prove that $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$

$(xy+yz+zx)(zx+xy+yz)\geq( x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy})^2$

by Cauchy Inequality.

Yes, it's correct.

Also, SOS helps: $$\sum_{cyc}(xy-x\sqrt{yz})=\frac{1}{2}\sum_{cyc}(xz+yz-2z\sqrt{xy})=\frac{1}{2}\sum_{cyc}z(\sqrt{x}-\sqrt{y})^2\geq0.$$

$$xy+yz+zx = {xy+yz\over 2} + {yz+zx\over 2} + {zx+xy\over 2} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}$$ By AM-GM Inequality. Done!