Problem 1439

1. Let $D$ be the midpoint of side $AB$ in triangle $ABC$. This implies that $AD = DB$.
2. Let $E$ and $F$ be points on segments $AC$ and $BC$ respectively. We need to show that the area of triangle $DEF$ is less than the sum of the areas of triangles $ADE$ and $BDF$.
3. Define $x = \frac{CE}{AC}$ and $y = \frac{CF}{BC}$. This means $E$ divides $AC$ in the ratio $x:1-x$ and $F$ divides $BC$ in the ratio $y:1-y$.
4. The area of triangle $ABC$ can be denoted as $[ABC]$. Since $D$ is the midpoint of $AB$, the area of triangle $ADE$ is $\frac{1}{2} \times [ADE]$ and the area of triangle $BDF$ is $\frac{1}{2} \times [BDF]$.
5. The area of triangle $DEF$ can be expressed in terms of $x$ and $y$. We need to show that:
$$[DEF] < [ADE] + [BDF]$$
6. Using the given sketch, we can express the areas in terms of $x$ and $y$. The area of triangle $DEF$ can be written as:
$$[DEF] = [ABC] \left(1 - \frac{1}{2}(1-x) - \frac{1}{2}(1-y) - xy\right)$$
7. The areas of triangles $ADE$ and $BDF$ can be written as:
$$[ADE] = [ABC] \left(\frac{1}{2}(1-x)\right)$$
$$[BDF] = [ABC] \left(\frac{1}{2}(1-y)\right)$$
8. Summing the areas of $ADE$ and $BDF$:
$$[ADE] + [BDF] = [ABC] \left(\frac{1}{2}(1-x) + \frac{1}{2}(1-y)\right)$$
9. We need to show that:
$$1 - \frac{1}{2}(1-x) - \frac{1}{2}(1-y) - xy < \frac{1}{2}(1-x) + \frac{1}{2}(1-y)$$
10. Simplifying the inequality:
$$1 - \frac{1}{2} - \frac{1}{2} + \frac{x}{2} + \frac{y}{2} - xy < \frac{1}{2} - \frac{x}{2} + \frac{1}{2} - \frac{y}{2}$$
$$1 - 1 + \frac{x}{2} + \frac{y}{2} - xy < 1 - \frac{x}{2} - \frac{y}{2}$$
$$\frac{x}{2} + \frac{y}{2} - xy < 1 - \frac{x}{2} - \frac{y}{2}$$
$$0 < xy - x - y + 1$$
$$0 < (1-x)(1-y)$$
11. Since $0 < (1-x)(1-y)$ holds for $0 < x < 1$ and $0 < y < 1$, the inequality is satisfied.