Intersection of a closed convex cone with the non-negative orthant

Question

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 $n=2$.

Answer 1

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 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.