RMM - Triangle Marathon 901 - 1000 · 901 – 1000 . Proposed by Daniel Sitaru – Romania,Vasile...

ROMANIAN MATHEMATICAL MAGAZINE

Founding EditorDANIEL SITARU

Available onlinewww.ssmrmh.ro

ISSN-L 2501-0099

RMM - Triangle Marathon 901 - 1000

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RMM

TRIANGLE

MARATHON

901 – 1000

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Proposed by Daniel Sitaru – Romania,Vasile Mircea Popa-Romania

Mehmet Sahin-Ankara-Turkey,Bogdan Fustei-Romania

Murat Oz-Turkey,Thanasis Gakopoulos-Greece

Muhammad Ozcelik-Turkey,Mustafa Tarek-Cairo-Egypt

Marian Ursărescu – Romania,Adil Abdullayev-Baku-Azerbaijan

Nguyen Van Nho-Nghe An-Vietnam

Seyran Ibrahimov-Maasilli-Azerbaijan

George Apostolopoulos-Messolonghi-Greece

Nguyen Van Canh-Vietnam

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Solutions by

Daniel Sitaru – Romania,Marian Ursărescu –

Romania,Soumava Chakraborty-Kolkata-India,Serban

George Florin-Romania, Rovsen Pirguliyev-Sumgait-

Azerbaijan,Seyran Ibrahimov-Maasilli-Azerbaijan,Shafiqur

Rahman-Bangladesh,Fotini Kaldi-Greece,Lahiru

Samarakoon-Sri Lanka,Tran Hong-Dong Thap-Vietnam,Ravi

Prakash-New Delhi-India,Bogdan Fustei-Romania

Myagmarsuren Yadamsuren-Darkhan-Mongolia

Mohamed Alhafi-Aleppo-Syria

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901. In the following relationship holds:

√ + √ + √√

+√

+√

≤

Proposed by Adil Abdullayev-Baku-Azerbaijan Solution by Soumava Chakraborty-Kolkata-India

In any ∑√ ∑√

≤

LHS = + ∑ + ∑ = + ∑ +

≤ + √ + = + √∑ + ∑

= + ( − )

= + ( + + ) − = + − +

∴ ≤( )

+ − +

Now, RHS = ≥( )

⋅ = = ∴ ≥( )

(1), (2) ⇒ it suffices to prove: ≥ √ − +

⇔( − )

≥( − + )

⇔ ( − ) ≥( )

( − + )

∵ ≥ ∴ (3) ⇒ it suffices to prove: ( − + ) ≤( )

( − )

Now, LHS of (4) ≤ + + ≤?

− +

⇔ − + ≥?

⇔ ( − )( − ) ≥?

→ true (Euler) (Proved)


+ + + + + ≤+ +

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Proposed by Bogdan Fustei – Romania

Solution 1 by Tran Hong-Vietnam Using the inequality:

− + − + − ≥ + + −

⇔+ +

≥ + + ++ +

− ;

⇔ ≥ + + + (*)

More, we have:

+ + ≤( ) +

=∑( + )

=∑ ( − )

=( + + ) −

=+ −

⇒ + + ≤ (**)

Form (*) and (**) we have

+ ≤∑

Proved.

Solution 2 by Soumava Chakraborty-Kolkata-India

+ ≤( ) ∑

(1)⇔ ∑ − ≥( )

∑

LHS of (2)= ∑ − = ∑ = ∑ = ∑

= =( )

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RHS of (2) ≤ ∑ = ∑ + = ∑ + = ∑ = ∑ ≤ ∑

∵ < ≤ − < < =( )

LHS of (2)⇒ (2) is true (Proved)

903. In ∆ the following relationship holds: +−

++−

++−

= + +

Proposed by Mehmet Sahin-Ankara-Turkey

Solution by Daniel Sitaru-Romania

+− =

( , , )

+ −−

=+ −−( , , )

=−

( − ) ∙ −( , , )

=( , , )

= −( , , )

= −( , , )

∙ =( , , )

904. In ∆ the following relationship holds:

+ + = + + , −

Proposed by Bogdan Fustei-Romania


= ∙ = = =

= = ( − ) =( − )( − )( − )

( − ) =

= ( − )( − ) =( − )( − )

=

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=( − )

∙( − )

=

905. In ∆ , − ′ , ⊥ , ⊥ , ⊥ ,

= , = , = . Prove that:

∙+

∙+

∙= ∙ ( + + )



− ′ → = = =⏞ ,

=+ +

=( + + )

→ = + +

=( , , )( , , )

∙= ∙ ( + + ) =

( , , )

=∙ + +

∙ ( + + ) = ∙ ( + + )

906. In ∆ the following relationship holds: ++ +

++ +

++ =

( + )



++ =

( , , )

+

− + −=

( , , )

+∙

( − )( − )− − =

( , , )

=( − ) ∙

( − ) =( , , )

−−

( , , )

=

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= + ∙ −( , , )

= + ∙+

= ∙+ +

=( + )

907. Prove that: + =

Proposed by Mohamed Ozcelic-Turkey

Solution 1 by Serban George Florin-Romania

, T sin =°, = = (1)

, T sin = = ,⇓

= = (2)

+ =

= °, = °, = ° − , = ( °− ) , = ,

= − , ( − + ) = , ≠

− ( − ) + = , + − =

° =− + √

, ° = − ° =+ √

° =+ √

, ° = ( ° − °) = ° =+ √

° = ( ⋅ °) = − ° = −√ −

=√ +

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° = ( ° − °) = ° =√ +

From (1) ⇒ = ⇒ = √ , = √

From (2) ⇒ = ⇒ = √ ⋅√

= √√

+ = |: , + = ⇒− + √

+ =+ √

√ +,

− √+ =

+ √

√ +,

− √=

+ √

√ +, − √ √ +

= + √ ⇒ − √ + √ = + √

+ √ − √ − = + √

+ √ = + √ true

Solution 2 by Rovsen Pirguliyev-Sumgait-Azerbaijan

By the sine theorem in ⇒°

=°⇒ = °

°= ° (1)

In ⇒°

=°⇒ = °

°

We verify the equality + = .

° + =°

° , ° + = °

It is known that ° = √ , ° = √ ,

we have √ − + = − √ . Q.E.D.

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Solution 3 by Seyran Ibrahimov-Maasilli-Azerbaijan

= =

= ⇒ = ⋅ = ⋅ √ (1)

= ⇒ = = √

√ (2)

+ =( ) − √

+ = ⋅− √

=?

=( )

⋅+ √

+ √= ⋅

− √(=)

(proved)


= ⇒ =( )

= = ⇒ =( )

(1), (2) ⇒ − = − =

=− − +

− =−

+ =

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=− −

+ =( − )( + )

( + )

=( − )

=− √ −

√ −=

− √×

− √=

⇒ + = (Proved)

908.

Prove that: = +

Proposed by Muhammad Ozcelik-Turkey Solution 1 by Soumava Chakraborty-Kolkata-India

www.ssmrmh.ro ∠ = ∠ = °

∴ = = (say)

From , = = = ⇒ =( )

From , =( )

Also, from , = + − ⇒

⇒ − = − ⋅ (using (1))

= ( − ) = ⋅ = ⇒

⇒ √ − = =? = (from (2))

⇔ =? ⇔ − =?

⇔ − − =? ⇔

⇔ + − =?

⇔ + − =?

⇔ =? − ± √

=

= ±√ → true ∵ = √ ∴ = +

(proved)


In ⇒ = ⋅ °

In ⇒ = = ⋅ °

In ⇒°

=°⇒ = °

° (1)

By cosine theorem in , we have:

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= + ° − ⋅ ⋅ ° ⋅ °

= + ⋅ ( ° − ° ⋅ °)

=( )

+°

° ⋅+ °

−° + °

=

= + °°⋅ ° = + °⋅ °

°, we prove that °⋅ °

°= ⇔

⇔ ° ° = ° ⇔ ⋅ ⋅ ( ° − °) = ° ⇔

⇔ ° − ° = ° ⇔ ° − ° = ⇔ ° − ° = (2)

Then (1)(2) ⇒ = + . Q.E.D.

909. In ∆ , − , ⊥ , ⊥ , ⊥ ,

= , = , = . Prove that:

+ + = + +

Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania

− → = = =⏞

= =( , , )

∙

− ∙ −=

( , , )

( − )( − ) =( , , )

= − ( + ) +( , , )( , , )

= (− + + + ) =

= ( + ) = ( − + + ) = ( − ) =( , , )

=−

=( , , )

=( , , )

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910.

=?

Proposed by Murat Oz-Turkey Solution 1 by Rovsen Pirguliyev-Sumgait-Azerbaijan

− + + + + + + ° = ° ⇒ + = ° (1)

= ⋅ ⋅ =°⋅ ( )

( )⋅ (2)

using (1) we have: =°⋅ °

( ° )⋅

( ° )

( ° − ) ⋅ ( ° − ) = ⋅

( ° − − ° + ) − ( ° − + ° − ) = −

− = ; √ ( − °) = ; − ° = ° ⇒ = °

Solution 2 by Shafiqur Rahman-Bangladesh + + + − + + ° + = ° ⇒ + = °

Now, from , & ⇒

( + ) ⋅ = ⋅ ( − ) ⋅ °

⇒ ⋅ = ⋅ ( − ) ⇒ − = − ( − ) ⇒

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⇒ = − ⇒ = = ° − ∴ = °


Firstly, + + = ° ⇒ + + + + + ° = ° ⇒ + = °

Now, ( )

=( )

⇒ √ =( )

( ). Again,

°= ⇒ √ =

( )

(1), (2) ⇒ =( ) ( ). Again, = ⇒ =

( )

(3), (4) ⇒ ( − ) = ⇒ − ( − ) = −

⇒ − ( − ) = ⇒ − ( + ) ( − ) = ⇒

⇒ ( ° − ) = (∵ + = °). Now, > ⇒ + > 2 ⇒ < 45° ⇒

⇒ < ° ⇒ − > − ° ⇒ ° − > −22 ⋅ °. Also, °− < 45°

∴ −°

< 45° − 3 < 45° ∴ ( °− ) = ⇒ ° − = ⇒ = °

911.

