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For any graph $G$ on $n$ vertices, not containing a $4$-cycle, prove that $$E(G)\le{n\over 4}\left(1+\sqrt{4n-3}\right)$$

Can someone tell me how to even approach?Not containing a 4-cycle means it can contain both $3$ cycles and $5$ cycles , right? How to handle such stuff in Ramsey Theory? An answer will be appreciated..

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    you likely mean **graph** $G$ not **group** $G$? – 2017-02-02
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    @gt6989b OOps.. sorry for the typo – 2017-02-02
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    For every couple of distinct and non-adjacent vertices $u,v$, $N(u)\cap N(v)\leq 1$ has to hold. If you assume that $G$ is regular, what is its maximum degree? If you assume that $G$ is bipartite, what is the maximum number of edges? – 2017-02-02
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    @JackD'Aurizio Can you please elaborate?What is $N$ ? I am a beginner, so it would help much if you wrote an answer.. – 2017-02-02
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    @Qwerty: $N$ stands for **N**eighbourhood: $N(u)$ is the set of vertices joined with $u$ through some edge. I meant $|N(u)\cap N(v)|\leq 1$ before. – 2017-02-02
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    This is an instance (luckily, a solvable one) of a famous open problem: https://en.wikipedia.org/wiki/Zarankiewicz_problem – 2017-02-02
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    This is really a Turan question, not a Ramsey one – 2018-06-28

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Let $X=\{(x,\{y,z\}):y\neq z,xy,xz\in E\}$. We will double count $X$. First, by fixing $x\in V$, we see that $$|X|=\sum_{x\in V}|\{\{y,x\}:(x,\{y,z\})\in X\}|=\sum_{x\in V}{\deg(x)\choose 2}.$$ On the other hand, we can fix the pair $\{y,z\}$. Now, notice that $\{x:(x,\{y,z\})\in X\}=N(y)\cap N(z)$ where $N(y)$ denotes the neighborhood of $y$. Now, if $|N(y)\cap N(z)|\geq 2$, then $G$ would have a copy of $C_4$, which is not the case, so $|N(y)\cap N(z)|\leq 1$ for all $y\neq z\in V$. Therefore, $$|X|=\sum_{\{y,z\}\in{V\choose 2}}|N(y)\cap N(z)|\leq \sum_{\{y,z\}\in {V\choose 2}}1={n\choose 2}.$$ Now, the function $f:\mathbb{R}\to\mathbb{R}:x\mapsto{x\choose 2}$ is convex, so we can apply Jensen's inequality to find $$ {n(n-1)\over 2}={n\choose 2}\geq |X|=\sum_{x\in V}{\deg(x)\choose 2}\geq n{{1\over n}\sum_{x\in X}\deg(x)\choose 2}=n{2{|E|\over n}\choose 2}=n{2{|E|\over n}(2{|E|\over n}-1)\over 2}.$$ Solving this equaltion for $|E|$, we see that $$|E|\leq{n\over 4}\bigl(1+\sqrt{4n-3}\bigr)$$