Algebra Iranian Olympiad Problem

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.

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.