= + Proposed by Mohamed Ozcelic-Turkey

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In ⇒°

=°⇒ = °

° (1)

In ⇒°

=°⇒ = °

° (2)

(1) = (2) ⇒ ° ° = ⋅ ° ° ⇒ = ° (3)

In ⇒°

=°⇒ = °

°

∢ = ∢ ⇒ = ⇒ = °°

(4)

By the cosine theorem in , we have:

= +°

° − ⋅°

° ⋅ ° =

= + ° °°− °

°=( )

+ ⋅ ° ⋅ ° °°− °

° (*)

Now we prove that: ° ⋅ ° °°− °

°= (5)

° ° ⋅° − ° °

° = ° ° ⋅

⋅− ° − °− °

° = ° ° ⋅°

° =

= ° ° ° =° + ° √ −

=

= ⋅ √ ⋅ √ = , using (5) in (*) ⇒ Q.E.D.

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From , = ⇒ =( )

.

From , = ⇒ =( )

(1), (2) ⇒ = ⇒ =( )

Also, from , = ⇒ =( )

& from , = ⇒ =( )

(3), (4) ⇒ = ⇒ =( )

From , = + − ⇒ − =( )

−

=−

=

=( )

( − ) =

=( − )

=

=( − − + )

=

=( ) ( − − + )

=

=+ ( − + )

=+

=?

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⇔ + =? = = − ⇔ + =? − ⇔

⇔ + − =? ⇔ =? ±√ = ±√ → true

∵ = √ ⇒ = + (Proved)

912. In ∆ , − , ⊥ , ⊥ , ⊥ ,

= , = , = . Prove that: + +

+ + =+ +

+ + = −


− → = = =⏞

+ ++ + =

+ ++ + =

+ ++ + =

( , , )

=

= −( , , )

= −( . . )

= ∙( − )

= −

913. In acute , = , – circumcentre,

– incentre, – orthocentre. Prove that:

[ ] = ⋅ | |

Proposed by Daniel Sitaru – Romania Solution 1 by Soumava Chakraborty-Kolkata-India

( [ ]) =( − ) ( − ) ( − )

Let = , = , =

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Then, ( [ ]) = ( )( )( )( ) = ∑ ∑

= ∑ ∑ ∑ = ∑ ∑ → (1)

= = + − − − = + + − , = = − and

= = + + − ∴ = + + −

∴ from (1), ( [ ]) =

− {( + + − ) + ( − ) + ( + + − ) } + ( + + − )

= → (a)

Now, ( − ) ( − ) ( − ) = ∑ ( + ) − (∑ )− (∑ ) + (∑ + ∑ ) − → (2)

Now, ∑ + = ∑ ∑ − = ∑ (∑ ) − ( ) −

= (∑ )(∑ ) − ( − − ) − → (2a)

Again, − (∑ ) = − { + ( − − )} =

= − − ( − − ) → (2b)

Also, − ∑ = − { + (∑ )(∑ − ( ))} =

= − − − ( )( )

= −(∑ ) ( ∑ ) + ( + + ) − → (2c)

Moreover, (∑ + ∑ ) = {∑ ( − )} = ⋅ (∑ )−

= ( + + ) − → (2d)

(2), (2a), (2b), (2c), (2d) ⇒ ( − ) ( − ) ( − )

= − − ( − − ) − ( − − ) +

+ ( + + ) − ( )

= − + − + − − − − − → (3)

∴ ( ) ( ) ( ) = (from (3)) → (b)

(a), (b) ⇒ ( [ ]) = ( ) ( ) ( )

= ( − ) + ( − ) +

+ ( − ) =

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{ ( − ) + ( − ) + ( − )}−

− { ( − ) + ( − ) + ( − )} ( = )

= −

− { ( − ) + ( − ) + ( − )} +

+ { ( − ) + ( − ) + ( − )}

= − +

+ ( − + − + − )

= ( − + − + − ) ( = , = , = )

= ( − ) + ( − )− ( + )( − )

= ( − )( + − − ) = ( − )( − )( − )

=( )

( − )( − )( − )

Now, − = − =

=( − ) − ( − ) + ( − )( + + )

=( ) ( − )( + + )( + − )

Similarly, − =( ) ( )( )( ) &

− =( ) ( − )( + + )( + − )

(1), (a), (b), (c)⇒ | | = ⋅|( )( )( )| ( )( )( )⋅ ( )

=

=|( − )( − )( − )|

⋅ =|( − )( − )( − )|

∴ | | =|( − )( − )( − )|

⋅=

|( − )( − )( − )|

= [ ] (Proved)

Solution 2 by Marian Ursărescu-Romania

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=− − −

=

=− − −

=

= −− ( − ) ( − )

= − −( − ) ( − )

= ( − )( − )− −

= ( − )( − )( − ) (1)

From (1) ⇒

[ ] = | ( − )( − )( − )|

= ⋅|( + )( − )( − )|

=− + − + − +

= (2)

But = (3). From (2)+(3) ⇒

[ ] = (4)

Now, from Sondat theorem ⇒

[ ] = |( − )( − )( − )| = ⋅ |( − )( − )( − )|

=− + + + + +

= ⋅ ⋅ (5)

But = (6)

From (5)+(6)⇒ [ ] = (7)

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From (4)+(7) relationship its true.

914. In ∆ , , , -circumradii of ∆ , ∆ , ∆

, , -excenters, −Bevan’s point. Prove that:

=−

, = , + + =[ ] − [ ]


= , = , =

∢( ) = − ,∢( ) = − ,∢( ) = −

= = = , = , =

∑ = ∑ = − = , ∏ = = =

= = ( − ) = − =

= − ∙ =−

=[ ]− [ ]

915.

Prove:

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( )( )( ) =

Proposed by Thanasis Gakopoulos-Greece Solution by Daniel Sitaru-Romania

( ) = − =

( ) = ⋅ = ⋅ = ⋅

= ⋅ = ⋅ = ⋅

= , = , =

⋅ ⋅ = ⋅ ⋅ =

= = ⋅ = =

916. In , – incenter, , , ′ - internal bisectors,

, , ∈ ℕ∗

Find: = + +

Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India

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Via angle-bisector theorem, on ,

′′ = ⇒ ′ =

+⇒ = + ⇒ = − + ⇒ =

( )

+

Via angle-bisector theorem on , =( )

=( )

Similarly, =( )

& ==( )

∵ , , ∈ ℕ∗, so, let, = , = & = (using (a), (b), (c))

Where , , ∈ ℕ∗ − { } ∵ , , > 1

∴ + =( )

, + =( )

& + =( )

(i)+(ii)+(iii)⇒ ∑ =( )

+ +

If , , ≥ , then (iv) ⇒ ∑ ≥ ∑ , which is impossible, ∴ at least one among

, , must be < 3 ⇒ at least one among , , = (∵ , , ∈ ℕ − { })

Case 1) = , , ≥ (i)⇒ + =( )

(ii)+(iii) =( )

= + =,

( + ) ⇒ ≥ + → impossible

⇒ at least one of , < 3 ⇒ at least one of , = (∵ , ≥ )

Case 1a) = (and of course, = )

(i), (ii)⇒ + = & + = ⇒ ( + )− ( + ) = ( − ) ⇒ =

& using + = , we get = ∴ = =

Case 1b) = (& of course = )

(i), (iii)⇒ + = & + = ⇒ ( + )− ( + ) = ( − ) ⇒ = &

using + = , we get = ∴ = =

Case 2) = , , ≥

(ii)⇒ + =( )

∴ using (3) & (i)+(iii), we get

= + ≥,

( + ) ⇒ ≥ + , which is impossible ⇒ at least of

, < 3 ⇒ at least one of , = (∵ , ≥ & ∈ ℕ∗)

Case 2a) = (& of course = )

(i), (ii)⇒ + = & + =

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⇒ ( + )− ( + ) = ( − ) ⇒ = & using + = , we get,

= ∴ = =


(ii), (iii) ⇒ + = & + =

⇒ ( + ) − ( + ) = ( − ) ⇒ = & using + = ,

we get = ∴ = = . Case 3) = , , ≥

(iii)⇒ + =( )

∴ using (4) & (i)+(ii), we get = + ≥,

( + ) ⇒

⇒ ≥ + , which is impossible ⇒ at least one of , < 3 ⇒

at least one of , = (∵ , ≥ & ∈ ℕ∗)

Case 3a) = (& of course = ). This is same as case 1b) ∴ = = .


This case is same as case 2b) ∴ = =

Combining all cases, we conclude = = .

∴ = = = = = = = = ∴ = + + = (answer)

917.

Find “ ”.

