By symmetry, we only need to prove $2(OP+OR) \le AD+BC$. It follows from these facts: Let the angle bisectors of $\angle AOB$ and $\angle COD$ be $OK$ and $OL$ ($K,L$ lie on $AB,CD$, respectively). Then $OP \le OK$, $OQ \le OL$. Let $M,N$ be the midpoint of $AB,CD$, then $MN=|\overrightarrow{MN}|=|\tfrac{1}{2}(\overrightarrow{AD}+\overrightarrow{BC})| \le \tfrac{1}{2}(AD+BC)$. In a triangle $XYZ$, the projection of the midpoint of $YZ$ on the angle bisector of $\angle YXZ$ lies outside the triangle (or on the side $YZ$). (Obvious.) In this configuration, $|\text{the projection of } MN \text{ on the line } KOL| \ge KL$. Therefore, $2(OP+OR) \le 2(OK+OL)=2KL \le 2 \cdot |\text{the projection of } MN \text{ on the line } KOL| \le 2MN \le AD+BC$.