Example 10 Determine the smallest real number $M$ such that the inequality
$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}$$

holds for all real numbers $a, b, c$. (47th IMO Problem)
Solving because
$$\begin{array}{l}
a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)= \\
a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left[\left(c^{2}-b^{2}\right)+\left(b^{2}-a^{2}\right)\right]= \\
\left(a^{2}-b^{2}\right)(a b-c a)+\left(b^{2}-c^{2}\right)(b c-c a)= \\
a(a+b)(a-b)(b-c)+c(b+c)(b-c)(b-a)= \\
(a-b)(b-c)[a(a+b)-c(b+c)]= \\
(a-b)(b-c)\left[\left(a^{2}-c^{2}\right)-c(a-c)\right]= \\
(a-b)(b-c)(a-c)(a+b+c)
\end{array}$$

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,
$$\left(\frac{(c-b)+(b-a)}{2}\right)^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2}$$

i.e.,
$$3(c-a)^{2} \leqslant 2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]$$

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 equations (4) and (5), we get
$$\begin{array}{l}
|(a-b)(b-c)(a-c)(a+b+c)| \leqslant\left|\frac{(c-a)^{3}}{4}(a+b+c)\right|= \\
\frac{1}{4} \sqrt{(c-a)^{6}(a+b+c)^{2}} \leqslant \\
\frac{1}{4} \sqrt{\left(\frac{2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right)^{3}(a+b+c)^{2}}= \\
\frac{\sqrt{2}}{2}\left(\sqrt[4]{\left(\frac{\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right)^{3}(a+b+c)^{2}}\right)^{2}
\end{array}$$

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

Equality holds in inequality (6) 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, we get
$$\begin{array}{l}
a=\left(1-\frac{3 \sqrt{2}}{2}\right) b \\
c=\left(1+\frac{3 \sqrt{2}}{2}\right) b
\end{array}$$

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