Find the maximum $r$ such that $(\beta_i,\,\beta_j)<0$, $\;1\leq i
Let $\mathcal{L}$ be an $n$-dimension real Euclid space. Find the maximum $r$ such that there are $r$ vectors $\beta_1,\cdots,\,\beta_r$ in $\mathcal{L}$ satisfying
$$
(\beta_i,\,\beta_j)<0,\quad 1\leq i < j  \leq r.
$$
I have tried to consider the case that $\beta_1,\cdots,\;\beta_r$ are dependent and the case that $r=2,\,3$, however, for technical reason, I couldn't get any hint from trying these.
 A: Since the inner products are only considered pairwise let us restrict to unit vectors.
Let $u_1,\ldots,u_m$ be unit vectors such that
$$
(u_i,u_j)\leq \delta<0,\quad 1\leq i<j\leq r.
$$
Then
$$
0\leq \left|\sum_{i=1}^m u_i\right|^2=(\sum_{i=1}^m u_i,\sum_{j=1}^m u_i)=\sum_{j=1}^m |u_i|^2+2\sum_{i<j} (u_i,u_j)
$$
$$
\leq m + m(m-1) \delta
$$
giving $$m\leq 1-\frac{1}{\delta}.$$
Edit: The result I quoted is from Bollobas' little white book Combinatorics, Set Systems and Hypergraphs.
There is a counterpart result which shows if a set $S$ of $n+r$ nonzero (not necessarily unit) vectors in $n$ dimensions satisfy
$$
(x_i,x_j)\leq 0,\quad 1\leq i<j\leq n+r
$$
then $r\geq n,$ and if $r=n,$ then the space $\mathbb{R}^n$ has an orthonormal basis such that for all $1\leq i\leq n,$
$$
S=\{\lambda_1 a_1,\mu_1 a_1, \lambda_2 a_2, \mu_2 a_2,\ldots,\lambda_n a_n, \mu_n a_n\}
$$
where $\mu_i<0<\lambda_i.$ If the inner products are strictly negative, and we let $\delta\rightarrow 0,$ I think it can be argued by continuity that we end up in this case. Thus, really the smallest $\delta$ must be $-1/2n$ or something like that.
Another way of thinking of it, is if the inner product $(u_i,u_j)$ is arbitrarily near zero and negative, that will force some $(u_i,u_j')$ to tend to be positive. So a fourth moment sum might make this idea clearer, but I have no time now.
