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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
www.ssmrmh.ro
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)
902. In the following relationship holds:
+ + + + + ≤+ +
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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
Solution by Daniel Sitaru-Romania
= ∙ = = =
= = ( − ) =( − )( − )( − )
( − ) =
= ( − )( − ) =( − )( − )
=
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=( − )
∙( − )
=
905. In ∆ , − ′ , ⊥ , ⊥ , ⊥ ,
= , = , = . Prove that:
∙+
∙+
∙= ∙ ( + + )
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
− ′ → = = =⏞ ,
=+ +
=( + + )
→ = + +
=( , , )( , , )
∙= ∙ ( + + ) =
( , , )
=∙ + +
∙ ( + + ) = ∙ ( + + )
906. In ∆ the following relationship holds: ++ +
++ +
++ =
( + )
Proposed by Bogdan Fustei-Romania
Solution by Daniel Sitaru-Romania
++ =
( , , )
+
− + −=
( , , )
+∙
( − )( − )− − =
( , , )
=( − ) ∙
( − ) =( , , )
−−
( , , )
=
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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)
Solution 4 by Soumava Chakraborty-Kolkata-India
= ⇒ =( )
= = ⇒ =( )
(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)
Solution 2 by Rovsen Pirguliyev-Sumgait-Azerbaijan
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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⇒ = − ⇒ = = ° − ∴ = °
Solution 3 by Soumava Chakraborty-Kolkata-India
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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Solution 1 by Rovsen Pirguliyev-Sumgait-Azerbaijan
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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Solution 2 by Soumava Chakraborty-Kolkata-India
From , = ⇒ =( )
.
From , = ⇒ =( )
(1), (2) ⇒ = ⇒ =( )
Also, from , = ⇒ =( )
& from , = ⇒ =( )
(3), (4) ⇒ = ⇒ =( )
From , = + − ⇒ − =( )
−
=−
=
=( )
( − ) =
=( − )
=
=( − − + )
=
=( ) ( − − + )
=
=+ ( − + )
=+
=?
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⇔ + =? = = − ⇔ + =? − ⇔
⇔ + − =? ⇔ =? ±√ = ±√ → true
∵ = √ ⇒ = + (Proved)
912. In ∆ , − , ⊥ , ⊥ , ⊥ ,
= , = , = . Prove that: + +
+ + =+ +
+ + = −
Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania
− → = = =⏞
+ ++ + =
+ ++ + =
+ ++ + =
( , , )
=
= −( , , )
= −( . . )
= ∙( − )
= −
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:
=−
, = , + + =[ ] − [ ]
Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania
= , = , =
∢( ) = − ,∢( ) = − ,∢( ) = −
= = = , = , =
∑ = ∑ = − = , ∏ = = =
= = ( − ) = − =
= − ∙ =−
=[ ]− [ ]
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,
= ∴ = =
Case 2b) = (& of course = )
(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) ∴ = = .
Case 3b) = (& of course = )
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
918. In the following relationship holds:
+ − = + + +
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:
⋅ ⋅⋅ ⋅ = ⋅ ⋅
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Lahiru Samarakoon-Sri Lanka
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⋅ ⋅⋅ ⋅
= ⋅ ⋅ . Consider, , = =
So, similarly, = , = . From, ,
=− −
=+
=
Similarly, = , =
=⋅ ∏
∏ =
= × ( ) ∏ ⋅ ∏
∏ =∏ ⋅ ∏
⋅ ∏ ⋅ ∏= ⋅ ⋅
(proved)
922. In the following relationship holds:
+ ++ + + + + =
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
923. In the following relationship holds:
≥ +
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:
⋅ + ⋅ + ⋅ ≥ ( + + )
Proposed by Daniel Sitaru – Romania
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)
From (3)+(4) we must show:
+ + ≥ ( + + ) ⇔
⇔ + + ≥ + + true.
Solution 2 by Soumava Chakraborty-Kolkata-India
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)
925. In the following relationship holds:
+ − > 4 Proposed by Marian Ursărescu – Romania
Solution by Ravi Prakash-New Delhi-India + − =
= + + ( + − ) = + ( + ) + =
= + + + + − + =
= ( + ) + ( − ) + ≥ =
926. In the following relationship holds:
+ + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution 1 by Soumava Chakraborty-Kolkata-India
≥
= ≥(∑ )
(∑ ) ≥√ (∑ )
(∑ ) =√ (∑ )
∑ ≥( )
≥√ (∑ )∑ ( − )
≥√ (∑ )
√ √ ∑( − )=√ (∑ )√
=∑
≥?