Proposed by Mohamed Ozcelik-Turkey

Solution by Fotini Kaldi-Greece

Ceva ⇒ ⋅( )

⋅ = ⇒ ( ) = ⋅ ⇒

( + )= ⋅

⋅⋅ ⇒

( + )= ⋅

⋅⋅ ⇒

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⋅⋅ =

⋅=

+=

⇒( + )

= ⇒ = , ( ) =( + )

, ( ) > 0


+ − = + + +

Proposed by Daniel Sitaru – Romania

Solution by Soumava Chakraborty-Kolkata-India

= ( ) = ( + + )

= { ( − ) + }

= { ( − ) − ( + )}

= ⋅ = =( )

Now, + − = ( + − ) =

= { ( − ) + ( + )}

= ⋅ =

=∏

⇒ + − =

= =( )

(where = ∏ )

Similarly, =( )

& =( )

(a)+(b)+(c)⇒ ∑ −∑

=( ) ∑ − = ∑

=∑ − − − −

=( − − ) − ( − − − )

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= ⋅ = =

⇒ + − = ( ∏ ) + ∑

(proved)

919.

Prove that: − − =

Proposed by Muhammad Ozcelik-Turkey Solution by Rovsen Pirguliyev-Sumgait-Azerbaijan

Using formula: = − , we have: in ⇒ = ,

⇒ =+ −

= − = = = (1)

⇒ = , ⇒ =

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= = = (2)

− + = − = =

(using formula = ) Q.E.D.

920.

, , collinears

Prove that: + =

Proposed by Mohamed Ozcelik-Turkey Solution by Rovsen Pirguliyev-Sumgait-Azerbaijan

∢ = ° ⇒ =

∢ = ∢ = ° ⇒ = (1)

∢ = °,∢ = ° ⇒ = = (2)

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⇒( )

=

∢ = °, = ° ⇒ =° (3)

In ⇒°

=°⇒ = ° (4)

+ = ⇒ + ° = ° ⇒ + ° = °

Considering that ° = √ and ° = √ , we have:

+

⋅ √ −= ⋅

+ √

+− √

=+ √

++ √

=+ √

921. pedal triangle of – incenter in , , , –

circumradii of

, , , , , – circumradii in , , . Prove that:

⋅ ⋅⋅ ⋅ = ⋅ ⋅


Solution by Lahiru Samarakoon-Sri Lanka

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⋅ ⋅⋅ ⋅

= ⋅ ⋅ . Consider, , = =

So, similarly, = , = . From, ,

=− −

=+

=

Similarly, = , =

=⋅ ∏

∏ =

= × ( ) ∏ ⋅ ∏

∏ =∏ ⋅ ∏

⋅ ∏ ⋅ ∏= ⋅ ⋅

(proved)


+ ++ + + + + =

Proposed by Mustafa Tarek-Cairo-Egypt

Solution 1 by Daniel Sitaru-Romania

∑

∑ + =∑ ( − )( − )( − )( − )

⋅∑ ( + )

=

=√

⋅⋅ ∑ −

∑ ( + ) =√

⋅∑ −

∑ ( − )=

=√

∙√

∙∑ −

∑ −= = = = =

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Solution 2 by Tran Hong-Dong Thap-Vietnam

Let = ∑ = ∑ = ⋅ ∑

(Because = )

Let = + + = ∑ + = ∑

= = ⋅ = ⋅ =

⇒ = ⇒ Proved


≥ +

Proposed by Bogdan Fustei – Romania Solution by Soumava Chakraborty-Kolkata-India

=+

=( + )

=

= ( + ) = ( + ) =⏞( ) +

By Bogdan Fustei (see proof below):

( − ) ≥ +

− ≥ +

− ≥ +

( + − )≥ +

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( + )≥⏞( )

+

By ( ), ( ): ∑ ≥ ∑ + ∑

( − ) = − = ⋅ − =

=( − )

=+ −

=+

− ≥

≥ + ⇔+

≥( )

+

, ≤( )

− + ( . . , . . ),

≤( )

( + ) ( ),

+≥( )

√ ( ) ⎭⎪⎪⎬

⎪⎪⎫

Now, ∑ ≤ ∑ ( − ) ≤ √ √ √ = √ ≤( )

∑ ⇒

⇒ ≤( ) +

(i) ⇒ in order to prove (1), it suffices to prove: ∑ ≥ ∑ + ∑ ⇔

⇔ ∑ ≥( )

∑ + ∑ . Now, LHS of (2) ≤( )

√ (∑ ) + (∑ ) ≤

≤( ),( )

√ ( + )( + + )

+ − + =

=( + )( + + ) + ( − + )

Again, LHS of (2) ≥( )

√

(m), (n) ⇒ in order to prove (2), it suffices to prove:

√ ≥( + ) + ( + ) − ( − )

⇔ ≥ + +

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+ ( + ) − ( − ) ⇔ { ( − ) − ( + ) } ≥ ⇔

⇔ ≤( )

− −

Now, LHS of (3) ≤ + + ≤?

− − ⇔

⇔ − − ≥?

⇔ ( − )( + ) ≥?

→ true ∵ ≥ ⇒

⇒ (3) is true ⇒ (2) is true (proved)

924. Let ′ ′ ′ be the pedal triangle of – incentre in . Prove

that:

⋅ + ⋅ + ⋅ ≥ ( + + )


Solution 1 by Marian Ursărescu – Romania

= ⇒ = + ⇒ = +

= = ⇒ = ⇒ we must show:

⋅ + + ≥ ( + + ) (1)

Let ≤ ≤ ⇒ ≥ ≥ and ≥ ≥ ⇒

From Cebyshev’s inequality ⇒

+ + ≥ + + + + (2)

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

+ + + + ≥ ( + + ) (3)

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But + + = + + + + + ≥ (4)


+ + ≥ ( + + ) ⇔

⇔ + + ≥ + + true.


Angle-bisector theorem ⇒ = ⇒ = ⇒ = ⇒ =( )

Again, angle-bisector theorem ⇒ = ⇒ = ( ) =( )

. Similarly,

=( )

& =( )

(a), (b), (c) along with ≥ ( − ), etc ⇒

≥( ) ( − ) ( + )

= ( − )( − ) =

= ( − + ) =(∑ )− + ( )

=

= = . Also, ≤( )

,

( ∑ + ∑ ) =

=− −

(i), (ii) ⇒ it suffices to prove: − ( − ) ≥ − ( + ) ⇔

⇔ − ( − ) + ( + ) ≥( )

.

Now, of (2) ≥ ( − ) + ( + ) ≥?

⇔

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⇔ ( − ) + ( + ) ≥( )

.

Now, LHS of (3) ≥ {( − )( − ) + ( + )} &

RHS of (3) ≤ ( + + )

The last 2 inequalities ⇒ in order to prove (3), it suffices to prove:

( − )( − ) + ( + ) ≥ + + ⇔

⇔ − + ≥ ⇔ ( − )( − ) ≥ → true (Euler) ⇒ (3) is true

(Done)


+ − > 4 Proposed by Marian Ursărescu – Romania

Solution by Ravi Prakash-New Delhi-India + − =

= + + ( + − ) = + ( + ) + =

= + + + + − + =

= ( + ) + ( − ) + ≥ =


+ + ≥



≥

= ≥(∑ )

(∑ ) ≥√ (∑ )

(∑ ) =√ (∑ )

∑ ≥( )

≥√ (∑ )∑ ( − )

≥√ (∑ )

√ √ ∑( − )=√ (∑ )√

=∑

≥?

⇔

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⇔ + + ≥?

⇔ ≥?

− .

Now, ≥ − ≥?

− ⇔ ≥?

⇔ ≥?

→ true (Euler)

(proved)


= ⋅ ( ) ≤ √ ⋅ ( ) = ( − ) ⇒ we must show:

( − )+

( − )+

( − )≥ ⇔

√+

√+

√≥ ( )( )( ) (1)

Now, let − = , − = , − = ⇒ + + = and

= + , = + and = + (2)

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

≥ (3)

But + ≥ √ and + ≥ (4)

From (3)+(4) we must show: ∑ ⋅√( )√

≥ ⇔

∑ ≥( )

(5)

But ∑ ⋅ ∑( + ) ≥ ⇔ ∑ ≥( )

⇒ (5) its true.


++

++

+≥

√



+ = + ≥⏞( + + )

+ ( + + ) = + =

= + ≥⏞√

( + ) ≥⏞√

+=

√

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928. In the following relationship holds: ( + )

+( + )

+( + )

≥ ( + + )√

Proposed by Bogdan Fustei-Romania Solution by Soumava Chakraborty-Kolkata-India

By Bogdan Fustei, ≥( )

( + ). Proof of (1):

( + ) = + = + =+

=

=∏

= =( )

. Using (a), (1) ⇔ ( ) ≥ ⇔

⇔( + )

≥ ⋅ ⋅( − )

= ( − ) ⇔ ( + ) ≥ ( + − ) ⇔

⇔ ( + − ) ≥ → true ⇒ (1) is true. Now, ≥( )

( + )( )

=( )

⋅ ( + ) ⋅ = √ ⋅ ( − )

( + ) = √ ⋅+ −( + ) =

= √ − ⇒ ≥( )

√ − . Similarly,

≥( )

√ − & ≥( )

√ −

(i)+(ii)+(iii) ⇒ ∑ ≥ √ ∑ −∑ = √ ∑ − ∑( )( )∏( )

= √∑

−∑ + ∑

+ ∑ ( − ) =

= √+ +

−+ + +

( + + − ) =

= √( + + )( + + ) − ( + + )

( + + )

= √{ − ( − ) + ( + )( + )− ⋅ ( + )}

( + + )

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= √− ( − ) + ( + )

( + + ) ≥?

√ ⇔

⇔{ − ( − ) + ( + )}

( + + ) ≥?

