Prove that if $diam(G) = 3$ for a regular graph G, then $diam(\bar{G}) = 2.$ 
Prove that if $diam(G) = 3$ for a regular graph G, then $diam(\bar{G}) = 2.$

I know that without using regularity one can show that if $diam(G) \geq 3, \text{then } diam(\bar{G})\leq 3$. How do I use regularity to show that it must be exactly 2?
 A: Suppose $G$ is a graph with $\operatorname{diam}(G)\gt2$ and $\operatorname{diam(\overline G})\gt2;$ I will show that $G$ is not regular.
More precisely, let $x,y,v,w$ be vertices with $d_G(x,y)\gt2$ and $d_{\overline G}(v,w)\gt2;$ I will show that $\deg(v)+\deg(w)\gt\deg(x)+\deg(y),$ the degrees being computed in $G.$
Clearly $x,y,v,w$ are four distinct vertices, $vw\in E(G),$ and $xy\in E(\overline G).$ Since $d_{\overline G}(v,w)\gt2,$ at least one of the edges $xv,xw$ is in $E(G);$ without loss of generality, we assume that $xv\in E(G).$ It readily follows that $yv\in E(\overline G),\ yw\in E(G),$ and $xw\in E(\overline G).$
Let $S=V(G)\setminus\{x,y,v,w\}.$ Let $m$ be the number of edges (of $G$) between $S$ and $\{v,w\},$ and let $n$ be the number of edges between $S$ and $\{x,y\}.$ Then $\deg(v)+\deg(w)=m+4,$ and $\deg(x)+\deg(y)=n+2.$ Since each vertex in $S$ is joined (by an edge of $G$) to at least one vertex in $\{v,w\}$ and to at most one vertex in $\{x,y\},$ we have $m\ge n;$ it follows that $\deg(v)+\deg(w)=m+4\gt n+2=\deg(v)+\deg(w).$
