Let $A$ be a finite set in an abelian group. A rich seam of questions in additive combinatorics concerns the relationship between the sizes of $A$ under simple arithmetic operations. For example, consider the doubling constant $\sigma=\abs{A+A}/\abs{A}$ and the difference constant is $\delta=\abs{A-A}/\abs{A}$. Ruzsa [Ru96] proved that $\delta\leq \sigma^2$. The exponent $2$ here is the best possible; this is demonstrated by taking $A$ to be the lattice points in a $d$-dimensional simplex, whence $\sigma\approx 2^d$ and $\delta\approx \binom{2d}{d}$. (This example originated in work of Freiman and Pigaev [FrPi73].)
The Plünnecke-Ruzsa inequalities imply a similar converse inequality $\sigma\leq \delta^2$. Despite a wealth of examples and constructions for similar-looking inequalities (many of them by Ruzsa, see e.g. Chapter 2 of his lecture notes), it was unknown whether $\sigma\leq \delta^c$ for some $c<2$ is possible.
The new construction of Lin, Li, and the Hyra AI research model [LiLi26] proves that the exponent cannot be improved past $2$. The previous record for this problem was achieved by Penman and Wells [PeWe13], who constructed arbitrarily large $A$ for which
\[\sigma\geq \delta^{1.0305\cdots}.\]
This was done by producing a single finite example (of size $67$) and using the tensor power trick of replacing $A\subseteq G$ by $A^d\subseteq G^d$ and noting that the quantities $\sigma$ and $\delta$ are replaced by $\sigma^d$ and $\delta^d$.
For arbitrarily large $K$ there exists $A$ (a finite set in some abelian group) such that $\abs{A+A}\gg K\abs{A}$ and $\abs{A-A} \ll K^{1/2}\abs{A}$.
The description in [LiLi26] is longer than it needs to be (in part because of their desire to construct an example in $\mathbb{Z}$, which is redundant via the machinery of Freiman isomorphisms, and in part because of the tracking of explicit constants, which is not needed). I'll present a simplified digest of the idea. Roughly speaking, the largeness of $A+A$ comes from the largeness of $S+S$, where $S$ is a Sidon set. To prevent $A-A$ also blowing up we 'twist' $S$ with a set $B$ where $\abs{B+B}$ is much larger than $\abs{B-B}$. Finally, since we are concerned not just with $\abs{A+A}$ and $\abs{A-A}$ but their size relative to $\abs{A}$, we throw in some large group component to make sure $\abs{A}$ grows suitably also.
Let $G$ and $H$ be two abelian groups, of sizes $L$ and $K$ respectively. Let $S\subset H$ be a Sidon set of size $\asymp K^{1/2}$ (this is easily constructed in a number of ways). Let $B\subseteq G$ be a set to be chosen later, and let
\[A=(G\times \{0\})\cup (B\times S).\]
The size, doubling constant, and difference constant of $A$ are easily estimated:
\[\lvert A\rvert\asymp L+\lvert B\rvert K^{1/2}\asymp L,\]
say, provided $\lvert B\rvert\ll L/K^{1/2}$. Secondly, since $\lvert S+S\rvert \gg K$,
\[\frac{\lvert A+A\rvert}{\lvert A\rvert}\gg \frac{\lvert B+B\rvert\lvert S+S\rvert}{L}\asymp K,\]
provided $\abs{B+B}\gg L$. Finally, using the trivial $\abs{S-S}\leq K$,
\[\frac{\lvert A-A\rvert}{\lvert A\rvert} \ll \frac{L+\lvert B-B\rvert K+L\lvert S\rvert}{\lvert A\rvert}\asymp K^{1/2}\]
provided $\lvert B-B\rvert \ll LK^{-1/2}$.
This set $A$ therefore provides the required example, as long as we can find some suitable $B\subseteq G$ with $\abs{B+B}\gg \lvert G\rvert$ and $\lvert B-B\rvert\ll \abs{G}K^{-1/2}$ (note the second condition automatically implies $\abs{B}\ll \abs{G}K^{-1/2}$ also). This can be provided by taking any fixed candidate $B'\subseteq G'$ in which $B'+B'=G'$ and $B'-B'\neq G'$ and blowing up using the tensor power trick, so taking $B=(B')^d$. Then $\abs{G}=\lvert G'\rvert^d$ and $\lvert B-B\rvert=(\frac{\lvert B'-B'\rvert}{\lvert B'+B'\rvert})^d\abs{G}\asymp K^{-1/2}\abs{G}$, say, and $K$ can be made arbitrarily large by taking $d\to \infty$. Examples of such $B'$ can be found via computation; [LiLi26] uses the example
\[B=\{0,1,2,4,5,9\}\subseteq \bbz/12\bbz.\]
Note that one could try to use $B$ itself to get a construction with large doubling constant and small difference constant; the advantage of the 'twisted' construction given above is that we don't need to worry at all about what the size of $\lvert B\rvert$ is!
