<s>[INST] Prove that $(\sum x^2)^3\ge9\sum x^4yz$ 
Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.

The $pqr$ method doesn't seem possible because the power is too high.
$$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$
Then expand the expression to get
$$\sum x^6+3\sum\left(x^4y^2+x^2y^4\right)+6x^2y^2z^2\ge9\sum x^4yz.$$
I wanted to use SOS but cannot find the weight of three squares, my progress:
$$3\sum x^4(y-z)^2=3\sum\left(x^4y^2+x^2y^4\right)-6\sum x^4yz.$$
Whats left is $\displaystyle\sum x^6+6x^2y^2z^2-3\sum x^4yz$. I have trouble dealing with it.
 [/INST] This inequality is true for any real variables.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ may be negative and $xyz=w^3$.
Thus, we need to prove that:
$$(9u^2-6v^2)^3\geq9w^3(27u^3-27uv^2+3w^3)$$ or $f(w^3)\geq0,$ where $f$ is a concave function.
But, a concave function gets a minimal value for an extremal value of $w^3$, which by uvw happens for equality case of two variables.
Since our inequality is symmetric,  even degree and homogeneous, it's enough to assume that $y=z=1$ (for $y=z=0$ the equality is obvious), which gives $$(x^2+2)^3-9x(x^3+2)\geq0$$ or
$$(x-1)^2(x^4+2x^3-2x+8)\geq0,$$ which is true because
$$x^4+2x^3-2x+8=(x^2+x-1)^2+x^2+7>0.$$
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