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I need help in proving this inequality->

Assuming q to be any positive integer prove that ( 1+ log(q) ) d(q) $\phi^{-2} (q) $ $\leq$ $ q^{-5/3}$ . where d(q) means number of divsor of q and $\phi(q) $ = number of integer less than q which are co-prime to q.

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I am unable to deduce (39) on page 6 of research paper and this inequality will be used there. The motive of asking the question and putting the bounty is to prove (39) .I hope that I am not wrong in thinking that the above mentioned inequality will lead to proving (39) ?

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The inequality asked for is equivalent to $\phi(n) \geq cn^{5/6}\sqrt{d(n)\log n}$ and this follows from known estimates on $\phi(n)$ and $d(n)$. For example, $\phi(n)$ is asymptotically at least $\dfrac{n}{\log \log n}$, whereas the divisor function is (asymptotically) upper-bounded by $n^{\epsilon}$ for every $\epsilon>0$. The required bound also follows from $\varphi(n)d(n) \geq cn^2$.

For proofs of the three bounds stated above, see: Lower bound for $\phi(n)$: Is $n/5 < \phi (n) < n$ for all $n > 1$? and https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/ and Lower bound of Euler phi function times sum of divisors

Aravind
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  • my inequality is $\phi_(q) \geq q^{5/6} {(d(q) +d(q) log(q) ) }^{0.5}$ how is that equivalent to your identity, I don't get it? Can you please explain? –  May 09 '20 at 12:48
  • Because $\sqrt{d(n)+d(n)\log n)}=\Theta(\sqrt{d(n)\log n})$. – Aravind May 09 '20 at 13:40
  • I think this should be used -> d(n) = o$ ( n^{\delta} ) $ for every $\delta$ >0 . So, LHS n / loglog(n) >0 and on RHS and $\epsilon $ tending to 0 come s due to small o notation. So, LHS always greater than or equal to LHS. Am I right? Can you please verify? –  May 12 '20 at 20:53
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    Yes, you are right, $d(n)=o(n^{\delta})$ for every $\delta>0$, and we need to use that. – Aravind May 13 '20 at 03:07
  • did you got bounty points? –  May 13 '20 at 07:48
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    did you got bounty? can you reply? –  May 13 '20 at 14:38
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    I think you have!! –  May 13 '20 at 14:39