<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] If we write $x+y+z=1$ (we can do that) then we have to prove $$1+9xyz\geq 4(xy+yz+zx)$$ Clearly one of $x,y,z$ is smaller than $4/9$, we can assume $y<4/9$. 
If we put $z=1-x-y$ we get $$0\geq (9y-4)x^2+x(9y^2-13y+4)-(2y-1)^2=:f(x)$$
Since the discirminant for $f(x)$ is: $$\Delta= (9y-4)y(3y-1)^2$$
is also nonegative for all $y<4/9$ we are done.
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