Prove inequality for $xyz=1$ I am having trouble to prove an intermediate step $(xy+xz+yz)^2 \geq 3xyz(x+y+z)$ for $x,y,z\geq0$. 

Using that $xyz=1$ we see that the inequality we want to prove is equivalent
to
$$(1/x + 1/y+1/z)^2 \geq 3xyz(x+y+z) \fbox {1}  $$
Set $x=1/x,y=1/y,z=1/z$, and notice that the restraint for the new variables stays the same i.e. $xyz=1$ again.
Now we substitube to  $\fbox{1}  $ and get$$ (  x+y+z  )^2\geq 3xyz(1/x+1/y+1/z)$$
So,
 $$ (  x+y+z  )^2\geq 3xy+3yz+3zx$$
By expanding and canceling and multiplying by 2 we get
$$2(x^2+y^2+z^2) \geq 2(xy+yz+zx)$$
But now by bringing everything to the left side and factoring we get
$$(x-y)^2+(y-z)^2 + (z-x)^2\geq 0$$
And we condlude.
