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.

Note that
$(x-1)^2 \geq 0 \implies x^2+1 \geq 2x $
$(y-1)^2 \geq 0 \implies y^2+1 \geq 2y $
$(z-1)^2 \geq 0 \implies z^2+1 \geq 2z $
Multiplying these three inequalities you will get $(x^2+1)(y^2+1)(z^2+1)\geq 8xyz \implies \frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\ge8$
