This might be a stupid question, but I have been thinking and googling for some time now and I still cannot seem to find a response.
I am interested in how many extreme rays a finitely generated (pointed, full-dimensional, or any other expected constraint in order to get a positive answer) cone in $\mathbb{R}^n$ has to have in order for me to be sure that I can find two vectors on extreme rays such that their sum lies in the interior of the cone? For example, this seems to be 4 in the 3-dimensional space, but this might work only because in 2D this constant is 2.
It seems to me that it is at most n+1, but I might be biased by low-dimensional intuition. I tried to come up with a counterexample by expressing the cone as a system of inequalities, with no success for now.
Starting in dimension $4$, this is no longer possible. Let $m \in \mathbb N$ be arbitrary. We consider the cone in $\mathbb R^4$ with the extremal rays $$ \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}, \qquad \begin{pmatrix} \cos(i \pi / n) \\ \sin(i \pi /n) \\ 0 \\ 1 \end{pmatrix}, $$ with $i = 1,\ldots, m$. Then, no convex combination of two extremal rays lies in the interior of the cone.
What is behind: When you think of properties of convex cones in $\mathbb R^n$, then you can in most cases think of convex sets in $\mathbb R^{n-1}$. Indeed, often a set $C \subset \mathbb R^{n-1}$ can be identified with the conical hull of $C \times \{1\} \subset \mathbb R^n$, and often, both sets have similar properties. My above example corresponds to a cone in $\mathbb R^3$ having a regular $m$-gon as base.