Is it possible that both a graph and its complement have small connectivity? Let $G=(V,E)$ be a simple graph with $n$ vertices. The isoperimetric constant of $G$ is defined as 
$$
i(G) := \min_{A \subset V,|A| \leq \frac n2} \frac{|\partial A|}{|A|}
$$
where $\partial A$ is the set edges $uv \in E$, with $v \in A$ and $u \in A^c$. 

Is there a graph $G$ such that  $$i(G) + i(G^c) < \frac 12$$
  where $G^c$ denotes the complement of $G$?

 A: Since for comment is long, I write the contribution here:
Let $A$ and  $L$ be the adjacency matrix and Laplacian matrix of the graph $G$, respectively. Also, suppose $\mu_2$ denotes the second smallest eigenvalue of $L$ and $\lambda_2$ be the second largest eigenvalue of $A$. It is proved that $i(G)\geq \frac{\mu_2}{2}$ (see Theorem 4.1 of "Isoperimetric numbers of graphs" by Bojan Mohar). So, we have $$i(G)+i(G^c)\geq \frac{\mu_2(G)+\mu_2(G^c)}{2}$$.
So, if we have an example for your question, we must find graph $G$ such that $\mu_2(G)+\mu_2(G^c)<1$. By some spectral graph theory, it can be seen that finding such a graph is a difficult problem. But, finding the spectrum of a graph is much easier than finding its isoperimetric number. 
Also, it can be shown that this problem is related to the classification of the graphs based on their second largest eigenvalue of the adjacency matrix. Actually, if we can find a suitable graph such that:
$$\lambda_2(G)+\lambda_2(G^c)\leq \delta(G)-\Delta(G)+n-2$$,
Then we find an example; where $\delta(G)$ and $\Delta(G)$ shows the minimum and maximum degree of the graph $G$ with $n$ vertices.
$\textbf{Added later:}$ By the relation $\mu_2(G)+\mu_2(G^c)<1$ and some random graph testing, it seems that such graphs are so rare. There is not any example of such graph up to $6$ vertices. Also, it seems that we can prove $\mu_2(G)+\mu_2(G^c)\geq 1$, which means there is not such an example.
A: It is almost obvious. Let $i(A)=\frac{|\partial_G A|}{|A|}$ and $j(B)$ be the same ratio for a vertex set $B$ and the complement of $G$. Let $x=|A\cap B|, y=|A\cap B^c|, z=|A^c\cap B|, t=|A^c\cap B^c|$. The edges going between $A\cap B$ and $A^c\cap B^c$ are boundary edges for both $A$ and $B$ and so are the edges between $A^c\cap B$ and $A\cap B^c$. Thus, we have
$$
i(A)(x+y)+j(B)(x+z)\ge xt+yz
$$
If $i(A)+j(B)<1$, the left hand side is strictly less than $x+\max(y,z)$, which is at most $xt+yz$ if $t,y,z>0$. Note that if $t=0$, then $x=0$ because of the condition $|A|,|B|\le \frac n2$, so if $y,z>0$ in this case, we still have a contradiction. Finally, if, say, $y=0$, then $x>0$ (otherwise $A=\varnothing)$ and $t\ge \frac n2$ (because $|B|\le\frac n2$), so the RHS is at least $\frac n2$ while the LHS is less than $\max(|A|,|B|)\le \frac n2$ and we get a contradiction again. 
