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The positive-power length assertion is an asymptotic one: it requires $N$ sufficiently large in terms of the density. Read literally for every $N$, it fails when $N=2$, $\delta =1/2$ and $A=\{1\}$, whose triple sumset is a singleton. We prove the intended result with this necessary qualification and an explicit positive exponent.

Embed the interval into the cyclic group $G=\mathbb{Z}/M\mathbb{Z}$ with $M=8N$, and set $u=1_A$, $\alpha =|A|/M\ge \delta /8$. Since all integer triple sums lie in $[3,3N]$, this embedding introduces no collisions between distinct sums. Use normalized Fourier coefficients on a finite abelian group and normalized convolution on a finite group, and putThe function $F$ is nonnegative and has mean ${\alpha }^3$, so some $t_0$ satisfies $F(t_0)\ge {\alpha }^3$. We show that a whole translated Bohr set is contained in its positive support, hence in the triple sumset.

Put $\theta ={\alpha }^2/8$, $\rho =\alpha /4$, and define the large spectrumThe Parseval identity on a finite group gives ${\sum }_r|\hat{u}(r){|}^2=\alpha$, soAlso $|\hat{u}(r)|\le \alpha$, and therefore ${\sum }_r|\hat{u}(r){|}^3\le {\alpha }^2$. Using Fourier inversion and the convolution theorem on a finite group, for $b\in B$ we obtainConsequently $F(t_0+b)\ge {\alpha }^3/2>0$. This proves the translated Bohr neighborhood in a triple sumset:It remains to produce a long ordinary integer arithmetic progression in this translate; we supply the simultaneous approximation and the lifting argument. SetThus $d=|\Gamma |\le D$. Partition the $d$-dimensional unit cube into $Q^d$ boxes of side $1/Q$. The $Q^d+1$ pointscontain two in the same box by the pigeonhole principle. Their index difference gives an integer $q$ with $1\le q\le Q^d<M$ andIn particular $|e^{2\pi irq/M}-1|\le 2\pi /Q$. The identity $z^j-1=(z-1)(1+\cdots +z^{j-1})$ on the unit circle now shows thatAll residues $t_0+jq$ therefore lie in the triple sumset. Let $s_j\in [3,3N]$ be their actual integer representatives. Their successive differences are congruent to $q$ modulo $M$, and each lies in $[-(3N-3),3N-3]$, which is contained in $(-M/2,M/2)$. There is at most one integer in this interval in any specified residue class, so every difference $s_{j+1}-s_j$ is the same integer. It is nonzero because $0<q<M$. Thus the $s_j$ form an ordinary integer arithmetic progression of distinct elements, even though $M$ is composite. Reversing it if necessary makes the common difference positive.

For sufficiently large $N$, we have $Q\ge \frac{1}{2}{M}^{1/(D+1)}$ and $\rho \ge \delta /32$, whenceAbsorb this fixed prefactor by decreasing the exponent: withthe last lower bound is at least $N^{r_{\delta }}$ once $N$ is sufficiently large in terms of $\delta$. This proves a polynomial-length progression in $A+A+A$ by the polynomial-length progression in a dense triple sumset argument, without replacing the triple sumset by a fourfold difference set.