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 A.M.$\geq$G.M.$$(x^2+1)(y^2+1)(z^2+1)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\geq8\times \sqrt[8]{(x^2y^2z^2)\times(x^2y^2)\times(y^2z^2)\times(z^2x^2)\times(x^2)\times(y^2)\times(z^2)\times1}=8xyz$$
