<s>[INST] $(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem:

Let $x$, $y$, $z$ be real 
  numbers. Prove that
  $$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$

The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square mean), but I haven't found anything helpful. 
(To be more specific, my attempts looked like this :
$\frac{x+1}{2} \leq \sqrt{\frac{x^2+1}{2}}$)
I also take into consideration Cauchy-Buniakowsky-Scwartz or Bergström inequality, but none seems to help.
Some hints would be apreciated.
Thanks!
 [/INST] Its just a matter of factorization:
$$
(x^2+1)(y^2+1)(z^2+1)+8-2(x+1)(y+1)(z+1)=(x^2y^2z^2-2xyz+1)+\left(\sum_{cyc}x^2y^2-2xy+1\right)+\left(\sum_{cyc}x^2-2x+1\right)=(xyz-1)^2+\left(\sum_{cyc}(xy-1)^2\right)+\left(\sum_{cyc}(x-1)^2\right)≥0
$$
With equality only if $x=y=z=1$.
</s>