Proving a much stronger version of AM-GM for three variables

I will use the Buffalo method: By symmetry of the inequality, we can assume without loss of generality that $a\le b\le c$. So we can substitute $a=x,b=x+y,c=x+y+z$ for some non-negative reals $x,y,z$.

Putting this in our inequality and expanding results in

$$\color{green}{12 x^4 y^2 + 12 x^4 y z + 12 x^4 z^2 + 28 x^3 y^3 + 42 x^3 y^2 z + 54 x^3 y z^2 + 20 x^3 z^3 + 27 x^2 y^4 + 54 x^2 y^3 z + 87 x^2 y^2 z^2 + 60 x^2 y z^3 + 15 x^2 z^4 + 12 x y^5 + 30 x y^4 z + 60 x y^3 z^2 + 60 x y^2 z^3 + 30 x y z^4 + 6 x z^5} + \color{blue}{2 y^6 + 6 y^5 z - y^4 z^2 - 12 y^3 z^3 - y^2 z^4 + 6 y z^5 + z^6}\geq 0.$$

The green part is automatically greater or equal than $0$ because the $x,y,z$ are non-negative. Also, $$\color{blue}{2 y^6 + 6 y^5 z - y^4 z^2 - 12 y^3 z^3 - y^2 z^4 + 6 y z^5 + z^6}=y^6+(y-z)^2(y+z)^2(y^2+6yz+z^2)\geq0.$$

This achieves a proof.

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: $$4k^3-19k+16\geq3\sqrt[3]{4k^3\cdot8^2}-19k=\left(12\sqrt[3]{4}-19\right)k>0.$$