Condition on the inequality holds asymptotically

Question

$$(n-1) \left[\frac{1}{2}\log\left(\frac{1}{2p}\right)+\frac{1}{2}\log\left(\frac{1}{2(1-p)}\right)\right]>2\log(n)$$

where $p$ is a function of $n$.

How to find out on which condition of $p$ such that the inequality holds (asymptotically)?

From the following experiment, it seems when $p>\frac{1+\sqrt{\frac{2\log n}{n}}}{2}$ the inequality holds, and when $p<\frac{1+\sqrt{\frac{2\log n}{n}}}{2}$ the inequality does not hold.

I can manage to prove that this is true by plugging the form of $p$ as $\frac{1+\sqrt{\frac{a\log n}{n}}}{2}$ into the inequality, but I have no idea to prove it without assuming the form of $p$.

It would be much appreciated if someone could help me out. Many thanks in advance!

Answer 1

Assuming $n>1$, gouping terms, you look for the problem of $$-\log (p(1-p) )~~ \geq~~ \frac{4 \log (n)}{n-1}+2 \log (2)= A$$ When $n$ increases, the solution requires that $p$ just be above the root of $\text{(lhs - rhs)}=0$ which is closer an closer to $p=\frac 12$.

Expanding as a Taylor series, we have $$\text{(lhs - rhs)}=-\frac{4 \log (n)}{n-1}+\sum_{n=1}^\infty \frac {t^n} n \qquad \text{with}\qquad t=(2p-1)^2$$ that is to say $$\text{(lhs - rhs)}=-\frac{4 \log (n)}{n-1}-\log (1-t)=0$$ $$t=1-n^{-\frac{4}{n-1}}\qquad \implies\qquad p=\cdots$$

Edit

To better see my point, type in Wolfram Alpha

Plot[-2*Log[2]-(4*Log[10])/(10-1)-Log[p(1-p)],{p,0,1}]

and look where the expression is positive. Change the $10$ by the number you want.

Answer 2

Some thoughts.

The inequality is written as $$p(1 - p) < \frac14\mathrm{e}^{- \frac{4\ln n}{n - 1}}. \tag{1}$$

WLOG, assume that $p \ge 1/2$. (1) is equivalently to $$p > \frac12 + \frac12\sqrt{1 - \mathrm{e}^{- \frac{4\ln n}{n - 1}}}.$$

We can prove that, for all $n \ge 2$, $$1 - \mathrm{e}^{- \frac{4\ln n}{n - 1}} \le \frac{4\ln n}{n}.$$

Thus, if $p > \frac12 + \frac12\sqrt{\frac{4\ln n}{n}}$, then the original inequality is true.