One can show, that a Graph with at least $n$ vertices and $m$ edges, has at least $\dfrac{4m}{3n}(m-\dfrac{n^2}{4})$. I was wondering, about the best lowest bound of this, and the best upper bound of this ( is the best upper bound truly, the number of triangles formed in the complete graph , i.e $ \Delta(G) \choose 3$ ?
2026-03-25 15:42:14.1774453334
Number of Triangles in a graph with $n$ vertices and $m$ edges
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An upper bound is $\frac 13{m \choose 2}=\frac {m(m-1)}6$ because at the most you can choose any two edges and have one triangle, then each triangle gets counted $3$ times.