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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.

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    $\begingroup$ Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. $\endgroup$
    – Martin R
    Commented Jan 16, 2021 at 3:01
  • $\begingroup$ Your comment was quite right, I've made a pretty daft error. I've deleted the (incorrect) answer, I'll try and have another go later. $\endgroup$ Commented Jan 16, 2021 at 7:06

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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$.

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  • $\begingroup$ Thank you! Your last inequality without the expansion is just $$(1+x)(1+x+\cdots+x^{n-1}) \le n(1+x^n)$$ which looks very familiar. Is it a variant of some famous inequality, or it's just my imagination? $\endgroup$
    – Neat Math
    Commented Jan 17, 2021 at 1:01
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    $\begingroup$ @NeatMath I think it is just your imagination. Another way to write it that is more symmetric is $(1+x)/(1-x) \leq n ( 1+x^n) / ( 1 - x^n)$. $\endgroup$
    – Calvin Lin
    Commented Jan 17, 2021 at 14:34

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