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.

This is a proof that continues your SOS attempt, which is to prove
$$
\sum x^6+6x^2y^2z^2\geqslant3\sum x^4yz
$$
By noticing the following identity, this inequality is obviously true by Schur's inequality
$$
\sum(x^6+6x^2y^2z^2-3x^4yz)=\sum x^4(x-y)(x-z)+4(x-y)^2(y-z)^2(z-x)^2+2\sum yz(x-y)^2(x-z)^2+2xyz\sum x(x-y)(x-z)
$$
This proof is by brute force using the Triangle of Coefficient Method. Though I've found a way to simplify this inequality, I didn't managed to solve it that way.
Namely, by dividing both sides with $(xyz)^2$ , we obtain
$$
\sum\left(\frac{x^2}{yz}\right)^2+6\geqslant3\sum\frac{x^2}{yz}
$$
Let $a=(x^2)/(yz),~b=(y^2)/(zx),~c=(z^2)/(xy)$ , we have to prove that
$$
a^2+b^2+c^2+6\geqslant 3(a+b+c)
$$
for positive real numbers $a,b,c$ such that $abc=1$ .
I hope someone would continue my attempt.
Edit. This edition contains a new proof of this inequality using SOS Method by LasterCircle
$$
\text{LHS}-\text{RHS}=\frac12\sum\left[\sum x^4+2xy(x^2+xy+y^2)-2xyz\sum x\right](x-y)^2
$$
It's apparent that $\sum x^4+2xy(x^2+xy+y^2)-2xyz\sum x>0$ , the inequality is true.
Edit2. River Li has given a nice proof in the comment.
And here's a simpler SOS
$$
\sum(x^6-3x^4yz)+6x^2y^2z^2=\frac1{18}\sum\left(x(x^2-yz)\right)^2+\frac12\sum x^2(x^2-y^2)(x^2+y^2)+\frac12\sum{x^2y^2(x-y)^2}\geqslant0
$$
