Induced subgraph with radius rad(G)-1 Let $G$ be a simple connected graph with $rad(G)=r$. From all the induced subgraphs  of $G$ with radius $r$ let $H$ be this with the least number of vertices. I want to show that for every vertex that is not a cutvertex of $H$ we have $rad(H-u)=r-1$
 A: If we take a connected graph and add a vertex to it (along with at least on edge adjacent to that vertex) then the radius increases by at most $1$. This tells us if we remove a vertex from a graph the radius decreases by at most $1$.
Take a graph $G$ and consider the graph with the least number of vertices that has radius $r$. Call it $H$, if we remove a vertex from $H$ the we obtain a graph which has radius $r-1,r$ or possibly a radius larger than these two.
What happens if the radius is $r$? we obtain a contradiction to the minimality of $H$.
What happens if we obtain a radius larger than $r$? We can continue to remove vertices from the graph, the radius will reach $0$ at some point, but since the radius decreases by $1$ at the most this means at some point it will have had radius $r$. So we can start at $H$ and then remove some vertices until we get another graph of radius $r$. This again is a contradiction to the minimality of $H$.
So indeed $H-\{u\}$ has radius $r-1$ for any non cutvertex $u$.
