Sum of reciprocals of $\phi(n)$ Let $Q$ be the set of square-free integers less than some $n$ and let $\phi$ denote the Euler totient.
I am currently reading an article where it is mentioned that :
$$\sum_{q\in Q}\frac1{\phi(k)}\ge\ln(n)$$
I would like to understand where that inequality does come from. Any hint ?
Thanks in advance.
 A: Note, merely for curiosity's sake, that one may actually obtain the asymptotic formula
$$\sum_{n \leq x} \frac{\mu^2(n)}{\varphi(n)} = \log x + c + o(1),$$
where
$$c = \gamma + \sum_{p} \frac{\log p}{p(p-1)} = 1.332\ldots.$$
Therefore the desired inequality follows for all $x$ sufficiently large, and if one had explicit bounds for the error term you could potentially do a machine computation to check the remaining cases.
We don't need any of this, however. The key observation is that, for $n$ squarefree, we have
$$\frac{1}{\varphi(n)} = \frac{1}{n}\prod_{p|n} \frac{1}{1 - \frac{1}{p}} = \frac{1}{n}\prod_{p|n} \left(1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) = \sum_{\substack{m \geq 1 \\ \text{rad}(m) = n}} \frac{1}{m}.$$
Here $\text{rad}(m)$ denotes the squarefree radical of $m$, i.e.
$$\text{rad}(m) = \prod_{p|m} p.$$
Therefore
$$\sum_{n \leq x} \frac{\mu^2(n)}{\varphi(n)} = \sum_{n \leq x} \mu^2(n)\sum_{\substack{m \geq 1 \\ \text{rad}(m) = n}} \frac{1}{m} \geq \sum_{k \leq x} \frac{1}{k}.$$
This inequality is the key step, but we see almost immediately it is true, since if $k \leq x$ then $\text{rad}(k) \leq x$. At this point we are basically done, since from elementary geometric considerations we have
$$\sum_{k \leq x} \frac{1}{k} \geq \int_1^x \frac{1}{t}dt = \log x.$$
