How can I prove $\sum_{q=1}^{\infty}\frac{\mu(q)c_{q}(n)}{\varphi(q)^{3}}>C$ for positive odd n where C is a constant? I have to prove
$$\sum_{d|n}\frac{\mu(d)}{\varphi(d)^{2}}>C$$
for all positive odd $n$ where $C>0$ is a constant. (You can choose the constant $C$.)
I got
$$\sum_{d|n}\frac{\mu(d)}{\varphi(d)^{2}}=\prod_{p|n}\left(1-\frac{p}{\varphi(p)^{2}}\right)\geq\prod_{p}\left(1-\frac{p}{(p-1)^{2}}\right),$$
but I cannot prove the infinite product converges to a positive real number.
Can anyone help me?
*Edited
The first question was just a conjecture I made to prove another question, but it seems to be false.
The original question was to prove that
$$\sum_{q=1}^{\infty}\frac{\mu(q)c_{q}(n)}{\varphi(q)^{3}}>C$$
for all positive odd $n$ where $C>0$ is a constant you can choose. ($c_q(n)$ is Ramanujan's Sum.)
I got
$$\sum_{q=1}^{\infty}\frac{\mu(q)c_{q}(n)}{\varphi(q)^{3}}=\sum_{d|n}\frac{\mu(d)}{\varphi(d)^{2}}\sum_{(n,e)=1}\frac{\mu(e)^2}{\varphi(d)^{3}},$$
but cannot solve the problem at all.
You can use another idea if you want.
 A: I need new spectacles ;)
You wrote:
$$\sum_{d|n}\frac{\mu(d)}{\varphi(d)^{2}}=\prod_{p|n}\left(1-\frac{p}{\varphi(p)^{2}}\right)\geq\prod_{p}\left(1-\frac{p}{(p-1)^{2}}\right),$$
and "but I cannot prove the infinite product converges to a positive real number", and I focused on the infinite product you wrote, that converges to $0$.
But the equality in that is wrong. If $d$ is squarefree, then $\dfrac{\mu(d)}{\varphi(d)^2} = \prod\limits_{p\mid d}\frac{(-1)}{\varphi(p)^2}$ and hence
$$\sum_{d|n}\frac{\mu(d)}{\varphi(d)^{2}}=\prod_{p|n}\left(1-\frac{1}{\varphi(p)^{2}}\right)\geq\prod_{p > 2}\left(1-\frac{1}{(p-1)^{2}}\right),$$
and the latter product is easily seen to be positive since $\sum \frac{1}{n^2}$ converges.

Another way to obtain the existence of a positive $C$ such that
$$\sum_{q=1}^{\infty}\frac{\mu(q)c_{q}(n)}{\varphi(q)^{3}}>C$$
for all positive odd $n$, and a different, maybe easier to handle(?) expression for the sum is to use Kluyvert's expression for the Ramanujan sum, $c_q(n) = \sum\limits_{d\mid (q,n)} \mu\left(\frac{q}{d}\right)d$, and set
$$P = \prod_p \left(1 + \frac{1}{\varphi(p)^3}\right)$$
for brevity. Then one can rearrange (all sums are absolutely convergent because a rather crude estimate gives $\varphi(n) \geqslant C\frac{n}{\log n}$)
$$\begin{align}
\sum_{q=1}^{\infty}\frac{\mu(q)c_{q}(n)}{\varphi(q)^{3}}
&= \sum_{q=1}^\infty \frac{\mu(q)}{\varphi(q)^3}\left(\sum_{d\mid (q,n)} \mu(q/d)d\right)\\
&= \sum_{d\mid n}\sum_{k=1}^\infty \frac{\mu(kd)\mu(k)d}{\varphi(kd)^3}\\
&= \sum_{d\mid n} \sum_{\substack{k=1\\(k,d) = (1)}}^{\infty} \frac{\mu(d)d\mu(k)^2}{\varphi(d)^3\varphi(k)^3}\\
&= \sum_{d\mid n} \frac{\mu(d)d}{\varphi(d)^3} \sum_{\substack{k=1\\(k,d) = (1)}}^{\infty}\frac{\mu(k)^2}{\varphi(k)^3}\\
&= \sum_{d\mid n} \frac{\mu(d)d}{\varphi(d)^3} \prod_{p \not\mid d}\left(1 + \frac{1}{\varphi(p)^3}\right)\\
&= P \cdot \sum_{d \mid n} \frac{\mu(d)d}{\varphi(d)^3 \prod\limits_{p\mid d}\left(1+\frac{1}{\varphi(p)^3}\right)}\\
&= P \cdot \sum_{d \mid n} \frac{\mu(d)d}{\prod\limits_{p\mid d}\bigl(\varphi(p)^3+1\bigr)}\\
&= P\cdot \prod_{p\mid n} \left(1 - \frac{p}{\varphi(p)^3+1}\right).
\end{align}$$
The last product again is bounded below (for odd $n$) by a positive number because for $p \geqslant 3$ we have $\frac{p}{\varphi(p)^3+1} \leqslant \frac{3}{p^2}$ and hence
$$\prod_{p \geqslant 3} \left(1 - \frac{p}{\varphi(p)^3+1}\right) > 0.$$
