G=(V,E) a graph that that maintains this condition $\left |E \right |\geq \left | V \right |\geq \left |3 \right |$ must have a circle

Question

Condition : $\left |E \right |\geq \left | V \right |\geq \left |3 \right |$ and graph G=(V,E).

I need to proove it without using any graph theorems and lemmas(if so then I have to proove them).

I think that somehow I need to pay attention to the fact that there are always more vertex than edges(in a non circular graph)(although I am not really sure that I am allowed to use because it is a theorem from graph theory and I have to somehow proove it). I was thinking about making 2 sets: one for vertex and the other for edges. Then to use "Pigeonhole principle" also called "Dirichlet's drawer principle(by wiki)" and to show that there is some contradiction. Overall this is the direction I was thinking but I'm stumped now and don't have specific direction of how to continue this particular proof.

Answer 1

Hint: prove that a graph with no cycles must have some vertex $v$ meeting at most one edge.

Now considering the graph $G-v$ allows you to prove by induction that $|E|<|V|$ for any such graph.

Answer 2

Let's first note that any acyclic graph (one without "circles") has to have a node of degree 1, the proof of which is noted in this question.

Hence, for an acyclic graph, we can remove a node of degree 1, which is convenient since this drops the number of edges and vertices by 1. Hence, you can easily use induction to prove your hypothesis.