ √ (proved)


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1. ∑( + − ) ≥ ⋅ (∑( + − )) = ⋅ ( + + ) = ⋅

2. ∑( + − ) + ≥ + = + ≥⏞

√

≥ ⋅ √ ⋅ ⋅ + = ⋅ √ = √


In any ,∑( + − ) + ≥ √

= ( − ) + ( ) ≥ ( − ) + = ( + )

≥ √ ( ) (∵ ≥ √ and ≥ ) = √ (proved)

623. If ∈ ( ) then:

+ + ≥+ +



+ + ≥ ( + + ) (1)

= + + = + + = ⇔ = ≥ > >> > Chebysev


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≥( + + )( + + )

⇔ ≥ ( + + )

⋅≥

( )⇔ ( + + ) ≥ (2)

(1); (2) ⇒ + + ≥ = (*)

2) RHS: = …

= = ; ∑ ⋅

=∑ ( )

=

= ∑ − ∑ + + + ≤ − − ( + ) + =

=( )

=( )

=( )

= ≤ = = (LHS)

≥ ≥∑ ⋅


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)


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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)


°+

° °+

°>


Solution by Soumitra Mandal-Chandar Nagore-India

= ° = ( ° + °) =° + °

− ° ° →

→ = ° + ° ° + °

° + ° ° + ° >⏞( + + )

° + ° ° + ° =

= ( + + ) ≥⏞ √ =


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625. In ∆ the following relationshiop holds:

+ + −+ +

≤√

( + + )


Solution by Boris Colakovic-Belgrade-Serbia

+ + − + + =+ +

−+ +

+ + =

=( + + ) − ( + + )

( + + ) ≥( + + )− ( + +

( + + ) =

=( + + )

( + + ) ≥( + + )

( + + ) = + + ≤√

( + + )↔

( + + ) ≤ ( + + ) ; ≤ + +

≤ + + = + + ↔ ≥ − ( )

≥⏞ − ≥ − ; ≥ ↔ ≥

626. If in acute , , , – altitudes, - orthocentre, -

circumcentre then: + + ≥

Proposed by Mehmet Șahin – Ankara – Turkey

Solution by proposer


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| | = ⋅ ; = − ⋅ +

= − + ⋅ ; = − +

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


=+

, =+ −

∑ = ∑ ≤( )

,(1) (C. Mateescu – 2016)


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= ≥⏞( ) (∑ )

∑ =+

+ − =

=( + )

+ − ≥ ( + ) ≥ ( ) ↔

↔ ( + − ) ≤ ( + ) ∙ ( + )( )

( + − ) ≤⏞ ( + + + − ) =

= ∙ ( + ) ≤ ( + ) ∙ ( + ) ↔ ≤ ( + )( + )

≤⏞ + + ≤ + + ↔ ≥


+ + + + + ≥ √

Proposed by George Apostolopoulos-Messolonghi-Greece


+ = + ≥√

√ ≥⏞

=√

∙ √ =√

∙ √ ≥⏞√

∙ ≥⏞

≥√

√ =√ ∙ √

=√

≥ √ ∙ ↔

≥ ↔ ≥


+ ≥⏞√

( + ) = √ =


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= √ = √ ∙ ≥⏞ √ ∙√

= √ ∙ ∙ ≥⏞

≥ √ ∙ ∙ = √


≤



= and = . It is known that ∑ = +

( ) ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ≤

⇔ ( ) ⋅( )

⋅ ≤ ( )⋅

⇔ ≥ (?) + + = +

+ + = + ⇒ + + = +

⇒ + + ≤ + =

≤ + + ≤ ⇔ ≥ ⇔ ≥

Solution 2 by Rozeta Atanasova-Skopje

If < , , < then ≤ ≤ = ,


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then = ( )

≤ = ⋅ ⋅ =


∵ in any acute-angled , , , > ,∴ ≥ ⇒

≤ ( + + ) = +

= ∵ ≥ ⇒ ≤ ⇒ ≤ (1)

Also, ∵ is acute, ∴ < , , <

and hence, >

= ( )( )( )( )

= ( ) ≤ (using (1)) = = (proved)


+ ≥ + + ≥( − )

Proposed by Vadim Mitrofanov-Kiev-Ukraine


≤⏞ =( − )

= ∙∑ − ∑

=

=( − + + )

= ( + ) = +

≥⏞( − )( − )( − )

( ) = =

= ≥ ( − ) ( ) ↔ ( − ) ≥


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( − ) ≥⏞ − = ∙ =

= ∙ ≥⏞ ∙ =

631. If in ∆ , − ninepoint center then:

a. =

b. + + =

c. + + ≤



a. ∆ , − → = ( + ) − =

= ( + ) − ( − − − ) = − − + =

= − ++ +

= − ++ +

=+ + −

b. + + = + + =

c. + + = =

= ≤⏞ =

632. PAPACU’S INEQUALITY

In nonacute :

−,−

,−

≤

Proposed by Nicolae Papacu-Romania


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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:

≥



= ⇒( − )( − )

⋅( − )( − )

=


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⇒−

= ⇒ = = ⇒ =( − )

⇒ = ⋅ ⋅ ⇒ = . We need to prove,

≥ ⇔ ≥ √ . We know, = + + ⇒ = + ⇒ ≥ √

∴ ≥ (proved)


⋅ ⋅ = and = ⇒ ⋅ ⋅ = (1)

⋅ = ⇒ − ⋅ − = ⇒ = ( − )( − )

⇒ ( − ) ⋅ = ( − )( − )( − )

⇒ ( − ) ⋅ = = ⇒ − = ⇒ = (2)

⋅ = ⇒ = ⋅ ⇒ = ⋅ = ⋅( − )

From (2) = ⋅ = ⋅ =

⋅ =⋅

= =

⋅ = ⋅ ⋅ = ⋅

From (2) ⋅ = ⋅ = ⋅ (3)

= ⇒ = ⇒ =( )

(4)

+ ≥ √ ⇒ ≥ √ ⇒ ≤ ⇒ ≤

and form (4) ≤( )

= =

from (3) ⋅ = ≤ ⋅ = ⋅

≤ ⋅ = as desired.


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, = ⇒ ≥

+ =+

−= +

= + =+

=+

=−

=

∴+

= ⇒ − =

⇒ = ⇒ =

⇒ = ⇒ =

⇒ ∑ = ⇒ =( )

⇒ = (1)

(1) ⇒ it suffices to prove: ≥

⇔ ≥ ⇔ ≥

⇔ ≥ √ ⇔ ≤ ° ⇔ ≤ ° (i)

(a) ⇒ ≤ √ ⇒ ≤ √ ⇒ ≤ ° ⇒ (i) is true (proved)


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+ + ≥


Solution 1 by Rovsen Pirguliyev-Sumgait-Azerbaidian

+ + ≥ (1)

=+ −

, =+ −

, =+ −

⇒⊕

⇒ + + = , we prove that ≥ ⇒

⇒ + + ≥ , (where = , =( )

)

we have = ⋅⋅

= ( + + )

hence + + ≥ ( + + ) and this is true.


