97. In a right triangle $\triangle ABC$, where $C$ is the right angle vertex, the sides opposite to vertices $A, B, C$ are $a, b, c$, respectively. If the circle $K_{a}$ has its center on $a$ and is tangent to sides $b, c$; the circle $K_{b}$ has its center on $b$ and is tangent to sides $a, c$; and the radii of circles $K_{a}, K_{b}$ are $r_{a}, r_{b}$, respectively. Try to find the largest real number $p$ such that for all right triangles, the inequality $\frac{1}{r_{a}}+\frac{1}{r_{b}} \geqslant p\left(\frac{1}{a}+\frac{1}{b}\right)$ holds. (2009 Czech and Slovak Mathematical Olympiad)
97. Let $A^{\prime}$ be the point symmetric to $A$ with respect to $BC$, then $K_{a}$ is the incircle of $\triangle A^{\prime} A B$. Given that the perimeter of $\triangle A^{\prime} A B$ is $2(b+c)$ and the area is $a b$, we have $r_{a}=\frac{a b}{b+c}$. Similarly, $r_{b}=\frac{a b}{a+c}$.

Therefore,
$$\begin{array}{l}
p \leqslant \frac{\frac{1}{r_{a}}+\frac{1}{r_{b}}}{\frac{1}{a}+\frac{1}{b}}=\frac{a+b+2 c}{a+b}= \\
1+\frac{2 c}{a+b}=1+\frac{2 \sqrt{a^{2}+b^{2}}}{a+b}
\end{array}$$

When $a, b$ are positive numbers, by the AM-GM inequality, $\frac{2 \sqrt{a^{2}+b^{2}}}{a+b} \geqslant \sqrt{2}$. The equality holds when $a=b$.

Therefore, the maximum value of $p$ is $1+\sqrt{2}$.