15 points $P_i$ are placed within the unit sphere. Find the maximum possible value of $$\sum_{1\leq i \lt j \leq 15} \vert P_i P_j\vert^2$$ given that $x_i^2+y_i^2+z_i^2 \leq 1$
I saw this problem on a mathematics website and I was curious about this. Can you please provide me some books that may enlighten me about this topic? I think the problem asks to find the square of the lengths of line segments connecting $P_1$ to $P_2$, $P_2$ to $P_3$, $P_3$ to $P_4$ and so on (except $P_{15}$ to $P_1$?) I tried to solve this using a little knowledge about properties of vectors$$A.A = \vert A \vert^2$$ hence the distance between the vector connecting the two points is $$\vert P_j - P_i\vert$$ and $$\sum_{1\leq i \lt j \leq 15} \vert P_j - P_i\vert^2 =\sum_{1\leq i \lt j \leq 15} (P_j-P_i).(P_j-P_i) = \sum_{1\leq i \lt j \leq 15} P_j^2-2P_j.P_i+P_i^2$$ I don't know how to solve the last expression so I got stuck here.
The expression $\sum\limits_{1 \le i < j \le 15}|P_i-P_j|^2$ isn't a sum over the $14$ line segments $P_1 \to P_2 \to \cdots \to P_{15}$. It is a sum over the $105 = \frac{15 \times 14}{2}$ pair of points. Notice $$\begin{align} \sum_{1 \le i<j \le 15} |P_i-P_j|^2 &= \frac12 \sum_{1 \le i, j \le 15}|P_i - P_j|^2 = \frac12 \sum_{i=1}^{15}\sum_{j=1}^{15} (|P_i|^2 + |P_j|^2 - 2 P_i \cdot P_j )\\ &= 15 \sum_{j=1}^{15} |P_j|^2 - \left|\sum_{j=1}^{15}P_j\right|^2 \\ &\le 15 \sum_{j=1}^{15} |P_j|^2 \le 225 \end{align} $$ The maximum value of this sum is at most $225$. It is easy to find a configuration of $15$ points such that all $|P_j| = 1$ and $\sum\limits_{j=1}^{15} P_j = 0$ (e.g. a regular pentadecagon lying on the equator). For such a configuration, the value $225$ is achieved. This means the maximum value we seek is indeed $225$.
For the sum over the $14$ line segments, the maximum is $56 = 14\times 2^2$ and this value is achieved when we place the points alternatively over a pair of anti-podal points (e.g. $P_1 = P_3 = \cdots P_{15} = (0,0,1)$ and $P_2 = P_4 = \cdots = P_{14} = (0,0,-1)$).