Given the $ x+y+z =3$ Prove that $ x^3+y^3+z^3+6xyz \geq 27/4$

Question

Also, $x$,$y$ and $z$ are positive.

I have figured out that it would be enough to prove that LHS is bigger than $(27-\text{LHS})/3$, then by substituting 27 with $ (x+y+z)^3 $, it's enough to prove that $$x^3+y^3+z^3+6xyz \geq x^2(3-x)+y^2(3-y)+z^2(3-z).$$

And I'm stuck. I feel like this has to be provable somehow with rearrangement, since the right hand side is conveniently ordered (if $x>y$, then $3-x<3-y$). But the equality only holds when one of the variables is $0$ and the other two are $1.5$, which isn't the case for rearrangement.

Answer 1

Without loss of generality, we assume $z\ge \frac{x+y+z}{3}=1$. We temporarily fix $z$ and check how to arrange $x$ and $y$ to achieve a minimum. Now $y=3-x-z$.

So $f(x)=x^3+(3-x-z)^3+z^3+6x(3-x-z)z$, and the derivative of $f$ is $$f'(x) = 3x^2-3(3-x-z)^2 + 6(3-x-z) z-6xz=x(-3*2*(z-3) -6z-6z) + (-3(z-3)^2+6(3-z)z) = x(-18z+18) + 3(3-z)(9z-9) $$ and $f''(x)=-18z+18 \le 0$, which means $f(x)$ assumes its minimum at the boundary of $[0, 3-z]$, then either $x=0$ or $y=0$. Without loss of generality, we assume $y=0$, and now we want to prove $x^3+z^3\ge 27/4$ given $x+z=3$ and $x\ge 0$ and $z\ge 0$, which is much easier than the original problem.

Answer 2

We need to prove that $$\sum\limits_{cyc}(4x^3+8xyz)\geq(x+y+z)^3$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)+3xyz\geq0,$$ which follows from Schur.

Done!