For positive real numbers $a,b,c$ prove that $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$

Question

For positive real numbers $a,b,c$ prove that $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$

I think the $A.M-G.M$ inequality needs to be used here. Pretty sure we have to use the fact that
$\frac{a^6+b^6+c^6}{3}\ge a^2b^2c^2$
and that
$\frac{x^2+y^2}{2}\ge xy$
But I'm not sure in what way to use it exactly, I can't seem to get the sum of two squares in any way, so any help is appreciated.

Answer 1

$2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
$12+2a^6+2b^6+2c^6\ge 6ab+6ac+6bc$
We can write this as
$a^6+b^6+1+1+1+1+a^6+c^6+1+1+1+1+b^6+c^6+1+1+1+1\ge 6ab+6ac+6bc$
We also know that, from the $A.M-G.M$ inequality
$a^6+b^6+1+1+1+1\ge6\sqrt[6]{a^6b^6}=6ab$
$a^6+c^6+1+1+1+1\ge6\sqrt[6]{a^6c^6}=6ac$
$b^6+c^6+1+1+1+1\ge6\sqrt[6]{b^6c^6}=6bc$
Summing the three inequalities above we get
$a^6+b^6+1+1+1+1+a^6+c^6+1+1+1+1+b^6+c^6+1+1+1+1\ge 6ab+6ac+6bc$
$=12+2a^6+2b^6+2c^6\ge 6ab+6ac+6bc$
$=2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
So we've proved the original inequality.

Answer 2

Hint: $a^6+b^6+1+1+1+1\geqslant 6ab$. What happens if you add three such AM-GMs?