$\square$ Example 10 Determine the smallest real number $M$ such that the inequality $\mid a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}\right.$ $\left.-c^{2}\right)+c a\left(c^{2}-a^{2}\right) \mid \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}$ holds for all real numbers $a, b, c$. (47th IMO Problem)
Solving because
$$\begin{aligned}
& a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right) \\
= & a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left[\left(c^{2}-b^{2}\right)+\left(b^{2}-a^{2}\right)\right] \\
= & \left(a^{2}-b^{2}\right)(a b-c a)+\left(b^{2}-c^{2}\right)(b c-c a) \\
= & a(a+b)(a-b)(b-c)+c(b+c)(b-c)(b-a) \\
= & (a-b)(b-c)[a(a+b)-c(b+c)] \\
= & (a-b)(b-c)\left[\left(a^{2}-c^{2}\right)-c(a-c)\right] \\
= & (a-b)(b-c)(a-c)(a+b+c)
\end{aligned}$$

Therefore, the problem is reduced to finding the smallest real number $M$ such that the inequality
$$|(a-b)(b-c)(a-c)(a+b+c)| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .$$

This relation is symmetric with respect to $a, b, c$, so we can assume without loss of generality that $a \leqslant b \leqslant c$. Under this assumption,
$$|(a-b)(b-c)|=(b-a)(c-b) \leqslant\left[\frac{(b-a)+(c-b)}{2}\right]^{2}=\frac{(c-a)^{2}}{4} .$$

Equality holds if and only if $a-b=b-c$, i.e., $2 b=a+c$, and $a, b, c$ form an arithmetic sequence.
Also, since $\left[\frac{(c-b)+(b-a)}{2}\right]^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2}$, we have
$$3(c-a)^{2} \leqslant 2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right] .$$

Equality holds if and only if $a-b=b-c$, i.e., $2 b=a+c$, and $a, b, c$ form an arithmetic sequence.

From (1), we get
$$\begin{aligned}
& |(a-b)(b-c)(a-c)(a+b+c)| \leqslant\left|\frac{(c-a)^{3}}{4}(a+b+c)\right| \\
= & \frac{1}{4} \sqrt{(c-a)^{6}(a+b+c)^{2}} \\
\leqslant & \frac{1}{4} \sqrt{\left\{\frac{2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right\}^{3}(a+b+c)^{2}} \\
= & \frac{\sqrt{2}}{2}\left\{\sqrt[4]{\left[\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right]^{3}(a+b+c)^{2}}\right\}^{2} .
\end{aligned}$$

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

Equality holds in (2) if and only if $\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2}$ and $2 b=a+c$.

Substituting $b=\frac{a+c}{2}$ into the first equation, we get $(c-a)^{2}=18 b^{2}$, i.e., $c-a=3 \sqrt{3} b$, solving for
$$a=\left(1-\frac{3 \sqrt{2}}{2}\right) b, c=\left(1+\frac{3 \sqrt{2}}{2}\right) b \text {. }$$

By symmetry, $M$ attains its minimum value $\frac{9 \sqrt{2}}{32}$ if and only if $(a, b, c)$ is proportional to $\left(1-\frac{3 \sqrt{2}}{2}, 1,1+\frac{3 \sqrt{2}}{2}\right)$.