Maximum number of edges in a non-Hamiltonian graph I need to show that the maximum number of edges of non-Hamiltonian, simple graph, on $n$ vertices noted by $t(n,H_n)$ is $\binom{n-1}{2} + 1$. It's essential to show the upper and lower bounds for this question.

Now, I can see why I need to show the upper bound - for $t(n,H_n) &gt; \binom{n-1}{2} + 1$ there's a clique on $n-1$ vertices and one other vertex with at least 2 edges. Surely, a Hamiltonian cycle exist in such graph.

What about the lower bound, though? Am I need to prove that for every $t(n,H_n) \leq \binom{n-1}{2} + 1$ there is no Hamiltonian cycle? By negation, I seem to not advance anywhere, because I don't know of any necessary conditions for a Hamiltonian cycle to exist, beside the cut-vertex one. Can someone give me a clue on that?

Thanks in advance.
Found a lot easier proof for the harder part, using Ore's theorem: If $G$ is complete, there's a Hamiltonian cycle. So suppose it's not. We'll take some non-neighbor vertices $v,w$, and delete them from the graph. For the resulting graph $G&#39;$ - $|E(G&#39;)| \geq \binom {n-1}{2} + 2 - (d(v)+d(w))$. In the complete graph on $n-2$ vertices there are $\binom {n-2}{2}$ edges. So, lastly, we get that $$\binom{n-2}{2} \geq \binom{n-1}{2} + 2 - (d(v)+d(w))$$ and after some easy operations, we get that $$d(v)+d(w) \geq n$$ which is exactly Ore's condition for a Hamiltonian graph to exist.
First see that you can have a complete graph on n-1 vertices where the number of edges is n-1 C 2 and then you just need to consider how many edges you can add to a new incoming vertex such that the resulting graph is Non- Hamiltonian. Hamiltonicity of the complete graph implies that only one edge can be added .