Maximum edges in graph with no triangles is complete bipartite I know that Mantel's theorem says that the maximum number of edges in a graph with no 3-cycles is $\lfloor n^2/4\rfloor$.
How can I prove that a graph with no 3-cycles and the maximum number of edges is necessarily $K_{\lfloor n/2 \rfloor , \lceil n/2 \rceil}$?
I have tried assuming equality in the inductive proof of Mantel's theorem, so for any edge, its endpoints together are incident with EXACTLY k−1 other edges, so proving the inductive hypothesis provides equality rather than an upper bound, but I am lost on how to go further.

You can show that if $G$ has the maximum edges on $n$ vertices and no $3$-cycles, then for any three vertices $x,y,z$ if edge $xy$ exists, then either $xz$ or $yz$ exists. This implies you should be looking at complete bipartite graphs. 
From there show $K_{n/2,n/2}$  maximizes edges among complete bipartite graphs. (I'm not sure how to get ceiling and floor functions)
