An improved bound on $(A+A)/(A+A)$

We show that, for a finite set $A$ of real numbers, the size of the set $$\frac{A+A}{A+A} = \left\{ \frac{a+b}{c+d} : a,b,c,d \in A, c+d \neq 0 \right \}$$ is bounded from below by $$\left|\frac{A+A}{A+A} \right| \gg \frac{|A|^{2+1/4}}{|A / A|^{1/8} \log |A|}.$$ This improves a result of Roche-Newton.

In this paper we study the related question of establishing lower bounds on the size of the set In [2], Balog and Roche-Newton showed that, if A is a set of strictly positive reals, This is completely sharp, as shown by the set A = {1, 2, 3}.
In [8], Roche-Newton and Zhelezov conjectured that Shkredov [9] has made some progress in this direction, showing that In [7], Roche-Newton proved that which implies that It is known that (4) is implied by (1), for example by work of Li and Shen [5], but the reverse implication does not hold (even if the extra log −25 |A| term were removed). This paper gives the following improvement to (3) and (4).

Theorem 1. Let A be a finite set of real numbers. Then
Consequently, In broad outline, the proof of Theorem 1 is similar to the proof by Roche-Newton of (3). We combine ideas of Solymosi [10] and Konyagin and Shkredov [4], developed in work on the Erdős-Szemerédi sum-product conjecture, with an incidence bound and a probabilitistc argument. The key difference between this proof and that of Roche-Newton is that we apply the probabilistic method in a more flexible way, which leads to a simpler proof of a stronger result.

Acknowledgments
I would like to thank Abdul Basit, Oliver Roche-Newton, and Adam Sheffer for useful conversations related to the material in this paper.

Proof of Theorem 1
The remainder of the paper is dedicated to the proof of inequality (5) Theorem 1; given (5), it is trivial to derive (6).
In section 2.1 we describe the general setup for the proof and fix notation. In section 2.2, we give the general idea of the proof, and briefly elaborate on the key difference between this proof and that of Roche-Newton [7]. Sections 2.3, 2.4, and 2.5 comprise the main body of the proof.

Setup
We assume that all of the elements of A are strictly positive. This is without loss of generality; if at least half of the elements of A are positive, we consider these elements; otherwise, we multiply by −1 and then consider the positive elements.
Using a dyadic pigeonholing argument, we find a set P ⊂ A × A and a number |A| 2 / (2|A/A|) ≤ τ ≤ |A| such that P is contained in the union of lines through the origin (in R 2 ) that each contain exactly τ points, and In more detail, A × A is contained in the union of |A/A| lines through the origin, one for each ratio in A/A. Note that no more than |A| 2 /2 points in A × A are contained in lines through the origin that each contain fewer than τ 0 = |A| 2 / (2|A/A|) points. Note also that no line contains more than |A| points of A × A. We partition the numbers [τ 0 , |A|] into O(log |A|) dyadic intervals of the form [2 i , 2 i+1 ), and use the pigeonhole principle to show that there is a τ such that there are at least Ω(|A| 2 / log |A|) points (a, b) ∈ A × A that lie on lines through the origin that each contain between τ and 2τ points. We then choose τ arbitrary points on each of these lines, and define the resulting point set to be P .
, it suffices to show that, for the set S of lines through the origin that each contain at least one point of P + P , we have Once (8) is demonstrated, the proof will be complete.
. . λ |Λ| be the set of slopes of lines through the origin that each contain τ points of P . Note that Let M be an integer parameter that we will set later. For each 1 ≤ i ≤ ⌊|Λ|/2M⌋, let For the remainder of the proof, we work with an arbitrary i and set T = T i and U = U i . We relabel λ f i +j as λ j , so that λ 1 , λ 2 , . . . , λ M ∈ T and λ M +1 , λ M +2 , . . . , λ 2M ∈ U. With this relabeling, let P i be the set of points of P contained in the line with slope λ i .

Brief overview
The basic idea of the proof is to show that we can select points a ij ∈ P i for 1 ≤ i ≤ M and M + 1 ≤ j ≤ 2M so that the union of the sets a ij + P j determines many slopes. There are two basic parts of the proof. First, we use a proof based on a geometric incidence bound to show that, for fixed i, j, k, ℓ with (i, j) = (k, ℓ), there are many selections of a ij , a kℓ such that a ij + P j shares few slopes with a kℓ + P ℓ . Second, we combine this with a probabilistic argument to show that we can select the a ij so that the total number of slopes is large.
The key difference between this proof and the proof of Roche-Newton in [7] is that we choose a point a ij for each pair in T × U, where Roche-Newton chose a single representative for each line in T .

