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$$\left({1-x\over1+x}\right)^{1/2}\lt{\ln(1+x)\over \arcsin(x)}\lt1$$ 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.

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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, $$g(x)>\lim_{x\rightarrow0^+}g(x)=0.$$

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    $\begingroup$ (+1) the cancellation of $\frac1{1+x}$ in the second derivation worked out nicely. $\endgroup$ – robjohn Jan 25 '19 at 7:32
  • $\begingroup$ 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. $\endgroup$ – robjohn Jan 25 '19 at 7:36
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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 $$ \sqrt{\frac{1-x}{1+x}}\le\frac{\log(1+x)}{\arcsin(x)}\le1\tag7 $$

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

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