2005 Cyprus Seniors TST/Day 1/Problem 2

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