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Difference between revisions of "2005 Cyprus Seniors TST/Day 1/Problem 2"

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== Solution ==
 
== Solution ==
  
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Let <math>\alpha =y+z, \beta =x+z, \gamma =x+y</math>. Then the inequality becomes
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<math>2x(y+z)^2+2y(x+z)^2+2z(x+y)^2 \leq 3(x+y)(y+z)(z+x) </math>
  
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After expansion, it is equivalent to
  
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<math>x^2y+y^2x+x^2z+z^2x+y^2z+z^2y \geq 6xyz</math>
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Which is true by <math>AM-GM</math> inequality
 
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Latest revision as of 21:50, 12 October 2022

Problem

Given a triangle with sides $\alpha, \beta, \gamma$. Prove that: $\alpha^2 (\beta+\gamma-\alpha)+\beta^2 (\gamma+\alpha-\beta)+\gamma^2 (\alpha+\beta-\gamma)\leq 3 \alpha \beta \gamma$

Solution

Let $\alpha =y+z, \beta =x+z, \gamma =x+y$. Then the inequality becomes $2x(y+z)^2+2y(x+z)^2+2z(x+y)^2 \leq 3(x+y)(y+z)(z+x)$

After expansion, it is equivalent to

$x^2y+y^2x+x^2z+z^2x+y^2z+z^2y \geq 6xyz$

Which is true by $AM-GM$ inequality

See also

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