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A nice IMO 1983 inequality from a stronger inequality

Prove variant of triangle inequality containing p-th power for 0 < p < 1A generalization of IMO 1983 problem 6How prove this inequality generalized from 1969 IMO problem 6Stronger Inequality than Holder InequalityStronger than Nesbitt inequalityNice InequalityA beautiful triangle inequalityIMO 1966 Shortlist InequalityA couple of big-O questions about a problem from number theoryInequality from AMM problems section

7

2

$begingroup$

If you are interested in IMO $1983$ please see:
$$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]geqq b(a+b-c)(a-c)(c-b),$$
where $a,b,c$ are three side-lengths of a triangle.

If $c≠rm mida,b,c$, the inequality is true, obviously! If $c=rm mid a,b,c$, we have $$(a - c)(c - b)= 0 Leftrightarrow c= fracc^2+aba+b.$$
I tried to prove that
$$f(c)- fleft(fracc^2+aba+bright)=(a - c)(c - b)Xgeqq 0,$$
where $$f(c)= 3[a^2b(a - b) + b^2c(b - c) + c^2a(c - a)]- b(a + b - c)(a - c)(c - b),$$
but without success! I found this inequality by using discriminant and some coefficient skills.

Hope you try to solve the rest on your own! Good luck everybody! Thank you so much!

inequality contest-math substitution geometric-inequalities buffalo-way

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks $!$ @Eevee Trainer $,$ so much $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It isn't same stronger with Bernhard Leeb $'$ solution $!$ See $:$ artofproblemsolving.com/community/c6h1175790p5672603 $.$ But both are nice $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:14
  
  

  
  
  
  

add a comment | 

7

2

$begingroup$

If you are interested in IMO $1983$ please see:
$$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]geqq b(a+b-c)(a-c)(c-b),$$
where $a,b,c$ are three side-lengths of a triangle.

If $c≠rm mida,b,c$, the inequality is true, obviously! If $c=rm mid a,b,c$, we have $$(a - c)(c - b)= 0 Leftrightarrow c= fracc^2+aba+b.$$
I tried to prove that
$$f(c)- fleft(fracc^2+aba+bright)=(a - c)(c - b)Xgeqq 0,$$
where $$f(c)= 3[a^2b(a - b) + b^2c(b - c) + c^2a(c - a)]- b(a + b - c)(a - c)(c - b),$$
but without success! I found this inequality by using discriminant and some coefficient skills.

Hope you try to solve the rest on your own! Good luck everybody! Thank you so much!

inequality contest-math substitution geometric-inequalities buffalo-way

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks $!$ @Eevee Trainer $,$ so much $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It isn't same stronger with Bernhard Leeb $'$ solution $!$ See $:$ artofproblemsolving.com/community/c6h1175790p5672603 $.$ But both are nice $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 3:14
  
  

  
  
  
  

add a comment | 

$begingroup$

If you are interested in IMO $1983$ please see:
$$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]geqq b(a+b-c)(a-c)(c-b),$$
where $a,b,c$ are three side-lengths of a triangle.

If $c≠rm mida,b,c$, the inequality is true, obviously! If $c=rm mid a,b,c$, we have $$(a - c)(c - b)= 0 Leftrightarrow c= fracc^2+aba+b.$$
I tried to prove that
$$f(c)- fleft(fracc^2+aba+bright)=(a - c)(c - b)Xgeqq 0,$$
where $$f(c)= 3[a^2b(a - b) + b^2c(b - c) + c^2a(c - a)]- b(a + b - c)(a - c)(c - b),$$
but without success! I found this inequality by using discriminant and some coefficient skills.

Hope you try to solve the rest on your own! Good luck everybody! Thank you so much!

inequality contest-math substitution geometric-inequalities buffalo-way

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

$endgroup$

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

share|cite|improve this question

edited 2 days ago

HaiDangel

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

asked Mar 7 at 3:01

HaiDangelHaiDangel

695

1 Answer
1

active

oldest

votes

3

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 11:16
  

  
  
  
  

add a comment | 

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3

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 11:16
  

  
  
  
  

add a comment | 

3

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Yes $!$ Your inequality is also very nice $!$ I saw it on AoPS $!$
  $endgroup$
  – HaiDangel
  Mar 7 at 11:16
  

  
  
  
  

add a comment | 

$begingroup$

Consider three cases.

1. 
   $a=maxa,b,c$, $a=x+u+v,$ $b=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+3(2u^3+u^2v-uv^2+v^3)x+2u^3(u+2v)geq0;$$

2. 
   $b=maxa,b,c$, $b=x+u+v,$ $a=x+u$ and $c=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(4u^2-4uv+3v^2)x^2+(6u^3-5u^2v+5uv^2+3v^3)x+2u(u^3-uv^2+3v^3)geq0$$ and

3. 
   $c=maxa,b,c$, $c=x+u+v,$ $a=x+u$ and $b=x+v$, where $x>0$, $ugeq0$ and $vgeq0.$
   

Thus, $$3[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)]-b(a+b-c)(a-c)(c-b)=$$
$$=(3u^2-2uv+3v^2)x^2+(3u^3+6u^2v-2uv^2+3v^3)x+6u^3vgeq0$$ and we are done!

Actually, the following stronger inequality is also true.

Let $a$, $b$ and $c$ be sides-lengths of a triangle. Prove that:
$$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)geq b(a+b-c)(a-c)(c-b).$$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

$endgroup$

share|cite|improve this answer

edited Mar 7 at 4:42

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200

answered Mar 7 at 4:08

Michael RozenbergMichael Rozenberg

107k1895200