Let $a-b=x$ and $b-c=y$. So, we get $a-c=x+y$. Thus, we want to prove $xy(x+y) \leq 2$. Because $0 \leq a, b, c \leq 2$, we get that $ -2 \leq x, \: y, \: x+y \leq 2$. $|xy| \leq \frac{(x+y)^2}{4} \leq 1$ (because $(x+y)^2 \leq 4$) So, $xy(x+y) \leq |xy| \cdot |x + y| \leq |x+y| \leq 2$ and we are done!

See also: https://artofproblemsolving.com/community/c6h3330877p30835292

Beautiful and correct

adityaguharoy wrote: See also: https://artofproblemsolving.com/community/c6h3330877p30835292 mudok wrote: biao wrote: Let $a,b,c$ be real number satisfy $a+b+c=6$ and $a^2+b^2+c^2=14$.Prove that $$(a-b)(b-c)(c-a) \leq 2$$ WLOG $c\ge b\ge a$ $0\le (a-1)^2+(b-2)^2+(c-3)^2=a^2+b^2+c^2 -(2a+4b+6c)+14=$ $=28 - (2a+4b+6c)=2(2+a-c) \ \ \Rightarrow \ \ c-a\le 2$ $(b-a)(c-b)(c-a)\le 2\cdot (b-a)(c-b)\le 2\cdot \frac{1}{4}(b-a+c-b)^2=\frac{1}{2}(c-a)^2\le 2$

WLOG $a \geq b \geq c$. Then $(a-b)(b-c)(a-c) \leq (2-b)(b)(2)$, and the result follows.