Example 6- Given a plane $\pi$ and a point $P$ on the plane as well as a point $Q$ not on the plane, find all points $R$ on the plane $\pi$ such that $\frac{Q P+P R}{Q R}$ is maximized.
Solution: Let point $R$ be the desired point, as shown in Figure $13-6$, denote $\angle Q P R=\alpha, \angle P Q R=\beta, \angle Q R P=\gamma$. Then, by the Law of Sines, we have $\frac{P R}{\sin \beta}=\frac{P Q}{\sin \gamma}=\frac{Q R}{\sin \alpha}$. Since $\alpha+\beta+\gamma=\pi$, we get $\frac{Q P+P R}{Q R}=\frac{\sin \beta+\sin \gamma}{\sin \alpha}=\frac{2 \sin \frac{\beta+\gamma}{2} \cos \frac{\beta-\gamma}{2}}{\sin \alpha}=\frac{\cos \frac{\beta-\gamma}{2}}{\sin \frac{\alpha}{2}}$.

First, fix $\alpha$. Clearly, when $\cos \frac{\beta-\gamma}{2}=1$, that is, $\beta=\gamma, P R=P Q$, $\frac{Q P+P R}{Q R}$ reaches its maximum, with the maximum value being $\frac{1}{\sin \frac{\alpha}{2}}$. Since $\sin \frac{\alpha}{2}(0<\alpha<\pi)$ increases as $\alpha$ increases, the maximum value of $\frac{1}{\sin \frac{\alpha}{2}}$ occurs when $\alpha$ is the smallest. Clearly, $\alpha$ is the smallest when $P R$ is the projection of $P Q$. Therefore, take a point $R$ on the projection of $P Q$ such that $P R=P Q$, then $R$ is the point that maximizes $\frac{Q P+P R}{Q R}$.