# Trisecting $3$-fold sumsets: is the middle part always thick?

Here is a truly minimalistic and seemingly basic question which should have a simple solution (I hope it does).

Let $$A$$ be a finite set of integers with the smallest element $$0$$ and the largest element $$l$$. The sumset $$C:=3A$$ resides in the interval $$[0,3l]$$, and we let $$C_1:=C\cap[0,l]$$, $$C_2:=C\cap[l,2l]$$, and $$C_3:=C\cap[2l,3l]$$.

Is it true that $$|C_2|\ge\frac12\,(|C_1|+|C_3|)$$, for any choice of the set $$A$$?

Computations seem to suggest that the answer is in the affirmative.

No. Take $$A = \{0,1,\ldots,9,10,20,30,\ldots,90,100,200,300,\ldots,900,1000\}$$. Then $$|C_1|=1001$$, $$|C_2|=272$$ and $$|C_3|=29$$.

A smaller counterexample in the same spirit is $$\{0,1,2,3,4,5,10,15,20,25,50,75,100\}$$, with sizes $$101, 53, 13$$.

• Great! Do you have a counterexample for $|C_2|\ge\min\{|C_1|,|C_3|\}$?
– Seva
May 29, 2021 at 17:41
• Not at the moment, no. For all I know your milder inequality might be true. May 29, 2021 at 18:06
• Just for the record, a counterexample to the "weak" version: $A=\{0, 1, 3, 4, 12, 15, 18, 19, 20\}$. Here $|C_1|=|C_3|=18$, but $|C_2|=17$.
– Seva
May 30, 2021 at 6:27
• Mysteriously, I count sizes 21, 20, 21 from your counterexample, but yes, it is still a counterexample to the weak version. May 30, 2021 at 9:03
• You are right, 21/20/21. Interestingly, all examples that I found satisfy $|C_2|=|C_1|-1=|C_3|-1$. I wonder whether the weak inequality holds if we redefine $C_1:=C\cap[0,l-1]$ and $C_3:=C\cap[2l+1,3l]$.
– Seva
May 30, 2021 at 9:43