# Trying to show that $\phi(n) > c_1 \frac{n}{\text{log log }n}$ for some constant $c_1 > 0$

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.

• Is the formula you give one that you already know or one that you are conjecturing? – Laertes Oct 18 '14 at 17:44
• one that I already know. – Pablo Oct 18 '14 at 17:47
• 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. – Laertes Oct 18 '14 at 17:53
• See the answer here and here. – Dietrich Burde Oct 18 '14 at 18:47
• $c_1={1+\epsilon \over e^\gamma}$ 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. – Laertes Oct 19 '14 at 17:02

The best theorem to look up is, I think, the following effective lower bound for $\phi(n)$
Theorem: For all $n>2$ we have $$\frac{\phi(n)}{n}>\frac{1}{e^{\gamma}\log \log n+\frac{3}{\log \log n}}.$$
• If I prove this result, can I then say that $\frac{\phi(n)}{n} > \frac{1}{\text{loglog }n}\left(\frac{1}{e^\gamma+3}\right)$? – Pablo Oct 19 '14 at 10:55