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.
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 $$(u_0,u_d)=<u_0, u_1, ..., u_i, ..., u_d > (0\leq i \leq d)$$ a shortest path from some node $u_0$ to some node $u_d$ such that its lenght is $d$. Let's add to $G$ an edge $e$.
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 $0<i<d$ ); it's easy to see that these different nodes are consecutive in $(u_0,u_d)'$, that is they replace a substring of $<u_0, u_1, ..., u_i, ..., u_d >$. Indeed, they form a new shortest path between some $u_j$ and $u_k$ using the new edge $e$ (because this new path $(u_j,u_k)$ wasn't the shortest before adding $e$); if there were not-consecutive $u_i$s in $(u_0,u_d)'$, then $e$ would be used more than one time, and $(u_0,u_d)'$ would not be a shortest path.
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 $l/2$.