Question

Show that $(x^{2}y+y^{2}z+z^{2}x)(x{y}^2+y{z}^2+z{x}^2)>9x^2{y}^2{z}^2$.

Answer 1

This is only true when $x,y,z$ are positive numbers

Apply $AM\ge GM$ for $x^{2}y,y^{2}z,z^{2}x$

$\Rightarrow \frac{x^{2}y+y^{2}z+z^{2}x}{3}\ge xyz\to$ eqn 1

Apply $AM\ge GM$ for $x{y}^2,y{z}^2,z{x}^2$

$\Rightarrow \frac{x{y}^2+y{z}^2+z{x}^2}{3}\ge xyz\to$ eqn 2

Multiply corresponding sides of eqn 1,2

$(x^{2}y+y^{2}z+z^{2}x)(x{y}^2+y{z}^2+z{x}^2)\ge 9x^2{y}^2{z}^2$