Let $F = \sup d(x,y)$ where $x ,y \in A$.
I came across a problem where it was given that if $ \|a\| < F$ , $ \|b\| < F$. Then $ -F^2/4 < a \bullet b < F^2 $ for any $a,b \in A$. I know the upper bound comes from the Cauchy Schwartz inequality. Can someone tell me the proof of the lower bound. Thanks
This fails in one dimension. Let $a = 2/3$ and $b = -2/3$ and $F = 1$.