Question

If x, y, z $>0$, then prove that $x^3{y}^3+y^3{z}^3+z^3{x}^3\ge 3x^2{y}^2{z}^2$.

Answer 1

We know that A.M $\ge$ G.M

Consider the terms $x^3{y}^3,y^3{z}^3,z^3{x}^3$

Arithmetic mean $=\frac{x^3{y}^3+y^3{z}^3+z^3{x}^3}{3}$

Geometric mean $=\sqrt[3]{3\left(x^3{y}^3\right)\left(y^3{z}^3\right)(z^3{x}^3})$

$=\sqrt[3]{3x^6{y}^6{z}^6}$

$=x^2{y}^2{z}^2$

$A.M\ge G.M$

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

$\Rightarrow x^3{y}^3+y^3{z}^3+z^3{x}^3\ge 3x^2{y}^2{z}^2$