<s>[INST] 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
 [/INST] This is SOS's proof
Since $$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
$$=\sum \Big[{\dfrac {1}{140}}\, \left( 11a+11b-70c \right) ^{2}+{\dfrac {1944
}{35}}\,ab+{\dfrac {999}{140}}\, \left( a-b \right) ^{2}\Big](a-b)^4 \geqslant 0$$
So we are done.
For the  best constant, assume  $c=\min\{a,b,c\}$$,$ $$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\Big(1+\dfrac{2}{\sqrt{3}}\Big)(a-b)^2(b-c)^2(c-a)^2$$
$$=\dfrac{1}{3} \left( 2\,a+2\,b-c \right) c \left( {a}^{2}-ab-bc+{b}^{2}-ac+{c}
^{2} \right) ^{2}+$$
$$+\frac19 A\cdot \left[  \left( \sqrt {3}-1 \right) {c}^{2
}- \left( \sqrt {3}-1 \right)  \left( a+b \right) c-{a}^{2}+ab+\sqrt {
3}ab-{b}^{2} \right] ^{2} \geqslant 0,$$
where $$\text{A}= \left( b-c \right) ^{2}+(2\,\sqrt {3} +2)\left( a-c \right)  \left( b-c
 \right)  + \left( a-c
 \right) ^{2} \geqslant 0. $$
Done.
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