= =+ −

⋅

= ( + − ) = =− −

≥

⇔( − − )

≥ ⇔ ( − − ) ≥

LHS of (1) ≥ ( − − )( − ) ≥?

⇔ ( − − )( − ) ≥?

⇔ ( − ) ≥?

( − )( + ) (2)

LHS of (2) ≥ ( − )( − ) ≥?

( − )( + )

⇔ − + ≥?

⇔ ( − )( − ) ≥?

true ∵ ≥ (Euler)


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( + )( + ) ≥


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 + + ≥ √


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Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian

( + )≥

( + )( + )( + )≥ √

√ ≥ √ ; ⋅ ≥ √ ∴ ≤ √


In any , for , , > , we have the following:

( ) + ( ) + ( ) ≥ √ ; ≥ ( )( )( )( ) (A-G)

≥ ( ⋅ ) (∵ ( + )( + )( + ) ≥ − )

≥ ⋅ ( )√

(∵ ≥ by Euler and ≥√

by Mitrinovic)

=√

= ⋅√

( ) = √ (proved)


+ √ + √≤


Solution 1 by Boris Colakovic-Belgrade-Serbia

+ √ + √ ≥ √ + √ = √ =

+ √ + √≤

+ √ + √≤ + + =

+ +=

= ⋅ =

Solution 2 by Geanina Tudose-Romania

+ √ ⋅ + √ = + √ ⋅ + √ ⋅ +


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= + √ ⋅ ( + ) ≥ + √ ⋅ √ =

Hence ∑√ √

≤ ∑ = = =⋅

⋅ ⋅=


√ + √ + = + √ ( + ) +

≥ + √ √ + =

∴+ √ + √

≤ ( + + ) = = ( ) =

Solution 4 by Serban George Florin-Romania

+ √ + √≤

+ √ + √ = √ ⋅ √ + √ + √ ⋅ √ ⋅ √ + √ + √

≥ √ ⋅ √ ⋅ √ ⋅ √ ⋅ √ ⋅ √ ⋅ √ = √ ⋅ √ ⋅ =

= √ ⋅ ( ) = √ ⋅ √ = ≥

∑√ √

≤ ∑ = ⋅ = ⋅ = = (A)


+ + ≥ ( − − )

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


= + + ≥+ ++ + =

( + + )+ +


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≥ ( + + ) ⋅ ( + + ) ⋅ ⋅

+ + = ⇒ ( + + ) ≤ ( + + )

≤ ( + + ) ⇒ + + ≥

Also, ≥ ( + + ) ⋅ ⋅ ⋅ ⋅ ; ≥ ( + + ) ⋅ ⋅ ⋅

= ⋅ , ≥ , + + = ( − − )

⇒ ≥ ⋅ ⋅ ( − − ), ⋅ ⋅ ⇒ ≥ ( − − )

Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian

∑ ≥ ∑ ≥ ∑∑

≥ ∑ = ∑ = ( − − )

+ + = ; + + = + +

= − − − = − −


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


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Weitzenbock’s inequality ⇒ ∑ ≥ √

∴ ∑ ≥ √ and of course ∑ ≥ √

639.

Proposed by Lelia Nicula-Romania



∙ ≥⏞ ≥⏞ ∙ ≥

≥ ≥⏞ ∙ √ =√

≥

≥⏞ ≥ − ↔ ≥ − + ↔

≥ + ↔ − + − ≥ , = ≤⏞ , − ≤

↔ − + − ≤ ↔ ( − )( − ) ≤

640.

Proposed by Hung Nguyen Viet-Hanoi-Vietnam


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+ − + + − ≤⏞

+ − ≤ + + = ( )

√ = ( + − ) = ( + − ) ≤⏞( )

≤ ∙ = = = ; √ ≤


( + + ) + + ≥ √


Solution 1 by Athanasios Mplegiannis-Greece

From cosines law:

= + −= + −= + −

⇒( )

+ + = ( + + ) ⇒

⇒ + + ≥ ⋅ √ ⋅ ⋅ ⇒

⇒ + + ≥ ⇒

⇒ ( + + ) ≥ ⇒

⇒ + + ≥ √ √ ⇒

⇒ ( + + ) + + ≥ √

Equality holds for = = ⇒ equilateral.


Given inequality ⇔


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⇔ ( + + ) ≥ ⋅ (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)


( + − ) + ( + − )≥ ( + − )( + − ) ⇒

≥ ( + − )( + − ) similarly,

≥ ( + − )( + − ) and ≥ ( + − )( + − )

( ) ≥ ( + − ) = ( ) ⇒


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⇒+ +

≥ ( )

≥ √

642. In ∆ , − Lemoine’s point.Prove that:

++

++

+≤

+ ++ +



= ∑ − (∑ ) =∑ −(∑ ) =

=( + ) −

(∑ ) = (∑ ) → = ∑

+ = + + + ≤⏞ + +


+ + ≤√ ( + )

√ + √+



= = ; = = ; = =

∴ given inequality ⇔ √√ √

( + ) + √√ √

+ √√ √

+ ∑ ≥ ∑ (1)

Now, √√ √

≤ √√ √

⇔ √ + ≤ + √


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⇔ ( − ) + √ √ − √ ≤ ⇔ √ − √ √ + √ + √ ≤

⇔ √ ≤ √ ⇔ ≤ (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)


+ + + ( + ) ≥√

+ +



Let = + + , = + + and =

+ + = , + + = + + and =

∴ + ( + ) = − + + − = −

= − = ( − − ), we will prove

( − − )( + + ) ≥ √

⇔ ( − − )( + + ) ≥ ∴ ≥ √

⇔ − ( + ) ≥ ⇔ ≥ ( + ) +

We know, ≥ − now we will prove,

( − ) ≥ ( + ) + ⇔ − + ≥


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⇔ ( − )( − ) ≥ , which is true

∴ + ( + ) ≥√

+ +


1. + + ≥

∑ ( + )−

2. + + ≥ ; 3. ≥ √

⇒ + ( + ) ≥( )

( + ) − =

= − − = − =

= − ≥( )

( − ) =

= ⋅ = ≥( )

√ ⋅ ≥( ) √

+ + ≥√

+ +


= + − + ( − )

= + ∑ − ∑ + ∑ − = ∑ (1)

Now, ∑ ≥ √ + ∑ (Hadwiger-Finsler) ≥ √

(Ionescu – Weitzenbock) ⇒ ∑ ≥ √

⇒ √∑

≤ √√

= (2)

(1), (2) ⇒ it suffices to prove: ≤ ∑

⇔ ( − − ) ≥ (3)

Now, LHS of (3) ≥ ( − )( − ) ≥

⇔ − + ≥ ⇔ ( − )( − ) ≥

→ true ∵ ≥ (Euler) ⇒ (3) is true


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+ + + ≤



=∑⋅ ⋅

∏≤∑

∏=

∑ ( − )( − )

( − )( − )( − )⋅

=

=

∑ ( − )( − )

= ⋅ ⋅ ( − )( − ) =

= − ( + ) + − + =

= ( − − − + ) = ( − ) =−

≤−

; + = − + =


,∑ ≤ ; = ∑( )( )

≤ ∑ ∑( )( )

(CBS)

