10. (15 points) As shown in Figure 2, given that the side length of square $ABCD$ is 1, a line passing through vertex $C$ intersects the rays $AB$ and $AD$ at points $P$ and $Q$ respectively. Find the maximum value of $\frac{1}{AP} +$ $\frac{1}{AQ} + \frac{1}{PQ}$.
10. From $\triangle P B C \backsim \triangle P A Q \Rightarrow \frac{B C}{A Q}=\frac{B P}{A P}$ $\Rightarrow \frac{1}{A Q}=\frac{A P-1}{A P} \Rightarrow \frac{1}{A P}+\frac{1}{A Q}=1$.
Since $1=\frac{1}{A P}+\frac{1}{A Q} \geqslant 2 \sqrt{\frac{1}{A P} \cdot \frac{1}{A Q}}$, therefore, $A P \cdot A Q \geqslant 4$.
Also, $\frac{1}{2} P Q \cdot A C \geqslant S_{\triangle A P Q}=\frac{1}{2} A P \cdot A Q \geqslant 2$, thus $\frac{1}{P Q} \leqslant \frac{A C}{4}=\frac{\sqrt{2}}{4}$.
Hence, $\frac{1}{A P}+\frac{1}{A Q}+\frac{1}{P Q} \leqslant 1+\frac{\sqrt{2}}{4}$.
When $A P=A Q$, the equality holds.
Therefore, the maximum value of $\frac{1}{A P}+\frac{1}{A Q}+\frac{1}{P Q}$ is $1+\frac{\sqrt{2}}{4}$.