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
 A: 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.
A: We write inequality we have
$$4(a^3+b^3+c^3-3abc) \geqslant 9 |(a-b)(b-c)(c-a)|,$$
or
$$2(a+b+c)[(a-b)^2+ (b-c)^2+ (c-a)^2] \geqslant 9 |(a-b)(b-c)(c-a)|.$$
It's easy to check $a + b \geqslant |a-b|,$ now using the AM-GM inequality, we have
$$2(a+b+c)[(a-b)^2+ (b-c)^2+ (c-a)^2] $$
$$\geqslant \left(|a-b|+|b-c|+|c-a|\right) \left[(a-b)^2+ (b-c)^2+ (c-a)^2\right]$$
$$ \geqslant 3 \sqrt[3]{\left | (a-b)(b-c)(c-a)\right |} \cdot 3 \sqrt[3]{(a-b)^2(b-c)^2(c-a)^2}.$$
$$ =9 \left | (a-b)(b-c)(c-a)\right | .$$
Done.
SOS proof. We have
$$4(a^3+b^3+c^3-3abc) - 9(a-b)(b-c)(c-a) = \sum b \left[(2a-b-c)^2+3(a-b)^2\right] \geqslant 0.$$
Note. The best constant is
$$a^3+b^3+c^3-3abc \geqslant \sqrt{9+6\sqrt{3}} \cdot \left | (a-b)(b-c)(c-a)\right |.$$
A: 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:
$$64t^3+97=64t^3+2\cdot\frac{97}{2}\geq3\sqrt[3]{64t^3\cdot\left(\frac{97}{2}\right)^2}>129t.$$
A: 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.