=−

∏( − ) = = ≤−

⇔ ∑ ≤ − + ⇔ ≤ − + (1)

Now, Gerretsen ⇒ of (1) ≤ + + ≤ − +

⇔ ≥ ⇔ ≥ → true (Euler) ⇒ (1) is true (proved)


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646. If in , - incentre, , , circumradii of , ,

with circumcentres , , then:

+ + ≥



= ° −

is a quadrilateral. Using Ptolemy theorem we get,

| | = ⋅ + ⋅ ⇒ | | = + ; | | = ⋅ + ⋅ ⇒ | | = +

| | = ⋅ + ⋅ ⇒| |

= ⇒ + + ≥ (Nesbitt)


< + (from )


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For an equilateral , = + ∴ combining above 2 situations,

≤ + = + ⇒ ≥ (1)

Now, ⋅ ⋅ = ⋅ ⋅ ⇒ = ⋅ ⋅ = =( )

(1), (2) ⇒ ≥ = = =

= − + = +

∴ ≥ (a)Similarly, ≥( )

& ≥( )

(a)+(b)+(c)⇒ ≥ ∑ ≥

647. If in : = then:

√ ≤ + +



In : = ( ) , = ( )


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=( − )( − )

, =( − )( − )

⋅ = ⋅ ⋅ ⇔

⇔( − )

⋅( − )

= ⋅( − )( − )

⋅( − )( − )

; =( − )

= ⇒ + + = ⇒ = + ≥ √ ⇒ ≥ √ (1)

Using the cosinus theorem, we get;

= + − ⋅ and + = ; + + =

+ = − ⇒ = − − ⋅

⇒ ( + ) = ⇒ + − = ⇒ ⋅ =

⇒ = ⇒ = √ ⋅√

(2)

From (1) and (2) ⇒ ≥ √ ⇒ ( ) ≤ °

√ ⋅ ⋅ ≤ + + = ⇒ ≤√

⇒ ( ) = °


⋅ = ⋅ ⋅ ⇔

⇔ = ⋅ = − ⋅ − = ( − )( − ) =

= ( − )( − ) =−

⇔ =−

⇔ = ⇔ = ⇒

⇒ = + (*)⇒ ≥ (1) ⇔ ≥ (Assure)

≥−

= ⋅ + ⋅ =

= ⋅ + ⋅ = ⋅ ⇔ ≥ (II)


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(I), (II) ⇒ ≥( )

= ≥( )

⇒ = (3)

√ ⋅ ⋅ =( )

√ ⋅ ⋅ = ⋅ √ ⋅ ⋅√

= = + =(∗)

+ +


− = ⇒ = (1)

⇒ − =

⇒ = = ≤ ⇒ ≤ ⇒ ≤ ⇒ ≤ ⇒ ≤ (a)

Now, √ ≤ + +

⇔ √ ( ) ≤ ⇔ √ ≤

⇔ √ ≤⏞( )

=+

+−

(1) ⇒ = − ⇒ = =( )

(i), (2) ⇒ given inequality ⇔ √ ≤

⇔ ≤√⇔ ≤ ⇔ ≤ true by (a)

648. If in :

= , = , = then:

≤√


Solution 1 by Rajeev Rastogi-India

Given = ⇒ =

= = ( ) ⇒ = . Similarly, =


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= ⇒ ⋅ = ( )( )( ) (1)

Now, AM ≥ GM ≥ ( − )( − )( − ) ⇒ ( )( )( ) ≤√

= √


+ = ( + )

= + − =+

−

⇒ =−

+

⇒−

+=

−

+⇒ =

⇒ = ≤√


= ⇒ = ⇒ = ⇒ = (1)

Similarly, =( )

and =( )

; (1)×(2)×(3)⇒ =

=( − )( − )( − )

≤?

⇔ ≥ ( − )( − )( − )

⇔ ≥ ( − )( − )( − ) = ⇔ ≥ (4)

Now, Gerretsen ⇒ ≥ − ≥ − = ⇒ (4) is true


++

++

+≥ +

( − )( − )



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Solution by Myagmarsuren-Yadamsuren-Darkhan-Mongolia

In : ( )√

≤ + + ≤ ( )√

(*) ⇒ ≥ √( )

⋅ ∑ ≥ (**)

(**)⇒ ∑ = ∑( )

= ∑ ( − ) = [∑ ( − + )] =

= ⋅ − + =

= ⋅ [ − ( − − ) + ( − − )] =

= ⋅ [ − + ( + ) + − ( + )] =

= ⋅ ( + − − ) = ( − ) ⇒ ∑ = ⋅ ( − ) (***)

(**); (***) ⇒ ≥ √( )

⋅ ( − ) ≥ , ≥ √ = √ ≥ (****)

(****)⇒ : ≥ √ ; : ≥√

650. If in , , , - internal bisectors,

∈ ( ), ∈ ( ), ∈ ( ) then:

+ + ≤ ( + )



Let = , = , = , = , = , =

Angle – bisector theorem ⇒ = → (1)

Stewart’s theorem ⇒ + = ( + ) (where = )→ (2)

Now, = √ ( − ) = ( )( )( )

→ (3)

(2) ⇔ + = ( )( )( )

+ (from (1), (3))


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⇒ ( + ) − ( + ) + ( + + )( + − ) =

⇒ = ( ) ± ( ) ( ) ( )( )( )

→ (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)

= − − + + + + + + − −

= −( − ) − ( + ) + ( + ) + ( + ) + +

≤ ( + )( + − + − ) + + (∵ ( − ) )

= ( + )( − ( − ) ) + + ≤ ( + ) + + (∵ ( − ) )

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

⇒ ≤ ( + )( + ) ∴ ( + ) ( + ) ≤ ( + )( + )


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⇒ ⋅

( )( ) (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)


( − ) ( − )+

≥ ( + )



[( + ) ⋅ ( + )]+ ≥

ö ∑( + ) ⋅ ( + )∑ + ∑ =

(∑ + ∑ )∑ + ∑

= ; = =( + )

+ =( + + )

+ ≥


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≥( + ) ⋅

+ = = ⋅ ( + + ) =

= ⋅ ( + + ) ≥ ( + ) = ( + )


= ∑ ( ) ( ) ≥ ∑ ∑∑

(Bergström)

=(∑ + )∑ + ∑ =

( + + + )− − =

( + + )− − ≥

≥ ( + ) ⇔ + ( + ) + ( + ) ≥

≥ ( + ) − ( + )( + )

⇔ + ( + )( + ) ≥ ( + ) → (1)

Now, LHS of (1) ≥ ( − ) + ( + )( + ) (Gerretsen)

≥ ( + ) ⇔ ( − ) + ( + )( + ) ≥

⇔ ( − ) + ( + )( + ) ≥ → (2) Now, LHS of (2) ≥ ( − )( − ) + ( + )( + )(Gerretsen) ≥

⇔ ≤ ( − )( − ) + ( + )( + ) → (3)

Now, LHS of (3) ≤ ( + + ) (Gerretsen)

≤ ( − )( − ) + ( + )( + )

⇔ − + ≥ ⇔ ( − )( − ) ≥

→ true ∵ ≥ (Euler) (proved)


++

++

+≤

√⋅

( − )



