For all nonnegative real numbers $x,y$ and $z$, prove that $\dfrac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$ 
For all nonnegative real numbers $x,y$ and $z$, prove that $$\frac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$$

It seems like AM-GM works here. We have $\dfrac{(x+y+z)^2}{3} \geq \dfrac{(2\sqrt{xy}+z)(2\sqrt{xz}+y)(2\sqrt{xy}+z)}{3}$. Then I get stuck and don't know what to do next.

Using AM-GM inequality we have:
$\displaystyle \sum_{\text{cyclic}} x\sqrt{yz}\leq \displaystyle \sum_{\text{cyclic}}\dfrac{x(y+z)}{2}=\displaystyle \sum_{\text{cyclic}} xy\leq \dfrac{2\displaystyle \sum_{\text{cyclic}} xy + \displaystyle \sum_{\text{cyclic}} x^2}{3}=\dfrac{\displaystyle \left(\sum_{\text{cyclic}} x\right)^2}{3}$
