Consider $P$ a convex polytope in $\mathbb R^d$, and $x\in P$ a vertex of $P$. Let $N_P(x)$ be the set of "neighboring" points of $x$. A vertex $y\in P, y\neq x$ is a neighbor of $x$ if $(x,y)$ is an edge (1-dimensional face) of $P$. Mathematically, $(x,y)$ is an edge if and only if there exists $c\in\mathbb R^d$, $c\neq 0$, such that $c^\top (y-x)=0$ and $c^\top (z-x)<0$ for all vertices $z$ different from $x,y$.
Question: Now suppose for some non-zero $c\in\mathbb R^d$, $c^\top (y-x) < 0$ for all $y\in N_P(x)$. Prove that $c^\top(z-x)<0$ as well for all other vertices of $P$.
My attempt (partial solution): when the vertex $x$ is not over-specified (i.e., there are exactly $d$ facets intersecting at $x$), it is not difficult to prove that each facet corresponds to an edge, and therefore $c$ must be in the normal cone of $x$. However, I do not know how this argument could be extended to over-specified $x$.