7. Through a fixed point $P$ inside $\angle A$, draw a line intersecting the two sides at $B, C$, such that $\frac{1}{P B}+\frac{1}{P C}$ is maximized.
(Draw $A D \perp B C$ at $D$, let $A D=h$, the areas of $\triangle A B P$ and $\triangle A C P$ be $S_{1}, S_{2}$, and the area of $\triangle A B C$ be $S$,
$$
\begin{array}{l}
\angle B A P=\alpha, \angle C P A=\beta \cdot \frac{1}{P B}+\frac{1}{P C}=\frac{h}{2} . \\
\left(\frac{1}{S_{1}}+\frac{1}{S_{2}}\right)=\frac{h S}{2 S_{1} S_{2}}=\frac{h \sin A}{A P^{2} \sin \alpha \sin \beta} \leqslant \frac{\sin A}{P A \sin \alpha \sin \beta} .
\end{array}
$$

Therefore, when $P A \perp B C$, the equality holds, and the desired value is maximized.)