
> 
> Theorem. Fix $\epsilon > 0$; for sufficiently large n, any graph with n vertices and $\epsilon \binom{n}{2}$ edges contains many (nondegenerate) cycles of length 4.
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The proof is simple; put an indicator variable $\delta\_{x, y}$ for each pair of vertices corresponding to whether or not there is an edge there; then start with


$n^8 \epsilon^4 = (\sum \delta\_{x, y})^4$


and apply Cauchy-Schwarz twice; finally, note that there are $O(n^3)$ "degenerate 4-cycles".


A basic corollary of this is the following fact:



> 
> Corollary. Any graph with girth at least 5 and n vertices has $o(n^2)$ edges.
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This seems like it should be possible to prove without resorting to "analytic" machinery like Cauchy-Schwarz; indeed, it seems like it should be weak enough to prove almost by arguing "locally." But none of the obvious lines of reasoning seem to provide a proof.


Is it possible to get a good bound on the density of large-girth graphs without using Cauchy-Schwarz or equivalent? 



Combinatorial proof that large-girth graphs are sparse?
It's known more specifically that any graph with girth ≥ 5 has $O(n^{3/2})$ edges — see e.g. [Wikipedia on the Zarankiewicz problem](http://en.wikipedia.org/wiki/Zarankiewicz_problem).


Here's a combinatorial proof. Suppose that graph $G$ has $\ge kn^{3/2}$ edges for a sufficiently large constant $k$. As long as there are vertices with degree smaller than some appropriate constant times $\sqrt n$ one can remove them and get a smaller graph with the same property of having at least $kn^{3/2}$ edges, so eventually one can reach a state where every vertex has degree at least $\Omega(\sqrt n)$. Once this happens, there are $O(n^2)$ possible pairs of neighbors that a vertex might have, and each vertex has $\Omega(n)$ pairs of neighbors, so some pair of neighbors appears twice causing the graph to have a 4-cycle.


