The question is:
Proof that if $c(n-1, 2) + 1 < |E|$ then $G = (V, E)$ is hamiltonian (Use Ore theorem)
My solution is:
Since $c(n-1, 2) + 1 < |E|$ we find $n^2-3n+4 < \sum \limits_{v\in G}deg(v)$. If $G$ is not hamiltonian then there are $u,v$ belonging to $V$ with $u + v < n$ Then $n^2-4n+5 < deg(v)$...
Who do I progress from here?
Choose arbitrary non-adjacent vertices $u, v$, and consider the graph $H = (V', E')$ formed by deleting the vertices $u, v$ from $G$. Observe that $|V'| = n - 2$. Also, the assumed inequality requires
$$|E'| \geq {n - 1\choose 2} + 2 - \text{deg}(u) - \text{deg}(v).$$
But since a $K_{n - 2}$ graph has ${n - 2 \choose 2}$ vertices and surely has a Hamiltonian cycle, we also require
$$|E'| \leq {n-2 \choose 2}.$$
With some algebra, these two inequalities imply $\deg(u) + \deg(v) \geq n$, which is the criterion necessary to apply Ore's Theorem.