"Theorem (Ore; 1960) Let $G$ be a simple graph with $n$ vertices.
If $$\deg v + \deg w ≥ n$$ for every pair of non-adjacent vertices $v$, $w$, then $G$ is Hamiltonian."
From this theorem, one could obtain the following corollary:
Let $G$ be a simple graph and $v_1$ and $v_n$ non-adjacent vertices such that $\deg v_1 + \deg v_n \geq n$. Consider $G'$ a supergraph obtained by joining an edge $e$ between $x$ and $y$ to the graph $G$. Then $G$ is hamiltonian iff $G + e$ is hamiltonian.
To prove this corollary, could I say that if $G+e$ is hamiltonian and $G$ is not, then $\deg v_1 + \deg v_n < n$ in $G$, but that'd contradict the initial assumption that $\deg v_1 + \deg v_n \geq n$ ? But maybe it could happen that $\deg x' + \deg y' < n$ not necessarily for the vertices $x,y$ that appear in our assumption?
EDIT: Suppose that $G+e$ is hamiltonian and $G$ is not. Let $C$ be an hamiltonian cycle in $G+e$. Now if $C$ doesn't use the edge $e$, we're done. Suppose that it doesn't happen. Now we have $P=C-e=(v_1,\dots,v_n)$ is a path in $G$.For any $v_i$ neighbour of $v_n$, we can´t have that $v_{i+1}$ is a neighbour of $v_1$ as we could construct $C=(v_{i+1},\dots,v_n,v_i,v_{i-1}, \dots, v_1, v_{i+1})$ is an hamiltonian cycle in $G$; but if so, $\deg v_1 \leq (n-1)-\deg v_2$ contraditing our initial statement. Could this be considered as an valid argument?