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.
 A: 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 $w > 0.$
A: You can do the famous "proof by definition" as follows:
$a,b,c\in (0, \pi)$ such that
$$x = \tan\frac a2, y = \tan\frac b2, z = \tan\frac c2$$
and then you will get:
$$\dfrac{(x^2+1)(y^2+1)(z^2+1)}{8xyz} = \dfrac{1}{\sin a\sin b\sin c}\geq 1$$
A: By A.M.$\geq$G.M.$$(x^2+1)(y^2+1)(z^2+1)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\geq8\times \sqrt[8]{(x^2y^2z^2)\times(x^2y^2)\times(y^2z^2)\times(z^2x^2)\times(x^2)\times(y^2)\times(z^2)\times1}=8xyz$$
A: 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 $(x^2+1)(y^2+1)(z^2+1)\geq 8xyz \implies \frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\ge8$
A: by Huygen's inequality $$\frac{(1+x^2)(1+y^2)(1+z^2)}{xyz}\ge \frac{({1+\sqrt[3]{x^2y^2z^2})}^3}{xyz}={\left(\frac{1}{\sqrt[3]{xyz}}+\sqrt[3]{xyz} \right)}^3\ge 2^3$$
