Proving monotonicity of a function $g$ Let $r\ge 1$. For $r-1\leq x \le r+1$ we define $f(x)=\arccos\left(\frac{x^2 + r^2 - 1}{2 r x}\right)$. Now let $g:[1,\infty)\to\mathbb{R}$ be given by
$$g(r)=\frac{\int_{r-1}^{r+1}{r(f(x))^2\,dx}}{\int_{r-1}^{r+1}{ f(x)\,dx}}.$$
I want to show that $g$ is a monotonically decreasing function.
Notes

*

*Numerically, this looks to be almost certainly true.

*I have already proved that $g$ has some interesting properties; for instance $g(1)=\pi-2$ and $\lim_{r\to \infty} g(r)=\frac{8}{3\pi}$. But ideally I'd like to show that the function decreases monotonically between these two values.

*I've tried looking at the derivative - but it seems too ghastly to be useful!

 A: $\def\c{\cos\phi}\def\L{\operatorname{\cal L}_r}\def\F{\operatorname F}\def\E{\operatorname E}\def\f{\rho_r}$
Not a proof but a way to simplify the problem.
Let $x=\sqrt{r^2-\sin^2\phi}-\cos\phi$. Then:
$$
\frac{x^2+r^2-1}{2xr}=\frac{r^2-\sin^2\phi+\cos^2\phi-2\cos\phi\sqrt{r^2-\sin^2\phi}+r^2-1}{2r\left(\sqrt{r^2-\sin^2\phi}-\cos\phi\right)}=\sqrt{1-\frac{\sin^2\phi}{r^2}}
$$
so that:
$$\begin{align}
\int_{r-1}^{r+1}\arccos\left(\frac{x^2+r^2-1}{2xr}\right)dx
&=\int_0^\pi\arcsin\left(\frac{\sin\phi}r\right)\,d\left(\sqrt{r^2-\sin^2\phi}-\cos\phi\right)\\
&=-\int_0^\pi\frac{\sqrt{r^2-\sin^2\phi}-\cos\phi}{\sqrt{1-\frac{\sin^2\phi}{r^2}}}
\frac{\cos\phi}rd\phi\\
&=\int_0^\pi\frac{\cos^2\phi}{\sqrt{r^2-\sin^2\phi}}d\phi\\
&=2\int_0^{\pi/2}\frac{\cos^2\phi}{\sqrt{r^2-\sin^2\phi}}d\phi\\
&\stackrel{\sin\phi=r\sin u}=2\int_0^{\arcsin\frac1r}\sqrt{1-r^2\sin^2 u}\;du.\\
%&=2\left[r\operatorname{E}\left(\frac1r\right)-\left(r-\frac1r\right)\operatorname{K}\left(\frac1r\right)\right],
\end{align}$$
Similarly:
$$\begin{align}
\int_{r-1}^{r+1}\left[\arccos\left(\frac{x^2+r^2-1}{2xr}\right)\right]^2dx
&=4\int_0^{\pi/2}\frac{\cos^2\phi}{\sqrt{r^2-\sin^2\phi}}\arcsin\left(\frac{\sin\phi}r\right)d\phi\\
&=4\int_0^{\arcsin\frac1r}u\sqrt{1-r^2\sin^2 u}\;du.\\
%&\stackrel{\sin\phi=r\sin u}=2\\
%&=2\left[r\operatorname{E}\left(\frac1r\right)-\left(r-\frac1r\right)\operatorname{K}\left(\frac1r\right)\right],
\end{align}$$
Introducing:
$$
\f(x)=\sqrt{1-r^2\sin^2 x};\quad \L[\psi(x)]=\int_0^{\arcsin\frac1r}\psi(x)dx
$$
the condition that the function $g(x)$ is decreasing can be written as:
$$
\frac d{dr}\left(\frac{r\L[x\f(x)]}{\L[\f(x)]}\right)<0,
$$
which is equivalent to
$$
\frac{\L\left[\dfrac x{\f(x)}\right]}{\L[x\f(x)]}-\frac{\L\left[\dfrac 1{\f(x)}\right]}{\L[\f(x)]}>1.\tag1
$$
Further introducing:
$$
f(x,r)=\frac{\F(x\arcsin\frac1r\,|\,r^2)}{\arcsin\frac1r},\quad
e(x,r)=\frac{\E(x\arcsin\frac1r\,|\,r^2)}{\arcsin\frac1r},
$$
where $\F(x)$ and $\E(x)$ are the elliptic integrals of the first and second kind, respectvely, the inequality $(1)$ can be written in explicit form as:
$$\frac
{f(1,r)-\int_0^1 f(x,r)\,dx}{e(1,r)-\int_0^1 e(x,r)\,dx}
-\frac{f(1,r)}{e(1,r)}>1.\tag2
$$
The expression $(2)$ looks not very complicated but it is not obvious how to prove it. Yet it can be easily shown that the left-hand-side tends to $1$ as $r$ goes to infinity.
