Algebra Iranian Olympiad Problem If:
$x^2+y^2+z^2=2(xy+xz+zy)$
 and $x,y,z \in R^+$
Prove:
$\frac{x+y+z}{3} \ge \sqrt[3]{2xyz}$
I tried my best to solve this thing but no use.
Hope you guys can help me.Thanks in advance.

If we put $a=\sqrt{x}$ and $b=\sqrt{y}$, the degree two equation (in $z$)
$x^2+y^2+z^2-2(xy+xz+yz)=0$ has two solutions, $(a-b)^2$ and $(a+b)^2$. By
cyclically permuting $x,y,z$, we may assume $z=(a+b)^2$. The inequality
to be shown is then equivalent to $(x+y+z)^3 \geq 54xyz$, or
$(a^2+b^2+(a+b)^2)^3 \geq 54(a^2b^2(a+b)^2)$. We are then done because
$$
(a^2+b^2+(a+b)^2)^3 -54(a^2b^2(a+b)^2)=2\Bigg((b-a)(2a+b)(a+2b)\Bigg)^2
$$
As guessed by CalvinLin, equality is reached exactly when
$(x,y,z)=(1,1,4)$ up to permutation.
