Sum of the reciprocals of radicals

You can get away with elementary analytic number theory. Consider the series $\sum_n\frac{1}{n^{\varepsilon}\rm{rad}(n)}$. It suffices to show that it converges. However, it can be written as a product of $$ S(p)=1+p^{-1-\varepsilon}+p^{-1-2\varepsilon}+\dots=1+p^{-1-\varepsilon}\frac 1{1-p^{-\varepsilon}}\le 1+p^{-1-\frac\varepsilon 2} $$ for all but finitely many $p$. Thus $\prod_p S(p)\le C\prod_p(1+p^{-1-\frac\varepsilon 2})\le\sum_n n^{-1-\frac\varepsilon 2}<+\infty$

First, notice that for any squarefree $m$ and any $\varepsilon>0$ we have

notice that

$$\sum_{n:\operatorname{rad}(n)=m} \frac{1}{n^\varepsilon}=m^{-\varepsilon}\prod_{p\mid m}(1-p^{-\varepsilon})^{-1}\ll_\varepsilon d(m)/m^\varepsilon,$$

thus, the series

$$r(s)=\sum_{n=1}^{+\infty} \frac{1}{n^s\mathrm{rad}(n)}$$

converges absolutely when $\mathrm{Re}\,s>0$. Now, using multiplicativity, one has

$$r(s)=\prod_p (1+p^{-s-1}+p^{-2s-1}+\ldots)=\prod_p (1+\frac{1}{(1-p^{-s})p^{1+s}}).$$

Next, notice that for positive $\varepsilon$ we have $1-2^{-\varepsilon}\gg \varepsilon$ and $1-p^{-\varepsilon}\geq \varepsilon$ for $p>2$ and $\varepsilon<1/6$. Therefore we deduce for any $\varepsilon>0$

$$r(\varepsilon)\ll \prod_p\left(1+\frac{1}{\varepsilon p^{1+\varepsilon}}\right)\leq \zeta(1+\varepsilon)^{1/\varepsilon}.$$

As $\zeta(1+\varepsilon)=\frac{1}{\varepsilon}+O(1)$, we finally obtain

$$r(\varepsilon)\ll \varepsilon^{-1/\varepsilon}.$$

Using Rankin trick we arrive at

$$\sum_{n\leq x} \frac{1}{\mathrm{rad}(n)}\ll x^\varepsilon \varepsilon^{-1/\varepsilon}.$$

Choosing $\varepsilon=\sqrt{\frac{\ln\ln x}{2\ln x}}$ we prove that

$$\sum_{n\leq x} \frac{1}{\mathrm{rad}(n)}\leq \exp(\sqrt{(2+o(1))\ln x\ln\ln x}),$$

which is a bit non-optimal by the answer of Don. (But at least we have the correct $\ln\ln$ asymptotics)

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).