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.
 A: 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, $$g(x)>\lim_{x\rightarrow0^+}g(x)=0.$$
A: 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
$$
\sqrt{\frac{1-x}{1+x}}\le\frac{\log(1+x)}{\arcsin(x)}\le1\tag7
$$
A: 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 $-\log$.
