If I were doing it inductively, I’d go in the other direction, removing edges instead of adding them. Take a graph G with n>0 edges, and remove an edge to get G’. The degree sum of G’ is two less than the degree sum of G (two vertices lose one degree, or one vertex loses two degree). Then induction shows that the degree sum is twice the edge count. There are probably simpler proofs, but having been primed by yours, this is the one that comes to mind.
I feel like being completely formal is the sort of thing that you learn to do at the beginning of your math education, and then gradually move away from it. But you move to a higher class of non-rigor than you started from, where you’re just eliding bookwork rather than saying things that don’t necessarily work. E.g. here I’ve omitted the inductive base case, because I consider it obvious that the base case works, and the word “induction” tells me the shape of the argument without needing to write it explicitly.
If I were doing it inductively, I’d go in the other direction, removing edges instead of adding them. Take a graph G with n>0 edges, and remove an edge to get G’. The degree sum of G’ is two less than the degree sum of G (two vertices lose one degree, or one vertex loses two degree). Then induction shows that the degree sum is twice the edge count. There are probably simpler proofs, but having been primed by yours, this is the one that comes to mind.
I feel like being completely formal is the sort of thing that you learn to do at the beginning of your math education, and then gradually move away from it. But you move to a higher class of non-rigor than you started from, where you’re just eliding bookwork rather than saying things that don’t necessarily work. E.g. here I’ve omitted the inductive base case, because I consider it obvious that the base case works, and the word “induction” tells me the shape of the argument without needing to write it explicitly.