Question

If $x,y,z,u>0$ , then find the minimum value of $\frac{x^6+y^6+2z^3+2u^3}{xyzu}$

Answer 1

Using A.M $\ge$ G.M
$\frac{x^6+y^6+z^3+z^3+u^3+u^3}{6}\ge {\left(x^6{y}^6{z}^6\right)}^{1/6}$
$\therefore \frac{x^6+y^6+2z^3+{2u}^3}{xyzu}\ge 6$