Any connected graph at most |V|+1 edges is 3-colourable

Question

I am attempting to prove the following:

Any connected graph at most $|V|+1$ edges is $3$-colourable.

My ideas so far have been induction on the number of vertices but I have only managed the basis of the induction, and then my idea for the more constructive proof was to use that this is a tree with at most two added edges (hence a cycle may have been created). If no edges are added, we just have a tree and this is clearly $3$-colourable. If one or two edges are added then a cycle will have been created and we know that even cycles can be coloured with $2$ colours and odd cycles with $3$.

I am struggling with both proofs and so I am looking for a nudge in the right direction for formalising what I have or a proof of the statement with some good explanation, please help.

Thanks in advance.

Answer 1

Since the graph is connected it has a spanning tree with $|V|-1$ edges. Every spanning tree is $2$-colorable (say with red and blue), hence you just have to prove you may re-insert back the two missing edges without altering too much the previous coloring. If a missing edge joins a red and a blue vertex, we do not need to alter the coloring. If just one edge joins a couple of vertices with the same color, we finish by changing the color of just one vertex (by making it green). If both missing edges join vertices with the same color, we need to perform a little case-by-case analysis, but that is simple, too.