Suppose I have a closed convex cone $C\subseteq \mathbb R^n$ and suppose that for every $x$ in the non-negative orthant $\mathbb R\_{0+}^n$ there is a $y\in C$ such that $x\cdot y>0$ (with the standard scalar product). Does it follow that intersection of $C$ with the non-negative orthant contains more than just the origin?


This is true for $n=2$.



Intersection of a closed convex cone with the non-negative orthant
Assume $C$ and $\mathbb{R}\_{\ge 0}^n$ can be (non-strictly) separated by a subspace of dimension $n-1$.   

Then a normal vector $x$ to that subspace lies in $\mathbb{R}\_{\ge 0}^n$; see e.g. [here](https://math.stackexchange.com/questions/106312/intersection-between-orthogonal-complement-of-a-subspace-and-a-set) for a proof.


But then by your assumption, there exists $y \in C$ on the same side of this subspace as $\mathbb{R}\_{\ge 0}^n$, contradiction.


