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Let $G$ be a simple graph of order $n$. If $G$ is triangle-free, then we know that there is a bipartite graph of the same order and the same size. So $G$ has size less than $n^2/4$.

Now if it has triangles, it might have more than $n^2/4$ edges. But then if we think of the graph as a smallest union of triangles and edges, then I guess that this union involves less than $n^2/4$ triangles or edges. I tried with several examples and my guess holds for them.

Is this a known result, or is it just false?

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What is order and size? What if $n$ is odd? $C_5$ has no bipartite graph with the same number of vertices. –  draks ... Nov 28 '12 at 22:02
If I understand correctly, the question basically asks if every $n$-vertex graph can be decomposed into at most $n^2/4$ triangles and edges. Turán's theorem implies every triangle-free graph has at most $n^2/4$ vertices, so any counter-example must have a triangle. –  Douglas S. Stones Jan 6 '13 at 0:04

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