prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$ prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$ if $x,y,z>0$
My try : dividing inequality by $\sqrt{xyz}$ and putting $\sqrt{x}=a,\sqrt{y}=b,\sqrt{z}=c$
we have to prove $$\sum_{cyc}\frac{ab}{c}\ge a+b+c$$
or
$$2\sum_{cyc}\frac{ab}{c}\ge 2(a+b+c)$$ using $\frac{ab}{c}+\frac{bc}{a}\ge 2b$ and similarly for others the proof can be completed.
Is it correct? Also i am looking for different proofs for this (possibly more simpler  ).Thanks

$x,y,z>0$ and that the inequality is a symmetric expression implies that we can take without loss of generality, an ordering $x\ge y\ge z \implies xy\ge zx\ge yz \implies \sqrt{xy}\ge \sqrt{zx}\ge \sqrt{yz}$
so that the sequences $\{\sqrt{xy},\sqrt{zx},\sqrt{yz}\}, \{\sqrt{xy},\sqrt{zx},\sqrt{yz}\}$, (i.e. the same sequence) are similarly sorted, so that, by the Rearrangement Inequality, we have
$$\sqrt{xy}\sqrt{xy}+\sqrt{yz}\sqrt{yz}+\sqrt{zx}\sqrt{zx}\ge \sqrt{xy}\sqrt{zx}+\sqrt{yz}\sqrt{xy}+\sqrt{zx}\sqrt{yz}$$ which is the required inequality.
