Dirac's Theorem

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

Proof
Let $$P=p_1p_2...p_k$$ be the longest path in $$G$$. If $$p_1$$ is adjacent to some vertex $$v$$ not in $$P$$, then the path $$vp_1p_2...p_k$$ 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 $$j$$ ($$1\leq j\leq k$$) 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 $$C=p_{j+1}p_{j+2}...p_{k}p_jp_{j-1}...p_2p_1p_{j+1}$$. 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.