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Suppose we're given a simple graph.

In Number of triangles in a graph based on number of edges there is a nice argument for $T \leq O(E^{3/2})$, where $T$ is the number of triangles, and E is the number of edges.

I am interested in determining the optimal constant in the equality, that is, the minimal possible $C$ such that $T \leq C E^{3/2}$.

One can follow the approach in the linked post and see that $C = \sqrt{2}$ is possible. (Some further denotions come from that post.) Indeed, suppose that some $A(v) > \sqrt{2E}$, then the number of edges adjacent to vertices in $A(v)$ is at least $$\frac 12 \sum_{w \in A(v)} d(w) \geq \frac 12 |A(v)| d(v) \geq \frac 12 |A(v)|^2 > E,$$

which yields a contradiction. We then have in fact $A(v) \leq \sqrt{2E}$, and, like in the linked post, we get $$T \leq E \cdot \sqrt{2E} = \sqrt{2}E^{3/2}.$$ The question is: do we know how to obtain better constants $C$? I am specifically interested whether $C=1$ would somehow work.

dnes
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This paper seems to prove the result that you want. Specifically, they show $C = \frac{\sqrt{2}}{3} \le 1$.

The paper then shows asymptotically, (i.e. as $n \to \infty$) the complete graph has $\frac{\sqrt{2}}{3}|E|^{3/2}$ many triangles. Thus this constant is sharp.