Pseudo-Tree , Proving $ 1 \leq \frac{|E|}{|V| - 1} \leq \frac{3}{2} $ Problem: Graph $ G = \langle V,E \rangle $ is called Pseudo-Tree if it is connected and also every edge of $ G $ is on one simple-circuit at-most. Prove that $ 1 \leq \frac{|E|}{|V| - 1} \leq \frac{3}{2} $
So I didn't know what to do at all. Is the best strategy to approach this problem would be induction on the number of vertices $ |V| = n $? ( Strong induction maybe? ), also what happens if I have only 1 vertex? ( In that case $ |V| - 1 = 0 $ the inequlity doesn't occur. But I don't fully understand why a such a case is not possible where $ |V|=1 $? ). I'd appreciate any guidance on this problem and thanks in advance for help!
 A: To answer your question, I came up with a counting argument first. You may be able to convert the reasoning below to an inductive proof, but there may be some details to keep track of. It is really observing Inequalities (1) and (2) below and putting them together that yields the desired result. In fact, a TLDR version of this proof in the following:
That each edge is in at most one cycle $\Rightarrow$ the cycles are all edge-disjoint [and have at least 3 edges each] $\Rightarrow$ removing at most a third of the edges can gives you a forest and so with $\le |V(G)|-1$ remaining edges $\Rightarrow$ desired result.
To elaborate: The circuits $C_1, C_2, \ldots, C_r$ of $G$ [for some integer $r$] are edge-disjoint [by the fact that each edge is on at most one circuit] and each circuit has 3 edges. So on the one hand, $$|E(G)| \ge \sum_{i=1}^r |C_i| \ge 3r.$$ This gives $$\frac{2|E(G)|}{3} = |E(G)|-\frac{E(G)}{3}
\le |E(G)|-r,$$ or in particular, $$(1) \quad \quad \quad \frac{2|E(G)|}{3} \le |E(G)| - r.$$
Now let $e_1,e_2, \ldots, e_r$ be such that $e_i\in C_i$ for each $i=1,\ldots, r$. Then $G \setminus \{e_1,\ldots, e_r \}$ has no cycles and thus is a forest. So removing from $G$ this set $\{e_1,\ldots, e_r\}$ of $r$ edges leaves a forest, in particular a graph with at most $|V(G)|-1$ edges. So on the other hand, $$(2) \quad \quad \quad |E(G)|-r \le |V(G)|-1.$$
So putting Inequalities (1) and (2) together yields
$$\frac{2|E(G)|}{3} \le |E(G)| - r \le |V(G)|-1,$$ or equivalently
$$\frac{2|E(G)|}{3} \le |V(G)|-1.$$ Finish using basic algebra.
