I have a trouble in proving the following:
If $G$ is a graph such that every vertex in it has degree greater than or equal to 3, then for any tree $T$ of $G$ the following is true:
$$|E(T)| \leq 2|E(G-T)|-3,$$
provided that $|E(G-T)| \geq 3.$
I tried to use the fact that the sum of degrees is equal to $2|E(G)|$ but couldn’t get a useful result.
Let $n_d$ be the number of vertices of degree $d$. Then $$|V(G)|=\sum_{d \ge 3} n_d$$ and $$2|E(G)|=\sum_{d \ge 3} d n_d.$$ So $$3|V(G)| = 3 \sum_{d \ge 3} n_d = \sum_{d \ge 3} 3 n_d \le \sum_{d \ge 3} d n_d = 2|E(G)|.$$ Subtract 3 from both sides to obtain $$3(|V(G)|-1) \le 2|E(G)|-3.$$ Now $|E(T)|=|V(G)|-1$ implies that $$3|E(T)| \le 2|E(G)|-3.$$ Finally, $|E(G-T)|=|E(G)|-|E(T)|$ yields $$|E(T)| \le 2|E(G-T)|-3.$$