$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ 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!

For real $x$,we have(or  Use Cauchy-Schwarz inequality) 
$$x^2+1\ge\dfrac{1}{2}(x+1)^2$$
the same we have
$$y^2+1\ge\dfrac{1}{2}(y+1)^2$$
$$z^2+1\ge\dfrac{1}{2}(z+1)^2$$
so
$$(x^2+1)(y^2+1)(z^2+1)\ge\dfrac{1}{8}[(x+)(y+1)(z+1)]^2$$
Use AM-GM inequality
$$\dfrac{1}{8}[(x+1)(y+1)(z+1)]^2+8\ge 2\sqrt{\dfrac{1}{8}[(x+)(y+1)(z+1)]^2\cdot 8}=2(x+1)(y+1)(z+1)$$
