Question

Prove that $(x+y)(y+z)(z+x)\ge 8xyz$ if $x,y,z>0$.

Answer 1

$LHS=x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y+2xyz$

Hence we need to prove

$x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y\ge 8xyz-2xyz$

A.M $\ge$G.M

$\frac{x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y}{6}\ge \sqrt[6]{6x^{2}y.x^{2}z.y^{2}x.y^{2}z.z^{2}x.z^{2}y}$

$\ge \sqrt[6]{6x^6{y}^6{z}^6}$

Hence $\frac{x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y}{6}\ge xyz$

$\Rightarrow x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y\ge 6xyz$