Example 5 Find the smallest real number $M$ such that
$$
\left|\sum a b\left(a^{2}-b^{2}\right)\right| \leqslant M\left(\sum a^{2}\right)^{2}
$$

holds for all real numbers $a, b, c$.
【Analysis】By symmetry, without loss of generality, let $a \geqslant b \geqslant c$.
Let $a-b=x, b-c=y$, then $x, y \geqslant 0$.
Let $a+b+c=s$.
By $\sum a b\left(a^{2}-b^{2}\right)=s \prod(a-b)$, then
$$
\left|\sum a b\left(a^{2}-b^{2}\right)\right|=|s x y(x+y)|,
$$

where, “川” represents the cyclic symmetric product.
$$
\begin{array}{l}
\text { and } 3 \sum a^{2}=\left(\sum a\right)^{2}+\left[\sum(a-b)^{2}\right] \\
=s^{2}+x^{2}+y^{2}+(x+y)^{2}, \\
\text { hence } \sum a^{2}=\frac{s^{2}+x^{2}+y^{2}+(x+y)^{2}}{3} \\
\Rightarrow\left|\sum a b\left(a^{2}-b^{2}\right)\right| \leqslant M\left(\sum a^{2}\right)^{2} \\
\Leftrightarrow|\operatorname{sxy}(x+y)| \\
\quad \leqslant M \cdot \frac{\left[s^{2}+x^{2}+y^{2}+(x+y)^{2}\right]^{2}}{3} .
\end{array}
$$

By the AM-GM inequality, we get
$$
\begin{array}{l}
{\left[s^{2}+x^{2}+y^{2}+(x+y)^{2}\right]^{2}} \\
\geqslant\left[s^{2}+2 x y+2 y x+\frac{(x+y)^{2}}{2}\right]^{2} \\
=\left[4 \sqrt[4]{s^{2} \times 2 x y \times 2 y x \times \frac{(x+y)^{2}}{2}}\right]^{2} \\
=16 \sqrt{2}|\operatorname{sxy}(x+y)| .
\end{array}
$$

Thus, taking $M=\frac{9 \sqrt{2}}{32}$, inequality (1) holds.
Furthermore, when $s=\sqrt{2} x=\sqrt{2} y=3 \sqrt{2}$, i.e., $a=\sqrt{2}+3$, $b=\sqrt{2}, c=\sqrt{2}-3$, equality in (1) is achieved.
In conclusion, the minimum value of $M$ that satisfies the condition is $\frac{9 \sqrt{2}}{32}$.