Problem 1686

1. Identify the areas:
Let $s_1 = \text{Area}(AKN)$, $s_2 = \text{Area}(BKL)$, $s_3 = \text{Area}(CLM)$, and $s_4 = \text{Area}(DMN)$. Let $s = \text{Area}(ABCD)$.

2. Draw diagonals and midlines:
Draw the diagonals $\overline{AC}$ and $\overline{BD}$. The segments $\overline{KL}$, $\overline{LM}$, $\overline{MN}$, and $\overline{NK}$ are midlines of $\triangle ABC$, $\triangle BCD$, $\triangle CDA$, and $\triangle DAB$ respectively.

3. Calculate the sum of areas:
Since $\overline{KL}$, $\overline{LM}$, $\overline{MN}$, and $\overline{NK}$ are midlines, each of the areas $s_1, s_2, s_3, s_4$ is exactly half of the area of the corresponding triangle formed by the diagonals. Therefore, we have:
$$s_1 + s_2 + s_3 + s_4 = \frac{s}{2}$$

4. Apply Jensen's inequality:
Jensen's inequality states that for a convex function $f$, the following holds:
$$f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) \leq \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$$
Here, we use the function $f(x) = \sqrt[3]{x}$, which is concave for $x \geq 0$. Applying Jensen's inequality to the function $f(x) = \sqrt[3]{x}$ for the areas $s_1, s_2, s_3, s_4$, we get:
$$\sqrt[3]{\frac{s_1 + s_2 + s_3 + s_4}{4}} \leq \frac{\sqrt[3]{s_1} + \sqrt[3]{s_2} + \sqrt[3]{s_3} + \sqrt[3]{s_4}}{4}$$

5. Substitute the sum of areas:
Since $s_1 + s_2 + s_3 + s_4 = \frac{s}{2}$, we have:
$$\sqrt[3]{\frac{\frac{s}{2}}{4}} \leq \frac{\sqrt[3]{s_1} + \sqrt[3]{s_2} + \sqrt[3]{s_3} + \sqrt[3]{s_4}}{4}$$
Simplifying the left-hand side:
$$\sqrt[3]{\frac{s}{8}} \leq \frac{\sqrt[3]{s_1} + \sqrt[3]{s_2} + \sqrt[3]{s_3} + \sqrt[3]{s_4}}{4}$$

6. Multiply both sides by 4:
$$4 \cdot \sqrt[3]{\frac{s}{8}} \leq \sqrt[3]{s_1} + \sqrt[3]{s_2} + \sqrt[3]{s_3} + \sqrt[3]{s_4}$$

7. Simplify the left-hand side:
$$4 \cdot \sqrt[3]{\frac{s}{8}} = 4 \cdot \frac{\sqrt[3]{s}}{\sqrt[3]{8}} = 4 \cdot \frac{\sqrt[3]{s}}{2} = 2 \cdot \sqrt[3]{s}$$

8. Conclude the inequality:
$$\sqrt[3]{s_1} + \sqrt[3]{s_2} + \sqrt[3]{s_3} + \sqrt[3]{s_4} \leq 2 \cdot \sqrt[3]{s}$$

The final answer is $\boxed{ \sqrt[3]{s_1} + \sqrt[3]{s_2} + \sqrt[3]{s_3} + \sqrt[3]{s_4} \leq 2 \cdot \sqrt[3]{s} }$