Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that 
$$\sqrt{x y z  } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution. 
Using AM GM inequality we have
 \begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y \sqrt{x z},\\
x z +y z \geq 2 z \sqrt{xz}.
\end{gather*}
Add and get 
$$
xy+xz+yz \geq  x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}.
$$
By condition 
$$
xyz \geq  x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}
$$
Dividing by  $\sqrt{xyz}$ we obtain the inequality.
Question.  Are there another ways  to prove it?

Here's a proof that avoids (or disguises) the AM-GM inequality.
Let's start by establishing a completely different inequality:  If $r,s\ge1$, then
$${1\over s}+r^2s+s\ge1+r+rs$$
The proof comes from first rewriting it in equivalent form as
$$(r-1)rs^2+(s-1)s\ge rs-1$$
and then letting $r=1+a$ and $s=1+b$ with $a,b\ge0$:
$$a(1+a)(1+b)^2+b(1+b)\ge a+b+ab$$
where we've expanded the right hand side but not the left.  However, since there are no negative signs on the left, it's easy to see that its expansion consists exclusively of non-negative terms, including an $a$, a $b$, and an $ab$ (in fact a $2ab$), hence cannot be less than $a+b+ab$.  This completes the proof.
Now to the problem about $x,y,z$.  The statement is symmetric in the three variables, so we can, without loss of generality, assume $0\lt x\le y\le z$.  This allows us to write $x=u^2$, $y=r^2u^2$ and $z=r^2s^2u^2$ with $u\gt0$ and $r,s\ge1$.  Then
$$\begin{align}
xyz\ge xy+yz+zx&\implies r^4s^2u^6\ge r^2u^4+r^4s^2u^4+r^2s^2u^4\\
&\implies r^2su^3\ge{u\over s}+r^2su+su= \left({1\over s}+r^2s+s\right)u\\
&\implies r^2su^3\ge(1+r+rs)u=u+ru+rsu\\
&\implies\sqrt{xyz}\ge\sqrt x+\sqrt y+\sqrt z
\end{align}$$
