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Ore's theorem, named after Øystein Ore, says that a finite simple graph with $n\ge 3$ vertices is Hamiltonian if every nonadjacent pair has degree sum at least $n$.

For a proof, enlarge a hypothetical counterexample to an edge-maximal non-Hamiltonian graph. Adding a missing edge $uv$ creates a Hamiltonian cycle, hence the original graph has a Hamiltonian path $u=v_1,\dots ,v_n=v$. If both $u{v}_{i+1}$ and $v{v}_i$ are edges, the path closes to a cycle by reversing its final segment. Thus the index sets of these two kinds of edges are disjoint in $\{1,\dots ,n-1\}$, giving $d(u)+d(v)\le n-1$, a contradiction. The degree condition is sufficient and need not follow from a lower bound of $n/2$ on every individual degree.

Suppose a graph satisfying Ore's theorem were not Hamiltonian. Extend it, on the same vertices, to an edge-maximal non-Hamiltonian graph $H$. Adding edges preserves the degree-sum hypothesis. The complete graph is Hamiltonian, so $H$ has a missing edge $uv$. By maximality, $H+uv$ has a Hamiltonian cycle, which must use the new edge. Removing it gives a Hamiltonian path in $H$, labelled $u=v_1,v_2,\dots ,v_n=v$.

If some $i$ has both edges $u{v}_{i+1}$ and $v{v}_i$, the path can be closed into the cyclecontradicting the choice of $H$. Thus the index sets $\{i:u{v}_{i+1}\in E(H)\}$ and $\{i:v{v}_i\in E(H)\}$ are disjoint subsets of $\{1,\dots ,n-1\}$. Their sizes are ${\mathrm{deg}}_{H}u$ and ${\mathrm{deg}}_{H}v$. Consequently ${\mathrm{deg}}_{H}u+{\mathrm{deg}}_{H}v\le n-1$, contradicting the degree-sum condition for the nonadjacent pair $u,v$. Hence the original graph is Hamiltonian.