An inequality between the diameter and the inverse dual degree of a tree

Let R(T ), D(T ) be respectively the radius and diameter of a nontrivial tree T and I(T ) = ∑ u∈V (T ) 1/d(u) be the inverse dual degree, where d(u) = ( ∑ v∈N(u) d(v))/d(u) for each u ∈ V (T ). In this note we prove that I(T ) ≥ { R(T ) + 1/3, if D(T ) is odd R(T ) + 5/6, if D(T ) is even, with equality if and only if T is a path of at least 4 vertices. This inequality strengthens a conjecture of Graffiti. Let G = (V (G), E(G)) be a simple, connected graph. The distance d(u, v) between two vertices u, v of G is the minimal length of a path from u to v in G. The diameter D(G) of G is the largest distance between any two vertices of G. The radius R(G) of G is minu∈V (G) maxv∈V (G) d(u, v). If S ⊆ V (G) and u ∈ V (G) \ S, then we denote d(u, S) = minv∈S d(u, v). The neighbours of u ∈ V (G) are vertices adjacent to u in G and the neighbourhood N(u) of u in G is the set of