Question: Prove that if the number of vertices in graph $G$ is $n \geq 2$ and the sum of $2$ degrees is at least $n - 2$, then graph $G$ has $2$ disjoint simple paths that their union completes the graph $G$ (Meaning their union is the graph) - a simple path can be of length $0$, with only one vertex.
My solution:
The strategy is to start from a random vertex in $G$, and then we will build a simple path $P$ such that $G \setminus P$ satisfies Ore's Theorem. and thus $G \setminus P$ can be 'covered' using only one simple path $P'$. So $P'$ and $P$ are disjoint and $P' \cup P = G$
Prove:
We will add to $G$ $2$ vertices which are neighbors, each one of them is a neighbor of each other vertex in $G$.
Because the bigger graph $G'$ has $n + 2 \geq 4$ vertices, and the sum of each $2$ different vertices is at least $n +2$, according to Ore's Theorem it has a Hamiltonian cycle. When we remove those vertices we added, we will get $2$ disjoint simple paths (Or even one) that cover $G$
I would appreciate your revision!
Thank you!
The proof you provide at the end of your question makes sense to me, but it's not clear to me how it matches the strategy you gave prior to it.
Depending on the level of the audience (for example if it is an undergraduate class in graph theory), it might be better to give some more detail about checking the degree condition, and how exactly you are finding the two disjoint simple paths upon removal of the addition of the two dummy vertices.