⋅ ⇔

⇔ + ( − ) + ( + ) − ( − ) −

− ( − )( + ) + ( + ) ≥?

≥ { + ( + ) + ( + )} = + ( + ) +

+ ( + ) ⇔ − ( − ) + ( + ) ≥( )

?

≥ ( + − ) + [ ( + ) + ( − )( + )]

∵ = ≥ ( − ), ∴ LHS of (2) ≥ ( − ) + ( + ) ≥

≥( )

?( + − ) + [ ( + ) + ( − )( + )]

Again, LHS of (3) ≥ ( − )( − ) + ( + ) ≥?

≥ ( + − ) + [ ( + ) + ( − )( + )] ⇔

⇔ ( − + ) + ( + ) ≥

≥( )

?[ ( + ) + ( − )( + )]. Now, LHS of (4) ≥

≥ ( − )( − + ) + ( + ) ≥?

≥ [ ( + ) + ( − ) ⋅ ( + )] ⇔

⇔ ( − + − ) + ( + ) ≥?

⇔

⇔ ( − + ) + ( + ) ≥( )

?

∵ − + = ( − )( − + ) + > 0

(as ≥ )

∴ LHS of (5) ≥ ( − )( − + ) + ( + ) &

RHS of (5) ≤ ( + + ) ∴ in order to prove (5), it suffices to

prove:

( − ) − + + ( + ) ≥ + + ⇔

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⇔ − + − − ≥ =

⇔ ( − ){( − )( + + ) + } ≥ → true

∵ ≥ ⇒ (5) is true (Hence proved)


+ + ≤√

+ +



≤√

≥ ⇔ ≤ ⇔ ≥ . Similarly, ≥ ⇔ ≤ ⇔ ≥ .

WLOG, we may assume ≥ ≥ . Then, ≥ ≥ & ≤ ≤

∴ ≤ ≤ ≤

≤ ∑ ⋅ √ = √ ∑ (proved)


+ + + + + ≥ −


+ =−

+ = ( + ) =

= ∙ ( − − ) ≥⏞ ( − − − ) =

= ( − ) = ( + − ) ≥⏞ ( + − ) = −

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≤( + )( + )( + )

≤

Proposed by Adil Abdullayev-Baku-Azerbaijan

Solution 1 by Lahiru Samarakoon-Sri Lanka

≤( + )( + )( + )

≤ ⇒ ( + ) ≥

∏( + ) ≥ × = × = but, ≥ √ , so,

( + ) ≥ ×

( + )( + )( + )≥

AM-GM ∑( ) ≥ ∏( + ) so, ∏( + ) ≤ × (∏( + ) =

= × ( + + ) , but ( ) ≥ ( + + )

≤ × ×[( + + )]

= ×

but, ≥ √ and ≥ (Euler)

≤ × × × × = , so, ∏( ) ≤

(it’s true)


In any we have ( + )( + )( + ) = (1)

(where = ). From (1) we must show: ≥ ⇔

( + + ) ≥ ⋅ (2)

From Mitrinovic inequality ≥ (3)

From (2) + (3) we must show: + + ≥ (4)

From Gerretsen inequality we have:

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≥ − ⇒ + + ≥ − (5)

From (4) + (5) we must show: − ≥ ⇔ ≥ ⇔ ≥ , true

because its Euler. From (1) we must show:

≤ ⇔ ( + + ) ≤ (6)

From Gerretsen inequality we have: ≤ + + ⇒ + + ≤

≤ + + ≤ + + = (7)

Form (6) + (7) we must show: ≤ ⇔ ≤ . But from Mitrinovic

inequality: ≤ and ≤ ⇒ ≤ ⋅ ⋅ = ⇒ ( ) its true.


° ≤ ( + + ) + ( + + ) + ( + + ) ≤ °



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

− − + +

+( + ) − + ( ) + ( + ) + ∑ . Now,

= = = ( − ) =

= − − =−

=( )

( − )

(1),(2)⇒ ∑( + + ) = − ( + ) + + ( + ) + + ( + ) +

+ − =( )

+ ( + ) + ≥ + + ≥√

+ √ = + √ = + √ =− √

= °

∵ ° = ° =− √

(3) ⇒ ∑( + + ) ≤ + + ⋅ √ = + √ = + √

= + √ =√

= ° = ° (Done)

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( − ) + ( − ) + ( − ) ≤+ +



( − ) ≤(∑ )

≤ ( − )( − )( − )−

=√

( − )=√

( − )

≤( ) √

( − ) =√

≥( ) (∑ )

= =

(1), (2) ⇒ it suffices to prove:

≥√

⇔ ≥ ⇔ ≥ → true (Euler) (Proved)


√−

+−

+−

≤ + + +

Proposed by Bogdan Fustei-Romania Solution by Daniel Sitaru-Romania

+ + + = +−

+−

+−

=

= +( − )

+( − )

+( − )

= ∙( , , )

=

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= √ ∙ ( − ) ∙( , , )( , , )

≥⏞ √ ∙ √ − ∙√( , , )

=

= √−

+−

+−


+ + + + + ≥ −


+ = +−

= ( − ) = − =

= + + − = + + − ≥⏞ + − = −


( + ) ≤ √


Firstly, ∑ = ∑ = ( + + ) =

= { ( − ) + } = { ( − ) − ( + )} =

= ⋅ = =( )

. Now, ∑ ( + ) =

= − = − =

= ⋅ ( − − ) − ∑ (by (1))

=( − − )

−( + − )

=( − − )

−

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−∑ (∑ − ) − ∑

=( ) ( − − )− (∑ )(∑ ) + + (∑ )

Numerator of (2) = ( − − )− (∑ )(∑ ) + +

+ + − = ( − − ) −

− + + = ( − − ) −

− + + − =

= ( − − ) − ( − − ) + ( − − ) + =

= − − + − − − + + + + =

= ( − − ){ + + (− − )} + =( )

(2), (3) ⇒ LHS = = ≤&

⋅ ⋅ √ = √ (Proved)

937. In ∆ the following relationship holds: + +

≥ + +

Proposed by Nguyen Van Nho-Nghe An-Vietnam


≥ + + ↔ ≥ + + ↔ + + ≥ ↔ + ≥

+ ≥⏞ − + ≥ ↔ ≥ ↔ ≥

938. In acute the following relationship holds:

( + ) ≤ ( + )

Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India Let = , = , = ; , , > 0. Then given inequality becomes:

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( + ) + ( + ) + ( + ) ≤ ( + + )( + )( + )( + ) ⇔

⇔ ( + − ) + ( + − ) + ( + − ) = ⇔

⇔ ( − ) + ( − ) + ( − ) ≥ → true (Proved)

939. Let be the pedal triangle of – incenter in . Prove

that:

⋅ + ⋅ + ⋅ ≥ ( + + )



Angle – bisector theorem ⇒ = ⇒ = ⇒ = ⇒ =( )

Again, angle-bisector theorem ⇒ = ⇒ = ( ) =( )

. Similarly,

=( )

& =( )

(a), (b), (c) along with ≥ ( − ), etc ⇒

≥( ) ( − ) ( + )

= ( − )( − ) =

= ( − + ) =(∑ )− + ( )

=

=( + + ) −

=( − + )

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Also, RHS ≤( )

,

( ∑ + ∑ ) =

(i),(ii) ⇒ it suffices to prove: − ( − ) ≥ − ( + ) ⇔

⇔ − ( − ) + ( + ) ≥( )

Now, LHS of (2) ≥ ( − ) + ( + ) ≥?

⇔

⇔ ( − ) + ( + ) ≥( )

.

Now, LHS of (3) ≥ {( − )( − ) + ( + )} &

RHS of (3) ≤ ( + + )

The last two inequalities ⇒ in order to prove (3), it suffices to prove:

( − )( − ) + ( + ) ≥ + + ⇔

⇔ − + ≥ ⇔ ( − )( − ) ≥ → true (Euler)⇒ (3) is true

(Done)


√ ++√ +

+√ +

≥√



√ +≥⏞

√ ∙ +=√

( , , )

−( , , )

=( , , )( , , )

=√ ( + + ) ≥

√( ) ↔ + + ≥

+ + ≥⏞ − + + ≥ ↔ ≥


+ + ≥


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Solution 1 by Bogdan Fustei-Romania Knowing the identity: + + = . The inequality from

enuciation becomes: + + ≥ ; ∑ = = = . So,

≥ ⇔ ≤ ⇔ ≤ (Euler)


=+ −

≥ , = ⇒

⇒ + − ≥

+ + −+ +

≥

( + + ) − ( + + ) ≥

= ( − )( − )( − ) (Heron)

( + + )− ( + + ) ≥

≥( + + )(− + + )( − + )( + − )

( + + )− ( + + ) ≥ [( + ) − ][ − ( − ) ]

− ≥ ( + ) − − ( − ) + ( − )

− ≥ + + − − − + + + −

− = −∑ + ∑ (true)

Solution 3 by Ravi Prakash-New Delhi-India

+ + = [ + + ] =

= [ + + ] =

= [ + + ] = [ ] = =

= ≥

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942. In acute ∆ , − , ⊥ , ⊥ , ⊥

,

= , = , = , − . Prove that:

+ + ≥



− ʹ → = = =⏞ ,

=+ +

=( + + )

→ = + +

=( , , )( , , )

= =( , , )( , , )

( , , )

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

≥⏞ ( + + − − )( + + − ( + ) ) =

= ( + ) ∙ ≥⏞∙ + ∙

=


( − )+

( − )+

( − )≥

−


( − )

( , , )

≥⏞( − − − )

+ + =( + )+ + ≥

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≥⏞( + )

+ + + += ≥

−↔ ( − )( + ) ≥


≥ +



= =

≥∑

≥ √ (∵ ( ) = ∀ ∈ ( , ) is concave as ( ) < 0)

= √ ∴ LHS ≥( )

√

Now, ∑ + ∑ = {( + + ) + ( + + ) + ( + + )}

≤ √ = √ ≤( )

(Proved)


≤ + + ≤−

Proposed by George Apostolopoulos-Messolonghi-Greece


First, we show: + + ≥ . From Bergström’s inequality, we have:

+ + ≥ ( ) = (1)

But in any we have: + + ≤ (2)

From (1)+(2) ⇒ + + ≥ (3)

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From (3) we must show: ≥ ⇔ ≥ , which its true, because its

Mitrinovic’s inequality. Second, we show: + + ≤ . We know:

≥ ≥ ⇒

⇒ ≤ ⇒ + + ≤ + + (4)

From (4) we must show:

+ + ≤ ⇔ ≤ (5)

But: + + = ( − − ) and = (6)

From (5)+(6) we must show: ≤ ⇔ − − ≤ − ⇔

⇔ ≤ + + , which its true, because its Gerretsen’s inequality.


