Belarusian Mathematical Olympiad

First we prove that the quadrilateral $SPQR$ is circumscribed and $O$ is the center of its incircle. The quadrilaterals $APOS$ and $BPOQ$ are cyclic since they have pairs of right angles, based on $AO$ and $BO$, respectively. In these circles $\angle OPS = \angle OAS$ and $\angle OPQ = \angle OBQ$. Wherein $\angle OAS = \angle OBQ$ in the circumcircle of $ABCD$. Therefore, $\angle OPS = \angle OPQ$ and $OP$ is the bisector of the angle $SPQ$. Similarly, $OQ$, $OR$ and $OS$ are the bisectors of angles $PQR$, $QRS$ and $RSP$, respectively. Hence point $O$ is equidistant from the sides of the quadrilateral $SPQR$, i.e. $SPQR$ is circumscribed and $O$ is its incenter.

Since $SPQR$ is circumscribed, $SP + QR = PQ + SR$, hence it is enough to prove the inequality $BD \ge PQ + SR$. The segments $BO$ and $DO$ are the diameters of the circumcircles of the quadrilaterals $BPOQ$ and $DSOR$, respectively. Therefore, $BO \ge PQ$ and $DO \ge SR$. Summing up these inequalities, we obtain the desired inequality.