= Solution
The <Bourgain bound for three-term-progression-free sets> states that, if $N=|G|$ is odd and $A\subseteq G$ contains no nonconstant three-term <arithmetic progression>, then
$$
|A|\ll N\left(\frac{\log\log N}{\log N}\right)^{1/2}.
$$
Here is how it follows from the standard <Bohr-set density increment lemma>. Begin with $B_0=G$ and relative density $\alpha_0=\alpha=|A|/N$. Whenever the lemma gives its second alternative, replace the current set by the denser translate inside the smaller regular Bohr set. The density changes by
$$
\alpha_{i+1}\geq(1+c\alpha_i)\alpha_i,
$$
so this can happen only $O(\alpha^{-1})$ times. Throughout the iteration the rank is $O(\alpha^{-1})$, the width is at least $\alpha^{O(\alpha^{-1})}$, and the elementary lower bound for the size of a Bohr set gives
$$
|B_i|\geq\alpha^{O(\alpha^{-2})}N.
$$
At the terminal stage the first alternative of the density-increment lemma holds. Combining it with the last display yields
$$
\alpha^{-2}\log(1/\alpha)\ll\log N.
$$
If $\alpha<1/\log N$, Bourgain's bound is already true. Otherwise $\log(1/\alpha)\ll\log\log N$, and rearranging proves
$$
\alpha\ll\left(\frac{\log\log N}{\log N}\right)^{1/2}.
$$
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