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.

Update: A simpler pqr
It suffices to prove that
$$\left(p^2-2q\right)^3\ge9r\left(p^3-3pq+3r\right).$$
Using $q^2 \ge 3pr$ and $pq \ge 9r$, it suffices to prove that
$$\left(p^2-2q\right)^3\ge 9 \cdot \frac{q^2}{3p}\cdot\left(p^3-3pq+3 \cdot \frac{pq}{9}\right)$$
or
$$(p^2 - 2q)^3 \ge q^2(3p^2 - 8q)$$
or
$$(p^2 - 2q)^3 + (2q)^3 \ge 3p^2 q^2$$
or (using $x^3 + y^3 = (x + y)[(x + y)^2 - 3xy]$)
$$p^2[p^4 - 3(p^2 - 2q)\cdot 2q] \ge 3p^2 q^2$$
or
$$p^2 (p^2 - 3q)^2 \ge 0$$
which is true.
We are done.