In any ,∑ ≤ √ ⋅ ( ); ≥ ( ) ∴ it suffices to prove:


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+ ≤( − )

⇔ + + − ≤( − )

⇔ (∑ ) ∑ ≤ ( ) ⇔ (∑ ) ⋅ ∑∏

≤ ( ) → (1)

Now, ∑ ( + )( + ) = ∑ (∑ + ) = ∑ + (∑ ) →(a)

Now, (∑ )(∑ ) = ∑ + ∑ ( − ) = ∑ + (∑ ) − (∑ )

⇒ ∑ = (∑ )(∑ ) + (∑ ) − (∑ ) → (b)

(a), (b) ⇒ ∑ ( + )( + ) = (∑ )(∑ ) + (∑ )

= + − + ( + + )

= + ( − − ) + ( + + )

= {( − − )( − − ) + ( + + )}

= { − ( + ) + ( + )( + )} → (c)

Again, ∏( + ) = + ∑ (∑ − )

= (∑ )(∑ )− = (∑ ) (∑ ) − ( ) −

= (( + + ) − ) −

= (∑ )( − ( − ) + ( + ) ) − → (d)

(c), (d) ⇒ (1) becomes: (∑ )[( + ){ − ( − ) + ( + ) }

− { − ( + ) + ( + )( + )}] ≥ ( + )

⇔ ( − ) − − − + ( + ) − − ≥

≥ ( + ) + (∑ ) → (2)

of (2) ≥ ∑ ( − ) − − − − +

( + ) − − ≥?

( + ) + ∑ → (3)

⇔ { ( − + ) + ( + )( − − )}

≥?

( + ) + (∑ ) → (3)


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RHS of (3) ≤ ( + ) + (∑ )( + + )

≤?

{ ( − + ) + ( + )( − − )}

⇔ (∑ ){ ( − + ) + ( + )( − − )} ≥?

( + ) → (4)

LHS of (4) ≥ ( − − ){( − )( − + ) +

( + )( − − )} ≥?

( + )

⇔ ( − ) − + + ( + ) − − − ( + ) ≥?

≥ ( + ) ( − ) − + + ( + ) − − → (5)

LHS of (5) ≥ ( − ){( − )( − + ) + ( + )( −

− ) − ( + )}

≥?

( + ){( − )( − + )} + ( + )( − − )

⇔ − + − + ≥ =

⇔ ( − )[( − ){( − )( + ) + } + ] ≥ → true ∵ ≥ (proved)


( + + ) + + ≤ ( + + ) + +


Solution 1 by Abdul Aziz-Semarang-Indonesia

1) ( + + ) ≤ ( + + ) + +

2) + + ≤ + + + +

Multiplying (1) and (2),

( + + ) + + ≤

≤ ( + + ) + +↙

+ +

≤

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


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then we have:

( + + ) + + ≤ ( + + ) + +


Let = , = , = , and consider

( + + )( + + )− ( + + )( + + )

= + + + + + + + + −

− + + + + ++ + + +

= ( + − − ) + ( + − − ) + ( + − − )

= ( − )( − ) + ( − )( − ) + ( − )( − ) ≥

∴ ( + + )( + + ) ≥ ( + + )( + + )


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)


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( + )≥


Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco

( + )≥

√ ( + )= ( + ) ≥

≥(∑( + ))

=+ +

=

Solution 2 by Dimitris Kastriotis-Athens-Greece

( + ) ≥ (*)

( + )=

( + ) ( + )≥(∗) ( + )

=( + )

≥ ( + ) + ( + ) + ( + )

= ( + + )

=( + + )

=

Solution 3 by Kunihiko Chikaya-Tokyo-Japan

( + )+

( + )+

( + )≥

( ) − ( + ) = ( ) ≥ , etc.

( + )+

( + )+

( + )≥ ( + ) + ( + ) + ( + )

= ( + + )( + + ) ≥ ( ⋅ + ⋅ + ⋅ )

= ( ) =


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=( + )

+≥

{∑( + )}

∑ +

≥∑ + ∵ + ≤

+ ≤ = =

655.

Prove that:

( ′) + ( ′) + ( ) ≥

Proposed by Abdilkadir Altintas-Afyonkarashisar-Turkey


( ′) = ( (∢ )) =+

=

= ≥ ∙ =


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656.


Design by Miguel Ochoa Sanchez


+ + + = + ( − ) = + ∙( − ) + ( + )

=

=( − ) + ( + )

≥( − )

↔ ( − − + ) ≤ ( + )

( − ) ≤⏞ ( + + )( − ) ≤ ( + ) ↔

↔ − + ≥ ↔ ( − ) ≥

657. , , : excenters

Prove that:

( ) + ( ) + ( ) ≤ √



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Prove that: ⋅ + ⋅ + ⋅ ≤ √

Lemma: = + ; = + ⋅ ; = + ⋅ ;

+ + = +

Applying Cauchy – Schwarz:

⋅ + ⋅ + ⋅ ≤ ( + + ) + +

= ( − − ) + ⋅ ≤

≤ ( + + − − ) + ( + ) (Gerretsen)

= ( + )[ + ( + )] ≤ ( + ) + + =

= ⋅ = √

658. If in , , , are Gergonne’s cevians, - inradius in

then: ≥( )




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In , , , are Gergonne’s cevians. If is the inradius of then:

≥( )

Lemma: = ⋅ ( ) ; = + − ⋅ ⋅ ( ) =

= ( − ) ( − ) = ⋅ ( − )

Similarly, = ⋅ ( − ); = ( − )

We have: = = ⋅

∑ ( )≥ ⋅

∑ ∑( ) (Cauchy – Schwarz)

= ⋅

⋅ ( )=

( )

≥( )

=( )

( )( )⇒

⇒ ≥( )

(Q.E.D.)


( + )+

( + )+

( + )≤

√

Proposed by Rovsen Pirguliyev-Sumgait-Azerbaidian


( + )=

⋅ ( + )⋅ = ( + )

= ⋅ ( + + ) = ( + ) + =

= ([∑ ⋅ ∑ − ( + + )] + [ (∑ ) − ( + + )]) =

= + − ⋅

= ( − − + + + ) − =

= [ ⋅ ∑ − ] = ( − + ) = ⋅ ( − + ) (*)


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(*) ⇒ ∑ ( ) = ⋅ ( − + ) ≤ ( − + ) ≤

√

≤√

⋅ ⋅ ( − + ) = √ ⋅


=( − )

=− +

= ∑ − ( ) + ∑ =( ) (∑ ) − . Also, ≥

( )

⋅ =

(1), (2) ⇒ it suffices to prove + ≥

⇔ ( + ) ≥ + + ⇔ ≤ + − (a)

Now, LHS of (a) ≤ + + ≤?

+ −

⇔ ≥?

⇔ ≥?

→ true (Euler)

660.




