Prove that $\bar{G}$ contains a triangle

Question

Prove that If $G$ doesn't contain a triangle then $\bar{G}$ contains one

$G$ is a graph of order at least 6, and $d_{G}(v)\geq 3$ where $v$ is a vertex of $G$

Any hints ?

Answer 1

Assuming $G$ is simple, then the set $N(v)$ (neighbours of $v$) is an independent set of $G$ since $G$ has no triangles. Therefore $\bar{G}$ has a triangle. In fact, any three vertices of $N(v)$ will make a triangle in $\bar{G}$ since $|N(v)| \ge 3$.