I strongly believe that the following assertion is true:
Suppose that $ v_1, \ldots, v_6 $ are $6$ points on $ S^1 \subset \mathbb{R}^2 $ satisfying $$ \sum_{i=1}^{6} \langle x,v_i \rangle ^2 = 3 $$ for any $ x \in S^1 $. Then for any $ x \in S^1 $, there exists $ v_i $ satisfying $$\langle x,v_i \rangle \geq \frac{\sqrt{3}}{2} .$$
Obviously $ v_1, \ldots, v_6 $ seem to be the vertices of a regular hexagon. But I am not sure whether they must be. To prove the assertion, I think I have to either show that they must be the vertices of a regular hexagon, or, prove directly without assuming that they are. Can you help me with the proof? Thanks.
This is not true (sorry for answering my own question).
For example $ v_1, v_2, v_3 $ are the vertices of an equilateral triangle and set $ v_4 = v_1, v_5 = v_2, v_6 = v_3 $. Then for some $ x \in S^1 $, we have $ | \langle x, v_i \rangle | < \frac{\sqrt{3}}{2} $ for every i.