3. Let $ABCD$ be a rectangle with an area of 2, $P$ a point on side $CD$, and $Q$ the point where the incircle of $\triangle PAB$ touches side $AB$. The product $PA \cdot PB$ varies as point $P$ changes. When $PA \cdot PB$ is minimized,
(1) Prove: $AB \geqslant 2BC$;
(2) Find the value of $AQ \cdot BQ$.


3. (1) Since $S_{\triangle A P B}=\frac{1}{2} S_{A B C D}=1$, we have $\frac{1}{2} P A \cdot P B \cdot \sin \angle A P B=1$, so $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$,
with equality holding only when $\angle A P B=90^{\circ}$.
This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$.
Thus, when $P A \cdot P B$ attains its minimum value, we should have $B C \leqslant \frac{1}{2} A B$, i.e., $A B \geqslant 2 B C$.
(2) Let the inradius of $\triangle P A B$ be $r$, then
$P A \cdot P B=(r+A Q)(r+B Q)=r(r+A Q+B Q)+A Q \cdot B Q$,
and since $P A \cdot P B=2 S_{\triangle A P B}, r(r+A Q+B Q)=S_{\triangle A P B}$,
we have $A Q \cdot B Q=S_{\triangle A P B}=1$.