TITLE: Sum of the reciprocals of radicals
QUESTION [16 upvotes]: Recall that the radical of an integer $n$ is defined to be $\operatorname{rad}(n) = \prod_{p \mid n } p$. 
For a paper, I need the result that
$$\sum_{n \leq x} \frac{1}{\operatorname{rad}(n)} \ll_\varepsilon x^{\varepsilon} \tag{$*$},$$
for all $\varepsilon > 0$. I have a proof of this using complex analysis and Perron's formula, but this seems a bit overkill given that I'm looking for a weak upper bound for a problem in elementary number theory.

Does anyone know of a short elementary proof of the bound $(*)$? Or better yet, a reference?

REPLY [9 votes]: de Bruijn studies this sum in "On the number of integers $\le x$ whose prime factors divide $n$", which was published in a 1962 volume of the Illinois J. Math; see
https://projecteuclid-org.proxy-remote.galib.uga.edu/euclid.ijm/1255631814
He proves there (see Theorem 1) that $$\sum_{n \le x} \frac{1}{\mathrm{rad}(n)} = \exp((1+o(1)) \sqrt{8\log{x}/\log\log{x}}),$$ as $x\to\infty$. Of course, this implies the $O(x^{\epsilon})$ bound you were after. However, his proof (which uses a Tauberian theorem of Hardy and Ramanujan) is not as elementary as some others that have been suggested here (but gives a more precise result).

REPLY [2 votes]: sIt seems that this argument hasn't been presented yet, so I might as well include it.
We can sort the integers $n \in [1, X]$ by their radicals, which is necessarily a square-free integer $m$. Thus we have
$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} = \sum_{\substack{m \leq X \\ m \text{ square-free}}} \frac{1}{m} \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1.$$
Now, $\text{rad}(n) = m$ if and only if $p | n \Rightarrow p | m$. If we write $m = p_1 \cdots p_k$, then 
$$\displaystyle \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1 = \#\{(x_1, \cdots, x_k) : x_i \in \mathbb{Z} \cap [0,\infty), p_1^{x_1} \cdots p_k^{x_k} \leq X/m\}.$$
The inequality defining the right hand side is equivalent to 
$$\displaystyle x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m),$$
and this is just counting integer points with non-negative entries bounded by a simplex, and it is easy to see that 
$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} \ll \frac{\log(X/m)}{\prod_{1 \leq i \leq k} \log(p_i)} \ll \log X.$$
EDIT: This last step is wrong, but it can be fixed. Indeed, we can arrange the $p_i$'s so that $p_1 < p_2 < \cdots < p_k$. It then follows from Davenport's lemma that
$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} = O \left(\sum_{i=0}^k \frac{(\log X/m)^{k-i}}{\prod_{1 \leq j \leq k-i} \log p_i} \right).$$
It then follows that 
$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} \ll \sum_{\substack{p_1 < \cdots < p_k \\ p_1 \cdots p_k \leq X}} \sum_{i=0}^k \frac{(\log X)^{k-i}}{\prod_{1 \leq j \leq k -i} p_i \log p_i}.$$
From here I think it is possible to get the bound $O_\epsilon(X^\epsilon)$, but it requires a somewhat more refined analysis on the interaction between the number of primes and the size of the primes.