The problem is this: If the smallest positive number $\lambda$, such that for any $17$ points on the plane $P_{1}, P_{2}, \ldots, P_{17}($ can overlap too), if the distance between any two of them does not exceed 1 , then $$\sum_{1 \leq i<j \leq 17}\left|P_{i} P_{j}\right|^{2} \leq \lambda \space \text{find this} \space \lambda$$
i have literally no clue how to optimize lambda the best possible, i can just think of large numbers which can be quite significant but not enough close to lamda minimum we want in the problem.
A hint:
Let us consider, in a first step, the extreme situation where the $A_k$s are the vertices of a regular polygon with $n=17$ vertices inscribed in a circle with radius $r=1/2$ (see figure).
This polygon, called a heptadecagon, is famous because it has attracted the attention of Gauss for reasons explained here.
In the case this configuration is optimal, we would have, for the sum of the squares of all mutual distances:
$$\lambda_{opt}=(nr)^2= (17/2)^2=\frac{289}{4}\tag{1}$$
This formula comes from a known formula $n^2$ for the sum of the squares of the lengths of all diagonals (including the sides) in any regular polygon with $n$ vertices inscribed in the unit circle : see formula 1 page 2 in this document.
In a second step, we have to slightly rectify our first guess. Indeed, the longest diagonal $A_1A_{9}$ of the heptadecagon with radius $1/2$ being equal to
$$a = \sin(8\pi/17)\approx 0.995734176295034$$
one shouldn't take $r=1/2$ in (1) but $r=1/(2a)$ for reaching the optimal (?) situation, once again without proof that the optimal $\lambda$ is obtained in this way.