Is there an "elementary" method to prove the following asymptotic bound ? $$\sum_{n>x}\frac{\mu^{2}(n)}{n\varphi(n)}=\mathcal{O}(\frac{1}{x}).$$ Here, $\varphi$ is the Euler totient function and $\mu$ is the Möbius function.
Using the inequalities $\varphi(n)\geq\sqrt{\frac{n}{2}}$ and $\mu^{2}(n)\leq 1$, $\forall n\geq 1,$ I can only get an error term of $\mathcal{O}(\frac{1}{\sqrt{x}})$.
Using Mertens' third theorem and the formula $\frac{\varphi(n)}{n}=\Pi_{p\vert n}(1-\frac{1}{p})$ I can only get a slightly better error term of $\mathcal{O}(\frac{\log x}{x})$.
Does someone have any idea on how to obtain the desired error term? I suspect I would have to use some result on the distribution of prime numbers.
Due to its Euler product, the LHS of $$ \sum_{n\ge 1} a_n n^{-s}=\frac{\sum_{n\ge 1} \frac{\mu(n)^2 }{n\varphi(n)}n^{-s}}{\sum_{n\ge 1}\frac{1}{n^2} n^{-s}}=\prod_p (1-p^{-s-2})(1+\frac{1}{p(p-1)}\frac{p^{-s}}{1-p^{-s}}) $$ converges absolutely at $s=-1$, so that $$\sum_{n> x}\frac{\mu(n)^2 }{n\varphi(n)}= \sum_{d \ge 1} a_d \sum_{m> x/d} \frac{1}{m^2}= \sum_{d \ge 1} a_d O(\frac{d}x)=O(\frac1x \sum_{d \ge 1} |a_d| d)=O(\frac1x)$$