In the answer to a different question titled "Is the Euler phi function bounded below?", one answer derives the fact that if $0<\delta<1$, then $\frac{\phi(n)}{n^{1-\delta}}$ attains its minimum at $n=2\cdot3\cdot\ldots\cdot p=p\sharp$, where $p$ is the largest prime satisfying $p-1\le p^{1-\delta}$. Thus, for a given prime $p$, we can set $\delta=1-\log_p (p-1)$ to find that $\phi(n)\ge\phi(p\sharp)(\frac{n}{p\sharp})^{\log_p (p-1)}$. Is there a way to use this to derive the well-known inequality $\phi(n)\ge \frac{n}{e^\gamma\log\log{n}+\frac{3}{\log\log{n}}}$? If not, how is this inequality derived? Thank you.
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1You want Rosser and Schoenfeld 1962. It is available for free online. See if I can find it. http://projecteuclid.org/euclid.ijm/1255631807 – Will Jagy Mar 30 '14 at 22:43
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possible duplicate of This question – Eric Naslund Apr 07 '14 at 13:09
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A complete derivation of the lower bound can be found in this answer.
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Eric Naslund
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How is this derived from the fact that $\phi(n)\geq \frac{n}{e^{\gamma}\log \log n}+O\left(\frac{n}{(\log \log n)^2}\right)$? – Pablo Oct 19 '14 at 11:21