# Prove that $k(x^6+y^6+z^6)+xyz(x^3+y^3+z^3)\geq0$

Let $x$, $y$ and $z$ be real numbers and $k=\frac{(2+\sqrt7)\sqrt[3]{3-\sqrt7}}{6}$. Prove that: $$k(x^6+y^6+z^6)+xyz(x^3+y^3+z^3)\geq0$$ The equality occurs here for $\frac{x}{\sqrt[3]{\sqrt7-3}}=y=z$.

I tried the following way.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.

Hence, the inequality $$\sum_{cyc}(kx^6+x^4yz)\geq\frac{k+1}{3(1-m+n)^2}\left(\sum_{cyc}(x^3+m(x^2y+x^2z)+nxyz)\right)^2$$ is a linear inequality of $w^3$ and it's enough to prove the last inequality for an extremal value of $w^3$, which happens for equality case of two variables.

It's obvious that we can assume $y=z=1$,

but I did not find values of $m$ and $n$, for which the last inequality would be true.