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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.

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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 $101, 53, 13$.

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  • $\begingroup$ Great! Do you have a counterexample for $|C_2|\ge\min\{|C_1|,|C_3|\}$? $\endgroup$
    – Seva
    Commented May 29, 2021 at 17:41
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    $\begingroup$ Not at the moment, no. For all I know your milder inequality might be true. $\endgroup$
    – Jukka Kohonen
    Commented May 29, 2021 at 18:06
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    $\begingroup$ 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 $|C_2|=17$. $\endgroup$
    – Seva
    Commented May 30, 2021 at 6:27
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    $\begingroup$ Mysteriously, I count sizes 21, 20, 21 from your counterexample, but yes, it is still a counterexample to the weak version. $\endgroup$
    – Jukka Kohonen
    Commented May 30, 2021 at 9:03
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    $\begingroup$ 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 $C_3:=C\cap[2l+1,3l]$. $\endgroup$
    – Seva
    Commented May 30, 2021 at 9:43

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