I remember reading an answer to a question similar to this one in MSE or MO, but I can't find it. I am looking for a proof or a reference.
Let $A\subseteq\mathbb{R}^n$ be a set of vectors such that no nonzero $v\in\mathbb{R}^n$ satisfies that for all $a\in A$, $\langle v,a\rangle\geq0$. Is every vector of $\mathbb{R}^n$ a positive linear combination of vectors of $A$ (that is, a sum $\sum_{i=1}^nk_ia_i$, with $k_i\geq0, a_i\in A$)? Also, can we always find a subset of $n+1$ elements of $A$ which satisfies the same condition?
If someone finds better tags for this question please add them.
Edit: As Andrew D. Hwang says in the comments, the second question about a subset of $n+1$ elements of $A$ is not necessarily true: we can just consider $A$ to be the union of the usual basis of vectors of $\mathbb{R}^n$ and their opposites.
After a few days I remembered this question and found an easy positive answer: it is enough to prove that, if $A$ is as in the question, then the convex closure $B$ of $A$ contains an open neighborhood of $0$.
Suppose that is not the case. The set $B$ is convex and $0$ is not in its interior, so there are two cases:
$0$ is in the boundary of $B$. Then there is a supporting hyperplane of $B$ at $0$ given by $\{x\in\mathbb{R}^n;\langle x,v\rangle=0\}$ for some $v$, which contradicts the statement of the question.
$0$ is not in the boundary of $B$. We still have supporting hyperplanes in this case, so a similar reasoning applies.