10. Find the smallest real number $M$ such that for all real numbers $a, b, c$, we have
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} \text {. }
$$
(IMO - 47 Problem)
10. Try to transform the left side of the given inequality into a form that can use the mean value inequality, and note the connection between this expression and $\left(a^{2}+b^{2}+c^{2}\right)^{2}$, thereby finding the minimum value of $M$.
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 $P(b)=P(c)=P(-c-b)=0$.
Then, 
$$
|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)|
$$
$$
=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)|.
$$

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 $a \leqslant b \leqslant c$, then
$$
|(b-c)(a-b)|=(b-a)(c-b) \leqslant\left[\frac{(b-a)+(c-b)}{2}\right]^{2}=\frac{(c-a)^{2}}{4},
$$

Equality holds if and only if $b-a=c-b$, i.e., $2 b=a+c$.
Note that $\left[\frac{(b-a)+(c-b)}{2}\right]^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2}$
$$
\begin{array}{l}
\Leftrightarrow 3(c-a)^{2} \\
\leqslant 2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right],
\end{array}
$$
Equality holds if and only if $2 b=a+c$. 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 value inequality,
$$
\begin{array}{l}
|(b-c)(a-b)(a-c)(a+b+c)| \\
\leqslant \frac{\sqrt{2}}{2}\left[\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\right]^{2} \\
=\frac{9 \sqrt{2}}{32}\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{array}
$$

Thus, $M \geqslant \frac{9 \sqrt{2}}{32}$, and 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 $a=1-\frac{3 \sqrt{2}}{2}, c=1+\frac{3 \sqrt{2}}{2}$. At this point, the equality in the original inequality holds.
Thus, $M_{\text {min }}=\frac{9 \sqrt{2}}{32}$.