Problem: If $G$ is a graph with $m$ edges and $n\geq 3$ vertices, where $2m \geq n^{2} - 3n + 6$, prove that $G$ is a hamiltonian graph.
(Hint: Consider the following theorem and corollary.)
Theorem: Let $G$ be a graph. Then $G$ is a hamiltonian graph if and only if its closure, denoted $\overline{G}$, is a hamiltonian graph.
Corollary: Let $G$ be a graph on $n \geq 3$ vertices. If the closure of $G$ is complete, then $G$ is hamiltonian.
Thoughts: Honestly, I'm stuck. The hint isn't very useful to me, and I'm not able to immediately pull out any ideas. If you could guide me on how to get started, that would be helpful and appreciated.
In $\overline G$, the inequality $n^2-3n+6\le 2m$ still holds (with possibly increased $m$). This means $$m\ge {n\choose 2}-(n-3). $$ Let $u,v$ be two vertices in $\overline G$. Then the graph lacks at least $(n-1-\rho(u))+(n-1-\rho(v))-1=2n-3-(\rho(u)+\rho(v))$ edges (the edges missing per vertex minus possibly double-counted $uv$). This must be $\le n-3$. It follows that $\rho(u)+\rho(v)\ge n$, i.e., $uv$ is an edge in $\overline G$.