$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ 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 all the equations of symmetrical type, the extreme value, if exists, is achieved when variables are equal.
It is not difficult to prove it. If you have $f(x,y,z)$ symmetrical over $x,y,z$ then all their derivatives are the same, so whatever condition is needed for $x$ is the same for $y$ and the same for $z$. This means that the extreme value, if exists, is for $x=y=z$.
$$(x^2+1)(y^2+1)(z^2+1) + 8 - 2(x+1)(y+1)(z+1)=0$$
$$x=y=z$$
$$(x^2+1)^3 + 8 - 2(x+1)^3=0$$
$$x=y=z=1$$
For any other value, since we can calculate $x=y=z=2$ for example and find that the expression is greater than 0, it is
$$(x^2+1)(y^2+1)(z^2+1) + 8 - 2(x+1)(y+1)(z+1)>0$$
All equations of this type can be solved this way as long as you can calculate the extreme value.
(Usually this is not the intention regarding the solution, since this method is removing the flavor from the equations completely.)
