Dirac's Theorem

Theorem
If a connected graph $$G$$ has $$n \ge 3$$ vertices and the degree of each vertex is at least $$\frac{n}{2}$$, then $$G$$ is Hamiltonian.

Proof
Let $$P = p_1 p_2 \ldots p_k$$ be the longest path in $$G$$.

If $$p_1$$ is adjacent to some vertex $$v$$ not in $$P$$, then the path $$v p_1 p_2 \ldots p_k$$ would be longer than $$P$$, contradicting the choice of $$P$$.

The same argument can be made for $$p_k$$.

So both $$p_1$$ and $$p_k$$ are adjacent only to vertices in $$P$$.

Since $$deg(p_1)\geq\frac{n}{2}$$ and $$p_1$$ cannot be adjacent to itself, $$k\geq\frac{n}{2}+1$$.

Claim: There is some value of $$j$$ ($$1 \le j \le k$$) such that:
 * $$p_j$$ is adjacent to $$p_k$$, and
 * $$p_{j+1}$$ is adjacent to $$p_1$$.

Suppose that the claim is not true.

Then since all vertices adjacent to $$p_1$$ or $$p_k$$ lie on $$P$$, there must be at least $$deg(p_1)$$ vertices on $$P$$ not adjacent to $$p_k$$.

Since all the vertices adjacent to $$p_k$$ and $$p_k$$ itself also lie on $$P$$, the path must have at least $$deg(p_1)+deg(p_k)+1\geq n+1$$ vertices.

But $$G$$ has only $$n$$ vertices: a contradiction.

This gives a cycle $$C = p_{j+1} p_{j+2} \ldots p_{k} p_j p_{j-1} \ldots p_2 p_1 p_{j+1}$$.

Suppose $$G - C$$ is nonempty.

Then since $$G$$ is connected, there must be a vertex $$v\in G-C$$ adjacent to some $$p_i$$.

So the path from $$v$$ to $$p_i$$ and then around $$C$$ to the vertex adjacent to $$p_i$$ is longer than $$P$$, contradicting the definition of $$P$$.

Therefore all vertices in $$G$$ are contained in $$C$$, making $$C$$ a Hamilton cycle.