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Let $q$ an arbitrary integer. Is there any chance of getting a bound like $$\underset{d\mid q}{\sum}\frac{1}{\phi\left(q/d\right)^{2}}\ll\frac{1}{\phi\left(q\right)^{2}}?$$

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  • $\begingroup$ By $a \ll b$, you mean "There is a $C$ such that $a \leqslant C\cdot b$"? $\endgroup$ – Daniel Fischer Jan 7 '14 at 16:27
  • $\begingroup$ Doesn't $\phi(q)\to \infty$ as $q\to\infty$, while the LHS is at least $1$? $\endgroup$ – Pablo Rotondo Jan 7 '14 at 16:35
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    $\begingroup$ How did your question arise? $\endgroup$ – Pablo Rotondo Jan 7 '14 at 16:50
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Not sure if I understand the question, but if this helps at all: $$\sum_{d\mid n}\frac{1}{\phi(n/d)^2}=\prod_{p\mid n}(1+\frac{1}{\phi(p)^2}+\frac{1}{\phi(p^2)^2}+\frac{1}{\phi(p^3)^2}...+\frac{1}{\phi(p^{v_p(n)})^2})$$ $$=\prod_{p\mid n}(1+\frac{1}{(p-1)^2}(1+\frac{1}{p^2}+\frac{1}{p^4}+..\frac{1}{p^{2(v_p(n)-1)}}))$$ $$=\prod_{p\mid n}(1+\frac{1-p^{-2v_p(n)}}{(p-1)^2(1-p^{-2})})$$

$$\sum_{d\mid n}\frac{1}{\phi(n/d)^2}<\prod_{p}(1+\frac{1}{(p-1)^2(1-p^{-2})})<3.4$$

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