<s>[INST] 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?
 [/INST] You could use this famous inequality: $$xy+yz+xz\geq \sqrt{3xyz(x+y+z)}$$ (which can be easily proved by AM-GM).
Using that inequality we have $$xyz\geq xy+yz+zx\geq \sqrt{3xyz(x+y+z)},$$
from where is $$\sqrt{xyz}\geq \sqrt{3(x+y+z)}=\sqrt{(1+1+1)(x+y+z)}$$
By Cauchy-Schwartz inequality, we have
$$\sqrt{(1+1+1)(x+y+z)} \geq (\sqrt{x}+\sqrt{y}+\sqrt{z})$$ 
and finally we can conclude $$\sqrt{xyz}\geq\sqrt{x}+\sqrt{y}+\sqrt{z}.$$
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