⇔
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⇔ + + ≥?
⇔ ≥?
− .
Now, ≥ − ≥?
− ⇔ ≥?
⇔ ≥?
→ true (Euler)
(proved)
Solution 2 by Marian Ursărescu-Romania
= ⋅ ( ) ≤ √ ⋅ ( ) = ( − ) ⇒ we must show:
( − )+
( − )+
( − )≥ ⇔
√+
√+
ó ( )( )( ) (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.
927. In ∆ the following relationship holds:
++
++
+≥
√
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
+ = + ≥⏞( + + )
+ ( + + ) = + =
= + ≥⏞√
( + ) ≥⏞√
+=
√
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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)
929. In the following relationship holds:
+ + ≤√
+ +
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
≤√
≥ ⇔ ≤ ⇔ ≥ . Similarly, ≥ ⇔ ≤ ⇔ ≥ .
WLOG, we may assume ≥ ≥ . Then, ≥ ≥ & ≤ ≤
∴ ≤ ≤ ≤
≤ ∑ ⋅ √ = √ ∑ (proved)
930. In ∆ the following relationship holds:
+ + + + + ≥ −
Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania
+ =−
+ = ( + ) =
= ∙ ( − − ) ≥⏞ ( − − − ) =
= ( − ) = ( + − ) ≥⏞ ( + − ) = −
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931. In the following relationship holds:
≤( + )( + )( + )
≤
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)
Solution 2 by Marian Ursărescu-Romania
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.
932. In the following relationship holds:
° ≤ ( + + ) + ( + + ) + ( + + ) ≤ °
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Soumava Chakraborty-Kolkata-India
( + + ) = + + + + + =( )
− − + +
+( + ) − + ( ) + ( + ) + ∑ . Now,
= = = ( − ) =
= − − =−
=( )
( − )
(1),(2)⇒ ∑( + + ) = − ( + ) + + ( + ) + + ( + ) +
+ − =( )
+ ( + ) + ≥ + + ≥√
+ √ = + √ = + √ =− √
= °
∵ ° = ° =− √
(3) ⇒ ∑( + + ) ≤ + + ⋅ √ = + √ = + √
= + √ =√
= ° = ° (Done)
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933. In the following relationship holds:
( − ) + ( − ) + ( − ) ≤+ +
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
( − ) ≤(∑ )
≤ ( − )( − )( − )−
=√
( − )=√
( − )
≤( ) √
( − ) =√
≥( ) (∑ )
= =
(1), (2) ⇒ it suffices to prove:
≥√
⇔ ≥ ⇔ ≥ → true (Euler) (Proved)
934. In ∆ the following relationship holds:
√−
+−
+−
≤ + + +
Proposed by Bogdan Fustei-Romania Solution by Daniel Sitaru-Romania
+ + + = +−
+−
+−
=
= +( − )
+( − )
+( − )
= ∙( , , )
=
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= √ ∙ ( − ) ∙( , , )( , , )
≥⏞ √ ∙ √ − ∙√( , , )
=
= √−
+−
+−
935. In ∆ the following relationship holds:
+ + + + + ≥ −
Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania
+ = +−
= ( − ) = − =
= + + − = + + − ≥⏞ + − = −
936. In the following relationship holds:
( + ) ≤ √
Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India
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
Solution by Daniel Sitaru-Romania
≥ + + ↔ ≥ + + ↔ + + ≥ ↔ + ≥
+ ≥⏞ − + ≥ ↔ ≥ ↔ ≥
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:
⋅ + ⋅ + ⋅ ≥ ( + + )
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
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)
940. In ∆ the following relationship holds:
√ ++√ +
+√ +
≥√
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
√ +≥⏞
√ ∙ +=√
( , , )
−( , , )
