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

Without loss of generality suppose that $a\geq b\geq c$. Then,
$$
6\cdot\left(\frac{a^3+b^3+c^3}{3}-abc\right)=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
$$
Note that
$$
a+b+c\geq (a-c)+(b-c)=2(a-b)+(b-c)
$$
and
$$
(a-b)^2+(b-c)^2+(c-a)^2=2(a-b)^2+2(b-c)^2+2(a-b)(b-c).
$$
Denote $x=a-b$ and $y=b-c$, due to our assumption $x$ and $y$ are nonegative. Then, we need to prove that ($\sqrt{(a-b)^2(b-c)^2(c-a)^2}=xy(x+y)$)
$$
(2x+y)(2x^2+2y^2+2xy)\geq 6\cdot\frac{3}{4}xy(x+y),
$$
or
$$
4(2x+y)(x^2+xy+y^2)\geq 9xy(x+y)
$$
or
$$
4(2x^3+3x^2y+3xy^2+y^3)\geq 9xy(x+y)
$$
or
$$
8x^3+3x^2y+3xy^2+4y^3\geq 0,
$$
which is obvious.
