Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$ Let $x$,$y$ and $z$ are positive and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$$
Prove that: $$xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$$
The things I have done so far
$$3\geq \sum \limits_{cyc}\frac{1}{x}\geq \frac{9}{\sum \limits_{cyc}x}\Rightarrow \sum \limits_{cyc}x\geq 3$$
Then, I tried to use AM-GM and Cauchy -Schwarz but without success.
 A: We need to prove that 
$ xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}\geq x+y+z $
$\Leftrightarrow \frac{xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}}{xyz}\geq \frac{x+y+z}{xyz}$
$\Leftrightarrow \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}$
$\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}$
$\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}} \geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}$
We have: $3\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Rightarrow 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}$
Now, we need to prove:
$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) $
Using Cauchy's inequality, we have: 
$\frac{1}{x^{2}}+\frac{2}{\sqrt{x}}=\frac{1}{x^{2}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}\geq 3\sqrt[3]{\frac{1}{x^{2}}.\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}}}=\frac{3}{x}$
Similar, we have:
 $\frac{1}{y^{2}}+\frac{2}{\sqrt{y}}\geq \frac{3}{y};\frac{1}{z^{2}}+\frac{2}{\sqrt{z}}\geq \frac{3}{z}$
So, $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) $