[ ] = √ + √ + √ √ + √ − √ =


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= + − + √ − − + √ = − ( + − ) =

= + + − − − ≤ ( − ) ↔

+ + − − − ≤ ( + − ) ↔ − − ≤

↔ ( − ) ≥


( + )( + )( + ) ≥ +( − )−



( + )( + )( + ) ≥ +( − )−

⇔ ( + )( + )( + )( − ) ≥ ( − ) + ( − )

we should prove that it is − ≥

= ( + + + )( + )( − )

= ( + + + + + + + )( − )

= [ ( + ) + ( + ) + ( + ) + ]( − )

= [ ( − ) + ( − ) + ( − ) + ]( − )

= [ ( + + )− ( + + ) + ]( − )

we know it is + + = − −

where , , , are area, circumradius, inradius, semiperimeter, respectively.

= [ ⋅ ( − − ) − ( − − ) + ⋅ ]( − )

= ( + + )( − )

= ( − ) + − + −

= ( − ) − − + ; = ( − ) + ( − )

= ⋅ ( − ) + ( − ) = − + −

= − −


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− = ( − ) − − + − ( − − ) ≥?

⇒ ( − ) ≥?

− − ⇔ ( − ) ≥ − −

≥ . Let’s use Gerretsen Inequality:

Gerretsen Inequality − ≤ ≤ + +

− −− ≤ ≤ + +

⇔ − − ≤ − + − + −

⇔ + − − ≥ ⇔ ( − )( − + ) ≥ as desired ∴.


In : ( + )( + )( + ) ≥ + ( )

1) ( + )( + )( + ) = ∑( + ) + =

= ⋅ − + = −

2) ∑ ⋅ ∑ − ≥ + ( ; ; )

⋅ − ≥ ( ; ; )

⋅ [ + + − ] = ( − + ) =

3) ( )

≤ ⋅ ⋅ ( − ) = ( − ) =

≥ (Assure) ( − + ) ≥ ( − )

− + ≥ −

≥ − ⇒ ≥ − ≥ − ; ≥ (Euler)


In any ,∏( + ) ≥ + ( ). Given inequality ⇔

+ ( − ) ≥ +( − )− ⇔ ≥ +

( − )−

⇔ ( + + ) ≥ + ( ) ⇔ − + ≥ ( ) (1)

Now, > ⇔ ( − ) > ⇔ > → true as


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≥ (by Euler) > ∴ > → (2)

∴ of (1) ≥ − = ( − ) = ( )

≥ ( ) (using (2)) ⇒ (1) is true (proved)

662. Let , , be the sides of extouch triangle for ∆ . Prove that:

+ + ≥



+ + =−

= − =

= − = − =

= − ∙ = − ≥⏞ − =

663. Let be the excentral triangle of ∆ with semiperimeter . If

is the inradii of ∆ then:

√ ≥


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[ ] = = = ≤ √

=[ ][ ] ≥

[ ]√

= ∙√

= ∙√

√ ≥ = = =

664. Let be the contact triangle of .

If = [ ], = [ ], = [ ], = [ ] then:

+ + + ≥ ⋅−



Let be the contact triangle of . If = [ ], = [ ], = [ ],

= [ ], then, + ∑ ≥( )

⋅

= = = ∴ (1) ⇔ ∑ ≥ ⋅ −

= − = ⋅ ⇔ ∑ ≥ ⋅ → (2).

Now, ∑ ≥ =[ ] [ ]

=


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= =

( )⇒ (2) is true (proved)


+ + + + + ≥



Let = + + . Using Bergström inequality

≥ ( ) = (1)

with Cauchy – Scwarz inequality ( + + ) ≤ + +

⇒ + + ≤ + + = ( + + )

⇒ + + ≤ ⋅ = (2)

Also, + ≥ ⋅ + ⋅ = , proof is completed.


In any ,∑ +∑

≥

+ ∑ ≥ö

∑ + ∑ = ∑

≥ ∑( + ) = ( − − ) + ( + + )

= − − ≥?

⇔ − − ≤?

⇔ ≤?

+ + → (1)

Now, LHS of (1) ≤ + + ≤?

+ +

⇔ − − ⇔ ( + )( − ) ≥ → true (Euler) (proved)


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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.



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Let is incenter of . =

| |⇒ | | =

| | = = ⋅ = ⋅( − )( − )

= ( − )( − )

= | | = [( − )( − ) + ( − )( − ) + ( − )( − )]

= ( − ⋅ + + + ) = ( − + + + ) =+

≥ ⇒ ≤ ⇒ ≤ ; ≥ ⇒ ≥ as desired.


=−

=( − )

( − ) =( − )

= ( − ) = ⋅−

= ⋅∑( − )( − )

∏( − ) =

= ⋅ ∑∏( )

=⋅∏( )

= ⋅ ( ) = (*)

1) ∑ = ≤ =

2) ∑ = ≤ =


+ + ≥ ( + )



+ + ≥?

( ); = + + = + + ≥ ( ) (1)

+ + = + ; = ; + + = ⋅ + ⋅ + ⋅


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= ⋅ − + − + −

= ⋅( − )( − ) + ( − )( − ) + ( − )( − )

( − )( − )( − )

= ⋅ (− ( + + ) + ( + + ) + )

= (− ⋅ + ⋅ ( − − ) + ⋅ ) = ( − )

= ( − ) = ( − ) (2)

From (1) and (2): ≥ ( )( )

≥( )

⇔ ( + ) ≥ ( − )

⇔ ( )( )

≥ ≥ ( − ) (Gerretsen’s inequality)

⇔ ( + ) ≥ ( − )( − )

⇔ + ⋅ + ⋅ + ≥ ( − − + )

⇔ + + + − + − ≥

⇔ − + − ≥ ⇔ − + − ≥

⇔ ( − )( − + ⋅ ) ≥ ⇔ ≥ (Euler) ∴

Solution 2 by Serban George Florin-Romania

= ⋅ ( − ) = ⋅ ⋅ ( − ) ≥ö

≥ ⋅( + + )∑ ( − ) = ∑ − ∑ = − ( − − )

≥ − + + = ( + )

∑ ≥⋅ ⋅( )

=⋅( )

q.e.d.


= ( − )


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={∑ ( − )( − )}( − )( − )( − ) =

∑{ ( − ( + ) + )}⋅

= − ( − ) +

= − + + − ( )

= − − = ( + ) −

= {( + + )( + ) − }

= { + ( + ) } =( ) + ( + )

(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:

<√



Let be the nine – point center. We know that the radius of the nine-point circle =

and = −∑ (1)


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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:

+ + ≤



≤⏞ ∙ ≤

≤ ∙ + + ∙ + + ≤⏞ ∙ =

= ∙ = ∙∙ ∙[ ] =

= ∙∏( )

= ∙ ∙ ∙ = ≤⏞ ∙ =


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⋅ ≤ + + ≤ −



In any , ≤ ∑ ≤ −

≥ ( − ) etc ∴ ∑ ≤ ∑ = ∑ ( )

= − − + ( − ) = − − + = − −

=∑( − )( − )

∏( − ) − = ( − ( + ) + )−

=− ( ) + + +

− =( − )

≤ −

Also, ∑ ≥∑

=∑

≥ (Done)


+ + ≤+

++

++



= × = ( ) etc ∴ given inequality becomes:

( + ) ≥ { ( − ) + ( − ) + ( − ) }

Let − = , − = , − = ⇒ = ∑

∴ = + , = + , = + ( , , > ). Then (1) becomes:


