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Trying to show that $\phi(n) > c_1 \frac{n}{\text{log log }n}$ for some constant $c_1 > 0$ where $\phi(n)$ is the euler phi function.

I was wondering if I could use something like

\begin{align} \phi(n)\geq \frac{n}{e^{\gamma}\log \log n}+O\left(\frac{n}{(\log \log n)^2}\right) \end{align}

to show this, but I'm not sure. Could anyone maybe give me some advice or tips on how to do this? I have quite a few Analytical Number Theory texts and notes with me, so even suggestions to some theorems would be great. Thanks.

Pablo
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  • Is the formula you give one that you already know or one that you are conjecturing? – Avi Oct 18 '14 at 17:44
  • one that I already know. – Pablo Oct 18 '14 at 17:47
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    Then your result follows, since the first term dominates all other terms as $n \to \infty$, so that there always exists some $\epsilon$ such that $\phi(n)\geq \frac{n}{e^{\gamma}\log \log n}(1+\epsilon)$ where $\lim_{n \to \infty} \epsilon=0$ and for all $n>N$ for some $N$, $\epsilon$ is monotonic decreasing, from which the result follows. – Avi Oct 18 '14 at 17:53
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    See the answer here and here. – Dietrich Burde Oct 18 '14 at 18:47
  • @Laertes, if I needed to choose a constant $c_1$ that has to hold for all $n$, what would it be w.r.t. $\epsilon$? – Pablo Oct 19 '14 at 11:57
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    $c_1={1+\epsilon \over e^\gamma}$ for the highest value of $\epsilon$ for any $n$. One value that appears to work for $c_1$ is $1\over 9$, although I haven't rigorously tested that. – Avi Oct 19 '14 at 17:02
  • perfect! Laertes your help has been invaluable, so I just let $c_1 = \text{lim }sup_{\epsilon}\left(\frac{1+\epsilon}{e^\gamma}\right)$. – Pablo Oct 19 '14 at 19:18

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The best theorem to look up is, I think, the following effective lower bound for $\phi(n)$

Theorem: For all $n>2$ we have $$ \frac{\phi(n)}{n}>\frac{1}{e^{\gamma}\log \log n+\frac{3}{\log \log n}}. $$

Dietrich Burde
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