On an estimation involving Euler totient function Let $m$ an arbitrary integer. I would like to know if exists an estimation of $$\underset{\left(q,\, m\right)>1}{\underset{q>1}{\sum}}\frac{1}{\phi\left(q\right)\phi\left(q/\left(q,\, m\right)\right)}$$ or something similar, where $(q, m)$ is the g. c. d. of $q$ and $m$. 
 A: \begin{align*}
\sum_{\substack{q>1 \\ (q,m)=1}} \frac{1}{\phi(q)\phi\big(q/(q,m)\big)} &= \sum_{\substack{d\mid m \\ d>1}} \sum_{\substack{q>1 \\ (q,m)=d}} \frac{1}{\phi(q)\phi(q/d)} \\
&\le \sum_{\substack{d\mid m \\ d>1}} \sum_{r=1}^\infty \frac{1}{\phi(rd)\phi(r)} \\
&\le \sum_{\substack{d\mid m \\ d>1}} \frac1{\phi(d)} \sum_{r=1}^\infty \frac{1}{\phi(r)^2} < C \sum_{d\mid m} \frac1{\phi(d)},
\end{align*}
where $C = \sum_{r=1}^\infty \frac{1}{\phi(r)^2}$. (The first inequality is because we've dropped the condition $(r,m/d)=1$; the second inequality comes from the elementary $\phi(ab) \ge \phi(a)\phi(b)$.)
As Alexey commented, this last sum is $\ll \log\log m$, although seeing so takes a few steps. Simplest perhaps is to note that $f(m) = \sum_{d\mid m} \frac1{\phi(d)}$ is a multiplicative function of $m$ satisfying $f(p^k) \le \frac p{p-1} \frac{p^2+2}{p^2}$ for every prime power $p^k$, whence
$$
f(m) = \prod_{p^k\|m} f(p^k) \le \prod_{p^k\|m} \frac p{p-1} \prod_{p^k\|m} \frac{p^2+2}{p^2} < \frac m{\phi(m)} \prod_p \frac{p^2+2}{p^2};
$$
it's a standard estimate that $\frac m{\phi(m)} \ll \log\log m$, while the last product converges to some constant.
It would even be possible to get an explicit evaluation of the original sum, via:
\begin{align*}
\sum_{\substack{q>1 \\ (q,m)=1}} \frac{1}{\phi(q)\phi\big(q/(q,m)\big)} &= \sum_{\substack{d\mid m \\ d>1}} \sum_{\substack{r\ge1 \\ (r,m/d)=1}} \frac{1}{\phi(rd)\phi(r)}  \\
&= \sum_{\substack{d\mid m \\ d>1}} \frac1{\phi(d)} \sum_{r=1}^\infty \begin{cases}
\frac{\phi(d)}{\phi(rd)\phi(r)}, &\text{if } (r,m/d)=1, \\ 0, &\text{if } (r,m/d)>1.
\end{cases}
\end{align*}
For a given $m,d$, the inner summand is a multiplicative function of $r$, and so the inner sum can be converted into its Euler product. That would leave a sum of the form $\sum_{d\mid m} g(d)$ for some multiplicative function $g$ (or something similar), which can be evaluated as an explicit multiplicatve function of $m$.
