China Hong Kong Mathematical Olympiad

Let $A, B, C, D, E, F$ be $6$ distinct points on the circle such that
$$\overline{AB} = \overline{BC} = \overline{CD} = \overline{DE} = \overline{EF} = \overline{FA}.$$
Let $O$ be the centre of the circle. Note that any two points in the same sector among $AOB, BOC, COD, DOE, EOF, FOA$ have distance at most $1$. Let $n_1, n_2, \dots, n_6$ be the number of points in the $6$ sectors respectively. It suffices to show
$$\binom{n_1}{2} + \binom{n_2}{2} + \dots + \binom{n_6}{2} \ge 2001.$$
Indeed, since the binomial function $\binom{x}{2}$ is convex, by Jensen's inequality, we have
$$\sum_{k=1}^{6} \binom{n_k}{2} \ge 6 \left( \frac{n_1+n_2+\dots+n_6}{6} \right)^2 \frac{1}{2} = 6 \binom{35}{2} = 3570 > 2001.$$