Given a finite group $G$ with $N$ elements what are the "smallest" possible subsets $A,B$ of $G$ such that $G=AB$ (ie. every element of $G$ is a product of an element in $A$ and an element in $B$)?

We have of course $\sharp(A)\sharp(B)\geq N$ but can one always find two subsets $A,B$ of size roughly $\sqrt N$ with the property $G=AB$? Since I do not know the optimal answer a good definition of "roughly" should be part of an answer (one can surely not do better than $A,B$ of size $O(1)+\sqrt N$ but I guess this is much to optimistic. It is perhaps more reasonable to hope for $\sqrt N$ times something bounded).

The optimal answer $A,B$ of size (at most) $1+\sqrt N$ works for cyclic groups: take $A=\mathbb N\left\lceil \sqrt N\right\rceil\cap \{0,\dots,N-1 \}$ and $ B= \{ 0,1,\dots,\left\lfloor \sqrt{N} \right \rfloor \} $ in $\mathbb Z/N\mathbb Z= \{0,\dots,N-1\}$.

On the other hand, if $G$ has a (not-necessarily normal) subgroup $H$ of size roughly $\sqrt{N}$ one can consider $A=H$ and take for $B$ coset-representantives.

Combining the two constructions we get a nearly optimal result for abelian groups.

Is there for example a sequence of "bad" groups $G_i$ such that, say $$\left((1+\sharp(A_i))(1+\sharp(B_i))-\sharp(G_i)\right)\Big/\sqrt{\sharp(G_i)}$$ is unbounded for every (sequence of) subsets $A_i,B_i\subset G_i$ such that $G_i=A_iB_i$? (The existence of such a sequence would of course the hope for the existence of $A,B$ of size $O(1)\sqrt{\sharp(G)}$ (with $AB=G$) in every finite group $G$.