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

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:
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geqslant(2+3 \sqrt{2}) a b c \text {. }
$$

Without loss of generality, let $c=1$, and 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
$$
\begin{aligned}
& a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
= & \sin ^{2} \theta(\cos \theta+1)+\cos ^{2} \theta(1+\sin \theta)+\sin \theta+\cos \theta \\
= & 1+(\sin \theta+\cos \theta)(\sin \theta \cdot \cos \theta+1) \geqslant 1+2 \sqrt{t}(t+1) \\
\geqslant & 1+2 \sqrt{2} t(t+1)\left(\text { since } t \leqslant \frac{1}{2} \text {, then } \sqrt{2} t \leqslant \sqrt{t}\right) \\
= & \left(2 \sqrt{2} t^{2}+\frac{\sqrt{2}}{2}\right)+2 \sqrt{2} t+\left(1-\frac{\sqrt{2}}{2}\right) \geqslant 2 \sqrt{2} t+2 \sqrt{2} t+2\left(1-\frac{\sqrt{2}}{2}\right) t \\
= & (2+3 \sqrt{2}) t=(2+3 \sqrt{2}) \sin \theta \cdot \cos \theta=(2+3 \sqrt{2}) a b c .
\end{aligned}
$$

Thus, (*) is proven, and the equality holds when $\theta=\frac{\pi}{4}$.
Therefore, the maximum value of $k$ is $2+3 \sqrt{2}$.
Note: (*) can also be proven with a simpler method:
$$
\begin{aligned}
& a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
= & c\left(a^{2}+b^{2}\right)+a\left(\frac{c^{2}}{2}+b^{2}\right)+b\left(\frac{c^{2}}{2}+a^{2}\right)+\frac{c}{2} \cdot c(a+b) \\
\geqslant & 2 a b c+\sqrt{2} a b c+\sqrt{2} a b c+\frac{c}{2} \cdot \sqrt{a^{2}+b^{2}} \cdot 2 \sqrt{a b} \geqslant(2+3 \sqrt{2}) a b c .
\end{aligned}
$$