Let $G$ be a graph and $e$ be an edge of the graph $G$ such that the subgraph $G\setminus e$
is connected. The subgraph $G\setminus e$ is the subgraph of $G$ obtained by deletion of the edge $e$ of $G$.
Assume that $G$ has $n$ vertices. 
Is it true that $\lambda(G)-\lambda(G\setminus e)\geq \frac{1}{n}$?
Here $\lambda(G)$ denotes the maximum eigenvalue of adjacency matrix of $G$.



Difference of the maximum eigenvalue of a graph with the one of one-edge-deleted subgraph
It's always useful to test these questions on actual examples. The largest eigenvalue of the cycle $C\_n$ is 2 and the largest eigenvalue of the path $P\_n$ on $n$ vertices is $2\cos(\pi/(n+1))$. When $n=8$, this is 1.879385 and $2-1.8793852=0.120615 <1/8$.


In fact it is not hard to see that for large $n$ the difference $2-2\cos(\pi/(n+1))$ is
of order $\pi^2/(n+1)^2$, and so your lower bound is not even of the right order.


