Let $X$ be the intersection of $\overline{CR}$ and $\overline{BP}$. Also, set $\angle BAP = \alpha$ with $\alpha \geq 60^{\circ}$, $\angle ABP = \beta$. WLOG set a side of the square to be 1. First, since $\angle BAQ = \angle PPQ = 90^{\circ}$, $ABPQ$ is cyclic, so $\angle BAP = \angle BQP = \alpha$. Then $\angle PBQ = 90-\alpha$ and $\angle ABQ = \alpha+\beta-90$. So \[BR = \frac{BX}{\sin{\alpha}} = \frac{\sin{\beta}}{\sin{\alpha}}\] \[BP = BQ\sin{\alpha} = \frac{\sin{\alpha}}{\sin{(180-\alpha-\beta)}}\] It remains to show that $\frac{\sin{\alpha}}{\sin{(180 - \alpha-\beta)}} \geq \frac{\sin{\beta}}{\sin{\alpha}}$, or $\sin{\beta}\sin{(180-\alpha-\beta)} \leq \sin^2{\alpha}$. It's not hard to show that the maximum value of $\sin{m}\sin{(n-m)}$ is attained when $m = \frac{n}{2}$. (Just consider the value of $\sin{\left(\frac{m}{2}-\theta\right)}\sin{\left(\frac{m}{2}+\theta\right)}$ for $\theta>0$ in relation to $\sin^2{\left(\frac{m}{2}\right)}$.) It then follows that $\sin{\beta}\sin{(180-\alpha-\beta)} \leq \sin^2{\left(90-\frac{\alpha}{2}\right)}$. But $90 - \frac{\alpha}{2} \leq 60$. So \[\sin{\beta}\sin{(180-\alpha-\beta)} \leq \sin^2{\left(90-\frac{\alpha}{2}\right)} \leq \sin^2{\alpha}\] Equality holds when $\alpha = 60^{\circ}$ and $\beta = 60^{\circ}$, or when $\triangle ABP$ is equilateral.