17. Let $x$, $y$, $z$ be the lengths of the sides of a right triangle, with $z$ being the hypotenuse. When the inequality $x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)>k x y z$ always holds, find the maximum value of the parameter $k$, and indicate when equality holds.
17. (1) When $x=y$, the original inequality is $x^{2}(x+\sqrt{2} x)+x^{2}(\sqrt{2} x+x)+2 x^{2}(x+x) \geqslant \sqrt{2} k x^{3} \Leftrightarrow k \leqslant 2+3 \sqrt{2}$.
(2) Now we prove $k \leqslant 2+3 \sqrt{2}$, i.e., $x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y) \geqslant(2+3 \sqrt{2}) x y z$.

Assume without loss of generality that $z=1, x \leqslant y, x$ is the side opposite to the acute angle $\theta$, then $\theta \in\left(0, \frac{\pi}{4}\right], t=\sin \theta \cos \theta \leqslant \frac{1}{2}$.
Thus, $x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)=1+(\sin \theta+\cos \theta)(\sin \theta \cos \theta+1)$
$$
\begin{array}{l}
\geqslant 1+2 \sqrt{t}(t+1) \geqslant 1+2 \sqrt{2} t(1+t)(\sqrt{2} t \leqslant \sqrt{t}) \\
=\left(2 \sqrt{2} t^{2}+\frac{\sqrt{2}}{2}\right)+2 \sqrt{2} t+\left(1-\frac{\sqrt{2}}{2}\right) \geqslant 2 \sqrt{2} t+2 \sqrt{2} t+2\left(1-\frac{\sqrt{2}}{2}\right) t \\
=(2+3 \sqrt{2}) t=(2+3 \sqrt{2}) \sin \theta \cos \theta=(2+3 \sqrt{2}) x y z .
\end{array}
$$

The equality holds if and only if $\theta=\frac{\pi}{4}$, hence the maximum value of $k$ is $2+3 \sqrt{2}$.