Firstly, ∏( + ) = + ∑ ( − ) = + + − =( )

+ +

=∑

≤∑ ( + )

∵ ≤+

,

≤∑ + ( )

+ + + ∵ ≥+

,

=∑ +∏( + ) =

( ) (∑ + )( + + ) ∴ ≤

( ) (∑ + )( + + )

Now, ∑ = (∑ ) − {(∑ ) − ( )}

= ( − − ) + {( − − ) − ( + + ) } +

= { + ( + ) − ( + )} + ( )(− − ) +

=( )

+ ( + ) − ( + )

(1), (2) ⇒ ∑ ≤( )

≤(?)

⇔ ( − ) ( + + ) ≥?

+ ( + ) − ( + )

⇔ ( − ) + {( − )( + ) + ( + )} ≥( )

?+ ( + )

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Now, LHS of (3) ≥( )

{( − )( − ) + ( − )( + ) + ( + )}

and RHS of (3) ≤( )

( + + ) + ( + )

(i), (ii) ⇒ in order to prove (3), it suffices to prove: ( − )( − ) + ( − )( + ) + ( + )

− ( + + ) ≥ ( + )

⇔ ( − + ) ≥( )

( + )

∵ − + = ( − )( − ) + > 0 (as ≥ )

∴ LHS of (4) ≥ ( − )( − + ) ≥?

( + )

⇔ − + − ≥?

(where = )

⇔ ( − ){ ( − ) + + } ≥?

→ true ∵ ≥ ⇒ ∑ ≤

Now, ∑ ≥∑

≥ (∵ ≥ and ∑ ≤ + )≥?

⇔

⇔ ≥?

+ ⇔ ≥?

→ true (Euler) ∴ ≤ ∑ and this completes the

proof.

946. In acute ∆ the following relationship holds: −+

+−+

+−+

≥ °



( ) =−+

, ( ) =−

( + ) , ( ) =( + ) +

( + ) > 0, −

−+ =

−+ ≥⏞

+ += = ∙

− √

+ √=

= − √ = − √ = °

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+ + ≥ √


≥ √

Let ( ) = ;∀ ∈ , ; ( ) = > 0 ∴ ( ) = is convex on

, . WLOG, we may assume ≥ ≥ . Then, ≥ ≥ & ≥ ≥

∵ ( ) = ,

∴ ≥ ≥ √

∵ ( ) = ,

= √ (∑ ) ≥ √ (∑ ) = √ (proved)

948. In ∆ , − , − ,

− . Prove that:

[ ] + [ ] + [ ] ≥ [ ] ∙



[ ] =∙ ∙

=∙ ∙ ∙ ∙

=

= ∙∙

= ∙ ∙ = ∙ ∙ =

= [ ] ∙ ∙ ≥ [ ] ∙ ↔ ≥

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= − − ≥ ↔ ≥ +

≥⏞ − ≥ + ↔ ≥ ↔ ≥


+ + + + + ≥ −



+ =− + −

=( − )( − )

=− ( + ) +

=

= −−

+ ≥⏞ − + + ≥

≥ − + ∙ + = −


++

++

+≥

Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan


+ =+

= + = + ≥⏞

≥ ∙( + + )

( + + ) = ( + + ) =

www.ssmrmh.ro


+ + ≥√

Proposed by Nguyen Van Nho-Nghe An-Vietnam Solution by Daniel Sitaru-Romania

( , , )

=( , , )

≥⏞( + + )( + + ) = ≥⏞

≥∙ √ ∙

=∙ ∙ √ ∙

=√

=√


+ + + + + ≥



+ ≥⏞( + + )( + + ) =

∙ + + = + + ≥

≥⏞ + + + + = ( + ) ≥⏞+

=

953. In ∆ , − , ⊥ , ⊥ , ⊥ ,

= , = , = . Prove that:

√ ≤ + + ≤



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− → = = =⏞

=+ +

=( , , )

→( , , )

=

=( , , )( , , )

= =( , , )

( + + )≥

≥⏞∙ √

= √

=( , , )( , , )

= =( , , )

( + + )=

=∙ ( − − )

≤⏞( + + − − )

=

=( + )

≤⏞+

=∙

=


+ + ≥



≥⏞( + + )

+ +=

( + + )=

= ∙ ( + + ) ≥ ∙=


www.ssmrmh.ro

+( + + − )

≥

Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India ∑ −

=− − −

= ⇒∑ −

=( )

= ( )( )( )

(1) ⇒ given inequality ⇔ ∑ + ∏( ) ≥( )

∑( ) ( )

Now, + − = ( + − ) =

= { ( − ) + ( + )}

(∵ ( + ) = ; = − ( + ))

= > 0 (∵ is acute-angled). Similarly, + −

− > 0 & + − > 0. Let + − = ,

+ − = & + − (of course, , , > 0)

Then, = , = & = . Via above substitution & (2), given

inequality ⇔ ∑ ( ) + ∏( + ) ≥ ∑( + ) ( + ) ⇔ ∑ + ∑ +

+ + + + ≥

≥ ( + + + + + ) ⇔ + + +

+ ≥ + + + ⇔ ≥ ⇔

⇔ ∑ ≥ → true by A-G (proved)

956. In ∆ the folowing relationship holds:

+ + +√

≤ + +

Proposed by Nguyen Van Nho-Nghe An-Vietnam


www.ssmrmh.ro

: ( , ) → ℝ, ( ) = , ( ) = − , − →

++ +

≤+

( , , )( , , )

↔

↔ + ≤−

( , , )( , , )

↔

↔ +√

≤( , , ) ( , , )

957. If , > 0, + ≤ 1 then in ∆ the following relationship

holds:

+ + + + + ≥



+=

( + )≥⏞ ∙

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

= ( + )( + + ) ≥⏞ + + ≥⏞√

=√

= ∙√

≥

≥⏞ = ≥⏞ ∙ =


+ + ≥ + +

Proposed by Bogdan Fustei-Romania Solution by Daniel Sitaru-Romania

≥⏞ ( + ) = = =∙

=


www.ssmrmh.ro

≤ + + ≤

Proposed by George Apostolopoulos-Messolonghi-Greece Solution by Daniel Sitaru-Romania

=( − )( − )

≥⏞( − ) ( − ) ( − )

=∙

=

= ≥ ↔ ≥ ↔ ≥ ↔ ≥

=( − )( − )

≤⏞− + −

= ∙ ∙ =


+ + ≤

Proposed by Adil Abdullayev-Baku-Azerbaijan

Solution 1 by Bogdan Fustei-Romania

We know: ≥ and the analogs ⇒ ≥ + + (1)

≤ (and the analogs) ⇒ ≤ (and the analogs)

⇒ ∑ ≤ ∑ (2). From (1) and (2) we have inequality from enunciation:

Namely: + + ≤ Q.E.D.

Solution 2 by Myagmarsuren Yadamsuren-Darkhan-Mongolia

≥ ≥ ; ≥ ≥ ; ≥ ≥

+ + ≤ ( + + ) + +

≤ ( + ) + + = ( + ) ⋅ ≤

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≤ + ⋅ = ⋅ =


By Tsintsifas, ∑ ≤ ∑ = ∑ = ∑ ( ) = =

=− +

≤?

⇔ ≥( )

?− +

Now, RHS of (1) ≤ + + ≤?

⇔ − − ≥?

⇔

⇔ ( − )( + ) ≥?

→ (true) (Euler) (proved)


+ + ≥ √ ( + + )


≥ √

WLOG, we may assume ≥ ≥ we shall prove ≤ ≤

≤ ⇔ ( + ) ⋅+ −

≤ ( + ) ⋅+ −

⇔

⇔( + − )

( + ) ≤( + − )

( + ) ⇔ + − ( + ) ≤

≤ + − ( + ) ⇔ + + ( + ) ≤( )

+ + ( + )

Now, ( + ) ≤ ( + ) (∵ ≥ ) ⇒ ≥ ⇒

⇒ ≤( )

. Also, ( + ) ≤ ( + )(∵ ≥ ) ⇒

⇒ + ≤( )

+

(a)+(b)⇒(1) is true ⇒ ≤ . Similarly, ≤ ∴ ≤ ≤ . Also, ≥ ≥ ⇒

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⇒ ≤ ≤ ∴ ≥ ≥

≥∑

∵ ( ) = ∀ ∈ ( , )

≥ ∑ √ (∵ ∑ ≥ ∑ ) = √ (∑ ) (proved)


++

++

+≥ √



+ =+

= + = + =

= ( + ) = + ≥⏞( + + )

( + + ) ≥

≥⏞∙ √

= √

963. If in , ( ) ≥ ( ) ≥ ( ) then:

+ + ≤+ +



Given inequality ⇔ ≤ ⇔

⇔+ +

≤+ +

⇔

⇔ ( − ) + ( − ) + ( − ) ≥?

