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.

Notice that:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right).$$
The minimum of the function $f(w) = w + \frac{1}{w}$ is attained for $w=1$. More specifically, $f(1)= 2.$ See addendum section below for further details.
Then:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right) \geq 2 \cdot 2 \cdot 2 = 8.$$

Addendum
This addendum uses calculus tools, not just pre-calculus ones!
$w=1$ is the minimum of $f(w) : \mathbb{R}^+ \to \mathbb{R}^+.$
Indeed $f(w)$ is continuous for $w>0$. Moreover, notice that:
$$f'(w) = \frac{w^2-1}{w^2}.$$
Hence, $f'(w) = 0 \Rightarrow w = 1$.
Moreover:
$$f''(w) = \frac{2}{w^3} > 0$$  for all $w > 0.$
Finally, notice that:
$$\lim_{w=0^+} f(w) = +\infty,$$
and
$$\lim_{w=+\infty} f(w) = +\infty.$$
Then, $w=1$ is the unique minimum for $w > 0.$