=( , , )( , , )
=√ ( + + ) ≥
√( ) ↔ + + ≥
+ + ≥⏞ − + + ≥ ↔ ≥
941. In acute the following relationship holds:
+ + ≥
Proposed by Daniel Sitaru – Romania
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Solution 1 by Bogdan Fustei-Romania Knowing the identity: + + = . The inequality from
enuciation becomes: + + ≥ ; ∑ = = = . So,
≥ ⇔ ≤ ⇔ ≤ (Euler)
Solution 2 by Serban George Florin-Romania
=+ −
≥ , = ⇒
⇒ + − ≥
+ + −+ +
≥
( + + ) − ( + + ) ≥
= ( − )( − )( − ) (Heron)
( + + )− ( + + ) ≥
≥( + + )(− + + )( − + )( + − )
( + + )− ( + + ) ≥ [( + ) − ][ − ( − ) ]
− ≥ ( + ) − − ( − ) + ( − )
− ≥ + + − − − + + + −
− = −∑ + ∑ (true)
Solution 3 by Ravi Prakash-New Delhi-India
+ + = [ + + ] =
= [ + + ] =
= [ + + ] = [ ] = =
= ≥
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942. In acute ∆ , − , ⊥ , ⊥ , ⊥
,
= , = , = , − . Prove that:
+ + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
− ʹ → = = =⏞ ,
=+ +
=( + + )
→ = + +
=( , , )( , , )
= =( , , )( , , )
( , , )
= ( + + ) ∙ − ( + ) = ( − − )( − ( + ) ) ≥
≥⏞ ( + + − − )( + + − ( + ) ) =
= ( + ) ∙ ≥⏞∙ + ∙
=
943. In ∆ the following relationship holds:
( − )+
( − )+
( − )≥
−
Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania
( − )
( , , )
≥⏞( − − − )
+ + =( + )+ + ≥
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≥⏞( + )
+ + + += ≥
−↔ ( − )( + ) ≥
944. In the following relationship holds:
≥ +
Proposed by Bogdan Fustei – Romania
Solution by Soumava Chakraborty-Kolkata-India
= =
≥∑
≥ √ (∵ ( ) = ∀ ∈ ( , ) is concave as ( ) < 0)
= √ ∴ LHS ≥( )
√
Now, ∑ + ∑ = {( + + ) + ( + + ) + ( + + )}
≤ √ = √ ≤( )
(Proved)
945. In the following relationship holds:
≤ + + ≤−
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution 1 by Marian Ursărescu-Romania
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.
Solution 2 by Soumava Chakraborty-Kolkata-India
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: −+
+−+
+−+
≥ °
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
( ) =−+
, ( ) =−
( + ) , ( ) =( + ) +
( + ) > 0, −
−+ =
−+ ≥⏞
+ += = ∙
− √
+ √=
= − √ = − √ = °
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947. In acute the following relationship holds:
+ + ≥ √
Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India
≥ √
Let ( ) = ;∀ ∈ , ; ( ) = > 0 ∴ ( ) = is convex on
, . WLOG, we may assume ≥ ≥ . Then, ≥ ≥ & ≥ ≥
∵ ( ) = ,
∴ ≥ ≥ √
∵ ( ) = ,
= √ (∑ ) ≥ √ (∑ ) = √ (proved)
948. In ∆ , − , − ,
− . Prove that:
[ ] + [ ] + [ ] ≥ [ ] ∙
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
[ ] =∙ ∙
=∙ ∙ ∙ ∙
=
= ∙∙
= ∙ ∙ = ∙ ∙ =
= [ ] ∙ ∙ ≥ [ ] ∙ ↔ ≥
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= − − ≥ ↔ ≥ +
≥⏞ − ≥ + ↔ ≥ ↔ ≥
949. In ∆ the following relationship holds:
+ + + + + ≥ −
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
+ =− + −
=( − )( − )
=− ( + ) +
=
= −−
+ ≥⏞ − + + ≥
≥ − + ∙ + = −
950. In ∆ the following relationship holds:
++
++
+≥
Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan
Solution by Daniel Sitaru-Romania
+ =+
= + = + ≥⏞
≥ ∙( + + )
( + + ) = ( + + ) =
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951. In ∆ the following relationship holds:
