60th Belarusian Mathematical Olympiad

Since $ABCD$ is cyclic, $\angle ABD=\angle ACD$, $\angle BAC=\angle BDC$. For the area of the triangles $ABD$ and $ACD$ we have $$S(ABD)=AB\cdot BD\sin \angle ABD=S_1+S_4,$$ $$S(ACD)=AC\cdot CD\sin \angle ACD=S_3+S_4.$$ By condition, $AC\cdot CD>AB\cdot BD$, so $S(ACD)>S(ABD)$, which gives $S_3>S_1$.

Similarly, for the area of $BAC$ and $BDC$ we have $$S(BAC)=AB\cdot AC\sin \angle BAC=S_1+S_2,S(BDC)=BD\cdot CD\sin \angle BDC=S_3+S_2.$$ Since $S_3>S_1$, we have $S(BAC)>S(BDC)$, which gives $CD\cdot BD>AB\cdot AC$.