Return to Answer

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 $\sum_{i=1}^n p_i \log \frac{q_i}{p_i} \leq 0\qquad (\sum_{i=1}^n p_i \log \frac{p_i}{q_i} \geq 0)$

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

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 $\sum_{i=1}^n p_i \log \frac{q_i}{p_i} \leq 0\qquad (\sum_{i=1}^n p_i \log \frac{p_i}{q_i} \geq 0)$

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

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 rewrote 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 $\sum_{i=1}^n p_i \log \frac{q_i}{p_i} \leq 0\qquad (\sum_{i=1}^n p_i \log \frac{p_i}{q_i} \geq 0)$

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

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 $\sum_{i=1}^n p_i \log \frac{q_i}{p_i} \leq 0\qquad (\sum_{i=1}^n p_i \log \frac{p_i}{q_i} \geq 0)$

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

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 rewrote 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 $\sum_{i=1}^n p_i \log \frac{q_i}{p_i} \leq 0\qquad (\sum_{i=1}^n p_i \log \frac{p_i}{q_i} \geq 0)$

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

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 rewrote 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 $\sum_{i=1}^n p_i \log \frac{q_i}{p_i} \leq 0\qquad (\sum_{i=1}^n p_i \log \frac{p_i}{q_i} \geq 0)$

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