Are you sure is $\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)|$? I`m asking that because we have the inequality $x^{3}+y^{3}+z^{3}-3xyz\geq 4(x-y)(y-z)(z-x)$

I'm very sure! I don't see the contradiction with your inequality. Anyway, one can prove it easily using a brute-force method (which works fine here).

I think the most of you contestants took $x\leq y\leq z$ and $y=x+a$, $z=x+a+b$, where $a,b \geq 0$

pohoatza wrote: I think the most of you contestants took $x\leq y\leq z$ and $y=x+a$, $z=x+a+b$, where $a,b \geq 0$ Well, that would be an easy way to do it. As for the additional task, you need two things to complete it in a reasonable amount of time: calculus and a computer algebra program.

See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101916

PhilAndrew wrote: Let $x, y, z \ge 0$ be real numbers. Prove that: \[\frac{x^{3}+y^{3}+z^{3}}{3}\ge xyz+\frac{3}{4}|(x-y)(y-z)(z-x)| . \]

with chebyshev inequalities see here http://diendantoanhoc.net/forum/index.php?showtopic=31296

Zamfirmihai wrote:

$x^{3}+y^{3}+z^{3}-3xyz\geq 4(x-y)(y-z)(z-x)$

As $x+y\ge |x-y|, y+z\ge |y-z|, z+x\ge |z-x|,$ we have \[x+y+z\ge \frac{|x-y|+|y-z|+|z-x|}{2}\ge \frac{3\sqrt[3]{|(x-y)(y-z)(z-x)|}}{2}.\]Then, \[\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}\ge \frac{3\sqrt[3]{|(x-y)(y-z)(z-x)|^2}}{2}\]and so \[\frac{x^3+y^3+z^3-3xyz}{3}=\frac{(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)}{6}\ge \frac{3}{4}|(x-y)(y-z)(z-x)|.\]