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[{( + )( + )} {( + ) + ( + ) }] ≥

≥ ( + ) ( + ) ( + ) { ( + ) ( + ) }

⇔ + + + + +

+ + + + + +

+ + + + +

+ + + + +

+ + + +

+ + ≥( )

+

+ + + +

+ + +

= ( + ) ≥ ( + )( + )

≥ ( + ) = ( + ) ≥ ( + )( + )

≥ ∑ ( + ) = (∑ ) ⇒⇒ (∑ ) ≥ (∑ ) (a)

Again, ∑ = ∑( + ) ≥( )

∑ = (∑ )

⇒ ∑ ≥ (∑ ) (b). Also, ∑ + ∑ ≥ ∑

(c) ≥⏞( )

(∑ )

Also, (∑ + ∑ ) ≥ ∑ ≥( )

( ) (∑ )


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Also, (∑ + ∑ ) ≥ ∑ ≥( )

( )(∑ )

Lastly, (∑ + ∑ ) ≥ ∑ ≥( )

( )(∑ )

Now, (∑ ) = ( ∑ )

≥( )

⋅ + ∵ ≥ +

= +

Also, ∑ + ∑ = ∑ ( + )

≥ =

= ≥( )

+

Last by, (∑ + ∑ ) = {∑ ( + )}

≥ { } { ( + )} = +

Now, (∑ + ∑ ) = {∑ ( + )} ≥ (∑ )

= ( + ) ≥ ( + )( + )

≥( )

( + ) = ∑ +

Also, (∑ + ∑ ) ≥ ∑ ( + ) +

≥ +

= + ≥( )

( )+

In the end, ∑ + (∑ + ∑ ) +

+ + +

+ + + +


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≥( )

( + + + + + ) =

(a)+(b)+(c)+(d)+(e)+(f)+(g)+(h)+(i)+(k)+(l)+(m)⇒ (2) is true


−+ + ≥ −



RHS: ∑ ⋅ = ∑ ⋅ − =

= ⋅+

⋅ ≥ ⋅+

⋅−

=

= ⋅ ( + ) = ⋅ ⋅ + ⋅ = ⋅

∑ ≥ Assure ≥ ⋅ ; − ≥ ; ≥ (Euler) RHS


+ + ≤ + + ≤ + +



= ⋅ =( + − )

=( + − )

=

= ( + + )( + − )( + − )( + − ) ≤

≤+ +

∵ ≥ ( + − ) = ; =( − )

=

= ∑ ≥ ( )( )( ) since, ∏ ( − ) = = =


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= , we need to prove, ≥ ⇔ ≥ ⇔ ≥ , which is true

∴ ≥


= =⋅

= ≤ ⇔ ≤ ⇔

⇔ ≥ → true (Euler) ∴ ∑ ≤ ∑ . Now, ∑ ≥ ∏

=∑

= ≥ = ⇔ ≥ ⇔ ≥ ⇔

⇔ ≤ √ → true (Mitrinovic) ∴ ∑ ≤ ∑ (proved)


≤ ↔ ≤ ↔ ≤

= = ⋅ = ≤∑

≤ ↔

↔ ≤ ↔ ≤√

674. If in , = then:

( − )( − )( − )+ √ + √ + √

≥√



: = . Prove that: ∏( )∏ √

≥∏

; =∏ ⋅

= =


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:∏( − )

∏ +=

∏( + )

∏ +√

=∏( + )

⋅ ∏ +√

=

= ⋅∏( )⋅∏ √

= ⋅ ⋅ ∏( )∏ √

= ⋅ ∏( )∏ √

≥ = (Assure)

( + ) ≥ + √

( + ) + ≥ √ + √ +

( + ) + ( + ) ≥ √ + √


= ⇒ √ =√

. Similarly, √ =√

and √ =√

∴ given inequality becomes: ( )( )( )√ √ √

≥

⇔ ( + )( + )( + ) ≥ ( + )( + )( + )

(where = √ , = √ , = √ )

⇔ ∑ + ∑ ≥ (∑ ) + ∑ → (a)

Now, ∑ ∑ ≥ ∑ = ∑ → (1)

Again, ∑ ∑ = ∑ ≥ ∑ = (∑ ) → (2)

(1)+(2)⇒ (a) is true (Proved)


√ ++√ +

+√ +

≤−



√ +≤ ⋅ +


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≤ ⋅ = ⋅+ +

≤

≤ ⋅ = ⋅ =−

≤−

=−


∵ + ≥ ∴ ≤ √√

(1). Similarly, ≤( ) √

√& ≤

( )√

√

(1)+(2)+(3)⇒ ≤ ∑ √√

≤ ∑ √√

= ∑ ≤?

⇔ ( − ) ≥?

( − − ) ⇔ − ≥?

− −

⇔ ≤?

+ + (a)

LHS of (a) ≤ + + ≤?

+ +

⇔ − − ≥?

⇔ ( − )( + ) ≥?

→ true ∵ ≥ (Euler) ⇒ (a) is true (proved)

Proof 2

+ ≥( + )

⇒√ +

≤( )

√ +

Similarly, ≤( )

√ & ≤( )

√

(1)+(2)+(3)⇒ ≤ √ ∑ ≤?

⇔∑ ( + )( + )

( + )( + )( + ) ≤? −

⇔(∑ ) + ∑

+ ∑ ( − ) ≤? −

⇔(∑ ) + + (∑ −∑ )

( + + ) − ≤? −

⇔( − − ) +

( + + ) ≤? −

⇔( − − )

+ + ≤? −


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⇔ ( − )( + + ) ≥?

( − − ) (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)


−≤ + + ≤

−



=

( − )

=( − )

= ∙( + − )

=


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=+ −

≤⏞+ + + −

=−

=+ −

≥⏞− + −

=

=−

=−


−+

−+

−≥



− = ( − ) ≥⏞(∑ )

∑ − ∑ = − ∑ ≥⏞

≥− ∑ ∑

=−

= =


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:


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+ ( + ) ≤ ( + ) ( + ) ( + )

Proposed by Do Quoc Chinh-Vietnam


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:


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+ ( + ) ≤ ( + ) ⇔ ( + ) [ + ( + )] ≤

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


The inequality needs to prove to be equivalent to:

( + ) + ( + )+ +

( + ) + ( + )+ +

( + ) + ( + )+ ≥

≥ + + + . We will prove that: ( ) + ( ) ≥ + . Applying the

Cauchy-Schwarz and AM-GM inequality, we have:

( + )+ +

( + )+ =

( + )√( + ) +

( + )√( + )

≥( + )√ + ( + )√

( + ) + ( + )

=( + )√ + ( + )√

( + ) + ( + ) =( + )√ + ( + )√

( + − ) + ( + )

≥( + )√ + ( + )√

( + ) ( + − ) + ( + )=

( + )√ + ( + )√( + ) ( + + )


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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:

( + )+ +

( + )+ +

( + )+ ≥


+ +≤



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= ≤ ( − ); =

+ + =+ +

≤ ≤ ( − )( − )( − ) = ⋅ =

= ; LHS ≤ ≤ ⋅ ; ≤ ⇔ ≤ ⇔ ≤ √

≤√

⋅ ⋅ =√


We know, √ ≤ ≤ √ and ≥

∑ ≤ = ∏ ( + ) ⋅ ( − )

