Graph Theory Question Related to Domination number. Let G be a graph whose diameter is at least 3. Prove that the domination number of the complement of G is at most 2. 
I know that since the diameter of G is at least 3, the diameter of the complement of G is at most 3. However, this doesn't seem to be enough to prove the domination number of the complement of G is at most 2. Any suggestions?

Damned if I know. Let me try and follow my nose here. The diameter of $G$ is at least $3$, what does that mean? It means there are two vertices $u,v$ in $G$ such that $\operatorname d(u,v)\ge3$. And we want to show that the complement $\bar G$ has a dominating set containing at most $2$ vertices. Hmm. Maybe I can show that $\{u,v\}$ is a dominating set for $\bar G$?
Let's see. I want to try and show that each vertex $x\notin\{u,v\}$ is adjacent to $u$ or $v$ in $\bar G$. What if I assume for a contradiction that $x$ is adjacent to neither $u$ nor $v$ in $\bar G$? So $x$ is adjacent to both $u$ and $v$ in $G$.
OK, so back in $G$ I've got two vertices $u,v$ with $\operatorname d(u,v)\ge3$, and another vertex $x$ is adjacent to both $u$ and $v$. Is there any way to get a contradiction out of that??
Beats me. I give up.
