This is my first time asking a question on this site, as I believe my question is probably related to computer science (and possibly to the analysis of Boolean functions), and someone here might be able to help me.
Consider the group $\mathbb{Z}_2^n$ equipped with the following dot product: for $a,b\in \mathbb{Z}_2^n$, define $a\cdot b := a_1b_1+\dots+a_nb_n$, where $a=(a_1,\dots,a_n)$ and $b=(b_1,\dots,b_n)$.
Let $A \subseteq \mathbb{Z}_2^n$ with $|A| = a$, where $2^{n-1} < a < 2^n$. Define $r(A) := \left|\{(x,y,z) \in A^3 : x+y = z\}\right|$. Using discrete Fourier analysis, one can show that $r(A)$ can be written as the following character sum: $$ r(A) = \frac{1}{2^n} \sum_{\chi\in \widehat{\mathbb{Z}_2^n}} \sum_{x,y,z \in A} \chi(x+y+z) = \frac{1}{2^n} \sum_{\varepsilon \in \{0,1\}^n} \sum_{x,y,z \in A} (-1)^{\varepsilon \cdot (x+y+z)}. $$
Define $$S(A):=\sum\limits_{\varepsilon \in \{0,1\}^n} \sum\limits_{x,y,z \in A} (-1)^{\varepsilon \cdot (x+y+z)}.$$ My goal is to prove the following lower bound for $S(A)$: $$S(A)\geq 2^n\Big(2^n-2^{1+\lfloor \log_2(2^n-a)\rfloor}\Big)\Big(3a-2^{n+1}+2^{1+\lfloor \log_2(2^n-a)\rfloor}\Big).$$ Do you have any ideas on how to prove this?
