Existence of non obtuse angle of n+2 vectors in n-dimensional euclidean space. There are n+2 distinct vectors $v_1,v_2,v_3,\cdots ,v_{n+2}$ in n-dimensional euclidean space.
Prove that there must be a integer pair of $(i,j)$ which satisfies $1\leq i<j\leq n+2$, and $dot(v_i,v_j)\geq 0$.
i.e. There are at most n+1 vectors in n-d euclidean space, and any pair of them forms an obtuse angle.
 A: W.l.o.g assume that the vectors are normal and suppose for contradiction that all these $n+2$ vectors have pairwise dot product less than zero.
In the following I further assume that the dimension of the span of these vectors is $n$ but I think for a smaller rank we can reduce it to the full-rank case.
Let $w_1=v_1$ and $w_2=v_2-\langle w_1,v_2 \rangle w_1$. Observe that the set $w_1,w_2$ is orthogonal and the inner product of one of $w_1$ or $w_2$ with one of $v_3,v_4,\ldots,v_{n+2}$ is less than zero.
Now let $w_3=v_3-\langle w_1,v_3 \rangle w_1 - \langle w_2,v_3 \rangle w_2$. Again the set $w_1,w_2,w_3$ is pairwise orthogonal and any of their inner product with any of $v_4,\ldots,v_{n+2}$ is less than zero.
Repeating this process $n$ times, we get $n$ orthonormal vectors $w_1,w_2,\ldots,w_n$ and such that $v_{n+1}$ and $v_{n+2}$ have inner product less than zero with each of them. Thus $v_{n+1}$ and $v_{n+2}$ must both lie in the same quadrant and hence the angle between them is acute.
