Alternatively, draw in the diagonals $\overline{AC}$ and $\overline{BD}$. It is clear that $\overline{KL}, \overline{LM}, \overline{MN},$ and $\overline{NK}$ are midlines of $\triangle{ABC}, \triangle{BCD}, \triangle{CDA},$ and $\triangle{DAB}$ respectively; whence, $S_1 + S_2 + S_3 + S_4 = \frac{S}{2}$. Therefore the result follows by Jensen's inequality. In fact, we find the follow generalization $S_1^r + S_2^r + S_3^r + S_4^r \le 4 \cdot \left(\frac{S}{8}\right)^r$ for all $0 \le r \le 1$ and the corollary result $S_1^r + S_2^r + S_3^r + S_4^r \ge 4 \cdot \left(\frac{S}{8}\right)^r$ for all $r \le 0$ and $r \ge 1.$

$ (\frac{S_1}{S})^{\frac{1}{3}}=(\frac{A(ABD)}{S}\cdot\frac{AK}{AB}\cdot\frac{AN}{AD})^\frac{1}{3}\leq\frac{1}{3}(\frac{A(ABD)}{S}+\frac{AK}{AB}+\frac{AN}{AD})$ (where $ A(XYZ)$ equals to the area of $ XYZ$) Summing these 4 inequalities gives the desired result.

As a plus, the equality holds when $ABCD$ is a parallelogram and $K,L,M,N$ are midpoints.