Given that quadrilateral $ABCD$ is a square, $P$ is a point on side $CD$ (point $P$ does not coincide with the vertices), and the extension of $AP$ intersects the extension of $BC$ at point $Q$. Let the inradii of $\triangle ABQ$, $\triangle PAD$, and $\triangle PCQ$ be $r_{1}$, $r_{2}$, and $r_{3}$, respectively.
(1) Prove that $\frac{r_{1}^{2}}{r_{2} r_{3}} \geqslant 4$, and determine the position of point $P$ for which the equality holds;
(2) If $AB=1$, find the range of $r_{1}^{2}+r_{2}^{2}+r_{3}^{2}$.
Solution: (1) As shown in Figure 3, since
$$
\begin{aligned}
& \mathrm{Rt} \triangle A B Q \backsim \\
& \mathrm{Rt} \triangle P D A \backsim \\
& \mathrm{Rt} \triangle P C Q, \\
& \text { therefore, } r_{1}: r_{2}: r_{3} \\
= & A B: P D: P C .
\end{aligned}
$$

And $A B=C D=P D+P C$, hence
$$
r_{1}=r_{2}+r_{3} \geqslant 2 \sqrt{r_{2} r_{3}},
$$

which means $\frac{r_{1}^{2}}{r_{2} r_{3}} \geqslant 4$.
Equality holds if and only if $r_{2}=r_{3}$, i.e., when $P$ is the midpoint of $C D$.
(2) From (1), we have $r_{1}=r_{2}+r_{3}$, so,
$$
\begin{array}{l}
r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=\left(r_{2}+r_{3}\right)^{2}+r_{2}^{2}+r_{3}^{2} \\
=\left(r_{2}+r_{2} \cdot \frac{P C}{P D}\right)^{2}+r_{2}^{2}+\left(r_{2} \cdot \frac{P C}{P D}\right)^{2} \\
=\left(r_{2}+r_{2} \cdot \frac{1-P D}{P D}\right)^{2}+r_{2}^{2}+r_{2}^{2}\left(\frac{1-P D}{P D}\right)^{2} \\
=r_{2}^{2}\left[\frac{1}{P D^{2}}+1+\left(\frac{1}{P D}-1\right)^{2}\right] \\
=r_{2}^{2} \times 2\left(\frac{1}{P D^{2}}-\frac{1}{P D}+1\right) \\
=\left(\frac{1+P D-A P}{2}\right)^{2} \times 2\left(\frac{1}{P D^{2}}-\frac{1}{P D}+1\right) \\
=\frac{\left(1+P D-\sqrt{1+P D^{2}}\right)^{2}}{2}\left(\frac{1}{P D^{2}}-\frac{1}{P D}+1\right) .
\end{array}
$$

Let $\angle D A P=\theta\left(0<\theta<\frac{\pi}{4}\right)$, then
$$
\begin{array}{l} 
r_{1}^{2}+r_{2}^{2}+r_{3}^{2} \\
= \frac{1}{2}\left(1+\tan \theta-\frac{1}{\cos \theta}\right)^{2}\left(1+\cot ^{2} \theta-\cot \theta\right) \\
= \frac{1}{2} \cdot \frac{(\cos \theta+\sin \theta-1)^{2}}{\cos ^{2} \theta} . \\
\frac{\sin ^{2} \theta+\cos ^{2} \theta-\sin \theta \cdot \cos \theta}{\sin ^{2} \theta} \\
= \frac{1}{2} \cdot \frac{(\cos \theta+\sin \theta-1)^{2}(1-\sin \theta \cdot \cos \theta)}{(\sin \theta \cdot \cos \theta)^{2}} .
\end{array}
$$

Let $\cos \theta+\sin \theta=t$, then $1<t<\sqrt{2}$, and $\sin \theta \cdot \cos \theta$ $=\frac{t^{2}-1}{2}$. Therefore,
$$
\begin{array}{l}
r_{1}^{2}+r_{2}^{2}+r_{3}^{2} \\
=\frac{1}{2} \cdot \frac{(t-1)^{2}\left(1-\frac{t^{2}-1}{2}\right)}{\left(\frac{t^{2}-1}{2}\right)^{2}}=\frac{3-t^{2}}{(t+1)^{2}} .
\end{array}
$$

This is a strictly decreasing function of $t$, reaching its upper bound $\frac{1}{2}$ when $t=1$, and its lower bound $3-2 \sqrt{2}$ when $t=\sqrt{2}$.
In summary, the range of $r_{1}^{2}+r_{2}^{2}+r_{3}^{2}$ is $\left(3-2 \sqrt{2}, \frac{1}{2}\right)$.