Bounding shared slopes
For a point p ∈ R 2 , denote by r(p) the slope of the line passing through p and the origin. For a i ∈ P i and a k ∈ P k , let The proof of the following lemma is based closely on the work of Roche-Newton [7].
, with at least one of i = k and j = ℓ. The number of pairs (a i , a k ) ∈ P i × P k such that E(a i , P j , a k , P ℓ ) ≥ K is bounded from above by Proof. For each pair of points (a, λ i a), (b, λ k b) in P i × P k , we define the curve For any set Q of points, denote by π x (Q) the projection of Q onto the x-axis; in other words, π x (Q) = {a : ∃y((a, y) ∈ Q)}.
Note that (x, y) ∈ l ab is equivalent to r((a+x, λ i a+λ j x)) = r((b+y, λ k b+ λ ℓ y)). Hence, to prove the lemma, it suffices to show that there are no more than τ 4 /K 3 + τ 2 /K curves from L that each contain at least K points of P.
To this end, we use the following result of Pach and Sharir [6].
Theorem 3. Let L be a family of curves and let P be a set of points in the plane such that 1. any two distinct curves from L intersect in at most two points of P, 2. for any two distinct points p, q ∈ P, there exist at most two curves L which pass through both p and q. Then, for any k ≥ 2, the set L k ⊂ L of curves that contain at least k points of P satisfies the bound Since |P| = τ 2 , Lemma 2 will follow directly from this theorem. It remains to show that L satisfies the hypotheses of Theorem 3.
We will first check that two distinct curves of L intersect in at most two points.
Hence, we conclude From equation (17), we get the quadratic equation Either there are at most two values of y which give a solution to this quadratic, or all of the coefficients are zero. Suppose that we are in this degenerate case. If δ 1 = 0, then the fact that the coefficient of y 2 is zero implies that a = a ′ . Otherwise, δ 4 = 0, and so the fact that the constant term is zero implies that a = a ′ . Combining the fact that a ′ = a with the fact that the linear term is zero, we conclude that δ 2 δ 3 − δ 1 δ 4 = 0, which contradicts (14).
Hence, l ab and l a ′ b ′ intersect in at most two points. Now, we show that there are at most two curves of L that pass through any pair of points in P.
If l xy ∈ L is a curve that that passes through (a, b) and (a ′ , b ′ ), then Rearrange these equations into the form These equations are the same as equations (15) and (16) with δ 4 interchanged with δ 1 , and δ 3 interchanged with δ 2 . Since the conditions on these quantities are symmetric with regard to the pairs δ 1 , δ 4 and δ 2 , δ 3 , we apply the previous argument to show that there are at most two solutions to this system of equations.

Choosing representatives
For each pair (i, j) with 1 ≤ i ≤ M and M + 1 ≤ j ≤ 2M, choose a ij ∈ P i uniformly at random. Note that we have chosen M 2 points. For each pair a ij , a kℓ of chosen points, let X(i, j, k, ℓ) be the event that where B is a parameter that we will fix later. Applying Lemma 2, we find that, for each quadruple (i, j, k, ℓ) with at least one of i = j and k = ℓ, we have In addition, note that X(i, j, k, ℓ) depends on X(i ′ , j ′ , k ′ , ℓ ′ ) only if either (i, j) = (i ′ , j ′ ) or (k, ℓ) = (k ′ , ℓ ′ ). Hence, X(i, j, k, ℓ) is independent of all but at most 2M 2 other events.
We apply the following version of the Lovász Local Lemma to bound the probability that at least one of the events occurs -see Corollary 5.1.2 in [1].
Theorem 4. Let X 1 , X 2 , . . . , X n be events in an arbitrary probability space. Suppose that each event X i is mutually independent from all but at most d of the events X j , with i = j. Suppose also that the probability of the event X i occuring is at most p, for all 1 ≤ i ≤ n. Finally suppose that Then, with positive probability, none of the events X 1 , X 2 , . . . , X n occur.
Using (23), we apply this lemma with p = c 1 (τ 2 B −3 + B −1 ) and d = 2M 2 , and conclude that, as long as there is a positive probability that none of the events X(i, j, k, ℓ) occur.

Combining the bounds
Let Q be the union of a ij + P j over all the a ij . Note that Q is the union of M 2 distinct sets, each containing τ points. Let R be the slopes of lines through points of Q. By inclusion-exclusion, |R| ≥ M 2 τ − E(a ij , P j , a kℓ , P k ), where the sum is over all 1 ≤ i, k ≤ M and M + 1 ≤ j, ℓ ≤ 2M such that at least one of i = k and j = ℓ holds. Since each E(a ij , P j , a kℓ , P ℓ ) < B, we have |R| ≥ M 2 τ − BM 4 .
It is possible to satisfy this constraint with M = c 2 τ 1/8 , for an appropriate choice of c 2 . Note that the sets R ⊂ S obtained for different choices of T, U are disjoint. Since there are ⌊|Λ|/2M⌋ choices for T and U, we have by equation 25 that the number of lines through the origin that contain some point of P + P is at least |S| = |R| ⌊|Λ|/2M⌋ ≫ Mτ |Λ| ≫ τ 1+1/8 |Λ|.
Combining this with equation (7) and the observation that τ ≥ |A| 2 /|A/A|, we have inequality (8), and the proof is complete.