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  "body": "\n## Proof\n\nWe argue by contradiction, with an edge-maximal counterexample.\n\n1. Let $G$ be a finite simple [[graph|0JD5C82QMB]] with vertex set $V$ and $n = \\abs{V} \\ge 3$. Suppose that $\\deg(u) + \\deg(v) \\ge n$ for all distinct nonadjacent $u, v$.\n2. Suppose, for a contradiction, that $G$ is not [[Hamiltonian|090WPKFABX]].\n3. Consider the graphs $H$ on the vertex set $V$ that contain every edge of $G$ and are not Hamiltonian. Choose one with the largest number of edges. Such an $H$ exists, because $G$ is one of these graphs and there are finitely many graphs on $V$.\n4. Each vertex has at least the same [[degree|0T0F6CCS9W]] in $H$ as in $G$, because $H$ contains all edges of $G$.\n5. The graph $H$ is not the [[complete graph|03WKB2H474]] on $V$. Indeed, for $n \\ge 3$ the complete graph has the [[Hamiltonian cycle|0HKXNG2RS1]] that visits the vertices in any order.\n6. So $H$ has two distinct nonadjacent vertices $u$ and $v$. They are also nonadjacent in $G$, so by Steps 1 and 4, $\\deg_H(u) + \\deg_H(v) \\ge n$.\n7. By the choice of $H$ in Step 3, the graph $H + uv$ is Hamiltonian. Its Hamiltonian cycle uses the edge $uv$, since otherwise it would be a Hamiltonian cycle of $H$.\n8. Remove the edge $uv$ from this [[cycle|07CT9RDY7Q]]. What remains is a [[path|0JDE56HWQ8]] $w_1, w_2, \\dots, w_n$ in $H$ that contains every vertex, with $w_1 = u$ and $w_n = v$.\n9. Let $A = \\set{i \\in \\set{1, \\dots, n-1} : u w_{i+1} \\text{ is an edge of } H}$. The neighbours of $u$ lie among $w_2, \\dots, w_n$, so $\\abs{A} = \\deg_H(u)$.\n10. Let $B = \\set{i \\in \\set{1, \\dots, n-1} : w_i v \\text{ is an edge of } H}$. The neighbours of $v$ lie among $w_1, \\dots, w_{n-1}$, so $\\abs{B} = \\deg_H(v)$.\n11. The sets $A$ and $B$ lie in $\\set{1, \\dots, n-1}$, which has $n - 1$ elements. So $\\abs{A \\cup B} \\le n - 1$. By Step 6, $\\abs{A \\cap B} = \\abs{A} + \\abs{B} - \\abs{A \\cup B} \\ge n - (n-1) = 1$. Let $i \\in A \\cap B$.\n12. Since $u$ and $v$ are not adjacent, $1 \\notin B$ and $n - 1 \\notin A$. Hence $2 \\le i \\le n-2$.\n13. Consider the closed walk $w_1, w_2, \\dots, w_i, w_n, w_{n-1}, \\dots, w_{i+1}, w_1$. Its edges are path edges, the edge $w_i w_n$ (as $i \\in B$) and the edge $w_{i+1} w_1$ (as $i \\in A$).\n14. This closed walk visits each of the $n \\ge 3$ vertices exactly once before it returns to $w_1$. So it is a Hamiltonian cycle of $H$.\n15. This contradicts the choice of $H$ as a non-Hamiltonian graph. Hence $G$ is Hamiltonian.\n\n## Idea\n\nAdd edges until one more edge would create a Hamiltonian cycle. Then two nonadjacent vertices $u, v$ are the ends of a Hamiltonian path. Their large degree sum forces a \"crossing\" pair of edges $u w_{i+1}$ and $w_i v$, and these two edges turn the path into a cycle.\n",
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