If $G$ is a graph on $n$ vertices in which every pair of non-adjacent vertices $v$ and $u$ satisfy, $\deg(v) + \deg(u) \geq n − 1$, then $G$ contains a Hamiltonian Path.
Attempt:
Form a new graph $H$ by adding a new vertex to $G$ that is adjacent to every vertex of $G$. Then in $H$, every two non-adjacent vertices $u, v \in V(G)$ satisfy
$$\deg_H (u) + \deg_H (v) = \deg_G ( u + 1) + \deg_G ( v + 1) ≥ n − 1+ 2 = n + 1 = |V(H )|,$$
and so all the edges between vertices of $G$ belong to the closure of $H$. Thus, the closure of $H$ is complete and it follows that $H$ must have a Hamiltonian cycle, call it $C$. So it follows that $C - w$ is a Hamiltonian path in $G$.