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I am reading that

$$\frac{\sqrt{x}-1}{\sqrt{x}+1} \leq \exp{\left(-\frac{C}{\sqrt{x}}\right)}.\tag{1}$$

Doing the graph, we can see that $\exp{\left(-\frac{C}{\sqrt{x}}\right)}$ upper bounds $\frac{\sqrt{x}-1}{\sqrt{x}+1}$ when $C$ is small enough. Could you please someone provide some analytic explanation on how we can get $(1)$ and what is the maximum value of $C$ to achieve the upper bound?

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2 Answers 2

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Only the case $x > 1$ is interesting, so we can substitute $y = 1/\sqrt x$. Then the inequality becomes $$ \frac{1-y}{1+y} \le e^{-Cy} $$ or $$ \ln \left( \frac{1-y}{1+y} \right) \le -Cy $$ for $0 < y < 1$. Using the Taylor series for the logarithm we get $$ \tag{*} \ln \left( \frac{1-y}{1+y} \right) = -2(y + \frac{y^3}{3} + \frac{y^5}{5} + \ldots) \le -2y. $$ It follows that the desired estimate holds with $\boxed{C=2}$. That is the best possible constant because $$ \lim_{y \to 0} \frac 1y \cdot \ln \left( \frac{1-y}{1+y} \right) = -2 \, . $$

Remark: An alternative method to obtain $(*)$ is $$ \ln \left( \frac{1-y}{1+y} \right) = - \int_0^y \frac{2}{1-t^2} \,dt \le - \int_0^y 2 \, dt = -2y \, . $$

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  • $\begingroup$ Up to linear terms in the initial inequality (before taking logarithms), the left side is $1-2y/(1+y) = 1-2y+O(y^2)$ and the right side is $e^{-Cy} = 1-Cy + O(y^2)$. So the desired inequality holds for small $y$ if $C < 2$, it does not hold for small $y$ if $C > 2$, and more care is needed if $C=2$. $\endgroup$
    – KCd
    Mar 13, 2021 at 19:56
  • $\begingroup$ @KCd: I have shown that it holds for all $y \in (0, 1)$ if $C \le 2$, and that is does not hold for $C > 2$ (using a limiting argument similar as yours). Is anything wrong with that? $\endgroup$
    – Martin R
    Mar 13, 2021 at 20:01
  • $\begingroup$ No. I was just pointing out that without have to make careful estimates, we can see that $C=2$ is the critical case, with larger and smaller $C$ being easy to figure out qualitatively (works for all small $y$ or fails for all small $y$ without specifying what “small” means precisely). $\endgroup$
    – KCd
    Mar 13, 2021 at 20:06
  • $\begingroup$ @KCd: Exactly, that was the idea. The Taylor series of the logarithm turned out to be a simple way to verify the inequality on the whole interval. There may be other methods. $\endgroup$
    – Martin R
    Mar 13, 2021 at 20:07
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    $\begingroup$ @Thoth: I am not sure what you are looking for, Taylor series and limits are analytical methods :) – I just added another way to obtain the estimate $(*)$, which is equivalent to your inequality $(1)$. $\endgroup$
    – Martin R
    Mar 13, 2021 at 20:33
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Observe that $\dfrac{\sqrt{x}-1}{\sqrt{x}+1} = \dfrac{1-\frac{1}{\sqrt{x}}}{1+\frac{1}{\sqrt{x}}}$. Thus if you set $u = \dfrac{1}{\sqrt{x}}$, then you want to prove: $e^{-Cu} \ge \dfrac{1-u}{1+u}$ for small $C > 0$. Consider: $f(u) = e^{-Cu} + \dfrac{u-1}{u+1}$ on $(0,\infty)$. Taking $1$st derivative of $f$: $f'(u) = -Ce^{-Cu} +\dfrac{1}{(u+1)^2} $. Your aim is to show: $f'(u) > 0$ on $(0,1)$ because if $u \ge 1$ then $f(u) > 0$ and the inequality you are trying to prove holds. But now $f'(u) > 0 \iff \dfrac{1}{\sqrt{C}}\cdot e^{\frac{Cu}{2}} \ge u+1$. Replace $C$ by $2C$ in this new inquality to get: $\dfrac{1}{\sqrt{2C}}\cdot e^{Cu} \ge u+1$. Now using the inequality: $e^{Cu} \ge 1+ Cu + \dfrac{C^2u^2}{2} \ge \sqrt{2C}u+\sqrt{2C}$. Consider $g(u) = 1+Cu+\dfrac{C^2u^2}{2}-\sqrt{2C}u - \sqrt{2C}$ on $(0,1)$. Then $g'(u) = C+C^2-\sqrt{2C}> 0 \iff C(1+C)^2 > 2$. Thus if you choose $C^{*} = 2C$ small $> 0$ and $C$ satisfy this condition then you have the inequality you want to show.

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