4.5.20 Find the smallest real number $M$ such that for all real numbers $a, b$, and $c$, we have
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
$$
Given $a=2-3 \sqrt{2}, b=2, c=2+3 \sqrt{2}$, we obtain $M \geqslant \frac{9 \sqrt{2}}{32}$. We now prove the inequality
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant \frac{9 \sqrt{2}}{32}\left(a^{2}+b^{2}+c^{2}\right)^{2}
$$

for all real numbers $a, b$, and $c$.
Since $a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)=-(a-b)(b-c)(c-a)(a+b+c)$, it suffices to prove that $|(a-b)(b-c)(c-a)(a+b+c)| \leqslant \frac{9 \sqrt{2}}{32}\left(a^{2}+b^{2}+c^{2}\right)^{2}$.
By symmetry, we may assume $a \leqslant b \leqslant c$. Then we have
$$
|(a-b)(b-c)|=(b-a)(c-b) \leqslant\left(\frac{(b-a)+(c-b)}{2}\right)^{2}=\frac{(c-a)^{2}}{4},
$$

and $\left(\frac{(c-b)+(b-a)}{2}\right)^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2}$, so $3(c-a)^{2} \leqslant 2\left[(b-a)^{2}+\right.$ $\left.(c-b)^{2}+(c-a)^{2}\right]$, thus
$$
\begin{array}{l}
|(b-c)(a-b)(a-c)(a+b+c)| \leqslant \frac{1}{4}\left|(c-a)^{3}(a+b+c)\right|= \\
\frac{1}{4} \sqrt{(c-a)^{6}(a+b+c)^{2}} \leqslant \frac{1}{4} \sqrt{\left(\frac{2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right)^{3}(a+b+c)^{2}} \\
\leqslant \frac{\sqrt{2}}{2}\left(\sqrt[4]{\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right)^{3}(a+b+c)^{2}}\right)^{2},
\end{array}
$$

By the mean inequality, we get
$$
\begin{aligned}
& |(b-c)(a-b)(a-c)(a+b+c)| \\
\leqslant & \frac{\sqrt{2}}{2}\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\right)^{2} \\
= & \frac{9 \sqrt{2}}{32}\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{aligned}
$$

Therefore, the minimum value of $M$ is $\frac{9}{32} \sqrt{2}$.