Using $$ \alpha(G) \ge \frac{n^2}{2\mathbb{E}(G) + n} $$ prove that any graph on n vertices with no triangles has at most $\frac{n^2}{4}$ edges.
$\alpha(G)$ is the maximum size of independent vertices in G.
Using $$ \alpha(G) \ge \frac{n^2}{2\mathbb{E}(G) + n} $$ prove that any graph on n vertices with no triangles has at most $\frac{n^2}{4}$ edges.
$\alpha(G)$ is the maximum size of independent vertices in G.
Let $G^{'}$ be a triangle free graph, and that $G$ is its complement.
From this we get that $\alpha(G) \le 2$ (Why?)
Finally we get $$\mathbb{E}(G) \ge \frac{n^2}{4} - \frac{n}{2}$$
After one more step, the proof is finished for $G^{'}$. (How?)