Quicker way of doing this inequality Prove that for $x \in (0,1)$
$$ \frac{\sqrt{1-x}}{\sqrt{1+x}} \leq \frac{\ln(1+x)}{\arcsin(x)} \leq 1$$
I've figured out this inequality by analyzing the derivatives. This took a rather long amount of time and I was wondering if there is a short trick to proving this. I've observed that the left hand side is the quotient of the logarithm and arcsin derivatives but I wasn't able to figure out how to make quick work of this problem with the use of that fact.
 A: METHOD $1$: Using Standard Inequalities That Can Be Obtained Without Calculus
For the right-side of the inequality of interest, we recall the inequalities
$$\log (1+x)\le x\,\,\dots ,x>-1 \tag 1$$
and 
$$\sin x\le x\,\,\dots,x>0 \tag 2$$

Note that $(1)$ can be established using only the limit definition of the exponential function and Bernoulli's Inequality and $(2)$ can be established using the geometric interpretation of the sine function.

Now, from $(2)$ we have for $x\in (0,1)$
$$x\le \arcsin x \tag 3$$
Putting $(1)$ and $(3)$ together yields
$$\frac{\log (1+x)}{\arcsin x}\le \frac{x}{x}=1$$
which is the right-side of the inequality of interest.  

For the left-side of the inequality of interest, we recall the inequalities
$$\log (1+x)\ge \frac{x}{x+1}\,\,\dots,-1<x \tag 4$$
and 
$$\sin x\ge x\cos x \,\,\dots,0 \le x \le \pi/2 \tag 5$$

Note that $(4)$ can be established using only the limit definition of the exponential function and Bernoulli's Inequality and $(5)$ can be established using the geometric interpretations of the sine and cosine functions.

Now, enforcing the substitution $x=\arcsin t$ in $(5)$ we have for $0\le t<1$
$$\sin \arcsin (t) \ge \arcsin(t) \cos \left(\arcsin (t) \right) \tag 6$$
which implies 
$$t \ge \arcsin (t) \sqrt{1-t^2} \tag 7$$
Therefore for $0\le x\le 1$, we have from $(7)$ 
$$\arcsin (x)\le \frac{x}{\sqrt{1-x^2}} \tag 8$$
Using $(4)$ and $(8)$ yields
$$\frac{\log(1+x)}{\arcsin (x)}\ge \left(\frac{x}{x+1}\right)\left(\frac{\sqrt{1-x^2}}{x}\right)=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$
which is the lef-side of the inequality of interest.

Putting the results together reveals 
$$\frac{\sqrt{1-x}}{\sqrt{1+x}}\le \frac{\log(1+x)}{\arcsin(x)}\le 1$$

METHOD $2$: Using Integral Representations Of The Log And Arcsine Functions
We have the integral representations of the logarithm and arcsine functions
$$\log x=\int_1^x \frac{1}{t}\,dt \tag 9$$
$$\arcsin x=\int_0^x\frac{1}{\sqrt{1-t^2}}\,dt \tag{10}$$
both of which are defined for $x>0$.  
Since both $1/x$ and $1\sqrt{1-x^2}$ are monotonically decrasing, it is trivial to establish the inequalities
$$\frac{x-1}{x}\le\log x\le x-1 \implies \frac{x}{x+1}\le\log (1+x) \le x \tag {11}$$
and 
$$x \le \arcsin(x)\le \frac{x}{\sqrt{1-x^2}} \tag{12}$$
From $(11)$ and $(12)$ we obtain
$$\frac{\sqrt{1-x}}{\sqrt{1+x}}=\frac{x}{(x+1)\frac{x}{\sqrt{1-x^2}}}\le \frac{\log (1+x)}{\arcsin x}\le \frac{x}{x}=1$$
as expected!
A: Actually, you can use the derivatives to prove this inequality. Let $x=\sin t$ for $0< t\le\frac{\pi}{2}$. Then first part is equivalent to
$$ \frac{\sqrt{1-\sin t}}{\sqrt{1+\sin t}}\le \frac{\ln(1+\sin t)}{t}. \tag{1} $$
Note $$  \frac{\sqrt{1-\sin t}}{\sqrt{1+\sin t}}=\tan(\frac{\pi}{4}-\frac{t}{2}). $$
Define
$$ f(t)=\tan(\frac{\pi}{4}-\frac{t}{2})- \frac{\ln(1+\sin t)}{t}. $$
Then it is easy to check
$$ f'(t)=\frac{-t^2-t \cos t+(1+\sin t) \log (1+\sin t)}{t^2 (1+\sin t)}. $$
Define
$$ g(t)=-t^2-t \cos t+(1+\sin t) \log (1+\sin t)$$ 
and then
$$ g'(t)=t (-2+\sin t)+\cos t \log (1+\sin t). $$ 
Noting that for $0< t\le\frac{\pi}{2}$, $\sin t\le t, \ln(1+\sin t)\le\sin t\le t$, we have $g'(t)\le 0$ or $g(t)$ is decreasing. Then for $0< t\le\frac{\pi}{2}$, $g(t)\le g(0)=0$ and hence $f'(t)\le0$ or $f(t)$ is decreasing for $0< t\le\frac{\pi}{2}$. Thus $f(t)\le f(0)=0$ for $0< t\le\frac{\pi}{2}$ and hence (1) is proved. Using the same way, you can prove the second part of the inequality.
