Problem

Source: China TST 2011 - Quiz 3 - D1 - P3

Tags: algorithm, combinatorics unsolved, combinatorics



Let $G$ be a simple graph with $3n^2$ vertices ($n\geq 2$). It is known that the degree of each vertex of $G$ is not greater than $4n$, there exists at least a vertex of degree one, and between any two vertices, there is a path of length $\leq 3$. Prove that the minimum number of edges that $G$ might have is equal to $\frac{(7n^2- 3n)}{2}$.