You can consider the orthogonal projections of P and Q onto AB.. Then its easy to show that it must be true for the projections, and thus true for the initial condition by triangle inequality.

Let's examine following figure. $AG =a, GM = b, MH = c, HB = d$ and $a + b + c + d = 2$. we will show $QK = GH = b + c \geq 1$ Because of $AQM \sim NQD$ and $BPM \sim NPC$, $JN = \frac {a^2}{b}$ and $ NI = \frac {d^2}{c}$. Therefore $b + c = \frac {a^2}{b} + \frac {d^2}{c}$ ... (1) Also by Cauchy - Schwarz $ \frac {a^2}{b} + \frac {d^2}{c} \geq \frac{(a+d)^2}{b+c}$ ... (2) Now, from (1) and (2), $b + c \geq \frac{(a+d)^2}{b+c}$ and hence we yields $b + c \geq a + d$. Since $a + b + c + d = 2$, we conclude that $b + c \geq 1$. * Equality holds when $PQ // AB$.

Attachments: