Separation from a closed convex cone

For a closed convex cone $C$ in a finite-dimensional real inner product space and $z\notin C$, there exists $y$ with $⟨y,z⟩<0$ and $⟨y,c⟩\ge 0$ for all $c\in C$. Here is a direct proof. A nearest point $c_0\in C$ exists by compactness after restricting to a sufficiently large ball. Differentiating squared distance along the segment towards $c\in C$ gives $⟨c_0-z,c-c_0⟩\ge 0$. Using $c=0$ and $c=2c_0$ shows $⟨c_0-z,c_0⟩=0$. Set $y=c_0-z\ne 0$: then $⟨y,c⟩\ge 0$ and $⟨y,z⟩=-\Vert y{\Vert }^2<0$. This is a conic form of the Hahn-Banach separation theorem.