Let $\mu, \phi, I$ denote mobius function, Euler-phi function and identity function, respectively.
Let $$f(n) := \frac{\mu^2(n)}{I(n)\phi(n)}$$
I follow some guide steps to estimate $\sum_{i=1}^\infty f(n)$ and $\sum_{n \leq x}\frac{1}{\phi(n)}$.
1) I show that ($*$ := Dirichlet convolution) $$\frac{1}{\phi} = \frac{1}{I}* f.$$
Since $I, \phi, \mu$ are multiplicative, $\frac{1}{\phi}, \frac{1}{I}*f$ are multiplicative. So it is enough to verify its value at prime power. A calculation show that $$\frac{1}{\phi}(p^a) = \frac{1}{I}*f (p^a)$$ for $p$ prime. So I am fine with this step.
2) I want to show that $\sum_{i=1}^\infty f(n) = O(1)$, or it the serie converges.
Sol. (Update : my new approach) $$\sum_{n=1}^\infty \frac{\mu^2(n)}{n\phi(n)} = \sum_{n=p_1p_2...p_k, \ p_i \neq p_j} \frac{1}{n(p_1 - 1)...(p_k - 1)}\\ = \sum_{n = 2p_1p_2...p_k, \\ p_i \neq p_j, p_m \neq 2} \frac{1}{n(p_1 - 1)...(p_k - 1)} + \sum_{n = p_1...p_k, \\ p_i \neq p_j, p_m \neq 2}\frac{1}{n(p_1 - 1)(p_2 - 1)...(p_k - 1)}.$$
For any prime $p > 2$, $$p^{1/2} < p-1 \ \ \ \ -----(1).$$ Then $$\sum_{n = p_1...p_k, \\ p_i \neq p_j, p_m \neq 2}\frac{1}{n(p_1 - 1)(p_2 - 1)...(p_k - 1)} \leq \sum_{n = p_1...p_k, \\ p_i \neq p_j, p_m \neq 2} \frac{1}{n^{3/2}}$$ so it is convergent. For the first sum $$\sum_{n = 2p_1p_2...p_k, \\ p_i \neq p_j, p_m \neq 2} \frac{1}{n(p_1 - 1)...(p_k - 1)}$$ I think it should be similar, but the argument is not exactly the same since $(1)$ does not hold.
Anyone has a better idea about this ?
3) From 2), for $n \geq 2$ $$\sum_{n \leq x} \frac{1}{\phi(n)} = O(\log x).$$ I think I will use 1) and use Dirichlet hyperbola method to do the estimation. But it does not go as I expected. Not much Idea what should I do.
Hint, suggestion, or help please ?