In Rt $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the median on the hypotenuse $A B$. Let the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ be $r$, $r_{1}$, and $r_{2}$, respectively, and $r_{1}+r_{2} \geqslant k r$. Find the maximum value of $k$.

In Rt $\triangle A B C$, $\angle C=90^{\circ}$, $C D$ is the median on the hypotenuse $A B$. Let the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ be $r$, $r_{1}$, and $r_{2}$, respectively, and $r_{1}+r_{2} \geqslant k r$. Find the maximum value of $k$.
Solve As shown in Figure 3, let $A B=c, A C=b, B C=a$. Then $r=\frac{a+b-c}{2}$.
Since $C D$ is a median, we have
$$
C D=A D=B D=\frac{c}{2} \text {. }
$$

Thus, the perimeter of $\triangle B C D$ is
$$
\begin{array}{l}
a+\frac{c}{2}+\frac{c}{2}=a+c . \\
\text { Hence } r_{1}=\frac{S_{\triangle B C D}}{\frac{1}{2}(a+c)}=\frac{a b}{2(a+c)} \\
=\frac{a b(c-a)}{2(a+c)(c-a)}=\frac{a c-a^{2}}{2 b} .
\end{array}
$$

Similarly, $r_{2}=\frac{b c-b^{2}}{2 a}$.
Then $r_{1}+r_{2} \geqslant k r$
$\Leftrightarrow \frac{a c-a^{2}}{2 b}+\frac{b c-b^{2}}{2 a} \geqslant k \cdot \frac{a+b-c}{2}$
$\Leftrightarrow \frac{b^{2} c-b^{3}+a^{2} c-a^{3}}{a b(a+b-c)} \geqslant k$.
Therefore, we only need to find the minimum value of $\frac{b^{2} c-b^{3}+a^{2} c-a^{3}}{a b(a+b-c)}$.
Let $\angle A=\alpha$. Then $a=c \sin \alpha, b=c \cos \alpha$.
Thus, $\frac{b^{2} c-b^{3}+a^{2} c-a^{3}}{a b(a+b-c)}$
$=\frac{\cos ^{2} \alpha-\cos ^{3} \alpha+\sin ^{2} \alpha-\sin ^{3} \alpha}{\sin \alpha \cdot \cos \alpha(\sin \alpha+\cos \alpha-1)}$.
Let $t=\sin \alpha+\cos \alpha\left(0<\alpha<\frac{\pi}{2}\right)$.
Then, $\sin \alpha \cdot \cos \alpha=\frac{t^{2}-1}{2}$.
Also, $1<t=\sin \alpha+\cos \alpha$
$=\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant \sqrt{2}$,
Thus, $\frac{\cos ^{2} \alpha-\cos ^{3} \alpha+\sin ^{2} \alpha-\sin ^{3} \alpha}{\sin \alpha \cdot \cos \alpha(\sin \alpha+\cos \alpha-1)}$
$=\frac{t+2}{t+1}=1+\frac{1}{t+1}$.
Let $f(t)=1+\frac{1}{t+1}$. Therefore, $f(t)$ is a monotonically decreasing function on $(1, \sqrt{2}]$.
Hence, $f(t)_{\min }=1+\frac{1}{1+\sqrt{2}}=\sqrt{2}$.
Therefore, $k_{\max }=\sqrt{2}$.