I've been trying to prove the following identity which has been used in a paper I'm currently reading:
$\displaystyle\sum_{q\leq Q}\frac{\mu(q)^2}{\phi(q)}\leq \frac{k}{\phi(k)}\displaystyle\sum_{q\leq Q, (q,k)=1}\frac{\mu(q)^2}{\phi(q)}$.
I tried to split the sum as $\displaystyle\sum_{q\leq Q}\frac{\mu(q)^2}{\phi(q)}=\displaystyle\sum_{r|k}\displaystyle\sum_{m\leq\frac{Q}{r},(m,\frac{k}{r}=1)}\frac{\mu(mr)^2}{\phi(mr)}$ but cannot proceed. Can anyone help?
Here $\mu$ and $\phi$ are the Mobius function and Euler's totient function respectively.
Hint1: any $q\leq Q$ is a product $mn$, where all prime factors of $m$ consists of prime factors of $k$, and $(n,k)=1$.
Hint2: note that $q\leq Q$ must be square-free because of $\mu(q)^2$. For $q=mn$ as in Hint1, we have $m|k$ and $\phi(q)=\phi(m)\phi(n)$
Applying the hints, we have $$ \begin{align} \sum_{q\leq Q} \frac{\mu(q)^2}{\phi(q)}&\leq \left(\sum_{m\leq Q, \ m|k} \frac{\mu(m)^2}{\phi(m)} \right)\left(\sum_{n\leq Q, \ (n,k)=1} \frac{\mu(n)^2}{\phi(n)}\right)\\ &\leq \left(\sum_{m|k} \frac{\mu(m)^2}{\phi(m)}\right)\left(\sum_{n\leq Q, \ (n,k)=1} \frac{\mu(n)^2}{\phi(n)}\right)\\ &=\left( \prod_{p|k} \left( 1 + \frac{1}{p-1} \right) \right) \left(\sum_{n\leq Q, \ (n,k)=1} \frac{\mu(n)^2}{\phi(n)}\right)\\ &=\left( \prod_{p|k} \frac p{p-1} \right) \left(\sum_{n\leq Q, \ (n,k)=1} \frac{\mu(n)^2}{\phi(n)}\right)\\ &=\frac{k}{\phi(k)} \left(\sum_{n\leq Q, \ (n,k)=1} \frac{\mu(n)^2}{\phi(n)}\right) \end{align} $$ as desired.