I'm trying to solve a problem, and I just need to prove the following lemma (Which I conjectured by myself):
$\forall \alpha >0, \exists \delta >0$ such that if $n$ is big enough, $G$ is a graph on $n$ vertices and $\alpha n^{2}$-far from being triangle-free (that is, you need to remove at least $\alpha n^{2}$ edges to have no triangles), then it has at least $\delta n^{3}$ triangles.
My first idea was to use Erdos-Simonovits Stability theorem (since $\alpha n^{2}$-far from being triangle-free implies $\alpha n^{2}$-far from being bipartite, which implies that $K_{3}(G)\geq\frac{n}{6}(e(g)+\alpha n^{2}-\frac{n^{2}}{4})$. Now it suffices to show that $e(G)\geq \frac{n^{2}}{4}$. Is this true?
Edit: I did not use the actual Erdos-Simonovits Stability theorem, but the first bound it is still true.
Note: For those who are curious, the original problem states: Prove that $\forall \alpha>0, n$ big enough, any $3-uniform$ hypergraph with at least $\alpha n^{2}$ edges has a set of six vertices with at least $3$ edges contained in those vertices.
This is the contrapositive of the triangle removal lemma, which states that
Usually this is proved using the regularity lemma. In particular, your suggested approach won't work, since (for many values of $\alpha$) there are graphs which are $\alpha n^2$-far from triangle-free but still have fewer than $\frac{n^2}{4}$ edges. For example, take a complete graph on about $2\alpha n$ vertices (this has about $2\alpha n^2$ edges, half of which need to be removed to make the graph triangle-free) and make all other vertices isolated. If it did work, the dependency $\delta = \frac{\alpha}{6}$ would be much stronger than known results, which suggests that it can't be easily fixed.