$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geqslant k a b c
$$

holds for all right triangles, and determine when equality occurs.
When $a=b$, i.e., $\triangle ABC$ is an isosceles right triangle, the original inequality becomes
$$
a^{2}(a+c)+a^{2}(c+a)+c^{2}(a+a) \geqslant k a^{2} c.
$$

Therefore,
$$
k \leqslant \frac{2 a^{2}(a+\sqrt{2} a)+2 a^{2} \cdot 2 a}{\sqrt{2} a^{3}}=2+3 \sqrt{2}.
$$

We conjecture that the maximum value of $k$ is $2+3 \sqrt{2}$. Below, we prove that
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geqslant(2+3 \sqrt{2}) a b c.
$$

Without loss of generality, let $c=1$, and the angle opposite side $a$ be $\theta$, then $\theta \in\left(0, \frac{\pi}{4}\right]$, and $t=\sin \theta \cos \theta \leqslant \frac{\sin ^{2} \theta+\cos ^{2} \theta}{2}=\frac{1}{2}$. Thus,
$$
\begin{aligned}
& a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
- & \sin ^{2} \theta(\cos \theta+1)+\cos ^{2} \theta(1+\sin \theta)+\sin \theta+\cos \theta \\
- & 1+(\sin \theta+\cos \theta)(\sin \theta \cos \theta+1) \\
\geqslant & 1+2 \sqrt{t}(t+1) \geqslant 1+2 \sqrt{2} t(t+1) \quad\left(\because t \leqslant \frac{1}{2}, \therefore \sqrt{2} t \leqslant \sqrt{t}\right) \\
= & \left(2 \sqrt{2} t^{2}+\frac{\sqrt{2}}{2}\right)+2 \sqrt{2} t+\left(1-\frac{\sqrt{2}}{2}\right) \geqslant 2 \sqrt{2} t+2 \sqrt{2} t+2\left(1-\frac{\sqrt{2}}{2}\right) t \\
= & (2+3 \sqrt{2}) t=(2+3 \sqrt{2}) \sin \theta \cos \theta=(2+3 \sqrt{2}) a b c.
\end{aligned}
$$

Thus, inequality (1) is proved, and the equality holds when $\theta=\pi$. Therefore, the maximum value of $k$ is $2+3 \sqrt{2}$.