+ + ≥√
Proposed by Nguyen Van Nho-Nghe An-Vietnam Solution by Daniel Sitaru-Romania
( , , )
=( , , )
≥⏞( + + )( + + ) = ≥⏞
≥∙ √ ∙
=∙ ∙ √ ∙
=√
=√
952. In ∆ the following relationship holds:
+ + + + + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
+ ≥⏞( + + )( + + ) =
∙ + + = + + ≥
≥⏞ + + + + = ( + ) ≥⏞+
=
953. In ∆ , − , ⊥ , ⊥ , ⊥ ,
= , = , = . Prove that:
√ ≤ + + ≤
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
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− → = = =⏞
=+ +
=( , , )
→( , , )
=
=( , , )( , , )
= =( , , )
( + + )≥
≥⏞∙ √
= √
=( , , )( , , )
= =( , , )
( + + )=
=∙ ( − − )
≤⏞( + + − − )
=
=( + )
≤⏞+
=∙
=
954. In ∆ the following relationship holds:
+ + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
≥⏞( + + )
+ +=
( + + )=
= ∙ ( + + ) ≥ ∙=
955. In acute the following relationship holds:
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+( + + − )
≥
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
Solution by Daniel Sitaru-Romania
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: ( , ) → ℝ, ( ) = , ( ) = − , − →
++ +
≤+
( , , )( , , )
↔
↔ + ≤−
( , , )( , , )
↔
↔ +√
≤( , , ) ( , , )
957. If , > 0, + ≤ 1 then in ∆ the following relationship
holds:
+ + + + + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
+=
( + )≥⏞ ∙
( + + )( + ) + ( + ) + ( + ) =
= ( + )( + + ) ≥⏞ + + ≥⏞√
=√
= ∙√
≥
≥⏞ = ≥⏞ ∙ =
958. In ∆ the following relationship holds:
+ + ≥ + +
Proposed by Bogdan Fustei-Romania Solution by Daniel Sitaru-Romania
≥⏞ ( + ) = = =∙
=
959. In ∆ the following relationship holds:
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≤ + + ≤
Proposed by George Apostolopoulos-Messolonghi-Greece Solution by Daniel Sitaru-Romania
=( − )( − )
≥⏞( − ) ( − ) ( − )
=∙
=
= ≥ ↔ ≥ ↔ ≥ ↔ ≥
=( − )( − )
≤⏞− + −
= ∙ ∙ =
960. In the following relationship holds:
+ + ≤
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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≤ + ⋅ = ⋅ =
Solution 3 by Soumava Chakraborty-Kolkata-India
By Tsintsifas, ∑ ≤ ∑ = ∑ = ∑ ( ) = =
=− +
≤?
⇔ ≥( )
?− +
Now, RHS of (1) ≤ + + ≤?
⇔ − − ≥?
⇔
⇔ ( − )( + ) ≥?
→ (true) (Euler) (proved)
961. In the following relationship holds:
+ + ≥ √ ( + + )
Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India
≥ √
WLOG, we may assume ≥ ≥ we shall prove ≤ ≤
≤ ⇔ ( + ) ⋅+ −
≤ ( + ) ⋅+ −
⇔
⇔( + − )
( + ) ≤( + − )
( + ) ⇔ + − ( + ) ≤
≤ + − ( + ) ⇔ + + ( + ) ≤( )
+ + ( + )
Now, ( + ) ≤ ( + ) (∵ ≥ ) ⇒ ≥ ⇒
⇒ ≤( )
. Also, ( + ) ≤ ( + )(∵ ≥ ) ⇒
⇒ + ≤( )
+
(a)+(b)⇒(1) is true ⇒ ≤ . Similarly, ≤ ∴ ≤ ≤ . Also, ≥ ≥ ⇒
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⇒ ≤ ≤ ∴ ≥ ≥
≥∑
∵ ( ) = ∀ ∈ ( , )
≥ ∑ √ (∵ ∑ ≥ ∑ ) = √ (∑ ) (proved)
962. In ∆ the following relationship holds:
++
++
+≥ √
Proposed by Mehmet Sahin-Ankara-Turkey
Solution by Daniel Sitaru-Romania
+ =+
= + = + =
= ( + ) = + ≥⏞( + + )
( + + ) ≥
≥⏞∙ √
= √
963. If in , ( ) ≥ ( ) ≥ ( ) then:
+ + ≤+ +
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
Given inequality ⇔ ≤ ⇔
⇔+ +
≤+ +
⇔
⇔ ( − ) + ( − ) + ( − ) ≥?
