It appears plausible to me that if we have a connected graph $G$ with diameter $d$ and we remove a non-cut edge $e$ from it, the diameter of the resulting graph $G_e$ will be at most $2d$.
By removing a non-cut edge I mean that the resulting graph is still connected.
I was not able to find a proof or a counter-example.
Any suggestion? Thanks.
EDIT:
I've tried to follow Perce's hint, and I'm quite sure I success in showing what I was looking for, but it was hard for me to explain. Then I ask for the correctness of my proof.
Let $G$ be a graph with a certain diameter $d$, and let $(u_0,u_d)=
By adding $e$, maybe, there will be a shorter path (u_0,u_d)' in the graph from $u_0$ to $u_d$ that differs from $(u_0,u_d)$ for some nodes $u_i$ (with $0 ); it's easy to see that these different nodes are consecutive in (u_0,u_d)', that is they replace a substring of $
EDIT2: fixed the following part of my solution
Hence, adding an edge we create at most one shorter route among the nodes in the path $(u_0,u_d)$, that in the best case is of lenght 1 (if we use $e$ itself). Now, it's easy to see that, by adding a an edge between two nodes in a path, we cannot reduce its lenght $l$ below $l/2$.