In an abelian group, is a certain inequality involving products of cardinalities of sum and differences of finite subsets (sumsets) true or not?

Question

For any two finite subsets $A,B$ of an abelian group, does the following ineqality hold?

$$|A+B|^2 |A-B|^2 \geq |A+A||A-A||B+B||B-B| \ $$

I’m interested in finding out if there are sumset inequalities that are sharper than the triangle inequalities and that show that some quantity related to sums and differences is smaller when the sums and differences are over the same sets.

Smaller examples don’t seem to work: $|A-B|^2 \geq |A-A||B-B|$ fails when $B = -A$ and $|A+A| < |A-A|$.

$|A+B|^2 \geq |A+A||B+B|$ doesn’t work when $B = -A$ and $|A-A| < |A+A|$.

The candidate given in the title combines the two and avoids these difficulties because of its symmetry. It also holds when $A$ is an arithmetic progression or a subspace (if such finite subspaces exist) and B is a random set.

Answer 1

I have been looking for possible counter-examples in $\left(\mathbb{Z}/n \mathbb{Z},+\right)$ for different values of $n$ and different subsets $A$ and $B$ using a SAGE program that I give below.

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I have found one modulo 7 with

$A = [0, 1, 3]$ and $B = [0, 3, 4, 5]$.

Indeed :

$A+B = [0, 1, 3, 4, 5, 6]$

$A+A = [0, 1, 2, 3, 4, 6]$

$A-B = A-A = B+B = B-B = [0, 1, 2, 3, 4, 5, 6]$

giving a LHS $(6 \times 7)^2$ less that the RHS $6 \times 7 \times 7\times 7$ in your formula.

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I have found another counterexample modulo 8 :

with $A = [0, 1, 2, 5]$ and $B = [0, 4, 5, 7]$.

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... and one modulo $12$ :

with $A = [0, 1, 3, 6, 9, 10, 11]$ and $B = [1, 7, 8, 10, 11]$.

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and also mod 14 with :

$A=[2,4,5,9,10,12]$ and $B=[0,3,5,6,7,9,12]$.

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Besides, your inequality is verified in almost all cases with, sometimes, an equality between the LHS and RHS.

Here is the SAGE program :

    from sage.misc.prandom import randrange
    n=6
    R=Integers(n); # integers mod n
    Li=list(R)
    X=Set([0,1,2,3,4,5]); # n elements  
    
    Y=X.subsets() # list of all the subsets of X
    ne=Y.cardinality()  # ne=2^n

    def selec(m) : # getting the m-th subset
        A=[];
        for x in Y[m] :
            A.append(Li[x])
        return list(A)

    def ol(L) : # opposite of a list
       mL=[] 
       for k in Li :
          mL.append(-k)
       return(mL)    
    
    def cs(A,B) : cardinality of the sum of 2 subsets 
       C=[]
       for a in A :
          for b in B :
             c=a+b
             if not(c in C) :
                C.append(c)
       return len(C)

    # Random selection of two subsets :
    m=randrange(ne);A=selec(m);show("A = ",A) 
    m=randrange(ne);B=selec(m);show("B = ",B)
    C=[cs(A,B),cs(A,ol(B)),cs(A,A),cs(A,ol(A)),cs(B,B),cs(B,ol(B))]
    show("Cardinalities of A+B, A-B, etc. : ",C)
    r=(C[0]*C[1])**2-C[2]*C[3]*C[4]*C[5];show("Diff, between LHS and RHS = ", r)