Generalization of the AM-GM inequality for three variables

Question

Theorem. Let $a,b,c$ be three non-negative real numbers. Then $$a^6+b^6+c^6\geq 3a^2b^2c^2+\frac12 (a-b)^2 (b-c)^2 (c-a)^2.$$

Remark. This Theorem is a generalization of the AM-GM inequality for three variables, which states $$a^6+b^6+c^6\geq3a^2b^2c^2.$$

I have found a proof of this Theorem (see my answer below), however, it is probably not the most elegant. More elegant proofs are welcome.

Answer 1

Note that \begin{eqnarray*} 2(a^6+b^6+c^6-3a^2b^2c^2)=(a^2+b^2+c^2)((a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2) \end{eqnarray*} Now use the Cauchy-Schwarz inequality for sums: \begin{eqnarray*} (a^2+b^2+c^2)((b^2-c^2)^2+(c^2-a^2)^2+(a^2-b^2)^2) \geq (a(b^2-c^2) +b(c^2-a^2)+c(a^2-b^2))^2. \end{eqnarray*}

Since the right-hand side equals $$(a-b)^2(b-c)^2(c-a)^2,$$ we are done.

Answer 2

Let $a,b,c$ be as in the question. Let $S:=a^6+b^6+c^6- 3a^2b^2c^2 -\frac12 (a-b)^2 (b-c)^2 (c-a)^2.$ Our inequality is $S\geq 0$.

By expanding and rearranging (note that there are $23$ terms after a full expansion!), one gets that (here $\sum_{\text{sym}}$ denotes the sum over all $6$ permutations of $(a,b,c)$) $$S=\frac12\cdot\left(\sum_{\text{sym}}a^6-\sum_{\text{sym}} a^3b^3+\sum_{\text{sym}} a^4b^2-\sum_{\text{sym}}a^4bc\right)+\sum_{\text{sym}} a^3b^2c-\sum_{\text{sym}}a^2b^2c^2.$$

By Muirhead's inequality, we have (as $(6,0,0)$ majorizes $(3,3,0)$; and $(4,2,0)$ majorizes $(4,1,1)$; and $(3,2,1)$ majorizes $(2,2,2)$) $$S\geq\frac12\cdot(0+0)+0=0$$ as desired.

Answer 3

Also, we have: $$\sum_{cyc}(a^6-a^2b^2c^2)-\frac{1}{2}\prod_{cyc}(a-b)^2=\frac{1}{2}\sum_{cyc}(a^2+b^2-ab-c^2)^2(a+b)^2\geq0.$$ The following inequality a bit of stronger.

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$a^6+b^6+c^6-3a^2b^2c^2\geq16(a-b)^2(a-c)^2(b-c)^2.$$