I am trying to lower bound the following function, $n \ge 3$ is a natural number:


$$l(n):=\frac{\log(n)}{\log(n)-\frac{1}{n}(\tau(n)\log(\tau(n))+(n-\tau(n))\log(n-\tau(n)))}$$


where $\tau(n)$ counts the number of divisors. Is this function exponential or polynomial in terms of $\log(n)$?


Thanks for your help.



Lower bound on a function of the number of divisors?
Tried to write a comment but got too long, not sure if correct or what you were looking for but hopefully is useful otherwise let me know and I will delete it.


What I did was to rewrite the denominator as follows
$\ln(n)-\frac{1}{n}(\tau(n)\ln(\tau(n))+(n-\tau(n))\ln(n-\tau(n)))\\
=\ln(2)+\ln(n/2)-((\tau(n)/n)\ln(\tau(n))+(1-\tau(n)/n)\ln(n-\tau(n)))\\
=\ln(2)+(\tau(n)/n)\ln(n/2)+(1-\tau(n)/n)\ln(n/2)+(\tau(n)/n)\ln(\frac{1}{\tau(n)})+(1-\tau(n)/n)\ln(\frac{1}{n-\tau(n)})\\
=\ln(2)+(\tau(n)/n)\ln(\frac{1/2}{\tau(n)/n})+(1-\tau(n)/n)\ln(\frac{1/2}{1-\tau(n)/n})\\
=\ln(2)+\sum\_{i=1}^2 p\_i\ln(q\_i/p\_i)\\
\leq\ln(2)
$


where the last inequality is because we have that $p\_1=\tau(n)/n, p\_2=1-\tau(n)/n,q\_1=q\_2=1/2$ then $\sum\_{i=1}^2 p\_i=1$ and $\sum\_{i=1}^2 q\_i=1$ and we can apply [Gibbs inequality](https://en.wikipedia.org/wiki/Gibbs%27_inequality) $\sum\_{i=1}^n p\_i \log \frac{q\_i}{p\_i} \leq 0\qquad (\sum\_{i=1}^n p\_i \log \frac{p\_i}{q\_i} \geq 0)$


Therefore we get that
$l(n):=\frac{\log(n)}{\log(n)-\frac{1}{n}(\tau(n)\log(\tau(n))+(n-\tau(n))\log(n-\tau(n)))}\geq\frac{\log(n)}{\log(2)}$


