Erken wrote: Find the max. value of $ M$,such that for all $ a,b,c>0$: $ a^{3}+b^{3}+c^{3}-3abc\geq M(|a-b|^{3}+|a-c|^{3}+|c-b|^{3})$ Let $ a\geq b\geq c.$ If $ b=c\rightarrow0^{+}$ then $ M\leq\frac{1}{2}.$ But $ a^{3}+b^{3}+c^{3}-3abc\geq \frac{1}{2}\cdot(|a-b|^{3}+|a-c|^{3}+|c-b|^{3})\Leftrightarrow$ $ \Leftrightarrow3(b+c)a^{2}-3(b+c)^{2}a+2b^{3}+3b^{2}c-3bc^{2}+4c^{3}\geq0$ and we obtain that $ \Delta=9(b+c)^{4}-12(b+c)(2b^{3}+3b^{2}c-3bc^{2}+4c^{3})=$ $ =-3(b+c)(5b+13c)(b-c)^{2}\leq0.$ Id est, $ M_{max}=\frac{1}{2}.$

Nice work. .Won't you try to solve another one: http://www.mathlinks.ro/viewtopic.php?p=902438#902438

Find the max. value of $ M$,such that for all $ a,b,c\geq 0$:$ a^{3}+b^{3}+c^{3}-3abc\geq M|(a-b)(a-c)(c-b)|$.

M

But arqady,do this: In triangle,prove that: \[a^2+b^2+c^2 \le 4\sqrt{3}S+|a^2-b^2|+|b^2-c^2|+|c^2-a^2|\] $a,b,c$ are sides of triangle

xzlbq wrote: In triangle,prove that: \[a^2+b^2+c^2 \le 4\sqrt{3}S+|a^2-b^2|+|b^2-c^2|+|c^2-a^2|\] $a,b,c$ are sides of triangle Let $a\geq b\geq c$. Hence, $a^2+b^2+c^2 \le 4\sqrt{3}S+|a^2-b^2|+|b^2-c^2|+|c^2-a^2|\Leftrightarrow$ $\Leftrightarrow a^2+b^2+c^2 \le 4\sqrt{3}S+2a^2-2c^2\Leftrightarrow b^2+3c^2-a^2\leq\sqrt{3\sum_{cyc}(2a^2b^2-a^4)}$. If $b^2+3c^2-a^2\leq0$ so our inequality is true. But for $b^2+3c^2-a^2\geq0$ after squaring of the both sides we need to prove that $(a^2-b^2)^2\leq3c^2(a^2-c^2)$. Since $a^2-c^2\geq a^2-b^2$, it remains to prove that $a^2-b^2\leq3c^2$, which is our assuming.

This is a fairly obvious inequality if you realize that $$2\left(a^3+b^3+c^3-3abc\right)= (a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right).$$ For $$\left(a^3+b^3+c^3-3abc\right)= (a+b+c)\left(a^2+b^2+c^2-ab-bc-ac\right).$$ It is clear that the maximum value is clearly not in $(1,1,1)$. Most likely it's $(2,0,0)$ since we were given non-negative numbers, then $M=\frac{1}{2}.$ W.L.O.G. let $a \geq b \geq c$ (changing variables doesn't change the meaning). $$(!) \ 2\left(a^3+b^3+c^3-3abc\right) \geq (a-b)^3+(b-c)^3+(c-a)^3,$$$$(!) \ (a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right) \geq (a-b)^3+(b-c)^3+(c-a)^3.$$This is easy to show. It is clear that $a \geq a-b, b-c, a-c.$ $$(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right) \geq a\left((a-b)^2+(b-c)^2+(c-a)^2\right) \geq (a-b)^3+(b-c)^3+(c-a)^3.$$