Prove that $\frac{\arccos(x)}{\sqrt{1-x}}$ is decreasing I want to show that $$f(x):=\frac{\arccos(x)}{\sqrt{1-x}}$$ is a decreasing function. I tried to differentiate but than I have to show that
$$\arccos(x)\sqrt{1+x} \leq 2 \sqrt{1-x}.$$
Everytime I tried something I ended up with a even more complex problem or in general something like finding roots of some nonlinear functions. E.g. I could transform the above inequality to 
$$\int_{-1}^x2\arccos(y)\mathop{}\!\mathrm{d}y +(1-x)\arccos(x)\leq 2\pi,$$
but it didn't helped me a lot.
Is there a way to prove analytically that the function $f$ is decreasing?
EDIT: The domain is $-1\leq x\leq 1$. This is valid, since
$$ \lim_{x\to 1^{-}}f(x)=\sqrt{2}.$$
 A: By $x=\cos y$ with $y\in(0,\pi)$ we obtain
$$g(y)=f(\cos y)=\frac{y}{\sqrt{1-\cos y}} \implies g'(y)=\frac{2-2\cos y-y\sin y}{2(1-\cos y)^\frac32}$$
which is positive, indeed we have that for $h(y)=2-2\cos y-y\sin y$


*

*$h(0)=0$

*$h'(y)>0$
therefore $g(y)$ in increasing and $f(x)$ is decreasing.
A: This was my attempt, though I might have missed something (Too long for a comment):
We have $$f(x)=\frac{\arccos(x)}{\sqrt{1-x}} \qquad x \in [-1;1]$$
Let's show that the first derivative is less than zero in the given domain.
$$f^{\prime}(x)= \dfrac{\arccos\left(x\right)}{2\left(\sqrt{1-x}\right)^3} - \dfrac{1}{\sqrt{1-x}\sqrt{1-x^2}}$$
Now,
$$f^{\prime}(x)< 0 \implies \dfrac{\arccos\left(x\right)}{2\left(\sqrt{1-x}\right)^3} < \dfrac{1}{\sqrt{1-x}\sqrt{1-x^2}}$$
We'll analyse the behaviour at the endpoint later, so for now, let us assume $\sqrt{1-x} \neq 0$. Then we multiply both sides and get
$$f^{\prime}(x)< 0 \implies \dfrac{\arccos\left(x\right)}{2\left(\sqrt{1-x}\right)^2} < \dfrac{1}{\sqrt{1-x^2}} \implies \dfrac{\arccos\left(x\right)}{2\left|{1-x}\right|} < \dfrac{1}{\sqrt{1-x^2}} $$
Now let $x = \cos z$. The range of cosine is exactly the set of permissible values for $x$ so we can make this substitution. This gives:
$$\dfrac{z}{2\left|{1-x}\right|} < \dfrac{1}{\sqrt{1-\cos^2 x}} \implies z|\sin z|<2|1-\cos z| $$
Since the positive case and the negative case yield the same results, we can drop the absolute value and consider the positive case:
$$z\sin z + 2\cos z < 2 $$
We note that if we let  $\space h(z) = z\sin z + 2\cos z\space $   then
$$ \sup_{z \in [-1,1]} h(z) = 2 \qquad \textrm{(which is attained when $z=0$)}$$
Which tells us that the inequality holds for all $z \neq 0$. 
Now, $z=0 \implies x = 1$. Taking the limit of the derivative as $x$ approaches $1$ gives a discontinuity (limit doesn't exist) which means it is not in the domain of the derivative. Besides, the first derivative test does not require that the derivative be defined at the endpoints.
Thus we have shown that
$$f'(x) < 0 $$ This proves that $f(x)$ is decreasing.
A: In the region $-1<x<1$
we set $x=\cos(t)$. With $t$ going from $0$ to $\pi$ we have $x$ going from $1$ to $-1$ in a bijective relation. Hence decreasing in $x$ means increasing in $t$.
With this substitution the function in question can be written as
$$f(x) = \frac{\operatorname{arccos}(x)}{\sqrt{1-x}} = g(t)=\frac{\operatorname{arccos}(\cos(t))}{\sqrt{1-\cos(t)}}  \\= \frac{t}{\sqrt{2\sin(\frac{t}{2})^2}}=\frac{\sqrt{2}}{\operatorname{sinc}\left(\frac{t}{2}\right)}$$
where $\operatorname{sinc}(z)=\frac{\sin{z}}{z}$.
Since  $\operatorname{sinc}(z)$ is decreasing in the interval $0<z<\pi$, its inverse is increasing. Hence $g(t)$ is increasing in the interval $0\le t \le \pi$ which means that $f(x)$ is decreasing in the interval $-1\le x \le 1$.
QED.
The region $x>1$
was not mentioned in the OP. But it can be treated similarly starting from
$$f(x>1) = \frac{\operatorname{arccosh}(x)}{\sqrt{x-1}}$$
Substituting $x=\operatorname{cosh}(t)$ so that $x \in [1,\infty)$ corresponds to $t \in [0,\infty)$ gives finally
$$f(x>1) = g(t) = \sqrt{2} \frac{\frac{t}{2}}{ \sinh (\frac{t}{2})}$$
which is obviously decreasing in $t$ and hence $fx)$ is decresing in $x$.
QED.
Discussion
The problem can be generalized. The function 
$$f(x,a) = \frac{\operatorname{arccos}(x)}{(1-x)^a}$$
is decreasing in the range $|x|\le 1$ if $a \le \frac{1}{2}$.
