consider the midpoints $M, N$ of $\overline{AC}, \overline{BD}$. As the distance $\overline{MN} \ge 0$, we have (using vectors), $4(\vec{A} + \vec{C} - \vec{B} - \vec{D})^2 \ge 0 \Rightarrow$ $4\vec{A}^2 + 4\vec{B}^2 + 4\vec{C}^2 + 4\vec{D}^2 + 8\vec{A} \cdot \vec{C} + 8\vec{B} \cdot \vec{D} \ge 8\vec{A} \cdot \vec{B} + 8\vec{B} \cdot \vec{C} + 8\vec{C} \cdot \vec{D} + 8\vec{D} \cdot \vec{A}$ $\Rightarrow 10\vec{A}^2 + 10\vec{B}^2 + 10\vec{C}^2 + 10\vec{D}^2 - 10\vec{A} \cdot \vec{B} - 10\vec{B} \cdot \vec{C} - 10\vec{C} \cdot \vec{D} - 10\vec{D} \cdot \vec{A} \ge 6\vec{A}^2 + 6\vec{B}^2 + 6\vec{C}^2 + 6\vec{D}^2 - 2\vec{A} \cdot \vec{B} - 2\vec{B} \cdot \vec{C} - 2\vec{C} \cdot \vec{D} - 2\vec{D} \cdot \vec{A} - 8\vec{A} \cdot \vec{C} - 8\vec{B} \cdot \vec{D}$ $\Rightarrow 5(\overline{AB}^2 + \overline{BC}^2 + \overline{CD}^2 + \overline{DA}^2) \ge 4(\overline{AP}^2 + \overline{BQ}^2 + \overline{CR}^2 + \overline{DS}^2)$

Problem. Let A, B, C, D be four arbitrary points in the plane, and let P, Q, R, S be the midpoints of the segments BC, CD, DA, AB. Prove that $4\left(AP^2+BQ^2+CR^2+DS^2\right)\leq 5\left(AB^2+BC^2+CD^2+DA^2\right)$. Solution. Here is a different way to write MysticTerminator's solution, without explicitely using vectors but with a formula for the length of a median of a triangle: Lemma 1. If $m_a$ is the length of the median from the vertex A of a triangle ABC, then $4m_a^2=2b^2+2c^2-a^2$. Applying Lemma 1 to the triangle ABC, whose median from the vertex A is the segment AP (in fact, P is the midpoint of the side BC of this triangle), we get $4\cdot AP^2=2\cdot AC^2+2\cdot AB^2-BC^2$. Similarly, $4\cdot BQ^2=2\cdot BD^2+2\cdot BC^2-CD^2$; $4\cdot CR^2=2\cdot AC^2+2\cdot CD^2-DA^2$; $4\cdot DS^2=2\cdot BD^2+2\cdot DA^2-AB^2$. Adding these four equations yields $4\cdot AP^2+4\cdot BQ^2+4\cdot CR^2+4\cdot DS^2=\left(2\cdot AC^2+2\cdot AB^2-BC^2\right)$ $+\left(2\cdot BD^2+2\cdot BC^2-CD^2\right)+\left(2\cdot AC^2+2\cdot CD^2-DA^2\right)+\left(2\cdot BD^2+2\cdot DA^2-AB^2\right)$. This simplifies to $4\left(AP^2+BQ^2+CR^2+DS^2\right)=4\cdot\left(AC^2+BD^2\right)+\left(AB^2+BC^2+CD^2+DA^2\right)$. Now, let M and N be the midpoints of the segments AC and BD, respectively. Then, the segment MN is the median from the vertex N of triangle ANC (since M is the midpoint of the segment AC); thus, by Lemma 1, we have $4\cdot MN^2=2\cdot AN^2+2\cdot CN^2-AC^2$. On the other hand, the segment AN is the median from the vertex A of triangle ABD (since N is the midpoint of the segment BD); hence, Lemma 1 yields $4\cdot AN^2=2\cdot AB^2+2\cdot DA^2-BD^2$. Similarly, $4\cdot CN^2=2\cdot BC^2+2\cdot CD^2-BD^2$. Adding these two equations, we get $4\cdot AN^2+4\cdot CN^2=\left(2\cdot AB^2+2\cdot DA^2-BD^2\right)+\left(2\cdot BC^2+2\cdot CD^2-BD^2\right)$ $=2\cdot\left(AB^2+BC^2+CD^2+DA^2-BD^2\right)$. Division by 2 yields $2\cdot AN^2+2\cdot CN^2=AB^2+BC^2+CD^2+DA^2-BD^2$, and thus $4\cdot MN^2=2\cdot AN^2+2\cdot CN^2-AC^2=AB^2+BC^2+CD^2+DA^2-BD^2-AC^2$ $=\left(AB^2+BC^2+CD^2+DA^2\right)-\left(AC^2+BD^2\right)$. Since $MN^2\geq 0$, we thus have $\left(AB^2+BC^2+CD^2+DA^2\right)-\left(AC^2+BD^2\right)\geq 0$, so that $AC^2+BD^2\leq AB^2+BC^2+CD^2+DA^2$. Thus, $4\left(AP^2+BQ^2+CR^2+DS^2\right)=4\cdot\left(AC^2+BD^2\right)+\left(AB^2+BC^2+CD^2+DA^2\right)$ $\leq 4\cdot\left(AB^2+BC^2+CD^2+DA^2\right)+\left(AB^2+BC^2+CD^2+DA^2\right)=5\left(AB^2+BC^2+CD^2+DA^2\right)$, and the problem is solved. darij