Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=\frac{1}{9}(a+b+c)^2(a-2b+c)^4\\ &+\frac{2}{3}(a+b+c)^2(a-2b+c)^2(b-c)(a-b)\\ &+\frac{1}{16}(a-b)^2(b-c)^2(a+4b+7c)(7a+4b+c)\\&\geqslant 0\end{align*}
However, this solution is too hard for me to find without computer. Could you help me with figuring out a better soltuion?
Thank you very much

Let $a=\min\{a,b,c\}$, $b=a+u,$ $c=a+v$ and $u^2+v^2=2tuv$.
Thus, by AM-GM $t\geq1$ and we need to prove that:
$$2(a+b+c)\sum_{cyc}(a-b)^2\geq9\sqrt{\prod_{cyc}(a-b)^2}$$ or
$$2(3a+u+v)(u^2+v^2+(u-v)^2)\geq9\sqrt{u^2v^2(u-v)^2},$$ for which it's enough to prove that
$$4(u+v)(u^2-uv+v^2)\geq9\sqrt{u^2v^2(u-v)^2}$$ or
$$16(u+v)^2(u^2-uv+v^2)^2\geq81u^2v^2(u-v)^2$$ or
$$16(t+1)(2t-1)^2\geq81(t-1)$$ or
$$64t^3-129t+97\geq0,$$ which is true by AM-GM:
$$64t^3+97=64t^3+2\cdot\frac{97}{2}\geq3\sqrt[3]{64t^3\cdot\left(\frac{97}{2}\right)^2}>129t.$$
