Prove that for positive reals $x,y,z$, $x^6+y^6+z^6 + 6x^2y^2z^2 \geq 3xyz(x^3+y^3+z^3)$. I am not sure if the inequality is true. My first attempt was to try AM-GM inequality in clever ways. I also tried Schur's inequality which gives
$$
x^6+y^6+z^6 + 6x^2y^2z^2 \geq (x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2)
$$
but it is not true that
$$
(x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2) \geq 3xyz(x^3+y^3+z^3)
$$
Edit: I found a variant and the solutions there work for this problem too: $a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$

The problem is equivalent to prove that
$a^2+b^2+c^2+6 \ge 3(a+b+c)$, $abc=1$
But the last inequality is true because
$$LHS - RHS = \frac{(a+b+c-3)^2+a^2+b^2+c^2+2abc+1 -2(ab+bc+ca)}{2} \ge 0$$
Note that the inequality $a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$ is well-known.
