Silk Road Mathematics Competition

Answer: $M=\frac{1}{2}$. The inequality $a^3+b^3+c^3-3abc\ge M(|a-b{|}^3+|b-c{|}^3+|c-a{|}^3)$ for $M=1/2$ implies from $$2(a^3+b^3+c^3-3abc)=(a+b+c)((a-b{)}^2+(b-c{)}^2+(c-a{)}^2)$$ and from obvious inequalities $$(x+y+z)(x-y{)}^2\ge |x-y{|}^3$$

The value $M=1/2$ is maximal, because for very large values of $a$ and very small values of $b$, $c$, the value of $a^3+b^3+c^3-3abc$ can be very close to $a^3$, and the value of $|a-b{|}^3+|b-c{|}^3+|c-a{|}^3$ can be very close to $2a^3$.