TITLE: Is the set $ AA+A $ always at least as large as $ A+A $?
QUESTION [111 upvotes]: Let $A$ be a finite set of real numbers. Is it always the case that $|AA+A| \geq |A+A|$?
My first instinct is that this is obviously true, and there is a one-line proof which I am foolishly overlooking. Can anyone provide one? Of course, any proof would be welcome! Any partial results would also be of interest.

REPLY [16 votes]: I believe there is an "energy" version of the conjectural inequality $|A+AA| \geq |A+A|$ which may explain why it was intuitive that there should be an "easy" proof of that inequality.  Namely:

Proposition Let $A$ be a finite collection of nonzero elements of a field $F$.  Let $a_1,a_2,a_3,a'_1,a'_2,a'_3$ be chosen uniformly and independently from $A$.  Then $$ {\mathbf P}( a_1 + a_2 a_3 = a'_1 + a'_2 a'_3 ) \leq {\mathbf P}( a_1 + a_2 = a'_1 + a'_2 ).$$

Informally, this asserts that $A+AA$ is "flatter" than $A+A$ in an $L^2$ sense, which leans toward $A+AA$ being larger in size than $A+A$, but does not imply it (as my counterexample in my other response shows).
The proof is basically Cauchy-Schwarz.  If one defines $E(A,B;C,D)$ to be the number of quadruples $a \in A, b \in B, c \in C, d \in D$ with $a+b=c+d$, then two applications of Cauchy-Schwarz give
$$ E(A,B;C,D) \leq E(A,A;A,A)^{1/4} E(B,B;B,B)^{1/4} E(C,C;C,C)^{1/4} E(D,D;D,D)^{1/4}$$
which imply in particular that
$$ {\mathbf P}( a_1 + a_2 b = a'_1 + a'_2 c ) \leq {\mathbf P}( a_1 + a_2 = a'_1 + a'_2 )$$
for any non-zero deterministic $b,c$.  Replacing $b,c$ by $a_3, a'_3$ and then taking expectations we obtain the claim.