Proving a much stronger version of AM-GM for three variables Here, @MichaelRozenberg stated the following inequality without proof:

Theorem. For all non-negative $a,b,c\in\mathbb R$,
  $$a^6+b^6+c^6-3a^2b^2c^2\geq16(a-b)^2(a-c)^2(b-c)^2.$$

In my answer below I give a complete brute-force proof. However, more elegant proofs are welcome.

I meant the following reasoning.
Let $a=\min\{a,b,c\},$ $b=a+u,$ $c=a+v$ and $u^2+v^2=2kuv.$
Thus, by AM-GM $k\geq1$ and
$$\sum_{cyc}(a^6-a^2b^2c^2)=\frac{1}{2}(a^2+b^2+c^2)\sum_{cyc}(a-b)^2(a+b)^2\geq$$
$$\geq\frac{1}{2}(u^2+v^2)(u^4+v^4+(u^2-v^2)^2)=(u^2+v^2)(u^4-u^2v^2+v^4).$$
Id est, it's enough to prove that:
$$(u^2+v^2)(u^4-u^2v^2+v^4)\geq16(u-v)^2u^2v^2$$ or
$$2k(4k^2-3)\geq16(2k-2)$$ or $$4k^3-19k+16\geq0,$$ which is true by AM-GM:
$$4k^3-19k+16\geq3\sqrt[3]{4k^3\cdot8^2}-19k=\left(12\sqrt[3]{4}-19\right)k>0.$$
