In the chapter on sieve methods, specifically the $\Lambda^{2}$-sieve, Iwaniec-Kowalski state that $$\sum_{d < \sqrt{D}}^{\flat} h(d) > \prod_{p | q} (1-g(p)) \log \sqrt{D},$$ where $h(d) = \prod_{p | d} g(p)/(1-g(p))$ for $d$ squarefree and $g(p) = 1/p$ for $p \nmid q$. I tried several approaches but could not justify this inequality. Instead, I obtained a rather different lower bound.
What I did is the following: Note that each squarefree $d$ can be written uniquely as $d = u v$, where $u \mid q$, $(v,q) = 1$, and $u,v$ are squarefree. Then we have $$\sum_{d < \sqrt{D}}^{\flat} h(d) = \sum_{u | q} \mu(u)^{2}h(u)\sum_{\substack{v < \sqrt{D}/u\\ (v,q) = 1}}^{\flat} \frac{1}{\varphi(v)} = \sum_{\substack{v < \sqrt{D}\\(v,q) = 1}}^{\flat}\frac{1}{\varphi(v)} \sum_{\substack{u | q\\u < \sqrt{D}/v}} \mu^{2}(u) h(u).$$ Next, restrict to $v < \sqrt{D}/q$, so that $q < \sqrt{D}/v$ and hence the condition $u < \sqrt{D}/v$ is satisfied for all $u \mid q$. The inner sum then becomes a clean convolution, which simplifies to $\prod_{p | q} (1-g(p))^{-1}$. Thus, we conclude $$\ge \prod_{p | q} (1-g(p))^{-1}\sum_{\substack{v < \sqrt{D}/q\\(v,q) = 1}}^{\flat}\frac{1}{\varphi(v)} > \prod_{p | q} (1-g(p))^{-1}\frac{\varphi(q)}{q}\log \sqrt{D}/q.$$ This seems to be the strongest bound I can obtain, assuming there are no mistakes above. Does anyone know how to recover the bound stated in the book?
