Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$

Question

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

Answer 1

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.

Answer 2

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: $$64t^3+97=64t^3+2\cdot\frac{97}{2}\geq3\sqrt[3]{64t^3\cdot\left(\frac{97}{2}\right)^2}>129t.$$

Answer 3

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 $$a^3+b^3+c^3-3abc \geqslant \sqrt{9+6\sqrt{3}} \cdot \left | (a-b)(b-c)(c-a)\right |.$$

Answer 4

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.