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.

It suffices to prove that $k(x^6+y^6+z^6) + xyz (x^3+y^3+z^3) \ge 0$ for $xyz = -1$.
Then, it suffices to prove that $k(x^2+y^2+z^2) - (x+y+z) \ge 0$ for $x, y > 0$ and $xyz = -1$.
Note that $x^2+y^2 \ge \frac{(x+y)^2}{2}$, $z^2 = \frac{1}{(xy)^2} \ge \frac{16}{(x+y)^4}$
and $-z = \frac{1}{xy} \ge \frac{4}{(x+y)^2}$. It suffices to prove that
$$k\Big(\frac{(x+y)^2}{2} + \frac{16}{(x+y)^4}\Big) - (x+y) + \frac{4}{(x+y)^2} \ge 0$$
for $x, y > 0$. It suffices to prove that
$k \ge \frac{2u^5 - 8u^2}{u^6 + 32}$ for $u > 0$.
Let $f(u) = \frac{2u^5 - 8u^2}{u^6 + 32}$.
We have $f'(u) = -\frac{2u(u+2)(u^2-2u+4)(u^6-24u^3+32)}{(u^6+32)^2}$.
$f'(u)=0$ has exactly two real solutions $u_1 = \sqrt[3]{12+4\sqrt{7}}$
and $u_2 = \sqrt[3]{12-4\sqrt{7}}$ on $(0, \infty)$.
On $[0,\infty)$, $f(u)$ attains its maximum at $u_1 = \sqrt[3]{12+4\sqrt{7}}$.
Note that $f(\sqrt[3]{12+4\sqrt{7}}) = \frac{\sqrt{7}-1}{24}\sqrt[3]{(12+4\sqrt{7})^2} = k$. We are done.