∵ ≥ ≥ ∴ ≥ ≥

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Let = + & = + + ( , ≥ ). Using the above substitution (1) ⇔

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

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

⇔ ( + + ) − ≥ ⇔ ( + ) ≥ → true ∵ , ≥ (Proved)


+ +

+ ++

+ +

+ ++

+ +

+ +≥



= + + =+ +

= = ; =

⇒+

+≥

∑ +

∑ +=

( + )+ =

= ( ) = = ≥ ⇒ ≥ |: ⇒ ≥ (Euler)

Solution 2 by Tran Hong-Vietnam Using Schwarz’s inequality we have:

≥+ +

+ += + +

+ + =+ +

= = ≥( )

Equality ⇔ = = .


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+ ++ ( − )( − )( − ) ≥

√



Mitrinovic ⇒ ≥ √ , which ⇒ it suffices to prove: ∏( ) (∑ ){∏( )}∑

≥( )

Now, ∏( + ) = + ∑ ( − ) = + + − =( )

+ +

(1) ⇒ LHS of (a) = ≥?

⇔ + ( + ) + ( + ) ≥?

+ ( + ) + ( + )( + )

⇔ ( + )( − ) ≥?

⇔ ≤( )

?− −

Now, LHS of (b) ≤ + + ≤?

− −

⇔ ( − )( + ) ≥?

→ true ∵ ≥ (Proved)


( + + )

( + + )+

( + + )≥ √

Proposed by Daniel Sitaru – Romania Solution 1 by Soumava Chakraborty-Kolkata-India

We shall first prove that:

∑

∑+

(∑ ) ∑ ≥( )

√ ⇔( ) + (∑ )

∑≥ √ ⇔

⇔ + ( ) + ⋅ ( ) ≥ ( ) ⇔

⇔ ( − − ) + ⋅ + ⋅ ( ) ⋅ ( − − ) ≥

≥ ⋅ ⋅ ( − − ) ⇔ − ( + ) −

− ( − − ) + − − − + +

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+ + + + ≥( )

Now, LHS of (1) ≥ ( − )

− ( − − ) + − − − +

+ + + + + − ≥?

⇔

⇔ ( − )− − − + − − − +

+ + + + + ≥( )

?

Now, LHS of (2) ≥ ( − )( − )

− ( − − ) + − − − + +

+ + + + & RHS of (2) ≤ ( + + )

∴ in order to prove (2), it suffices to prove:

⇔ ( − )( − ) + ( − − − ) + +

+ + + + ≥( )

Again, LHS of (3) ≥ ( − )( − )( − ) +

+ ( − − − ) + + + +

+ + & RHS of (3) ≤ ( + + )


( − ) − + + + + + + ≥

≥( )

Now, LHS of (4) ≥ ( − )( − )( − + ) +

+ + + + +

& RHS of (4) ≤ ( + + )


− + − + ≥ =

⇔ ( − ){( − )( + + ) + } ≥

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→ true ∵ ≥ ⇒ (4) is true ⇒ (i) is true

Applying (i) on a triangle with sides = , , , we get,

∑

∑+

∑ ⋅ ∑≥ √ ⇒

∑∑

+( ∑ ) ∑

≥ √

⇒∑

∑+

∑ ∑≥ √ ⇒

∑

∑+

(∑ )≥ √

(proved)


( + + )

( + + )+

( + + )+

( + + )+

( + + )≥

≥( + + ) + ( + + )

( + + ) ⋅⇒

We must show:

( + + ) ⋅ ( + + ) ( + + )

( + + ) ⋅≥ √ ⇔

⇔( + + ) ( + + )

≥ ⇒

( + + )( + + ) ≥ ⋅ (1)

But + + = ⇒ + + = + + (2)


( + + ) ⋅ + + ≥ ⋅ ⋅ ⋅ ⋅ ⇔

( + + ) + + ≥ (3)

From Cauchy inequality we have:

+ + ≥ ( + + ) ⇒

+ + ≥ ( + + ) (4)

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( + + ) ( + + ) ≥ ⇔

⇔ ( + + ) ≥ ⇔ + + ≥ , which its true.


+ + ≤



+ + ≤ − ( )

=∙

= = =

= ≤⏞ ∙ = ∙ =

968. In ∆ the following relationship holds: +

++

++

≤



+=

+=

+=

( + )= ( + ) =

= ( − ) = − =

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= ( ∙ ( − − ) − ( − − )) = ( + − ) ≥

≥⏞ ( − + − ) = ( − ) ≤ ↔

↔ − ≤ ↔ ( − )( − ) ≥


⎝

⎜⎜⎜⎛

+⎠

⎟⎟⎟⎞

≥


Let = , = , = . Using the substitution, given

inequality becomes: ∑ ≥ . WLOG, we may assume ≥ ≥ .

≥ ⇔ + ≥ + ⇔ ( − ) + ( − ) ≥ → true

∵ ≥

∴ ≥ . Similarly, ≥ ⇒ ≥ ≥

∴ + = + ≥ +

≥ =∑

=

= ⋅∑

∏= ⋅

∑=∑

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≥?

⇔ ≥( )

?

Now, ∑ = ∑ = ∑ ( )( )

≥( − )( − )

( + )

∵ <−

≤ − <−

<

=+ ( − )( − )

=∑ ( + )( − )( − )

∴ ≥( )∑ ( + )( − )( − )

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

= ( + )− ( + ) + ( + )

= ( + + ) + − ( − )

= ( + + )− ( − + )

= ( + + ) − ( ) + −

= ( + + ) − + ( − − ) − ( − − )

= ( ) = ⇒ ( + )( − )( − ) =( )

(1),(2)⇒ ∑ ≥ = ⇒ (a) is true ⇒ ∑ ≥ (Hence proved)

Now, ∑ = ∑⋅

≥ ∑

∵ < ≤ − < <

=+ −

= ( + )

= − = −

=+

( ) − =+

− ⋅ =+

−⋅

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= − = ⇒ (a) is true (Proved)

970. If is pedal triangle of – incentre of then: [ ][ ] ≤ + +

Proposed by Marian Ursărescu – Romania Solution by Soumava Chakraborty-Kolkata-India

Angle bisector theorem ⇒ = ⇒ = ⇒ = ⇒ =

Similarly, = , = , = , = , =

[ ] = ⋅ = ⋅ + ⋅ + ⋅ =( )

⋅ ( + )( + )

Similarly, [ ] =( )

⋅( )( )

& [ ] =( )

⋅( )( )

(1)+(2)+(3)⇒ − [ ] = ⋅ ∑ ( )∏( )

= ∑ ( )∑ ( )

= ⋅

=− ++ + ⇒ [ ] = −

− ++ + = + + ⇒

⇒[ ][ ] =

+ +≤

++ ⇔

+ +− ≤

+⇔

⇔ − + ≤ + ⇔ ≤ + + → true (Gerretsen) (Proved)


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+ + ≥√


Solution by Daniel Sitaru-Romania Known:

≥ ( )

≥ ( )

= = ≥⏞

≥ ∙ ∙ ≥⏞( ),( )

∙ ∙ = ≥

≥⏞ ∙ ∙ = ( ) = ≥√

↔

↔ ≥√

↔ ≥√

↔ ≤√

↔ ≤√

( )


+ + ≥ √ ⋅

Proposed by Daniel Sitaru – Romania Solution by Marian Ursărescu – Romania

+ + ≥ ⇒ We must show:

≥ √ ⇔

⇔ ≥ √ (1)

But = (2), = (3) and = ( ) (4).

From (1)+(2)+(3)+(4) we must show: ⋅ ≥ √ ( ) ⇔

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⇔ ≥ √ ( − ( + ) ) (5). But ≤ √ (6). From (5)+(6) we must

show:

≥ ( − ( + ) ) (7). From Gerretsen’s inequality: ≤ + + ⇒

⇒ − ( + ) ≤ (8). From (7)+(8) we must show:

≥ ⇔ ≥ true (Euler)


+ + ≤ ( + + − )

Proposed by Daniel Sitaru – Romania Solution 1 by Mehmet Sahin-Ankara-Turkey

= | | ⇒ | | =

= | | = | | + | | + | | ⇒ ≤ (| | + | | + | |) ≤

≤ + + ⇒

≤ ⋅ ( − )( − ) + ( − )( − ) + ( − )( − ) ⇒

≤ ⋅( − ) + ( − ) + ( − )

≤⋅

[ ( + + ) − ]

≤ ( + − )

≤ ( + − ) (1)

+ + − = + + − = ( + + ) − =

= + − = (2)

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From (1) and (2): | | + | | + | | ≤ ( + + − ) ∴

Solution 2 by Marian Ursărescu-Romania We must show ( + + ) ≤ ( + + − ) (1)

But form Cauchy’s inequality ( + + ) ≤ ( + + ) (2)

From (1)+(2) we must show: + + ≤ ( + + − ) (3)

But = (4). From (3)+(4) we must show:

≤ ( + + − ) ⇔

⇔ ⋅ ∑ ≤ ( + + − ) (5)

But ∑ = ⇒ ∑ = (6)

Now, + + − = − = = (7)

From (6)+(7)⇒ (5) its true.

