I was reading the proof of the Ore theorem from ProofWiki which says that if $G$ is a simple graph with $n\geq 3$ vertices and $d(v)+d(w)\geq n$ $\forall v,w \in V$ such that $v\neq w$ and $v, w$ are not neighbors, then $G$ is Hamiltonian Graph.
I understood the proof except this part:
By the Pigeonhole Principle, for some $i$ such that $2\le i\le n−1$, $v_i$ is adjacent to $v_1$, and $v_{i−1}$ is adjacent to $v_n$.
I think i didn't understand the explanation of the Pidgeonhole principle at ProofWiki. What is an explanation of why those vertices exist using the principle (or not) applied to this case? Also, wouldn't suppose to be required $n\geq4$ vertices instead of $3$?
Thanks in advance.
Consider the pairs of vertices $(v_{i-1}, v_i)$ for $2 \leq i \leq n-1$. There are $n-2$ of them. There are $d(v_1)$ edges between $v_1$ and some of the vertices $v_k$ for $2 \leq k \leq n-1$. In this way, $v_1$ touches $d(v_1)$ of the pairs $(v_{i-1}, v_i)$ ('touches' means $v_1$ is adjacent to $v_i$). Similarly $v_n$ touches $d(v_n)$ of the pairs $(v_{i-1}, v_i)$ (here 'touches' means $v_n$ is adjacent to $v_{i-1}$). Thus $d(v_1) + d(v_n) \geq n$ pairs are touched by $v_1$ or $v_n$. This is more than the total number of pairs, so by the pigeonhole principle, there is some pair $(v_{i-1}, v_i)$ that $v_1$ and $v_n$ both 'touch', as desired.