<s>[INST] Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$. 
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.

This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \ge 4(x + y + z)(xy + yz + zx)$
Furthermore, $$(x + y + z)^3 - 6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 9xyz$$
$$ = x^3 + y^3 + z^3 - 3(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 15xyz$$
In addition, $x^3 + y^3 + z^3 + 3xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$.
So now we need to prove that $6xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$, which I don't even know if it is true or not.
 [/INST] We need to prove that $$\sum_{cyc}(x^3+3x^2y+3x^2z+2xyz)+\sum_{cyc}3xyz\geq 4\sum_{cyc}(x^2y+x^2z+xyz)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
Your way is wrong because you took too strong estimation, that got a wrong inequality.
Here happens like the following.
Let we need to prove that $2>1$.
We know that $1<3$, but it does not say that $2>3$ can help because it's just wrong. 
The Schur's inequality we can proof so:
Let $x\geq y\geq z.$
Thus, $z(z-x)(z-y)\geq0$ and  $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)=\sum_{cyc}x(x-y)(x-z)\geq$$
$$\geq x(x-y)(x-z)+y(y-x)(y-z)=(x-y)^2(x+y-z)\geq0.$$
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