45.5 * If the inequality $a^{3}+b^{3}+c^{3}-3 a b c \geqslant M(a-b)(b-c)(c-a)$ holds for all non-negative real numbers $a, b, c$, find the maximum value of the real number $M$.


Suppose $M>0$, let $a=\min \{a, b, c\}, c \geqslant b$, and $b=a+c, c=a+y$, for $y \geqslant x \geqslant 0$, then
$$
\begin{aligned}
& a^{3}+b^{3}+c^{3}-3 a b c \geqslant M(a-b)(b-c)(c-a), \\
\Leftrightarrow & (a+b+c)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right) \geqslant 2 M(-x)(x-y) y, \\
\Leftrightarrow & (3 a+x+y)\left(x^{2}-x y+y^{2}\right) \geqslant M x y(y-x),
\end{aligned}
$$

Assume $y \neq 0$, if $a=0$ then
$f(t)=t^{3}+M t^{2}-M t+1 \geqslant 0$, where $t=\frac{x}{y}, 0 \leqslant t \leqslant 1$,
$$
f^{\prime}(t)=3 t^{2}+2 M t-M \text {, which implies } t_{\min }=\frac{-M+\sqrt{M^{2}+3 M}}{3} \text {, }
$$

Thus $\quad f\left(t_{\min }\right) \geqslant 0 \Leftrightarrow 2 M^{3}+9 M^{2}+27 \geqslant\left(2 M^{2}+6 M\right) \sqrt{M^{3}+3 M}$
$$
\leftrightarrow M^{4}-18 M^{2}-27 \leqslant 0 \leftrightarrow M \leqslant \sqrt{9+6 \sqrt{3}},
$$

Furthermore, $a^{3}+b^{3}+c^{3}-3 a b c \geqslant \sqrt{9+6 \sqrt{3}}(a-b)(b-c)(c-a)$ holds for all non-negative real numbers $a, b, c$, with equality when $a=0, b=\frac{-M+\sqrt{M^{2}+3 M}}{3}, c=1$, where $M=\sqrt{9+6 \sqrt{3}}$.
In summary, $M_{\max }=\sqrt{9+6 \sqrt{3}}$.