Example 10 (47th IMO Problem) Find the smallest real number $M$ such that for all real numbers $a, b, c$, we have
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
$$
Solve: First consider $P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right)$.
It is easy to see that $P(b)=P(c)=P(-c-b)=0$. Therefore, we have
$$
\begin{array}{c}
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)| .
\end{array}
$$

Thus, the original inequality is equivalent to $|(b-c)(a-b)(a-c)(a+b+c)| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}$.
By symmetry, assume without loss of generality that $a \leqslant b \leqslant c$, then we have $|(a-b)(b-c)|-(b-a)(c \cdots b) \leqslant$ $\left[\frac{(b-a)+(c-b)}{2}\right]^{2}=\frac{(c-a)^{2}}{4}$, and the equality holds if and only if $b-a=c-b$, i.e., $2 b=a+c$.
$$
\text { Also, }\left[\frac{(c-b)+(b-a)}{2}\right]^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2}
$$
$\Leftrightarrow 3(c-a)^{2} \leqslant 2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]$, and the equality holds if and only if $2 b=a+c$.
Therefore, $|(b-c)(a-b)(a-c)(a+b+c)|$
$$
\begin{array}{l}
\leqslant \frac{1}{4}\left|(c-a)^{3}(a+b+c)\right| \\
=\frac{1}{4} \sqrt{(c-a)^{6}(a+b+c)^{2}} \\
\leqslant \frac{1}{4} \sqrt{\left\{\frac{2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right\}^{3}(a+b+c)^{2}} \\
\leqslant \frac{\sqrt{2}}{2}\left\{\sqrt[1]{\left[\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right]^{3}(a+b+c)^{2}}\right\}^{2} .
\end{array}
$$

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

Thus, $M=\frac{9 \sqrt{2}}{32}$, and the equality holds if and only if $2 b=a+c$, and
$$
\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2} .
$$

Solving, we get $2 b=a+c,(c-a)^{2}=18 b^{2}$.
Taking $b=1$, we get $a=1-\frac{3 \sqrt{2}}{2}$ and $c=1+\frac{3 \sqrt{2}}{2}$.
Thus, when $(a, b, c)=\left(1-\frac{3 \sqrt{2}}{2}, 1,1+\frac{3 \sqrt{2}}{2}\right)$, the equality holds, hence $M=\frac{9 \sqrt{2}}{32}$.