How to maximize $\frac 1 {PC}+\frac 1 {PB}$? Assume $P$ is a point inside the angle $\hat A$. How to draw a line that intersects with lines $a$ and $b$ AND maximizes $\frac 1 {PC}+\frac 1 {PB}$ ?
Here is the picture:

EDIT:
The point $P$ is given. the only thing changeable is the angle of the line passing the point $P$.

I claim that $AP \perp BC$.
We set $X \in AB$ such that $AC \parallel PX$, and $Y \in AP$ such that $XY \parallel BC$.
Now, we have $$\triangle AYX \sim \triangle APB$$ $$\triangle PYX \sim \triangle APC$$
This gives us $$\frac{XY}{PB}=\frac{AY}{AP}$$ $$\frac{XY}{PC}=\frac{YP}{AP}$$
Summing them, we have $$\frac{1}{PB}+\frac{1}{PC}=\frac{1}{XY}$$
We need to minimize $XY$. Since $X$ is fixed and $Y$ is on $AP$, we need $XY \perp AP$, i.e. $AP \perp BC$.