974. If in , – Nagel’s point then:

⋅( ⋅ + ⋅ ) − ⋅

( , , )( , , )

≥


Solution by Soumava Chakraborty-Kolkata-India Let = , = & = . Then, given inequality becomes:

+ − + + − + + − ≥ ⇔ ( + − )( + − ) +

+ ( + − )( + − ) + ( + − )( + − ) ≥

≥ ( + − )( + − )( + − ) ⇔ + ≥( )

+

Now, ∑ + ≥ ∑ + ∑ & (∑ + ∑ ) ≥

Adding the last two inequalities, (1) is true (proved)

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975. If - circumcevian triangle of – incentre in then:

+ + ≥ √ ( + ) + ( + ) + ( + )



∵ is the orthocenter of & = , = & =

∴ = =( )

, =( )

& ( )

(∵ circumradius of = )

Also, ≤( )

√ ∑ = √ = √

(1), (2), (3), (4)⇒ it suffices to prove:

≥( )

√

Now, LHS of (5) = ∑ ≥ ∑( + )

∵ <−

≤ − <−

<

= + = ( + ) ≥ ⋅ = ≥?

√

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⇔ ≥?√ → true (proved)


( ° + ° + °)° + ° + ° > 108


∵ + + ≥ ,∀ , , ≥

∴ ° + ° + ° ≥ √ ° ° ° + ° °

> 4 √ ° ° + ° ° + ° °

= ( ° + ° + ° + ° + ° + °)

= −√

+ ° + ° + °

= −√

+ ° ° + ° −

>( )

−√

+√

° + ° −

= −√

+ √√ +

++ √

−

=− √ + √ + √ + + √ −

=+ √ + √ + √ − √

Also, ° < ° = √√

⇒°

>( ) √

√

& ∵ ° , ° < ° = √

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∴ ° + ° >( )

⋅√

=√

(1), (2), (3) ⇒ LHS

> 4+ √ + √ + √ − √ √ +

√+√

>√

√+ √ + √ + √ − √ √ +

√+√

> 120 > 108 (Proved)


( − )( − )+

( − )( − )+

( − )( − )≥ √



( − )( − )

( , , )

= − − − −( , , )

=

= ( − )( − )( , , )

=( , , )

∙ ≥⏞

≥∙

=∙

=∙

=

= ≥⏞√

=√

=√

= √

978. ADIL ABDULLAYEV’S REFINEMENT FOR TERESHIN’S

INEQUALITY

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In the following relationship holds:

≥+

+( − ) ( − − )


≥( ) +

+( − ) ( − − )

(1) ⇔ ≥ ( + ) + ( ) ⇔

⇔( + )

+( − )

− ≥ ⋅ ( + ) +( − ) ⋅

⇔( + ) −

+( − )

≥( + )

+( − )

( − )

⇔ ( − ) ≥( + )

− ( − ) ⇔

⇔ ≥( + )

− ( + ) + ⇔

⇔ − ( + )+ −

+ ≥

⇔ − ( + ) + ≥ ⇔

⇔ − ++

≥ ⇔

⇔ − ( − ) ++

≥ ⇔

⇔ − −+

+ ++

≥ ⇔

⇔+

−+

− ≥ ⇔

⇔+

− ≥ ⇔

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⇔ ( − ) ≥ → true ⇒ (1) is true (Proved)


− + − + − ≤ + + +



= + ⋅−

= +−

= + − =( )

= ⋅ − = − =−

≤ { ( − )}

= ∑ = (by (1)) (Proved)


+ + ≥


Solution by Daniel Sitaru-Romania Known: + + ≥ ( )

≥⏞( + + )

+ + ≥⏞( )

≥ = ≥⏞∙

=


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( + − ) ≥ ( + + )


Solution 1 by Marian Ursărescu – Romania

⋅ ≥ ⋅ ⋅ ( + + ) ⇔

⇔ ≥ ( + + ) ⇔

⇔ ∑ ≥ ( + + ) (1)

∑ ≥ √ (2)


√ ≥ ⋅ ( + + )

⇔ √ ≥ ( + + ) ⇔

⇔ ≥ ( + + ) ⇔

≥ ( + + ) (3)

But ≤ (4) From (3)+(4) we must show:

≥ ( + + ) ⇔ ≥ + + , which its true.

Solution 2 by Soumava Chakraborty-Kolkata-India WLOG, we may assume ≥ ≥ ∴ ≥ ≥ and + − ≥ + −

≥ + − ∴ ≥ ( + − ) ≥∑

≥?

⇔ ≥?

⇔ ( − − ) ≥? − ( + )

⇔ − − ≥?

− ( + )

⇔ ≤?

+ + ⇔ ≤( )

?+ +

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Now, LHS of (1) ≤ ( + + ) ≤?

+ + ⇔ ≥?

→ true

(Euler) (Proved)


− ≤ + ( + )


Solution 1 by Daniel Sitaru – Romania Known: ≤ + (1)

− =( )

( + )−⋅

=

= + − = + − =

= + − = = + +


= ⋅ =⋅

= ( − − )

=( )

− −

Now, ∑ = {(∑ ) −∑ } ≤( )

&( − + ) −

− ⋅ ⋅ ( − − ) = − + − + + =

= − +

(1)+(2)⇒LHS ≤ ( − + ) − ( − − ) =

= − + ≤?

+ + ⇔ ≥?

⇔ ≥?

→ true (Euler)

(Proved)

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+ + − − − < 1

Proposed by Daniel Sitaru – Romania Solution by Mohamed Alhafi-Aleppo-Syria

Let = , = then our inequality is:

+ + − − − < 1 ⇔( + − − )( − )

< 1 ⇔

⇔ ( − )( − )( − ) <

Note that: = = √√

, = = √√

But since , , are lenghts of sides of a triangle then = √ , = √ , = √ are

lengths of sides of a triangle too

Note that ( − )( − ) = − − = < ⋅ =

− = − =( − )

< =

So: ( − )( − )( − ) <

984. In , – incentre, , , – circumradii in

, , . Prove that:

− − ≤ + + ≤ − +

Proposed by Marian Ursărescu – Romania


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= ( )

= ( )= ( − ) (and analogous)

= ( − ) (and analogous) ⇒ + + = ⋅ ∑( − )

+ + = + − ( + + )

+ + = ⇒ = ( + + ) −

= ( + + ) −

+ + = [( + + ) − − ( + + ) + ]

+ + = [( + + )( + + − ) − + ]

+ + = [( + )( − )− + ]

+ + = ( − − + ) = ( − + )

= = . The inequality from enunciation becomes:

− − ≤− +

≤ − +

− − ≤ − + ⇒ ≤ + − + + =

= + + (Gerretsen’s inequality)

− + ≤ − + ⇒ − ≤ (Gerretsen’s inequality)

From the above the inequality from enunciation is proved.


− − ≤( ) ∑

≤( )

− +

From , ⋅ ⋅ = ⋅ ⇒ = = =

Similarly, = & =

∴ ∑ =∑

(using above 3 relations)

= = ( − ) = ( + − − )

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= −⋅

− + = −− −

− ( + )

=− + + − −

=( ) + −

(1) ⇒ (a) ⇔ + − ≥ − −

⇔ ≤ + + → true by Gerretsen ⇒ (a) is true

Also, (1) ⇒ (b) ⇔ + − ≤ − +

⇔ ≥ − → true by Gerretsen ⇒ (b) is true (Done)


(( − ) + ( − ) + ( − ) )≤ −


∑( − )≤

−

Given inequality ⇔ ∑ + − ≤ ( − ) ⇔

⇔ − ≤ ( − ) ⇔ ( + ) − − ≤

≤ ( − ) ⇔ ≥( )

( + )

Now, LHS of (1) ≥ ( − ) ≥( )

( + ) ⇔

⇔ − − ≥?

⇔

⇔ ( − )( + ) ≥?

→ true (Euler) ⇒ (1) is true (Proved)


( + + ) ≥ ( + + ) Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan


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Panaitopol’s inequality (1980): ≥ ( − )

= ∙ = ≥

≥⏞ ∙ ( − ) = ∙ ∙−

=

= ∙+ −

≥ ( + − ) ≥⏞

≥ ( + − ) ≥ ( + + ) ↔

+ − ≥ + + ↔ ≥ −

↔ ≥ − ( )

≥⏞ − ≥ − ↔ ≥ ↔ ≥


( + + ) ≥ √ Proposed by Daniel Sitaru – Romania

Solution 1 by Bogdan Fustei-Romania 1. In the following relationship holds:

( + + ) ≥ √

= → the area of the triangle having the sides , , ;

( + + ) ≥ √ ⋅ = √

We will write this inequality for , ,

( + + ) ≥ √ ⇔ ( ) ≥ √

≥ √ ⋅ ⇒ ≥ √ (Mitrinovic)

So, the inequality from enunciation is proved.


≥ √

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Applying the above inequality on a triangle with sides , , whose area of

course will be , we get,

≥ √ ⇒ ≥( )

√

Now, (∑ ) ≥ ∑ ≥( )

√ (Proved)


√ ≤ + + ≤ √ ⋅



= =( )

≥ =

= ≥?√ ⇔ ≥

( )

?

