TITLE: Proving that a graph of a certain size is Hamiltonian
QUESTION [6 upvotes]: For any graph with order $n \geq 3$, given that its size is
$$m \geq \frac{\left(n-1\right)(n-2)}{2} + 2,$$
show that the graph is Hamiltonian.
I know that if I can show that the degree sum of any two non-adjacent vertices is $\geq n$, then I'd be done.
Likewise, if I could show that the above somehow implied that the degree of every vertex in the graph is $\geq n/2$, I'd also be done. However, I cannot see how to get to either one of those given the information I have. I have been trying the following: assuming that there exists two non-adjacent vertices $u$ and $v$ whose degree sum is $\leq (n-1)$, then
$$2m = \sum d\text{(other vertices))} + d(u) + d(v) \leq \sum d(\text{other vertices)} + (n-1).$$
If I could show that this implied that
$$2m < \left(n-1\right)(n-2) + 4,$$
I would have a contradiction, thereby proving that the graph is Hamiltonian. However, I have not been able to show this, so I am thinking it is the wrong approach.

REPLY [4 votes]: Proof by induction on $n$. 
Let $G$ be a graph of order $n$ and size $m$ (given). Denote the maximum degree of a vertex of G by $\Delta$ and the average degree by $\delta$. Then
$$\delta = \frac{2m}{n} \geq \frac{(n-1)(n-2)+4}{n} \geq n-3+\frac{6}{n}$$
$\Delta \geq \delta$ and $\Delta\in \mathbb{Z}$, so $\Delta \geq n-2$. Let $v$ be a maximum degree vertex of G. 
$d(v)$ is $n-1$ or $n-2$.
Case 1: $d(v)=n-2$ 
$$e(G-v)= m-d(v)\geq \frac{(n-2)(n-3)}{2}+2$$
So, by induction hypothesis, $G-v$ contains a Hamiltonian cycle. Two adjacent vertices in this cycle are neighbors of $v$, so add v to the cycle and we are done.
Case 2: $d(v)=n-1$
Now the above doesn't quite hold, since $G-v$ contains 1 fewer edge than required. No problem! Let $H$ be $G-v$ with an arbitrary edge added (call this $jk$. By induction hypothesis, $H$ is Hamiltonian. If this cycle doesn't contain the added edge we are done, as in Case 1. Otherwise, deleting $jk$ gives a Hamiltonian path in $G-v$ from $j$ to $k$. $j\sim v$ and $k\sim v$, since $d(v)=n-1$, so we have a Hamiltonian cycle, as required.
Base case: easy.
Edit: A general tip 
I find it useful in problems like this to see how the numbers arise. Consider briefly why any fewer edges are insufficient.