Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0<p<1, n\in \mathbb N$

Question

Prove that $p+\frac{2p-1}{ \left(\frac{1-p}{p}\right)^n -1} \leq \frac{1}{2} - \frac{1}{2n}$, if $0<p<1, n\in \mathbb N$

This is from a recently closed question.

Notice that the fraction on the LHS is well defined. The denominator can be written as

$$\frac{1}{p^n} \left( (1-p)^n - p^n \right)=\frac{1-2p}{p^n} g(p)$$ where $g(p)$ is a polynomial of $p$ taking positive values. When $p=\frac 12$, the LHS becomes $p-\frac{1}{2^n g(\frac 12)}$.

My effort was shown here with variable substitution and taking derivatives. I'd like to know if there's a simpler, elementary proof without calculus. Thanks.

Answer 1

No calculus is required.

Using your substitution in your linked effort:

Let $x=\frac{1-p}{p}$ then $p=\frac{1}{1+x}, x>0$. The inequality becomes $$\frac{1}{1+x}+\frac{1}{1+x}\cdot \frac{1-x}{x^n-1} = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac1 2-\frac{1}{2n}$$ $$\iff g(x) = \frac{x-x^n}{(1+x)(1-x^n)} \le \frac 12 - \frac{1}{2n}\tag 1$$

From here, notice that

$$ \begin{array}{ll} g(x) & = &\frac{ x}{1+x} \times \frac{ 1 - x^{n-1 }} { 1 - x^n} \\ &=& \frac{x}{1+x} \times \frac{ 1 + x + \ldots + x^{n-2} } {1 + x + \ldots + x^{n-1} } \\ & =& \frac{ x + \ldots + x^{n-1}} { 1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n} } \\ & =& \frac{1}{2} - \frac{1}{2} \times \frac{ 1+x^n} {1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n}} \end{array}$$

So, it remains to show that

$$ 1 + 2x + 2x^2 + \ldots + 2x^{n-2} + 2x^{n-1} + x^{n} \leq n ( 1 + x^n). $$

This is obvious, since $ x^k + x^{n-k} \leq 1 + x^n$ for $ 0 \leq k \leq n$.
(EG by factoring, if you want to avoid convexity-calculus arguments ).

As you mentioned, equality holds iff $ x =1$ or $ n = 1$.