∵ ≥ ≥ ∴ ≥ ≥
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Let = + & = + + ( , ≥ ). Using the above substitution (1) ⇔
⇔ ( + + )( + ) + ( + ) + ( + + )(− − ) ≥ ⇔
⇔ ( + + )( + )− ( + + ) + ( + )− ( + + ) ≥ ⇔
⇔ ( + + ) − ≥ ⇔ ( + ) ≥ → true ∵ , ≥ (Proved)
964. In the following relationship holds:
+ +
+ ++
+ +
+ ++
+ +
+ +≥
Proposed by Daniel Sitaru – Romania
Solution 1 by Serban George Florin-Romania
= + + =+ +
= = ; =
⇒+
+≥
∑ +
∑ +=
( + )+ =
= ( ) = = ≥ ⇒ ≥ |: ⇒ ≥ (Euler)
Solution 2 by Tran Hong-Vietnam Using Schwarz’s inequality we have:
≥+ +
+ += + +
+ + =+ +
= = ≥( )
Equality ⇔ = = .
965. In the following relationship holds:
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+ ++ ( − )( − )( − ) ≥
√
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
Mitrinovic ⇒ ≥ √ , which ⇒ it suffices to prove: ∏( ) (∑ ){∏( )}∑
≥( )
Now, ∏( + ) = + ∑ ( − ) = + + − =( )
+ +
(1) ⇒ LHS of (a) = ≥?
⇔ + ( + ) + ( + ) ≥?
+ ( + ) + ( + )( + )
⇔ ( + )( − ) ≥?
⇔ ≤( )
?− −
Now, LHS of (b) ≤ + + ≤?
− −
⇔ ( − )( + ) ≥?
→ true ∵ ≥ (Proved)
966. In the following relationship holds:
( + + )
( + + )+
( + + )≥ √
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) ≤ ( + + )
∴ in order to prove (3), it suffices to prove:
( − ) − + + + + + + ≥
≥( )
Now, LHS of (4) ≥ ( − )( − )( − + ) +
+ + + + +
& RHS of (4) ≤ ( + + )
∴ in order to prove (4), it suffices to prove:
− + − + ≥ =
⇔ ( − ){( − )( + + ) + } ≥
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→ true ∵ ≥ ⇒ (4) is true ⇒ (i) is true
Applying (i) on a triangle with sides = , , , we get,
∑
∑+
∑ ⋅ ∑≥ √ ⇒
∑∑
+( ∑ ) ∑
≥ √
⇒∑
∑+
∑ ∑≥ √ ⇒
∑
∑+
(∑ )≥ √
(proved)
Solution 2 by Marian Ursărescu-Romania
( + + )
( + + )+
( + + )+
( + + )+
( + + )≥
≥( + + ) + ( + + )
( + + ) ⋅⇒
We must show:
( + + ) ⋅ ( + + ) ( + + )
( + + ) ⋅≥ √ ⇔
⇔( + + ) ( + + )
≥ ⇒
( + + )( + + ) ≥ ⋅ (1)
But + + = ⇒ + + = + + (2)
From (1)+(2) we must show:
( + + ) ⋅ + + ≥ ⋅ ⋅ ⋅ ⋅ ⇔
( + + ) + + ≥ (3)
From Cauchy inequality we have:
+ + ≥ ( + + ) ⇒
+ + ≥ ( + + ) (4)
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From (3)+(4) we must show:
( + + ) ( + + ) ≥ ⇔
⇔ ( + + ) ≥ ⇔ + + ≥ , which its true.
