Prove that every $n$-vertex graph with at least $\lfloor n^2/4\rfloor+1$ edges contains some edge in at least $(1/6-o(1))n$ triangles, and that this constant $1/6$ is best possible.
I have been thinking on this problem for a while and the following idea came to my mind. It is basically averaging argument since we need to show an existence of some edge which appears in $(1/6-o(1))n$ triangles.
My attempt: For any edge $uv\in E(G)$ consider the following set $N(u)\cap N(v)=\{x\in V(G): xu,xv\in E(G)\}$, i.e. this set represents all triangles where $uv$ is one of the sides. Let's apply averaging argument on the size of $N(u)\cap N(v)$, i.e. consider the sum $$\sum \limits_{uv\in E(G)}|N(u)\cap N(v)|=\sum \limits_{uv\in E(G)}|\{x\in V(G): xu,xv\in E(G)\}|=$$ $$=\frac{1}{2}|\{(u,v,x)\in V^3: uv,xu,xv\in E(G)\}|=3|\Delta_G|,$$ where $|\Delta_G|$ is the number of triangles in graph $G$. Here I am using the non-trivial fact that if $e(G)\geq \lfloor n^2/4\rfloor+1$, then $|\Delta_G|\geq \lfloor n/2\rfloor$. Therefore, $$\frac{1}{e(G)}\sum \limits_{uv\in E(G)}|N(u)\cap N(v)|\geq \frac{3|\Delta_G|}{e(G)}\geq \frac{3\lfloor n/2 \rfloor}{e(G)}.$$
Hence there is an edge $uv\in E(G)$ such that $|N(u)\cap N(v)|\geq \frac{3\lfloor n/2\rfloor}{e(G)}$. But of course it does not give the desired result.
I was wondering is it possible to modify this method to solve the problem?
I found this problem as a Lemma 2 in a paper of Paul Erdos "On a theorem of Rademacher-Turan".
Remark: I am gonna write the proof in my own words with more details. However, there was one moment in the proof which I did not understand at all so I'd be very grateful if someone can explain it to me.
Proof. Let $\{\{\alpha_1,\beta_1,\gamma_1\},\dots,\{\alpha_r,\beta_r,\gamma_r\}\}$ be a maximal system of disjoint triangles in our graph $G$. In other words, if we remove there $3r$ vertices, we obtain a triangle-free graph on $n-3r$ vertices and by Turan's theorem it has at most $\lfloor (n-3r)^2/4\rfloor$ edges.
Denote by $G(n,i)$ the graph $G-\sum \limits_{j=1}^{i-1}(\alpha_j+\beta_j+\gamma_j)$, i.e. we removed the vertices $\{\alpha_j,\beta_j,\gamma_j\}$ for $1\leq j\leq i-1$ from original graph $G$. Let $v^{(i)}(\alpha_i), v^{(i)}(\beta_i), v^{(i)}(\gamma_i)$ are the degrees of $\alpha_i,\beta_i,\gamma_i$ in $G(n,i)$.
We claim that for some $i\in \{1,\dots,r\}$ we should have $v^{(i)}(\alpha_i)+v^{(i)}(\beta_i)+v^{(i)}(\gamma_i)>\dfrac{5n}{2}-3i$. In the original proof of Erdos I just took $c_2=1/6$. If this is not true, then for any $i\in \{1,\dots,r\}$ we have $v^{(i)}(\alpha_i)+v^{(i)}(\beta_i)+v^{(i)}(\gamma_i)\leq \dfrac{5n}{2}-3i$. Since,
$$\left\lfloor \frac{n^2}{4}\right\rfloor+1-\sum \limits_{i=1}^{r}\left(v^{(i)}(\alpha_i)+v^{(i)}(\beta_i)+v^{(i)}(\gamma_i)\right)+3r\leq \left\lfloor \frac{(n-3r)^2}{4}\right\rfloor \Leftrightarrow$$ $$\left\lfloor \frac{n^2}{4}\right\rfloor+1\leq\sum \limits_{i=1}^{r}\left(v^{(i)}(\alpha_i)+v^{(i)}(\beta_i)+v^{(i)}(\gamma_i)\right)-3r+ \left\lfloor \frac{(n-3r)^2}{4}\right\rfloor.$$ hence $$\left\lfloor \frac{n^2}{4}\right\rfloor+1\leq\sum \limits_{i=1}^{r}\left(\frac{5n}{2}-3i\right)-3r+ \left\lfloor \frac{(n-3r)^2}{4}\right\rfloor.$$ However, one can show that $$\sum \limits_{i=1}^{r}\left(\frac{5n}{2}-3i\right)-3r+ \left\lfloor \frac{(n-3r)^2}{4}\right\rfloor\leq \left\lfloor \frac{n^2}{4}\right\rfloor,$$ which gives a contradiction.
Hence there is $i_0\in \{1,\dots,r\}$ such that $v^{(i_0)}(\alpha_{i_0})+v^{(i_0)}(\beta_{i_0})+v^{(i_0)}(\gamma_{i_0})>\dfrac{5n}{2}-3i_0.$ Then a simple computation shows that there are at least $3\lfloor n/6\rfloor$ vertices of $G(n,i_0)$ which are connected with more than one of the vertices of $\{\alpha_{i_0},\beta_{i_0},\gamma_{i_0}\}$. Therefore there are at least $\lfloor n/6 \rfloor$ of them which are connected with the same pair, i.e. $G(n,i_0)$, and therefore original graph $G$ contains an edge which appears in at least $\lfloor n/6\rfloor$ triangles. This is the proof of the statement.
Question. But I have one question which bothers me. Can anyone explain why are there at least $3\lfloor n/6\rfloor$ vertices of $G(n,i_0)$ which are connected with more than one of the vertices of $\{\alpha_{i_0},\beta_{i_0},\gamma_{i_0}\}$? I don't see any simple computation here as Erdos claims.