<s>[INST] 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.
 [/INST] You have:
\begin{align}
(x + y + z)^2 
  &=   x^2 + y^2 + z^2 + 2 (x y + x z + y z) \\
  &=   4 (x y + x z + y z) \\
  &=   4 \cdot 3 \cdot \frac{x y + x z + y z}{3} \\
  &\ge 12 \sqrt[3]{x^2 y^2 z^2} \\
x + y + z
  &\ge 2 \sqrt[3]{3 x y z} \\
\frac{x + y + z}{3}
  &\ge \sqrt[3]{\frac{8}{9} x y z}
\end{align}
</s>