Diameter of a Connected Graph Problem

*

*Prove that if $G$ is connected and $\text{ diam}(G) \geq 3$, then $\overline{G}$ is connected.

*Prove that if $\text{ diam}(G) \geq 3$, then $\text{ diam}(\overline{G}) \leq 3.$

*Prove that if $G$ is regular and $\text{ diam}(G) = 3$, then $\text{ diam}(\overline{G}) = 2$.

Approach
I tried to find relationships between the diameters of $G$ and $\overline{G}$, but all I could think of was the effect of adding edges to the graph reduces the diameter in most cases, but not necessarily.
Any help would be greatly appreciated!
 A: (1) Let $G$ be a connected graph. Suppose $\bar{G}$ (the complement of $G$) be disconnected. We shall prove that $G$ has $diam(G)\le 2$. (hence, proving the claim by contra-positive)
Let $C_1,C_2,\dots C_n$ be connected components of $\bar{G}$. Since, $\bar{G}$ is disconnected, there are at least two connected components.
In $G$, 


*

*there is an edge between any two vertices that belong to different
connected components of $\bar{G}$.

*For any two vertices belonging to a connected component of
$\bar{G}$, there is a path connecting them via a vertex belonging to
a different connecting component.


Hence, $diam(G)\le 2$. The case of $diam(G)=1$ occurs when $\bar{G}$ has no edges at all.
(2) Let $diam(G)\ge 3$. If $G$ is disconnected, then $\bar{G}$ is connected with $diam(\bar{G})=2$ (from 1). Suppose, G is connected and $diam(G)=k\ge 3$. Let $v_1$ and $v_{k+1}$ be two vertives of $G$ connected via shortest path of length $k$. Let $L_r:=\{v \mid \, \text{ length of the shortest path between } v_1 \text{ and } v \text{ is } r, 1\le r\le k-1\}$. Since, $v_{k+1}$ exists, $L_r\ne \emptyset \, ,\, \forall \, 1\le r\le k-1$.
In $\bar{G}$,


*

*$v_1$ is connected to $v_{k+1}$ via an edge.

*$v_1$ is connected to any element of $L_r$ where $2\le r\le k-1$ via
an edge.($v_{k+1}$ is connected to any element of $L_r$ where $1\le
    r\le k-2$ via an edge)

*$v_1$ is connected to any element of $L_1$ via $v_{k+1}$. ($v_{k+1}$
is connected to any element of $L_{k-1}$ via $v_{1}$)

*Let $1\le i\le k-1$ and $1\le j\le k-1$. If $\mid i-j \mid>1$, any
two vertices of $L_i$ and $L_j$ are adjacent. If $i-j=1$, any two
vertices of $L_i$ and $L_j$ are connected via $v_{L_i} \to v_1 \to
    v_{L_j}\to v_{k+1}$ where $v_{L_i}$ and $v_{L_j}$ are any two
vertices belonging to $L_i$ and $L_j$ respectively.


This proves, $diam(\bar{G})\le 3$
(3) Let $G$ be a regular graph with vertex degree $k$ and $diam(G)=3$. Using the same notation as in (2), let $v_1$ and $v_4$ be two vertices such that the shortest path between them has length $3$. Since, $G$ is regular with degree $k$, both $L_1$ and $L_2$ contain $k$ vertices each. There are $2k+2$ vertices and $k(k+1)$ edges. In, $\bar{G}$, a path from a vertex in $L_1$ to a vertex in $L_2$ on the lines of 4 of (2) has length $3$. Let $G'$ be the graph obtained by deleting the vertices $v_1$ and $v_4$ from $G$ (therefore, deleting all the edges incident on them too). 
Note that $\bar{G'}$ is a regular graph with degree $2k-1-(k-1)=k$ and $2k$ vertices. Suppose $diam(\bar{G'})\ge 3$, then we cannot partition the vertices of $\bar{G'}$ in the form $\{v_1,L_1,\dots,L_m,v_{m+1}\}$ for some $m\ge 3$ as we would require at least $2+\mid L_1\mid +\mid L_{m} \mid=2k+2$ vertices while we have only $2k$ vertices. Hence, we prove that $diam(\bar{G'})=2$ and thereby $diam(\bar{G})=2$. ( $diam(\bar{G})=1$ would mean that $G$ had no edges at all contradicting the fact that $diam(G)=3$)