= ∏ ( + ) ⋅ =⋅

( + + )( + + )−

= ( + + ) − = + + ≤⋅ ⋅+ +

= ( + ) = + ≤⋅

+ =

681. If in acute , - centroid, - incentre, - orthocenter then:

( + ) + ≥



For acute-angled , ≤( )

( + ) etc. Also, ∑ = ∑ =


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= ∏ = ⋅ =( )

. Moreover, ∑ = + ( − − ) =( )

= − − ; ≤∑

≤( )

≤ ( + ) = + =( )

= + ( − ) = + −∑

=

= + −( − − )

=+ − ( − − )

=+ + −

=( + + − )

∴ ≤( ) ( + + − )

=− +

+ + + − +

=( − + + + + − + )

=( )

; (a), (b) ⇒ it suffices to prove: + + − ≥

≥ + + − ⇔ ≤ − + ⇔

≤ − + (i)

Now, LHS of (i) ≤ + + ≤? − + ⇔

− + ≥?

⇔ ( − )( − ) ≥?

→ true ∵ ≥ (Euler) (proved)


+ ( − )( − ) ≥ √


Solution 1 by Mihalcea Andrei Stefan-Romania

LHS ≥√∑( − )( − ) + =

√∑ ( − − ) =

√( − ) =


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= √ . We will prove: √ ≥ √ ⇔ ≥ . By AM-GM ⇒

⇒ ( − )( − )( − ) ≤ = ⇒ ≤√

(1)

≥ √ . We will prove: ⋅ √ ≥ ⇔ ≥ √ ⇔ ≥ √ ⋅ ⇔

⇔ ≥ √ ⇔ (1) q.e.d.

Solution 2 by Myagmagsuren Yadamsuren-Darkhan-Mongolia

⋅ + ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅

= ⋅ + ≥ ⋅√

⋅ ⋅ ( + ) = √ ⋅ ( + + )

= √ ⋅ ( − )( − ) = √ ⋅ ⋅( − )⋅ ∏( − ) = √ ⋅ ≥ √ ⋅


+ ( − )( − ) ≥√

( + ) ( − )( − )

=√ − + −

( − )( − ) =√

( − + − )

=√∑ = √ ≥

√√ (proved)

683.


−≤ + + ≤

−



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= − ∙ = − = ∙( − )

=−

+ + =−

≤−

↔ ≤ ( − ) ↔ ≤

+ + =−

≥−

↔ ≥

684.

In the following relationship holds:

≤ + + + + + ≤−



+ ≥ ⇒ + ≤

= + + + + + ≤ + + =+ +

+ + = − − (well known)

≤+ +

=− −

⋅ =− −

=− −

Using Gerretsen Inequality we get

≤ = = proof is completed.


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+ ≤−

⇔ + + ≤+

⇔ + ≥+

⇔ ( + ) ( + ) ≥ ( + )( + )

⇔ ( + ) ( + ) ≥ +

⇔ ( + ){∏( + )} ≥ (∑ ){(∑ ) + ∑ } (1)

LHS of (1) ≥ ( + ) ⋅ (∑ )(∑ ) ≥?

(∑ )((∑ ) + ∑ )

⇔ ( + ) ≥?

+

⇔ ( + ) ≥?

⇔ ( + ) ≥?

+ ( + )

⇔ ( + ){ + ( + ) + ( + ) }

≥?

{ + ( + ) − ( + )} + ( + )

⇔ ( − ) + ( + )( + ) + ( + ) ( − ) ≥?

≥ + ( + )

LHS of (2) ≥ ( − ) ( − ) +

+ ( + )( + ) + ( + ) ( − )

RHS of (2) ≤ ( + + ) + ( + )

∴ it suffices to prove: ( − )( − ) + ( + )( + ) −

( + + ) − ( + ) +

+ ( + ) ( − ) ≥

⇔ ( − + ) + ( + ) ( − ) ≥ (3)


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∵ − + > 0

∴ LHS of (3) ≥ ( − )( − + ) +

+ ( + ) ( − ) ≥?

⇔ − + − ≥?

=

⇔ ( − )( − + ) ≥?

⇔ ( − ){( − )( − ) + } ≥?

→ true (Euler) ∵ ≥

∴ ∑ ≤

,∑ ≥ (proved)

685.

In acute the following relationship holds:

++

++

+<

− −



In any acute – angled ,∑ <

− −=

( − − )=∑

=(∑ )

=

∴ given inequality ⇔ ∑ < ∑ → (1)

Now, < ⇔ < + ⇔ ( − ) <

⇔ < ⇔ > ⇔ > → which is true


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∀ ∈ , . Hence, ∵ is acute – angled, ∴ < → (a)

Similarly, < → (b) and < → (c)

(a)+(b)+(c)⇒ (1) is true (Hence proved)

686.


( + ) + ( + ) + ( + ) ≤



LHS ≤ ∑( + ) ∑ = ∑ + ∑ ∑

= ( − − )( − − ) ≤?

⇔ ( − − )( − − ) ≤?

⇔ − ( + ) + ( + ) ≤?

⇔ + ( + ) ≤( )

?+ ( + )

Now, LHS of (1) ≤ ( + + ) + ( + )

≤?

+ ( + ) ⇔ ( − ) ≥( )

?( + )

Now, LHS of (2) ≥ ( − )( − ) ≥?

( + )

⇔ − + ≥?

⇔ ( − )( − ) ≥?

→ true

⇒ (2) is true (proved)


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( + ) + ( + ) + ( + ) ≤

≤ ( + ) ⋅ , ≤ ⋯, ≤ ⋯

( + ) ⋅ ≤ ( + )( + ) = ( + ) + ( + ) ⋅ =

= + ( + ) ⋅ = + ⋅ ( + )+ −

=

= + ⋅ ( + ) ( + − ) =

= +[ − + − + + ] +

+[ − + − + + ] ++[ − + − + + ]

=

= + ⋅ ( + + ) =

687.


( + )( + )( + )≥

+ ++ +



:∏ ( + )⋅ ⋅ ⋅ ≥

∑∑ ⋅

1) ∏( )⋅ ⋅ ⋅

= ∑ ⋅∑ = ( ) = (*)


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2) ≤ ≤ ( − )

≤ ≤ ( − )

≤ ≤ ( − )

=

=

=

∑∑ ⋅

≤ ∑ ( )∑ ⋅

=⋅ ∑

= ⋅⋅

= ⋅ ⋅⋅ ⋅

= (*)

=( )

≥( )


∏( + )=

+ ∑ ( + − )=

( + ) −

=( )

∴ (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:

√ + + ≤+ +




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1) =( ) ⋅

∑; … , …

) ∑ ≥ √ ( ) true

√ ⋅ =( )

√ ⋅⋅

∑ ⋅ = √ ⋅∑∑ ≤

( ) √ ⋅ ∑√ ⋅

=∑

689.


