This problem is indeed nice and hard, but it gets quickly solved when one knows what to look for. More specifically, see this Romanian TST problem from 2011: https://artofproblemsolving.com/community/c6h455245p2558775. That is not strong enough to imply the desired bound (at least not in an immediate manner), but the paper linked in @jukcter's comment for that problem does the job. Indeed, invoking the main result of the mentioned paper, the sum of the number of edges of all faces bounded by any given line in the configuration is bounded above by $9.5n$. Summing over all the lines, a bounded region with $m$ vertices gets counted $m$ times, while an infinite region with $m$ vertices gets counted $m+1$ times. In particular, a face with $m$ vertices contributes with at least $m^2$ to the total sum of edges of faces supported by lines in the configuration. However, that sum is bounded above by $9.5n\cdot n=9.5n^2<10n^2$, implying the problem. I wonder if the official solution is any different...

This problem is (10-11).7 and (8-9).8

The official solution is different - much simpler. But it only gives an estimate $10n^2-8n$.

fagot wrote: The official solution is different - much simpler. But it only gives an estimate $10n^2-8n$. Can you give a link to the official solution?thanks.

In the paper it is also proven that for $n+1$ lines, the sum of the ranks of the regions, bounded by some line $l$ is maximum $10n+2$ (so for $n$ lines, it's max $10n-8$, so this proof is also the official proof probably; the finish of the solution using this fact is mentioned in #2). I put it here for the sake of completeness. We consider $l$ to be horizontal line. Let $F$ be a face supported by $l$ such that $F$ lies above $l$. If $F$ is bounded we call the two edges of $F$ incident to the highest vertex of $F$ top edges (of $F$). If $F$ is unbounded, then the unbounded edge(s) of $F$ are top edges. The other edges of $F$ are divided into left edges and right edges (of $F$) according to whether they are in the left or right component of the boundary of $F$ minus $l$ and the top edges of $F$. If $F$ lies below $l$ we analogously define two bottom edges as well as left and right edges (of $F$). Note that every line in the arrangement may contain at most one left edge and at most one right edge of some faces in the zone of $l$ that lie above $l$ and similarly at most one left edge and at most one right edge of some faces in the zone of $l$ below $l$. This count gives the upper bound of $10n+2$ ($2n$ top edges, $2n$ bottom edge, $2n$ right edges, $2n$ left edges, and $2(n+1)$ edges on $l$).

This problem is appeared in the paper <The Power of Geometric Duality>, written by B. Chazelle, L. J. Guibas, D. T. Lee in 1985. It used the same idea as @VicKmath7 used in #6, so we can prove our 239MO problen by using the idea in the paper below; <The Power of Geometric Duality> Actually the bound $10n^2$ is not tight. We can prove the bound $9.5n^2-12.5n$ by using the result in the paper <The Zone Theorem Revisited>, written by Rom Pinchasi in 2011. This is also the paper @huricane mentioned in #2. You can find it's google drive file in his homepage. <The Zone Theorem Revisited> In this paper, Rom extended the result of Chazelle et al. and proved that for $n+1$ lines, the sum of the ranks of the regions, bounded by some line $L$ is less or equal to $9.5n-3$. He also proved that this bound is tight. So for $n$ lines it's bounded by $9.5n-12.5$, and we can achieve the extended bound $9.5n^2-12.5n$ directly by using the argument @huricane mentioned in #2.

This result is called Zone theorem. Seems like commonly known, important topic in computational geometry.