Question 7
Let $3n^2$ be the vertex number of a simple graph $G$ (integer $n \ge 2$). If the degree of each vertex is not greater than $4n$, there exists at least one vertex with degree $1$, and there exists a route with length not greater than $3$ between any two vertices. Prove that the minimum number of edges of $G$ is $\frac{7}{2}n^2 - \frac{3}{2}n$.
Remark. A route between two distinct vertices $u$ and $v$ with length $k$ is a sequence of vertices $u = v_0, v_1, \dots, v_k = v$, where $v_i$ and $v_{i+1}$, $i = 0, 1, \dots, k-1$, are adjacent. (posed by Leng Gangsong)
Remark. A route between two distinct vertices $u$ and $v$ with length $k$ is a sequence of vertices $u = v_0, v_1, \dots, v_k = v$, where $v_i$ and $v_{i+1}$, $i = 0, 1, \dots, k-1$, are adjacent. (posed by Leng Gangsong)