3. Let $a, b, c$ be the three sides of a right triangle, with $c$ being the hypotenuse. The maximum value of $k$ such that the inequality $a^{2}(b+c)+b^{2}(c+a)$ $+c^{2}(a+b) \geqslant k a b c$ holds for all right triangles is $\qquad$.
$$
\begin{array}{l}
3.2+3 \sqrt{2} \\
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
=\left(a^{2}+b^{2}\right) c+a\left(b^{2}+\frac{c^{2}}{2}\right)+b\left(\frac{c^{2}}{2}+a^{2}\right) \\
+\frac{c}{2} \cdot c(a+b) \\
\geqslant 2 a b c+\sqrt{2} a b c+\sqrt{2} a b c+\frac{c}{2} \sqrt{a^{2}+b^{2}} \\
\cdot 2 \sqrt{a b} \\
\geqslant(2+2 \sqrt{2}) a b c+c \cdot \sqrt{2 a b} \cdot \sqrt{a b} \\
=2+3 \sqrt{2}) a b c .
\end{array}
$$

When and only when $a=b=\frac{\sqrt{2}}{2} c$, the equality holds.
$$
\therefore k_{\max }=2+3 \sqrt{2} \text {. }
$$