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Let $$f(x)=\left(1-\frac{2x}{x+c}\right)^{-n}-\left(1+\frac{x}{c}\right)^{2n}$$ and $$g(x)=\left(1-\frac{2x}{c}\right)^{-n}-\left(1-\frac{x}{c}\right)^{-2n}$$ where $c>0$ and $n>0$ are constants.

I am wondering if $f(x)\leq g(x)$ for $0\leq x <c/2$ (I am actually interested in small positive $x$).

Clearly, this holds with equality when $x=0$ and I think the inequality holds for $0< x <c/2$. This is based on the numerical evaluations as well as the series expansion in Mathematica of $h(x)=g(x)-f(x)$ which yields:

$$h(x)=\frac{2 n x^3}{c^3}+\frac{3 \left(2 n^2+n\right) x^4}{c^4}+\frac{2 \left(4 n^3+5 n^2+3 n\right) x^5}{c^5}+\ldots$$

However, I am having trouble actually proving this. Any help?

M.B.M.
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    Use mathematica's hint. Try to expand $f-g$ and try to show all coefficients are nonnegative. – Kunnysan Aug 11 '13 at 04:26
  • The binomial theorem gives three of the four terms as power series in $x/c$ with closed form for the $n$-th term. The remaining one is $(1 + \frac{x}{c})/(1 - \frac{x}{c})$ all raised to the $n$th power, for which there is either a binomial coefficient identity to save the day, or something explicit enough that the question of positivity should be answerable. – zyx Aug 11 '13 at 05:51

1 Answers1

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Let's normalise things a bit first.

$$\left(1 - \frac{2x}{x+c}\right)^{-n} = \left(\frac{c-x}{c+x}\right)^{-n} = \left(\frac{1+\frac{x}{c}}{1-\frac{x}{c}}\right)^n.$$

So we have everything clearly as a function of $\frac{x}{c}$, whence we can without loss of generality assume $c = 1$. Then the inequality we want to show becomes

$$\begin{align} \left(\frac{1+x}{1-x}\right)^n - (1+x)^{2n} &\leqslant \frac{1}{(1-2x)^n} - \frac{1}{(1-x)^{2n}}\\ \iff (1-x^2)^n - (1-x^2)^{2n} &\leqslant \left(\frac{(1-x)^2}{1-2x}\right)^n - 1\\ \iff (1-x^2)^n\left(1 - (1-x^2)^{n}\right) &\leqslant \left(1+\frac{x^2}{1-2x}\right)^n - 1 \end{align}$$

Now, for $0 \leqslant x < \frac12$, we have $x^2 \leqslant \frac{x^2}{1-2x}$, and by Bernoulli's inequality, we have

$$(1+x^2)^n - 1 \geqslant n\cdot x^2.$$

On the other hand, also by Bernoulli's inequality, we have

$$1 - (1-x^2)^n \leqslant n\cdot x^2,$$

and since that is multiplied with $(1-x^2)^n \leqslant 1$, we conclude

$$\begin{align}(1-x^2)^n\left(1 - (1-x^2)^{n}\right) &\leqslant (1-x^2)^n\cdot n\cdot x^2\\ &\leqslant n\cdot x^2\\ &\leqslant (1 + x^2)^n - 1\\ &\leqslant \left(1+\frac{x^2}{1-2x}\right)^n - 1. \end{align}$$

Daniel Fischer
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  • Very nice answer -- so simple! I knew about the Bernoulli's inequality, just didn't see its application here... Thanks! – M.B.M. Aug 11 '13 at 23:11