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
$$S(BAC) = AB \cdot AC \sin \angle BAC = S_1 + S_2, \quad S(BDC) = BD \cdot CD \sin \angle BDC = S_3 + S_2.$$
Since $S_3 > S_1$, we have $S(BAC) > S(BDC)$, which gives $CD \cdot BD > AB \cdot AC$.