<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] Hint :
Due to homogeneization we have :
$$\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)$$
We need to show for $x\ge 1$ and WLOG $c\geq 1$
$$f(x)=\left(x^{2}+1^{2}+c^{2}\right)^{3}-9xc\left(x^{3}+1^{3}+c^{3}\right)\geq 0$$
The function is convex on $(0,\infty)$ because we have :
$$f''(x)=6(c^{6}+6c^{3}x^{2}+2c^{3}-18cx^{2}+5x^{4}+6x^{2}+1)>0$$
So we use the tangent line method we have :
$$f(x)\geq f(1)(x-1)+f(1)=(6 (c^3 + 2)^2 - 9 c (c^3 + 2) - 27 c)(x-1)+(c^3 + 2)^3 - 9 c (c^3 + 2)\geq 0$$
Because :
$$(6(c^{3}+2)^{2}-9c(c^{3}+2)-27c)\geq 0,(c^3 + 2)^3 - 9 c (c^3 + 2)=(c^3 + 2) (c - 1) (c^5 + c^4 + c^3 + 5 c^2 + 5 c - 4)\geq 0$$
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