This question is from Assignment 3 of following notes of Zeev Rudnick:http://www.math.tau.ac.il/~rudnick/courses/sieves2015.html
Let $f(n)=n^2$. (a) Show that $(f\star \mu)(n)=n^2 \prod_{p|n} (1-\frac{1}{p^2})$
(b) Show that $\sum_{n \leq z} \frac{\mu(n)^2} { (\mu \star f)(n)} \geq \zeta(2) +O(1/z)$.
I have done (a). But I am note able make any progress for (b).The LHS of (b) is S(z) in Lecture 10( Selberg Sieve).
Can you please help with this?
\begin{aligned} \sum_{n\le z}{\mu^2(n)\over(f*\mu)(n)} &=\sum_{n\le z}{\mu^2(n)\over n^2}\prod_{p|n}{1\over1-p^{-2}}=\sum_{n\le z}\mu^2(n)\prod_{p|n}{p^{-2}\over1-p^{-2}} \\ &=\sum_{n\le z}\mu^2(n)\prod_{p|n}\sum_{r\ge1}{1\over p^{2r}}=\sum_{n\le z}\mu^2(n)\sum_{\substack{m\ge1\\p|m\Rightarrow p|n}}{1\over m^2} \\ &=\sum_{m\ge1}{1\over m^2}\sum_{\substack{n\le z\\p|m\Rightarrow p|n}}\mu^2(n)=\sum_{m\le1}{1\over m^2}|S_{z,m}| \end{aligned}
where $|S_{z,m}|$ denotes the cardinality of the set
$$ S_{z,m}=\{n\le z:n\text{ squarefree}\wedge p|m\Rightarrow p|n\} $$
Since $\prod_{p|m}p\in S_{z,m}$ whenever $m\le z$, we conclude the last sum is bounded below by
$$ \sum_{n\le z}{\mu^2(n)\over(f*\mu)(n)}\ge\sum_{m\le z}{1\over m^2}=\zeta(2)+\mathcal O\left(\frac1z\right) $$
Hope this will address your concern!