967. In ∆ the following relationship holds:
+ + ≤
Proposed by Bogdan Fustei-Romania
Solution by Daniel Sitaru-Romania
+ + ≤ − ( )
=∙
= = =
= ≤⏞ ∙ = ∙ =
968. In ∆ the following relationship holds: +
++
++
≤
Proposed by Bogdan Fustei-Romania
Solution by Daniel Sitaru-Romania
+=
+=
+=
( + )= ( + ) =
= ( − ) = − =
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= ( ∙ ( − − ) − ( − − )) = ( + − ) ≥
≥⏞ ( − + − ) = ( − ) ≤ ↔
↔ − ≤ ↔ ( − )( − ) ≥
969. In the following relationship holds:
⎝
⎜⎜⎜⎛
+⎠
⎟⎟⎟⎞
≥
Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India
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)
971. In ∆ the following relationship holds:
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+ + ≥√
Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan
Solution by Daniel Sitaru-Romania Known:
≥ ( )
≥ ( )
= = ≥⏞
≥ ∙ ∙ ≥⏞( ),( )
∙ ∙ = ≥
≥⏞ ∙ ∙ = ( ) = ≥√
↔
↔ ≥√
↔ ≥√
↔ ≤√
↔ ≤√
( )
972. In the following relationship holds:
+ + ≥ √ ⋅
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)
973. In acute the following relationship holds:
+ + ≤ ( + + − )
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:
⋅( ⋅ + ⋅ ) − ⋅
( , , )( , , )
≥
Proposed by Daniel Sitaru – Romania
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:
+ + ≥ √ ( + ) + ( + ) + ( + )
Proposed by Daniel Sitaru – Romania
Solution by Soumava Chakraborty-Kolkata-India
∵ 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)
976. In the following relationship holds:
( ° + ° + °)° + ° + ° > 108
Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India
∵ + + ≥ ,∀ , , ≥
∴ ° + ° + ° ≥ √ ° ° ° + ° °
> 4 √ ° ° + ° ° + ° °
= ( ° + ° + ° + ° + ° + °)
= −√
+ ° + ° + °
= −√
+ ° ° + ° −
>( )
−√
+√
° + ° −
= −√
+ √√ +
++ √
−
=− √ + √ + √ + + √ −
=+ √ + √ + √ − √
Also, ° < ° = √√
⇒°
>( ) √
√
& ∵ ° , ° < ° = √
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∴ ° + ° >( )
⋅√
=√
(1), (2), (3) ⇒ LHS
> 4+ √ + √ + √ − √ √ +
√+√
>√
√+ √ + √ + √ − √ √ +
√+√
> 120 > 108 (Proved)
977. In ∆ the following relationship holds:
( − )( − )+
( − )( − )+
( − )( − )≥ √
Proposed by Bogdan Fustei-Romania
Solution by Daniel Sitaru-Romania
( − )( − )
( , , )
= − − − −( , , )
=
= ( − )( − )( , , )
=( , , )
∙ ≥⏞
≥∙
=∙
=∙
=
= ≥⏞√
=√
=√
= √
978. ADIL ABDULLAYEV’S REFINEMENT FOR TERESHIN’S
INEQUALITY
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In the following relationship holds:
≥+
+( − ) ( − − )
Proposed by Adil Abdullayev-Baku-Azerbaijan Solution by Soumava Chakraborty-Kolkata-India
≥( ) +
+( − ) ( − − )
(1) ⇔ ≥ ( + ) + ( ) ⇔
⇔( + )
+( − )
− ≥ ⋅ ( + ) +( − ) ⋅
⇔( + ) −
+( − )
≥( + )
+( − )
( − )
⇔ ( − ) ≥( + )
− ( − ) ⇔
⇔ ≥( + )
− ( + ) + ⇔
⇔ − ( + )+ −
+ ≥
⇔ − ( + ) + ≥ ⇔
⇔ − ++
≥ ⇔
⇔ − ( − ) ++
≥ ⇔
⇔ − −+
+ ++
≥ ⇔
⇔+
−+
− ≥ ⇔
⇔+
− ≥ ⇔
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⇔ ( − ) ≥ → true ⇒ (1) is true (Proved)
979. In the following relationship holds:
− + − + − ≤ + + +
Proposed by Bogdan Fustei-Romania
Solution by Soumava Chakraborty-Kolkata-India
= + ⋅−
= +−
= + − =( )
= ⋅ − = − =−
≤ { ( − )}
= ∑ = (by (1)) (Proved)
980. In ∆ the following relationship holds:
+ + ≥
Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan
Solution by Daniel Sitaru-Romania Known: + + ≥ ( )
≥⏞( + + )
+ + ≥⏞( )
≥ = ≥⏞∙
=
981. In acute the following relationship holds:
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( + − ) ≥ ( + + )
Proposed by Daniel Sitaru – Romania
Solution 1 by Marian Ursărescu – Romania
⋅ ≥ ⋅ ⋅ ( + + ) ⇔
⇔ ≥ ( + + ) ⇔
⇔ ∑ ≥ ( + + ) (1)
∑ ≥ √ (2)
From (1)+(2) we must show:
√ ≥ ⋅ ( + + )
⇔ √ ≥ ( + + ) ⇔
⇔ ≥ ( + + ) ⇔
≥ ( + + ) (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)
982. In the following relationship holds:
− ≤ + ( + )
Proposed by Bogdan Fustei – Romania
Solution 1 by Daniel Sitaru – Romania Known: ≤ + (1)
− =( )
( + )−⋅
=
= + − = + − =
= + − = = + +
Solution 2 by Soumava Chakraborty-Kolkata-India
= ⋅ =⋅
= ( − − )
=( )
− −
Now, ∑ = {(∑ ) −∑ } ≤( )
&( − + ) −
− ⋅ ⋅ ( − − ) = − + − + + =
= − +
(1)+(2)⇒LHS ≤ ( − + ) − ( − − ) =
= − + ≤?
