Prove that if the complement of a graph $G$ is disconnected, then we have the following inequality: $$ |\text{number of edges of}~ G| \leq\Delta(G)^2$$
Such that $\Delta(G)$ is the maximum degree in the main graph G.
My problem with this question is that I do not know how we can use the disconnectivity fact, Is there any similar problems that gives the lower or upper bound on edges when a graph or its complement is disconnected?
I'll use two claims to prove this:
Using that, we get: $$\vert E(G)\vert \leq \frac{\vert V(G)\vert}{2}\cdot\Delta(G)\leq \Delta(G)^2$$
Which is what we wanted to prove.
Proof of 1:
If $\bar G$ (the complement of $G$) is disconnected, it has more than one connected component, (so at least two).
Therefore, one of those connected components size is at most $\frac{\vert V(G)\vert}{2}$.
Denote it by $A$.
Now, let $v$ be some vertex in $G$.
$v$ is adjacent (in $G$) to every vertex $u\notin A$. Therefore $$\deg_G(v)\vert V(G)\vert-\vert A\vert\geq \frac{\vert V(G)\vert}{2}$$
and since $\Delta(G)\geq \deg_G(v)$, we get $\frac{\vert V(G)\vert}{2}\leq\Delta(G)$
Proof of 2:
$$2\vert E(G)\vert = \sum_{v\in V(G)} \deg(v)\leq\sum_{v\in V(G)} \Delta(G)=\vert V(G)\vert\cdot\Delta(G)\implies \\ \vert E(G)\vert\leq \frac{\vert V(G)\vert}{2}\cdot\Delta(G)$$