# prove the inequality $\left({1-x\over1+x}\right)^{1/2}\lt{\ln(1+x)\over \arcsin(x)}\lt1$ for $x\in (0,1)$

$$\left({1-x\over1+x}\right)^{1/2}\lt{\ln(1+x)\over \arcsin(x)}\lt1$$ for $$x\in(0,1)$$
My attempt:
For the upper bound, I took the derivative of $${\ln(1+x)\over \arcsin(x)}$$ and found out its a decreasing function over $$(0,1)$$. So I can found the upper bound less than 1 by taking the limit as $$x\to0$$. $$\lim_{x\to0}{\ln(1+x)\over \arcsin(x)}=\lim_{x\to0}\left({1-x\over1+x}\right)^{1/2}=1$$ The lower bound, in this case, should be $$\lim_{x\to1}{\ln(1+x)\over \arcsin(x)}={2\ln2\over\pi}$$. Instead the lower bound is $$\left({1-x\over1+x}\right)^{1/2}$$. How could I prove that? Noted that the term $$\left({1-x\over1+x}\right)^{1/2}$$ does appear when I was calculating the limit for the upper bound.

The right inequality.

Let $$f(x)=\arcsin{x}-\ln(1+x).$$

Thus, $$f'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x}=\frac{\sqrt{1+x}-\sqrt{1-x}}{(1+x)\sqrt{1-x}}=\frac{2x}{(1+x)\sqrt{1-x}(\sqrt{1+x}+\sqrt{1-x})}>0.$$ Thus, $$f(x)>\lim_{x\rightarrow0^+}f(x)=0.$$ The left inequality.

Let $$g(x)=\ln(1+x)-\sqrt{\frac{1-x}{1+x}}\arcsin{x}.$$

Thus, $$g'(x)=\frac{1}{1+x}-\left(\sqrt{\frac{1-x}{1+x}}\right)'\arcsin{x}-\sqrt{\frac{1-x}{1+x}}\cdot\frac{1}{\sqrt{1-x^2}}=$$ $$=-\left(\sqrt{\frac{2}{1+x}-1}\right)'\arcsin{x}>0.$$ Id est, $$g(x)>\lim_{x\rightarrow0^+}g(x)=0.$$

• (+1) the cancellation of $\frac1{1+x}$ in the second derivation worked out nicely. – robjohn Jan 25 '19 at 7:32
• I think the fact that $\frac{1-x}{1+x}$ is decreasing is pretty obvious, just as much as $\frac2{1+x}-1$, but that's up to the beholder. – robjohn Jan 25 '19 at 7:36

For $$0\lt x\lt1$$, $$\log(1+x)=\overbrace{\int_0^x\frac{\mathrm{d}t}{1+t}\le\int_0^x\,\mathrm{d}t}^{\normalsize t\ge0}=x\tag1$$ and $$\arcsin(x)=\overbrace{\int_0^x\frac{\mathrm{d}t}{\sqrt{1-t^2}}\ge\int_0^x\,\mathrm{d}t}^{\normalsize t\ge0}=x\tag2$$ Thus, we get $$\frac{\log(1+x)}{\arcsin(x)}\le\frac xx=1\tag3$$ Furthermore, for $$0\lt x\lt1$$, $$\log(1+x)=\overbrace{\int_0^x\frac{\mathrm{d}t}{1+t}\ge\int_0^x\frac{\mathrm{d}t}{1+x}}^{\normalsize t\le x}=\frac{x}{1+x}\tag4$$ and $$\arcsin(x)=\overbrace{\int_0^x\frac{\mathrm{d}t}{\sqrt{1-t^2}}\le\int_0^x\frac{\mathrm{d}t}{\sqrt{1-x^2}}}^{\normalsize t\le x}=\frac{x}{\sqrt{1-x^2}}\tag5$$ Thus, we see that $$\frac{\log(1+x)}{\arcsin(x)}\ge\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}}\tag6$$ Combining $$(3)$$ and $$(6)$$ yields $$\sqrt{\frac{1-x}{1+x}}\le\frac{\log(1+x)}{\arcsin(x)}\le1\tag7$$

• I don't really understand the inequality in (3) and (4) – clement Jan 25 '19 at 7:33
• @YibeiHe: I have expanded on them. Hopefully, it's clear now. – robjohn Jan 25 '19 at 7:39

Through the substitution $$x=\sin\theta$$ we just have to prove that $$\log(1+\sin\theta)<\theta,\qquad \theta\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}<\log(1+\sin\theta)$$ hold over $$(0,\pi/2)$$. The former inequality is trivial as a consequence of $$\log(1+\sin\theta)<\sin\theta<\theta$$.
The latter is a consequence of $$\log(1+\sin\theta)>\tan(\theta)\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}$$ which on its turn is equivalent to $$\log(1+\cos\theta)>\frac{1}{2}\left(1-\tan^2\frac{\theta}{2}\right)$$ or to $$\log\left(1+\frac{1-t^2}{1+t^2}\right) > \frac{1}{2}(1-t^2)$$ or to $$\log\left(\frac{2}{1+t}\right)>\frac{1-t}{2}$$ for $$t\in(0,1)$$, which is a consequence of the convexity of $$-\log$$.