√ ⋅ ≤ + + ≤√

⋅



≤ + + ≤ ( . − − )

√ ≤⏞ = ≤⏞ ∙ = ∙ =

= ∙ ∙ = ∙ ≤⏞ ∙ =

= ≤⏞ ∙√

=√



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∑ = ∑ ⋅ = = ; √ ≤ ≤ √ ⋅ ; √ ≤ ≤ √

) = ⋅ ⋅ ≥√

√

) = ⋅ ⋅ ≤ ⋅ ≤√

√⋅

⎭⎪⎬

⎪⎫

690.


≤√√

Proposed by Nguyen Van Nho-Nghe An-Vietnam


≤√√

↔ √ ≤ √ ↔ √ ≤ √ ↔ √ ≤

≥⏞ ∙ ∙ ≥⏞ ∙ √ = √



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+ + ++ +

≥+ +

+ +



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)


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692.


+ +≥

Proposed by Nguyen Van Nho-Nghe An-Vietnam


≥+

↔ ≥+

↔

( + )−≥ ( + )

( + ) −

↔ ( − ) ≥ ( − ) ↔ ( − ) ( − + )( + − ) ≥


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≥+

≥⏞ ∙ = ∙ =

↔∏

≥ ↔ + + ≥

693.


( + + ) ( + + ) ≥



Given inequality ⇔ ∑ + ∑ + ∑ + ∑ + ∑ ≥

Now, ∑ + ∑ + ∑ = ∑( + ) + ∑ ( + )

= ( + )( + ) ≥ √ √ =


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⇒ ( ∑ + ∑ + ∑ ) + ∑ + ∑ ≥ ∑ + ∑ ≥ ∑

≥ ⋅ √ ≥√

(proved)

694.


≤ + + ≤− +

Proposed by Marin Chirciu-Romania


In any , ≤( )∑ ≤

( ); ∑ = ∑ = ∑

=( − )

( − ) ( − ) =∑( − )

{ ( − )( − )( − )} =∑( − )

=( ) ∑( − )

( − ) = ( − + ) = + + − − +

= + + − −

∴ ( − ) =( )

+ + − ( )−

Now, ∑ = (∑ ) − ∑


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= − − ( ) =( )

− +

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) ≥ ( − ) + + + − ≥(?)

⇔ ≤( )

+ +


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Now, LHS of (ib) ≤ ( + + ) ≤(?)

+ +

⇔ − − ≥ ⇔ ( + )( − ) ≥ → true ∵ ≥ (Euler)

⇒ (i) is true (Done)


In : ≤ ∑ ≤

I) LHS: ⇒ = ; = ⇒ ∑ = ∑ ( ) = ∑ ( ) =

= ⋅ ∑ ( ) = ⋅ ∑ ( ) ≥ ⋅⋅

= = ⋅ ≥ (LHS)

II) RHS: ∑ ≤

= (I) ; ⋅ = − (II)

⋅ ⋅ =( );( ) −

⋅ ( − ) =

= − ( − ) = + − + −

= + − + −∑

=

= + ( − − ) − +∑

−⋅ ( − − )

=

= − + − − − + + +∑

=

= − + + ∑ = − + + =

= − − + + +

= − + + − − ++ +

=

= − ++ +

≤ − +


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+ + − +( + ⋯ )

≤ − +

+ +≤ + ⇔ ( + + ) ≤ ( + )

( + + ) ≤ ( − )( − )

( + + ) ≤ ( − )( + )

+ + ≤ + − ⇒ ≥ (Euler)

695.

In any triangle the following relationship holds:

−=

+ ( + )≥ { , ( + )√ }

Proposed by Lelia Nicula-Romania


+ ( + )≥ ↔ + ( + ) ≥ ↔ ( + ) ≥ ( )

+ ( + )≥⏞

( + ) + ( + )≥ ( + )√ ↔

↔ + + ≤ ( + )√ ↔ ( + )√ ≤ + +

( + )√ ≤⏞ ( + ) ≤ + + ↔ − − ≥ ↔

↔ − − ≥ ↔ ( − )( + ) ≥


Let = + + ; = ( )( ) ( )( ) ( )( )( )( )( )


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=(− + + ) + (− + + ) + (− + + )

= ( ) ( ) ( ) ( ) (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.


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+ +

+ +≤



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:


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∑

∑≥

∑⇒ ⋅ ∑

∑≥ ⇒ ∑

∑≥ = (Proved)


≥ ; ∑ ⋅ ∑ ≥

∑ ∑

∑≥ ⇔

∑

∑≥

∑≥

∑= =

697.


+ + ≥



In ,∑ ≥ ; ∑ =⋅

= ⋅ = ⋅

= ⋅ = ≥ ⋅∑

= (∑( + )) = + + = ( ) ≥ = (proved)



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= + ; … …

= ⋅ + = ⋅ + ⇒

∑ = ⋅ ∑ ⋅ ∑ − (*);

) ∑ =

) ∑ =

) = ⎭⎪⎬

⎪⎫

(**)

(*), (**) ⇒ ∑ = ⋅ ⋅ − = − = − ≥


≥ − and ≥

= ≥⏞∑

∑=

∑ −

∑=

∑

we need to prove, ≥ ⋅ ⇔ ≥ ⋅

⇔ ≥ ( + + ) ⇔ ≥ ( + + ) since, ≥

⇔ ≥ ( + ) we will prove, − ≥ ( + )

⇔ ( − ) ≥ , which is true. Hence proved.

698.

If in : = then: + ≥



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∵ = , ∴ given inequality ⇔ + ≥ ( )

⇔ − + ≥ ⇔ ( − )( − ) ≥ → true ∵ ≥ (Euler)

699.

If in , ≤ , , ≤ then:

( + + ) + + < ( + )



= ⋅ ⋅ ⋅ = ⋅

= = ( − ) = − = − −

=−

=( )

( − )

=⋅ ⋅

= = ( + )


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= + + = ⋅( + )

=( ) ( + )

(1) × (2) ⇒ = ( − )( + ) > ( + ) ⇔ ( − ) > ( + )

⇔ > → true ∵ ≥ >

700. In any acute the following relationship holds:

− , − , − ≥



WLOG, we may assume , , =

Now, ≥ ⇔ ≥ ⇔ ≥ ⇔

⇔ + − − ≥ + − −

(∵ + , + > )

⇔ ( − ) − ( + )( − ) − ( − ) ≥ ⇔

⇔ ( − ) − ≥ ⇔ ( − ) − − ≥ ⇔

⇔ − ≤ ⇔ ≤ ⇔ ≥ (1) (∵ , are acute angles)

Similarly, ≥ ⇔ ≥ (2)

(1), (2) ⇒ ≥ ⇒ ≥ ⇒ ≥ (3)


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∵ ≥ , ∴ it suffices to prove: ≥ (a)

∵ < , < ⇒ − < < ⇒ < ≤ (4)

(4) ⇒ ≥ ≥( )

⇒ (a) is true (Proved)


−= ⋅ + ⋅ = ⋅ +

−⋅ =

+⋅

Similary = ⋅

= ⋅

() = + ; + ; +

≥ ≥⇓

⇔ + ; + ; + = +


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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


RMM - Triangle Marathon 601 - 700 · RMM - Triangle Marathon 601 - 700. RMM TRIANGLE MARATHON 601...

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