Is there a graph with $n$ vertices and $n^2/4$ edges that isn't bipartite? Is there a graph with $n$ vertices and $\lfloor n^2/4\rfloor$ edges that isn't bipartite and contains no triangles ($K_3$)?
Rather, what I am asking is whether Mantel's Theorem implies that every graph on $n$ vertices with $\lfloor n^2/4\rfloor$ edges and contains no triangles must also be bipartite. 
EDIT: I apologize. I forgot to add that the graph must also not contain any triangles. Sorry, I realize that it was a trivial question before.
 A: Let $G=(V,E)$ be a (simple) triangle-free graph with $n$ vertices. I will show that $|E|\le\lfloor\frac{n^2}4\rfloor,$ with equality only when $G=K_{\lceil\frac n2\rceil,\lfloor\frac n2\rfloor}.$
Let $k$ be the independence number of $G,$ the maximum cardinality of an independent set of vertices. Since $G$ is triangle-free, every vertex $v$ has degree $d(v)\le k.$ Choose an independent set $S\subseteq V$ with $|S|=k,$ and let $T=V\setminus S.$
Since every edge has at least one endpoint in $T,$ we have
$$|E|\le\sum_{v\in T}d(v)\le\sum_{v\in T}k=k(n-k)\le\left\lceil\frac n2\right\rceil\left\lfloor\frac n2\right\rfloor=\left\lfloor\frac{n^2}4\right\rfloor.$$
If $|E|=\left\lfloor\frac{n^2}4\right\rfloor,$ then each inequality on the displayed line must be an equality.
If $|E|=\sum_{v\in T}d(v)$ then no edge can have both endpoints in $T,$ i.e., $G$ is a bipartite graph with partite sets $S$ and $T.$ If also $\sum_{v\in T}d(v)=\sum_{v\in T}k,$ then $G$ must be a complete bipartite graph $K_{k,n-k}.$ Finally, if $k(n-k)=\left\lfloor\frac{n^2}4\right\rfloor,$ then $\{k,n-k\}=\{\left\lceil\frac n2\right\rceil,\left\lfloor\frac n2\right\rfloor\}$ and $G=K_{\lceil\frac n2\rceil,\lfloor\frac n2\rfloor}.$
