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.
 A: 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.
A: As the equations are all homogenous, we'd add the condition that $x+y+z=1$. This gives us $ 1 = (x+y+z)^2 = 4 (xy + yz + zx) $. Let $C = xyz$, which is a positive number. We want to show that $ 0\leq C \leq \frac{1}{54}$.
Consider the cubic equation with roots $x, y, z$. It has the form $ X^3 - X^2 + \frac{1}{4} X - C$. For a cubic equation to have 3 real roots, it must have a non-negative discriminant, which gives us $ C-54C^2 \geq 0$ (courtesy of Wolfram, sorry I screwed this up earlier), or hence that $ 0\leq C \leq \frac{1}{54}$. Hence we are done.
A: 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}
A: It's my old problem. See here: https://artofproblemsolving.com/community/c6h193555
My solution.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition does not depend on $w^3$, 
which says that it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$.
Hence, the condition gives $z(z-4x)=0$.
For $z=0$ our inequality is obviously true and for $z=4x$ we get identity. 
Done!