Now, LHS of (a) = ⋅ ⋅ ≥ ⋅ ⋅ = ⋅ = ( × ) = ⇒

(a) is true ⇒ ∑ ≥ √

Also, ∑ =( )

∑ ≤ ∑ = ∑( + ) = + +

= ≤⋅

= ≤√

√= √ ⇒ ∑ ≤ √ (Done)


+ + ≥ (1)

But = and = (2)

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From (1)+(2) ⇒ + + ≥ (3)

But ≤ √ ⇒ ≥√

(4)

From (3)+(4)⇒ + + ≥√

=

=√

=√

= = √

Now from Cauchy’s inequality ⇒

+ + ≤ + + (5)

From (5) we must show: + + ≤ √ ⇒

+ + ≤ √ (6)

But + + = ( ) (7)

From (6)+(7)⇒ √ ≥ ( ) ⇔ √ ≥ ( + ) (8)

But ≥ √ ⇒ ≥ ( + ) ⇔ ≥ + ⇔ ≥ (true)

989. In ∆ , , , −circumradii of ∆ ,∆ ,∆ ,

, , -excenters, -Bevan’s point. Prove that:

+ + ≥


= , = , =

∢( ) = − ,∢( ) = − ,∢( ) = −

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= = = , = , =

= + ( − ) ∙( − )( − )

=

=( − )( − )( − )

+ ≥⏞ ∙( + + )

+ + + + + =

= ∙ =


+( + + − )

≥


In any acute-angled ,∑ + ∑ ≥ ∑

∑ −=

− − −=

⇒∑ −

=( )

( )( )( )

(1) ⇒ given inequality ⇔ ∑ + ∏( ) ≥( )

∑( ) ( )

Now, + − = ( + − ) =

= { ( − ) + ( + )}

(∵ ( + ) = & = − ( + ))

= > 0 (∵ is acute-angled)

Similarly, + − > 0 & + − > 0

Let + − = , + − =

& + − (of course , , > 0)

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Then, = , = & =

Via above substitution & (2) , given inequality ⇔ ∑ ( ) + ∏( + ) ≥

≥ ( + ) ( + ) ⇔ + + + + +

+ ≥ ( + + + + + ) ⇔

⇔ + + + ≥ + + +

⇔ ∑ ≥ ⇔ ∑ ≥ → true by A-G (proved)

991. If in , ( ) = then the following relationship holds:

√ + ≥

Proposed by Daniel Sitaru – Romania Solution by Marian Ursărescu – Romania

= ⇒ √ = ⇒ =√⇒ √ = ⇒ + ≥ ⇔ ≥ ⇔ ≥ (1)

But √ ≤ (2). From (1)+(2) we must show ≤ ⇔ + ≤ ⇔

⇔ + ≤ ⇔+ −

≤ ⇔

⇔ −−

≤ ⇔−

≤ ⇔

⇔ ≤ ⇔ ≤ , true with equality for = . i.e equilateral

992. In ∆ , , , −excenters, = ( ) −

, , -circumradii in ∆ , ∆ ,∆ . Prove that:

+ + = ( + + )

Proposed by Mehmet sahin-Ankara-Turkey

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= , = , =

∢( ) = − ,∢( ) = − ,∢( ) = −

= = = , = , =

= = =( − )( − )

=

= ( − )( − ) = ∙( − )( − ) =

= ∙ ( + ) = ( + + )

993. In , ≥ ≥ , + ≥ .

Prove that: − ≥ .

Proposed by Nguyen Van Canh-Vietnam

Solution by Tran Hong-Vietnam

We have: = ( )( )( ) ; Let ( ) = ( )( )( ),

⇒ ( ) =( + ) − + ( + )( − )

≥( + − )

=+ −

≥ ;

⇒ ( ) ≤+

=( − )( − )

= − ⋅( − )

≤ ;

⇔ ≤ ⇔ ≤ . (Proved)


+ + ≤ −


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≥ ( − ) = ⇒ ≥ ⇒ ≤ (and the analogs)

⇒ + + ≤ + +

( − )( − ) = − ( + ) + | ⋅ ⇒ ( − )( − ) =

= − ( + ) + ⋅ (and the analogs) ⇒ ∑ ( − )( − ) =

= ( + + )− ( + + + + + ) + ( + + ) =

= ( − − ) − ( + − ) + ⋅

+ + = ( − − )

+ + + + + = ( + − )

( − )( − ) = ( − − − − + + ) =

= ( − )

( − )( − ) = ( − ) = ( − ) = ( − )

= + + =+ +

=+ +

= ( − )( − ) (and the analogs); =

= ( − )( − ) =( − )

=( − )

=( − )

∑ = − (Q.E.D)

Finally, we have the inequality from enunciation: + + ≤ −


We have: ≥ ⇔ ≥ ( ) ⇒

≥ √ ( ) ⇒ ≥ ( − ) ⇒ we must show:

+ + ≤ − (1)

But in we have: + + = ( ) (2)

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From (1)+(2)⇒ + + = ( ) = − ⇒ relationship it’s true.


+ + ≥



LHS =( ) ∑

= = ( ) =

= { ( − ) − ( + )} =

= − { ( + ) ( − )} + ( ) =

= − ( + ) + ( + ) =

=( )

( + − − )

Similarly, =( )

( + − − ) &

=( )

( + − − )

(a)+(b)+(c) ⇒ ∑ = ( − ∑ ) = { − (− − )}

=( )

( + )

(1), (2)⇒ LHS = ( ) = = + (where = ∏ )

∵ is acute – angled, ∴ < ≤ ⇒ + ≥ ⇒ + ≥

≥ = (Proved)


+ + = (1)

But = + − ⇒ = ⇒

+ + = ( + + ) (2)

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From (1)+(2) we must show this: ≥ (3)

But ≤ ⇒ ≥ (4)

From (3)+(4) we must show: ≥ (5)

But + + = ( − − ) (6)

and = (7)

From (5)+(6)+(7) we must show this: ⋅ ≥ ⇔

≥ ⇔ − − ≥ ⇔ ≥ + (8)

But from Gerretsen’s inequality: ≥ − and from (8) we must show this:

− ≥ + ⇔ ≥ ⇔ ≥ (true)


≤ + + ≤ −

Proposed by George Apostolopoulos-Messolonghi-Greece


= − = − = ∙( − )

=−

−≥ ↔ − ≥ ↔ ≥

−≤ − ↔ ( − ) ≤ − ↔ − + ≥ ↔

↔ − − + ≥ ↔ ( − ) − ( − ) ≥ ↔

↔ ( − )( − ) ≥

997. In , the following relationship holds:

( + )( + )( + ) ≥ √ + √ − √

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Solution 1 by Tran Hong-Vietnam

√ + √ − √

= ⋅ ⋅ + ⋅ − ⋅

= ⋅ √ ⋅ + √ − √

≤ √ ⋅ + − √ ⋅ +

= + + √ ⋅ (1) ( + )( + )( + ) = ( + + + + + + + )

≥ + + ( ) ⋅ ( )

= + + √ ⋅ (2)

From (1) and (2) ⇒ Proved. Equality ⇔ = , = , = .

Solution 2 by Soumava Chakraborty-Kolkata-India LHS= + + (∑ + ∑ ) = + + ∑( + )

≥( )

+ + √ ⋅ = + + √

= + + √ ⋅

RHS= √ ′+ + − √ ⋅

=( )

√ + + − √

(1), (2)⇒ it suffices to prove: √ ⋅ ≥ √ ⇔

⇔ ⋅ ≥ ⇔ ≥ → true ∵ ≥ & (Proved)


+ + ≥ √ Proposed by Daniel Sitaru – Romania

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Solution by Lahiru Samarakoon-Sri Lanka

+ + ≥ √

( + + ) ≥ √ × √ ×

We have to prove, ≥ √

=∑

≥+ +

= √

So, it’s true.

∵ =


( + + ) ≥ Proposed by Adil Abdullayev-Baku-Azerbaijan

Solution 1 by Lahiru Samarakoon-Sri Lanka AM-GM

( + + ) ≥

So, ( + + ) ≥ but, ≥ ( − ). So,

≥ [ ( − ) ( − ) ( − ) ] = [ ]

But, ≥ √ . So, ≥ [ + ] =

Solution 2 by Mehmet Sahin-Ankara-Turkey In any triangle the following inequalities are valid:

≤ + + ≤ (1)

≥ ( − ) (2)

From (1) and (2): ( + + ) ≥ ⋅ ⋅ = ∴


≥ ⇒ ≥ ⇒ ≥( )

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Applying (1) on a triangle with sides , ,

(whose area = of course), we get

∑ ≥ ⇔ (∏ )(∑ ) ≥ (Proved)

1000. Prove that:

=√

√+

Proposed by Vasile Mircea Popa-Romania Solution by Lahiru Samarakoon-Sri Lanka

Let’s consider, √

= ⇔ =√

=√

+

−

√=

But, = −

∴ we have to prove,

−

√−

−

√=

√

So, − − − =

− × ⋅ + ×

− − × + =

− × = × =

Therefore, now we have to prove,

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− − =

Consider L.H.S,

L.H.S = − −

= − −

=+

⋅+

− −

L.H.S = + + + − −

= + + + − −

[∵ ( + ) + ( − ) = ]

= + + + − + − +

(∵ = − and = − )

= + − −

= ⋅ ⋅ + − +

= ⋅ −

= +

∵ = −

So, . . = =

= ⋅

= ⋅ … . ⇔ using similar way,

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L.H.S = = =

L.H.S. = R.H.S. Therefore, it’s true.

∴ =√

√+

(proved)

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

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru

RMM - Triangle Marathon 901 - 1000 · 901 – 1000 . Proposed by Daniel Sitaru – Romania,Vasile...

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