How to prove that a graph containing more paths of length 2 than $\binom{n}{2}$ (where n= #vertices) must contain a cycle of length 4.

Question

I need to prove that a graph that satisfies this equality also contains a cycle of length 4. $$\sum_{\upsilon\in V(G)}\binom{d(\upsilon)}{2} > \binom{n}{2}$$

The left hand side of this inequality represents the number of paths of length 2, and it is quite obvious that this number of paths becomes so large that it would be impossible for there not to be a cycle of length 4, though I don't know how to prove it mathematically. Any help is appreciated

Answer 1

Apply the pigeonhole principle.

There are $\binom n2$ pairs of vertices.

There more than $\binom n2$ paths of length $2.$