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.
We know that for non negative reals (in fact for all reals) we have $$a^2+b^2+c^2 \ge ab+bc+ca$$
Hence $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) \ge 3(xy+yz+zx)$$
and by applying the same trick on $a=\sqrt{xy}, b=\sqrt{yz}, c=\sqrt{zx}$ we have $$xy+yz+zx \ge y\sqrt{xz}+z\sqrt{yx}+x\sqrt{yz}$$
This proves the claim.
Using AM-GM inequality we have: $\displaystyle \sum_{\text{cyclic}} x\sqrt{yz}\leq \displaystyle \sum_{\text{cyclic}}\dfrac{x(y+z)}{2}=\displaystyle \sum_{\text{cyclic}} xy\leq \dfrac{2\displaystyle \sum_{\text{cyclic}} xy + \displaystyle \sum_{\text{cyclic}} x^2}{3}=\dfrac{\displaystyle \left(\sum_{\text{cyclic}} x\right)^2}{3}$
