Translated Bohr neighborhood in a triple sumset ($t_0+B(\Gamma ,\alpha /4)\subseteq A+A+A,|\Gamma |\le 64{\alpha }^{-3}$)

For a subset $A$ of density $\alpha$ in a finite cyclic group, put $F=1_A\ast 1_A\ast 1_A$ using normalized convolution on a finite group. Its average is ${\alpha }^3$, so choose $t_0$ with $F(t_0)\ge {\alpha }^3$. Retain the large spectrum where $|\widehat{1_A}|\ge {\alpha }^2/8$. The Parseval identity on a finite group gives at most $64{\alpha }^{-3}$ frequencies. Their phases change by at most $\alpha /4$ on the associated Bohr set, contributing at most ${\alpha }^3/4$ to the change of $F$. The discarded frequencies contribute at most another ${\alpha }^3/4$, since $\sum |\widehat{1_A}{|}^2=\alpha$. Therefore $F(t_0+b)\ge {\alpha }^3/2>0$ on the Bohr set, proving the containment.

For fixed positive density $\delta$, every sufficiently large $A\subseteq [1,N]$ of size at least $\delta N$ has a long arithmetic progression in its triple sumset. Embed it in $\mathbb{Z}/(8N)\mathbb{Z}$, where its density is at least $\delta /8$. A translated Bohr neighborhood in a triple sumset has bounded rank and fixed positive radius depending only on $\delta$. A torus pigeonhole principle argument supplies a nonzero step whose small multiples stay in the Bohr set, with polynomially many multiples. All resulting residues have representatives in $[3,3N]$, an interval shorter than half the modulus, so their successive ordinary integer differences agree and give an actual integer arithmetic progression. The positive constant prefactor is absorbed by reducing the exponent for sufficiently large $N$.