Question

If x, y, z$>0$, then prove that $(x+y+z)(x^3+y^3+z^3)\ge 9(xyz{)}^{\frac{4}{3}}$.

Answer 1

We know that for a finite number of positive real terms,

$A.M\ge G.M$

Consider the terms to be $x,y,z$

$A.M=\frac{x+y+z}{3},G.M=\sqrt[3]{3xyz}$

$\Rightarrow$ $\frac{x+y+z}{3}\ge \sqrt[3]{3xyz}...\left(1\right)$

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

$A.M=\frac{x^3+y^3+z^3}{3}$, $G.M=\sqrt[3]{3x^3{y}^3{z}^3}=xyz$

$\Rightarrow$ $\frac{x^3+y^3+z^3}{3}\ge xyz....\left(2\right)$

$\left(1\right)\times \left(2\right)$

$(x+y+z)(x^3+y^3+z^3)\ge 9{\left(xyz\right)}^{4/3}$