+ + ⇔ ≥?
⇔ ≥?
→ true (Euler)
(Proved)
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983. In the following relationship holds:
+ + − − − < 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
Solution 1 by Bogdan Fustei-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.
Solution 2 by Soumava Chakraborty-Kolkata-India
− − ≤( ) ∑
≤( )
− +
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)
985. In the following relationship holds:
(( − ) + ( − ) + ( − ) )≤ −
Proposed by Adil Abdullayev-Baku-Azerbaijan Solution by Soumava Chakraborty-Kolkata-India
∑( − )≤
−
Given inequality ⇔ ∑ + − ≤ ( − ) ⇔
⇔ − ≤ ( − ) ⇔ ( + ) − − ≤
≤ ( − ) ⇔ ≥( )
( + )
Now, LHS of (1) ≥ ( − ) ≥( )
( + ) ⇔
⇔ − − ≥?
⇔
⇔ ( − )( + ) ≥?
→ true (Euler) ⇒ (1) is true (Proved)
986. In ∆ the following relationship holds:
( + + ) ≥ ( + + ) Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan
Solution by Daniel Sitaru-Romania
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Panaitopol’s inequality (1980): ≥ ( − )
= ∙ = ≥
≥⏞ ∙ ( − ) = ∙ ∙−
=
= ∙+ −
≥ ( + − ) ≥⏞
≥ ( + − ) ≥ ( + + ) ↔
+ − ≥ + + ↔ ≥ −
↔ ≥ − ( )
≥⏞ − ≥ − ↔ ≥ ↔ ≥
987. In the following relationship holds:
( + + ) ≥ √ 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.
Solution 2 by Soumava Chakraborty-Kolkata-India
≥ √
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Applying the above inequality on a triangle with sides , , whose area of
course will be , we get,
≥ √ ⇒ ≥( )
√
Now, (∑ ) ≥ ∑ ≥( )
√ (Proved)
988. In the following relationship holds:
√ ≤ + + ≤ √ ⋅
Proposed by Mehmet Sahin-Ankara-Turkey
Solution 1 by Soumava Chakraborty-Kolkata-India
= =( )
≥ =
= ≥?√ ⇔ ≥
( )
?
Now, LHS of (a) = ⋅ ⋅ ≥ ⋅ ⋅ = ⋅ = ( × ) = ⇒
(a) is true ⇒ ∑ ≥ √
Also, ∑ =( )
∑ ≤ ∑ = ∑( + ) = + +
= ≤⋅
= ≤√
√= √ ⇒ ∑ ≤ √ (Done)
Solution 2 by Marian Ursărescu-Romania
+ + ≥ (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:
+ + ≥
Proposed by Mehmet Sahin-Ankara-Turkey Solution by Daniel Sitaru-Romania
= , = , =
∢( ) = − ,∢( ) = − ,∢( ) = −
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= = = , = , =
= + ( − ) ∙( − )( − )
=
=( − )( − )( − )
+ ≥⏞ ∙( + + )
+ + + + + =
= ∙ =
990. In acute the following relationship holds:
+( + + − )
≥
Proposed by Daniel Sitaru – Romania Solution by Soumava Chakraborty-Kolkata-India
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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Solution by Daniel Sitaru-Romania
= , = , =
∢( ) = − ,∢( ) = − ,∢( ) = −
= = = , = , =
= = =( − )( − )
=
= ( − )( − ) = ∙( − )( − ) =
= ∙ ( + ) = ( + + )
993. In , ≥ ≥ , + ≥ .
Prove that: − ≥ .
