Let $A_i$ have coordinates $(\sin {x_i},\cos{x_i})$ then the distance $A_iA_j$ is equal to $$\sqrt{(\cos x_j-\cos x_i)^2+(\sin x_j -\sin x_i)^2}=\sqrt{2-2\cos(x_j-x_i)}$$wich is always less or equal than $2$ since $-1\leq \cos x \leq 1$ Therefore $\sum_{1\leq i<j\leq k} d (A_i,A_j) \leq 2\dbinom{k}{2} =k(k-1) <k^2$ $\blacksquare$. Edit: @below right, I was expecting something sharper.

XbenX wrote: Let $A_i$ have coordinates $(\sin {x_i},\cos{x_i})$ then the distance $A_iA_j$ is equal to $$\sqrt{(\cos x_j-\cos x_i)^2+(\sin x_j -\sin x_i)^2}=\sqrt{2-2\cos(x_j-x_i)}$$wich is always less or equal than $2$ since $-1\leq \cos x \leq 1$ Therefore $\sum_{1\leq i<j\leq k} d (A_i,A_j) \leq 2\dbinom{k}{2} =k(k-1) <k^2$ $\blacksquare$. What was the need for coordinate bash shouldn't the largest distance be 2 as no distance can be larger than diameter?

There was a typo in the problem that is fixed now.

The following identity is true for any vectors $v_1,\dots,v_n\in\mathbb R^2$: $\sum_{1\leq i<j\leq n}||v_i-v_j||^2=n\sum_{i=1}^n ||v_i||^2-||\sum_{i=1}^n v_i||^2$ Hence: $\sum_{1\leq i<j\leq n}||A_i-A_j||^2=n\sum_{i=1}^n ||A_i||^2-||\sum_{i=1}^n A_i||^2=n^2-||\sum_{i=1}^n A_i||^2\leq n^2$.

Let the coordinates of $A_i$ be $(a_i,b_i)$, and observe that $$\sum_{1\le i<j\le k} ((a_i-a_j)^2 + (b_i-b_j)^2) = \left(k\sum_{1\le i \le k} (a_i^2+b_i^2)\right) - (a_1+a_2+\dots+a_k)^2- (b_1+b_2+\dots+b_k)^2 \le k^2$$Equality occurs when the vectors add up to zero.

Complex numbers also work Let $a_1,a_2 \dots a_n$ be the respective complex numbers of the points $A_1,A_2 \dots A_n$, such that $|a_1| = |a_2| = \dots = |a_n| = 1$ We have that: $$\sum_{1 \leq i < j \leq k}{d(A_i,A_j)^2} = \sum_{1 \leq i < j \leq k}{|a_i-a_j|^2} = \sum_{1 \leq i < j \leq k}{(a_i-a_j)(\overline{a_i}-\overline{a_j}}) = \left(k \sum{a_i \overline{a_i}}\right) - (a_1+a_2 \dots +a_n)\overline{(a_1+a_2 \dots +a_n)} = k^2-|a_1+a_2 \dots +a_n|^2 \leq k^2$$so we are done.