Notice
$$\begin{array}{l}
a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)= \\
(a-b)(b-c)(c-a)(a+b+c)
\end{array}$$

Thus, the original inequality is transformed into
$$|(a-b)(b-c)(c-a)(a+b+c)| \leqslant M\left(|a|^{2}+|b|^{2}+|c|^{2}\right)^{2}$$

Let $a-b=\omega, b-c=\omega^{2}, c-a=1, a+b+c=1$, where $\omega=\frac{-1+\sqrt{3} i}{2}$.
That is, $a=\frac{\omega}{3}, b=-\frac{2 \omega}{3}, c=1+\frac{\omega}{3}$. Substituting these values, we find
$$M \geqslant \frac{9}{16}$$

Next, we prove that for any $a, b, c \in C$, we have
$$|(a-b)(b-c)(c-a)(a+b+c)| \leqslant \frac{9}{16}\left(|a|^{2}+|b|^{2}+|c|^{2}\right)^{2}$$

Let $|a-b|=x,|b-c|=y,|c-a|=z,|a+b+c|=w$. Then
$$x^{2}+y^{2}+z^{2}+w^{2}=3\left(|a|^{2}+|b|^{2}+|c|^{2}\right)$$

Thus, the original proposition is transformed into
$$x y z w \leqslant \frac{1}{16}\left(x^{2}+y^{2}+z^{2}+w^{2}\right)^{2}$$

This is equivalent to
$$\sqrt{\frac{x^{2}+y^{2}+z^{2}+w^{2}}{4}} \geqslant \sqrt[4]{x y z w}$$

This is the well-known AM - GM inequality.
Therefore, the minimum value of $M$ is $\frac{9}{16}$.
Example 14.22 Try to find the minimum value of $M$, such that for any complex numbers $a, b, c$, we have
$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(|a|^{2}+|b|^{2}+|c|^{2}\right)^{2}$$