Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$ I found the following exercise reading my calculus notes:

If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$

I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.

by Huygen's inequality $$\frac{(1+x^2)(1+y^2)(1+z^2)}{xyz}\ge \frac{({1+\sqrt[3]{x^2y^2z^2})}^3}{xyz}={\left(\frac{1}{\sqrt[3]{xyz}}+\sqrt[3]{xyz} \right)}^3\ge 2^3$$
