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?

From your assumption, we obtain that
\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le 1.
\end{align*}
Since
\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} &=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z} +\frac{1}{y} + \frac{1}{z} \right)\\
&\ge\frac{1}{2}\left(\frac{2}{\sqrt{xy}} + \frac{2}{\sqrt{xz}} +\frac{2}{\sqrt{yz}} \right),
\end{align*}
that is,
\begin{align*}
\frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{xz}} +\frac{1}{\sqrt{yz}}\le 1,
\end{align*}
the required inequality follows immediately by multiplying $\sqrt{xyz}$ to both sides.
