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I'm trying to solve the following exercise from the book A Textbook of Graph Theory by R. Balakrishnan and K. Ranganathan

Show that for a simple bipartite graph, $m\leq \frac{n^2}{4}$

$m$ is the size of the graph.

$n$ number of vertices.

I was considering the following cases:

Let $X$ and $Y$ be a partition of the set of vertices of $V(G)$ and I tried to verify some cases accordingly to the cardinality of the each set. If $|V(G)|=n$ and $|X|=|Y|=\frac{n}{2}$ it is easy to see, but I got stuck trying to show the case when the cardinalities are $|X|<|Y|$

Any help will be greatly appreciated!

Thanks!

JimmyJP
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    If $x+y=n$ what's the biggest that $xy$ can be? You may have seen a problem like this in calculus class. What's the biggest rectangular field (area in square meters) you can enclose with $2n$ meters of fence? – bof May 07 '15 at 07:49
  • I solved this exercise thanks to your suggestion. Thank you bof! – JimmyJP May 07 '15 at 08:12

1 Answers1

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If your graph has $x$ vertices in one group and $y$ in the other, we have $m \leq m$.

By AM-GM inequality $$\sqrt{m} \leq \sqrt{xy} \leq \frac{x+y}{2}=\frac{n}{2}$$

N. S.
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