This same idea can be used more widely: in general, let $B\subseteq G$ and $S\subset H$ be any set of size $K$ which is dissociated. Let $l_1,l_2,k_1,k_2\in \bbz$ (not both zero). Let $A$ be defined as above with $\abs{A}\asymp L=\abs{G}$, valid provided $\abs{B}\ll L/K$. Since $S$ is dissociated, $\abs{l_1S-k_1S}\gg K^{l_1+k_1}$ and so, by the calculation above,
\[\frac{\abs{l_1A-k_1A}}{\abs{A}}\gg K^{k_1+l_1}\]
provided $\abs{l_1B-k_1B}\gg L$, and similarly
\[\frac{\abs{l_2A-k_2A}}{\abs{A}}\ll K^{k_2+l_2-1}\]
provided $\abs{l_2B-k_2B}\ll L/K$. Constructing $B$ as above, by taking any fixed example and taking arbitrarily large powers, yields the following.
Let $l_1,l_2,k_1,k_2\in \bbz$ be such that there exists some finite $B\subseteq G$ where $l_1B-k_1B=G$ and $l_2B-k_2B\neq G$. Then for arbitrarily large $K$ there exists $A$ (a finite set in some abelian group) such that
\[\abs{l_1A-k_1A}\gg K^{k_1+l_1}\abs{A}\]
and
\[\abs{l_2A-k_2A} \ll K^{k_2+l_2-1}\abs{A}.\]
For example, if we choose $B=\{0,1,3\}\subset \bbf_7$ then $B-B=\bbf_7$ but $B+B=\{0,1,2,3,4,6\}$, so this construction yields arbitrarily large $K$ with associated $A$ for which $\abs{A-A}\gg K^2\abs{A}$ and $\abs{A+A}\ll K\abs{A}$. This provides another construction which shows that $\delta \ll \sigma^2$ cannot be improved -- in fact this is superior to the simplex construction, which actually has $\delta \gg \frac{\sigma^2}{\sqrt{\log \sigma}}$ (rather than $\delta \gg \sigma^2$ as in this construction). This therefore answers (in the negative) a question of Ruzsa which asked whether $\delta \leq \sigma^2$ can be improved by some factor of the shape $(\log \sigma)^c$.
As another example, if one takes $B=\{0,1,2\}\subset \bbf_7$ then $B+B+B=\bbf_7$ but $B+B=\{0,1,2,3,4\}$, and so this construction yields arbitrarily large $K$ with associated $A$ for which
\[\abs{A+A+A}\gg K^3\abs{A}\]
and
\[\abs{A+A}\ll K\abs{A}.\]
Sets with this property were already constructed by Ruzsa, but this offers an alternative (and perhaps simpler) construction.
This raises the interesting question of characterising those quadruples $l_1,l_2,k_1,k_2\in \bbz$ for which there exists some finite $B\subseteq G$ where $l_1B-k_1B=G$ and $l_2B-k_2B\neq G$. This is clearly impossible if $l_2\geq l_1$ and $k_2\geq k_1$; this is probably the only obstruction. It is easy to construct such sets if $k_1+l_1>k_2+l_2$.
EDIT: In fact recent work by Kravitz [Kr26] proves that such quadruples are precisely those for which $\min(k_2,l_2)>\min(k_1,l_1)$ or $\max(k_2,l_2)>\max(k_1,l_2)$.
G. Freiman and V. P. Pigaev, The relation between the invariants $r$ and $t$, Kalinin. Gos. Univ. Moscow (1973), 172–174.
N. Kravitz, Inequalities among higher-order difference sets, or, remarks on a construction of Ruzsa, arXiv 2606.27087 (2026).
H. Lin and S. Li, Settling the optimal exponent relating sumsets and difference sets, arXiv 2607.27199 (2026).
D. Penman and M. Wells, On sets with more restricted sums than differences, Integers 13 (2013).
I. Ruzsa, Sums of finite sets. In Number Theory: New York Seminar, D.V. Chudnovsky, G.V. Chudnovsky and M.B. Nathanson (eds), Springer-Verlag, (1996), 281–293.