Ore's Theorem

Theorem
Let $$G = \left({V, E}\right)$$ be a simple graph of order $n \ge 3$ which is connected.

Suppose $$G$$ fulfils the following condition:


 * For each pair of non-adjacent vertices $$u, v \in V$$, $$\deg u + \deg v \ge n \qquad (1)$$.

Then $$G$$ is Hamiltonian.

Proof
Suppose it were possible to construct a graph that fulfils condition $$(1)$$ which is not Hamiltonian.

For a given $$n \ge 3$$, let $$G$$ be the graph with the most possible edges such that $$G$$ is non-Hamiltonian which satisfies $$(1)$$.

Although it does not contain a Hamilton cycle, $$G$$ has to contain a Hamiltonian path $$\left({v_1, v_2, \ldots, v_n}\right)$$.

Otherwise it would be possible to add further edges to $$G$$ without making $$G$$ Hamiltonian.

Since $$G$$ is not Hamiltonian, $$v_1$$ is not adjacent to $$v_n$$, otherwise $$\left({v_1, v_2, \ldots, v_n, v_1}\right)$$ would be a Hamilton cycle.

By $$(1)$$, we have that $$\deg v_1 + \deg v_n \ge n$$.

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 $$v_n$$.

But the cycle $$\left({v_1, v_2, \ldots, v_{i-1}, v_n, v_{n-1}, \ldots, v_1}\right)$$ is then a Hamilton cycle.

So $$G$$ is Hamiltonian after all.

Note
Not all Hamiltonian graphs fulfil the condition $$\deg v_1 + \deg v_n \ge n$$.

For example, the cycle graphs $$C_n$$ where $$n \ge 4$$ are all such graphs.

It was demonstrated in 1960.