Inequalities Of Polynomial [closed]

Let $\ (x,y,z) \ $ be positive real numbers such that $\ \ (x+y)(y+z)(z+x)=8 \ \ $ Prove That $x^3y^3+y^3z^3+z^3x^3+x^2y^2z^2-4xyz>=0$

Solution 1:

Muirhead Sols First homogenize, by multiplying by $2$ and replacing the coefficient of $xyz$ with the condition. \begin{align*} \sum_{sym}x^3y^3+2x^2y^2z^2-(x+y)(y+z)(z+x)xyz&\ge0\\ \sum_{sym}x^3y^3+2x^2y^2z^2-\sum_{sym}x^3y^2z-2x^2y^2z^2&\ge0\\ \sum_{sym}x^3y^3&\ge\sum_{sym}x^3y^2z \end{align*} which follows by Muirhead.

Schur Solution

$x^3y^3+y^3z^3+z^3x^3+3x^2y^2z^2\ge\sum_{sym}x^3y^2z.$ Then using this bound in the homogenized inequality we can simplify to $\sum_{sym}x^3y^2z\ge6x^2y^2z^2$ which follows by AM-GM