Prove that for positive reals $x,y,z$, $x^6+y^6+z^6 + 6x^2y^2z^2 \geq 3xyz(x^3+y^3+z^3)$.

Question

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: $a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$

Answer 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.