Indicating with $p$ and $q$ prime numbers, is it true that for $x\rightarrow\infty$ $$ \sum_{\substack{p\leq x \\ p\equiv 1 \mod{m}}}\prod_{q\mid\frac{p-1}{m}}\left(1+\frac1{q}\right)2^{\omega\left(\frac{p-1}{m}\right)} \ll \log\log x \sum_{\substack{p\leq x \\ p\equiv 1 \mod{m}}} \tau\left(\frac{p-1}{m}\right) \;, $$ where $\tau(n)$ is the divisor function and $\omega(n)$ counts the number of primes dividing $n$?
I tried in this way: first, $2^{\omega\left(\frac{p-1}{m}\right)} \leq \tau\left(\frac{p-1}{m}\right)$, but then using Mertens' third theorem $$ \prod_{q\mid\frac{p-1}{m}}\left(1+\frac1{q}\right) = \prod_{q\mid\frac{p-1}{m}}\left(1-\frac1{q^2}\right)\left(1-\frac1{q}\right)^{-1} \leq O(1)\prod_{q\leq\frac{p-1}{m}}\left(1-\frac1{q}\right)^{-1} =O\left( \log\left(\frac{p-1}{m}\right)\right) $$ and I can't retrieve the $\log\log x$ term. Am I doing something wrong? Any suggestion?
Note that $$\prod_{q\leq x}\left(1+\frac{1}{q}\right)\sim\log\left(x\right) $$ and $$\sum_{q\leq x}\left(1+\frac{1}{q}\right)\sim x+\log\left(\log\left(x\right)\right) $$ then $$\prod_{q\mid\left(p-1\right)/m}\left(1+\frac{1}{q}\right)\ll\sum_{q\mid\left(p-1\right)/m}\left(1+\frac{1}{q}\right)\leq\frac{3}{2}\omega\left(\frac{p-1}{m}\right) $$ and using the well-knonw bound $$\omega\left(a\right)\ll\log\left(\log\left(a\right)\right) $$ we have $$\frac{3}{2}\omega\left(\frac{p-1}{m}\right)\ll\log\left(\log\left(\frac{p-1}{m}\right)\right)\ll\log\left(\log\left(x\right)\right). $$