Example 5 Let $a, b, c$ be the lengths of the three sides of a right-angled triangle, and $a \leqslant b \leqslant c$. Find the maximum constant $k$, such that
$$
\begin{array}{l}
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
\geqslant \text { kabc }
\end{array}
$$
holds for all right-angled triangles, and determine when equality occurs.
When $a=b$, i.e., $\triangle ABC$ is an isosceles right triangle, the original inequality becomes
$$
\begin{array}{l}
a^{2}(a+\sqrt{2} a)+a^{2}(\sqrt{2} a+a)+2 a^{2}(a+a) \\
\geqslant \sqrt{2} k a^{3},
\end{array}
$$

which simplifies to $k \leqslant 2+3 \sqrt{2}$. We conjecture that the maximum value of $k$ is $2+3 \sqrt{2}$. We now prove:
$$
\begin{array}{l}
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
\geqslant(2+3 \sqrt{2}) a b c .
\end{array}
$$

Without loss of generality, let $c=1$, and let the angle opposite side $a$ be $\theta$, then $\theta \in \left(0, \frac{\pi}{4}\right]$. Let $t=\sin \theta \cdot \cos \theta \leqslant \frac{\sin ^{2} \theta+\cos ^{2} \theta}{2}=\frac{1}{2}$.

Thus, we have
$$
\begin{array}{l}
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
=\sin ^{2} \theta(\cos \theta+1)+\cos ^{2} \theta(1+\sin \theta) \\
+\sin \theta+\cos \theta \\
=1+(\sin \theta+\cos \theta)(\sin \theta \cdot \cos \theta+1) \\
\geqslant 1+2 \sqrt{t}(t+1) \\
\geqslant 1+2 \sqrt{2} t(t+1) \text { (since } t \leqslant \frac{1}{2} \text {, then } \sqrt{2} t \leqslant \sqrt{t} \text { ) } \\
=\left(2 \sqrt{2} t^{2}+\frac{\sqrt{2}}{2}\right)+2 \sqrt{2} t+\left(1-\frac{\sqrt{2}}{2}\right) \\
\geqslant 2 \sqrt{2} t+2 \sqrt{2} t+2\left(1-\frac{\sqrt{2}}{2}\right) t \\
=(2+3 \sqrt{2}) t \\
=(2+3 \sqrt{2}) \sin \theta \cdot \cos \theta \\
=(2+3 \sqrt{2}) a b c \text {. } \\
\end{array}
$$

Thus, inequality (3) is proven, and equality holds when $\theta=\frac{\pi}{4}$. Therefore, the maximum value of $k$ is $2+3 \sqrt{2}$.

Note: This problem first uses special values to guess the maximum value of $k$, and then proceeds to prove it. This method of "guessing and then proving" using special values is a common approach to solving such problems.

Inequality (3) can also be proven more succinctly:
$$
\begin{array}{l}
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
=c\left(a^{2}+b^{2}\right)+a\left(\frac{c^{2}}{2}+b^{2}\right) \\
\quad+b\left(\frac{c^{2}}{2}+a^{2}\right)+\frac{c}{2} \cdot c(a+b) \\
\geqslant 2 a b c+\sqrt{2} a b c+\sqrt{2} a b c+\frac{c}{2} \cdot \sqrt{a^{2}+b^{2}} \cdot 2 \sqrt{a b} \\
\geqslant(2+3 \sqrt{2}) a b c .
\end{array}
$$