<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] Since the expression is symmetric, WLOG, $ z \geq y \geq x$.
Let $ x = a, y = a + b, z = a+b+c$, where $a, b, c \geq 0$.
Then, expanding out $(x^2 + y^2 + z^2)^3 - 9 xyz(x^3+y^3+z^3)$ gives us, courtesy of Wolfram:
$9 a^4 b^2 + 9 a^4 b c + 9 a^4 c^2 + 28 a^3 b^3 + 42 a^3 b^2 c + 30 a^3 b c^2 + 8 a^3 c^3 + 42 a^2 b^4 + 84 a^2 b^3 c + 78 a^2 b^2 c^2 + 36 a^2 b c^3 + 12 a^2 c^4 + 30 a b^5 + 75 a b^4 c + 90 a b^3 c^2 + 60 a b^2 c^3 + 27 a b c^4 + 6 a c^5 + 8 b^6 + 24 b^5 c + 36 b^4 c^2 + 32 b^3 c^3 + 18 b^2 c^4 + 6 b c^5 + c^6.$
Since all the coefficients are non-negative, this is clearly non-zero, hence we are done.
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