A note I'm reading currently (the note itself deals with a novel way of estimating the number of square-free integers up to, say, $z$) uses without proof the following estimates:
(i) If $f(n)=n^2\prod_{p|n}(1-\frac{1}{p^2})$, then $\sum_{d|n}f(d)=n^2$
(ii) $\zeta(2)-\frac{1}{z}\leq \sum_{n\leq z}\frac{\mu(n)^2}{f(n)}\leq\zeta(2)$.
I can't figure out a way to obtain any of these estimates. Are they even related to each other in some way? I would appreciate any help. Thank you.
On
The first proposition can be proved using the same proof of $$\frac{\phi\left(n\right)}{n}=\prod_{p\mid n}\left(1-\frac{1}{p}\right) $$ (it can be found in the Apostol's book). Assume that $p_{1},\dots,p_{r} $ are the prime divisors of $n $. Then note that the product can be writte as $$\prod_{i=1}^{r}\left(1-\frac{1}{p_{i}^{2}}\right)=1-\sum\frac{1}{p_{i}^{2}}+\sum\frac{1}{p_{i}^{2}p_{j}^{2}}+\dots+\frac{\left(-1\right)^{r}}{p_{i}^{2}\cdots p_{r}^{2}} $$ where every sum is understood that we consider all the possibile product of distinct prime factors of $n$. So it is not difficul to see that $$ f\left(n\right)=n^{2}\prod_{i=1}^{r}\left(1-\frac{1}{p_{i}^{2}}\right)=\sum_{d\mid n}\frac{n^{2}}{d^{2}}\mu\left(d\right) $$ and so the claim $(i)$ follows using the Möbius inversion. About the proposition $(ii)$, the upper bound has been well explained by user1952009.
Update: 04/12/2017
Some ideas for the lower bound. Let $z \geq 2$ and $P\left(z\right)=\prod_{p\leq z}p$, $p$ prime number. Then we have $$\sum_{n\leq z}\frac{\mu^{2}\left(n\right)}{f\left(n\right)}=\sum_{\underset{{\scriptstyle n\mid P\left(z\right)}}{n\leq z}}\frac{1}{f\left(n\right)}\leq\sum_{n\mid P\left(z\right)}\frac{1}{f\left(n\right)}=\frac{1}{\prod_{p\leq z}\left(1-\frac{1}{p^{2}}\right)}$$ hence, taking $t>0$, we have $$\frac{1}{\prod_{p\leq z}\left(1-\frac{1}{p^{2}}\right)}-\sum_{n\leq z}\frac{\mu^{2}\left(n\right)}{f\left(n\right)}=\sum_{\underset{{\scriptstyle n\mid P\left(z\right)}}{n>z}}\frac{1}{f\left(n\right)}\leq\frac{1}{z^{t}}\sum_{\underset{{\scriptstyle n\mid P\left(z\right)}}{n>z}}\frac{n^{t}}{f\left(n\right)}$$ $$\leq \frac{1}{z^{t}}\prod_{p\leq z}\left(1+\frac{1}{p^{2-t}\left(1-1/p^{2}\right)}\right).$$ Unfortunately the choice $t=1$ doesn't work but maybe a clever choice can be done. Now I haven't much time but I will work on it. Furthermore it is possible to prove, using sieve theoretic arguments, that for any $\lambda >0$ we have $$\sum_{d\leq z}\frac{\mu^{2}\left(d\right)}{f\left(d\right)}\geq\frac{1}{\prod_{p<z}\left(1-1/p^{2}\right)\left(1+O\left(\exp\left(-\lambda+\left(\frac{C}{\lambda}+\frac{D}{\log\left(z\right)}\right)e^{\lambda}\right)\right)\right)}$$ where $C,D>0$ are suitable positive constants and $z \geq 2.$
If $h(n) = \frac{f(n)}{n^2} = \prod_{p | n} h(p)$ then $h(n)$ is multiplicative, so that $f(n)=h(n)n^2$ and $g(n) = \frac{\mu(n)^2}{f(n)}$ are multiplicative too.
Thus, since $g(p)=\frac{\mu(p)^2}{f(p)} = \frac{1}{p^2-1}$ $$\sum_{n=1}^\infty g(n)= \prod_p (1+\sum_{k \ge 1}g(p^k)) = \prod_p (1+\frac{1}{p^2-1}) = \prod_p \frac{1}{1-\frac{1}{p^2}} = \zeta(2)$$
i.e. $$\sum_{n \le x} g(n) \le \zeta(2)$$
Then for the lower bound, you have to show that $\sum_{n=k}^\infty g(n) \le \int_k^\infty \frac{dt}{t^2} = \frac{1}{k}$ $\color{red}{\text{I have no idea how}}$