# Why $\frac{\sqrt{x}-1}{\sqrt{x}+1} \leq \exp{\left(-\frac{C}{\sqrt{x}}\right)}$?

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

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?

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 $$\ln \left( \frac{1-y}{1+y} \right) = - \int_0^y \frac{2}{1-t^2} \,dt \le - \int_0^y 2 \, dt = -2y \, .$$

• 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 $C=2$.
– KCd
Mar 13, 2021 at 19:56
• @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? Mar 13, 2021 at 20:01
• 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).
– KCd
Mar 13, 2021 at 20:06
• @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. Mar 13, 2021 at 20:07
• @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 $(1)$. Mar 13, 2021 at 20:33

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.