Proposed by Nguyen Van Canh-Vietnam
Solution by Tran Hong-Vietnam
We have: = ( )( )( ) ; Let ( ) = ( )( )( ),
⇒ ( ) =( + ) − + ( + )( − )
≥( + − )
=+ −
≥ ;
⇒ ( ) ≤+
=( − )( − )
= − ⋅( − )
≤ ;
⇔ ≤ ⇔ ≤ . (Proved)
994. In the following relationship holds:
+ + ≤ −
Proposed by Mehmet Sahin-Ankara-Turkey
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Solution 1 by Bogdan Fustei-Romania
≥ ( − ) = ⇒ ≥ ⇒ ≤ (and the analogs)
⇒ + + ≤ + +
( − )( − ) = − ( + ) + | ⋅ ⇒ ( − )( − ) =
= − ( + ) + ⋅ (and the analogs) ⇒ ∑ ( − )( − ) =
= ( + + )− ( + + + + + ) + ( + + ) =
= ( − − ) − ( + − ) + ⋅
+ + = ( − − )
+ + + + + = ( + − )
( − )( − ) = ( − − − − + + ) =
= ( − )
( − )( − ) = ( − ) = ( − ) = ( − )
= + + =+ +
=+ +
= ( − )( − ) (and the analogs); =
= ( − )( − ) =( − )
=( − )
=( − )
∑ = − (Q.E.D)
Finally, we have the inequality from enunciation: + + ≤ −
Solution 2 by Marian Ursărescu-Romania
We have: ≥ ⇔ ≥ ( ) ⇒
≥ √ ( ) ⇒ ≥ ( − ) ⇒ we must show:
+ + ≤ − (1)
But in we have: + + = ( ) (2)
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From (1)+(2)⇒ + + = ( ) = − ⇒ relationship it’s true.
995. In the following relationship holds:
+ + ≥
Proposed by Mehmet Sahin-Ankara-Turkey
Solution 1 by Soumava Chakraborty-Kolkata-India
LHS =( ) ∑
= = ( ) =
= { ( − ) − ( + )} =
= − { ( + ) ( − )} + ( ) =
= − ( + ) + ( + ) =
=( )
( + − − )
Similarly, =( )
( + − − ) &
=( )
( + − − )
(a)+(b)+(c) ⇒ ∑ = ( − ∑ ) = { − (− − )}
=( )
( + )
(1), (2)⇒ LHS = ( ) = = + (where = ∏ )
∵ is acute – angled, ∴ < ≤ ⇒ + ≥ ⇒ + ≥
≥ = (Proved)
Solution 2 by Marian Ursărescu-Romania
+ + = (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)
996. In ∆ the following relationship holds:
≤ + + ≤ −
Proposed by George Apostolopoulos-Messolonghi-Greece
Solution by Daniel Sitaru-Romania
= − = − = ∙( − )
=−
−≥ ↔ − ≥ ↔ ≥
−≤ − ↔ ( − ) ≤ − ↔ − + ≥ ↔
↔ − − + ≥ ↔ ( − ) − ( − ) ≥ ↔
↔ ( − )( − ) ≥
997. In , the following relationship holds:
( + )( + )( + ) ≥ √ + √ − √
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Proposed by Daniel Sitaru – Romania
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)
998. In the following relationship holds:
+ + ≥ √ Proposed by Daniel Sitaru – Romania
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Solution by Lahiru Samarakoon-Sri Lanka
+ + ≥ √
( + + ) ≥ √ × √ ×
We have to prove, ≥ √
=∑
≥+ +
= √
So, it’s true.
∵ =
999. In the following relationship holds:
( + + ) ≥ 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): ( + + ) ≥ ⋅ ⋅ = ∴
Solution 3 by Soumava Chakraborty-Kolkata-India
≥ ⇒ ≥ ⇒ ≥( )
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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