I am trying to prove the following:
"Let $\mathscr{G}(V, E)$ be a connected, unweighted, undirected graph with no self-loops or multiple edges and $n$ nodes ($|V| = n$). Suppose that there is a unique node, $\bar{v}$, maximizing the closeness centrality, i.e.,
$$ C(\bar{v}) = \frac{n-1}{\sum_{u\in V}d(\bar{v}, u)} \gt C(v) = \frac{n-1}{\sum_{u\in V}d(v, u)} \forall v \neq \bar{v}$$
Where $d(v, u)$ is the geodesic distance between $v$ and $u$. Consider $\bar{v}$ and an arbitrary node $v \neq \bar{v}$. Let $p$ be the number of nodes that are closer to $\bar{v}$ than to $v$ ($d(u, \bar{v}) \lt d(u, v)$) and $q$ the number of nodes that are closer to $v$ than to $\bar{v}$ ($d(u,v) \lt d(u, \bar{v})$).
Prove that $p \gt q$."
I started with the fact that $\sum_{u\in V}d(v, u)-d(\bar{v}, u) \gt 0$. Then, I separated the sum as follows: one sum is over all nodes such that $d(v, u)-d(\bar{v}, u)$ is positive (set $P$); and the other is over all nodes such that $d(v, u)-d(\bar{v}, u)$ is negative (set $Q$). I am trying to prove that the amount of terms in the first sum is greater than the amount of terms in the second. I achieve this but only when:
$$d(u, v) = d(u, \bar{v}) + d(v, \bar{v}) \forall u \in P$$
$$d(u, \bar{v}) = d(u, v) + d(v, \bar{v}) \forall u \in Q$$
I want a general proof.
Any suggestion, counterexample, or the proof is appreciated.