How to prove that $f(x)-x f'(x)\neq 0$? I have this function:
$f(x)=\cosh ^{-1}\left(\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi  x)}{\left(a^2 x^2+1\right)^2}\right),$
where $0<x<\frac{1}{a}$ and $a$ is a positive real number. I want to prove that $\;f(x)-x f'(x)>0$, or at least to prove that $f(x)-x f'(x)\neq 0$. Is there any way to prove this?
 A: I hope and wish that you will receive simpler answers.
If you compose Taylor series around $x=0$ (I skip the intermediate steps)
$$A=\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi  x)}{\left(a^2
   x^2+1\right)^2}=1+2 \pi ^2 x^2+\left(\frac{2 \pi ^4}{3}-8 \pi ^2 a^2\right) x^4+O\left(x^6\right)$$
$$f(x)=\cosh ^{-1}(A)=2 \pi  x-4 \pi  a^2 x^3+O\left(x^5\right)$$
$$f(x)-x f'(x)=8 \pi  a^2 x^3+O\left(x^5\right)$$
Edit
Doing the same around $x=\frac 1a$, we have
$$A=1+2 a^2 \left(x-\frac{1}{a}\right)^2 \sinh ^2\left(\frac{\pi }{a}\right)+O\left(\left(x-\frac{1}{a}\right)^3\right)$$
$$f(x)=\cosh ^{-1}(A)=-2 \left(x-\frac{1}{a}\right) \left(a \sinh \left(\frac{\pi
   }{a}\right)\right)+O\left(\left(x-\frac{1}{a}\right)^3\right)$$
$$f(x)-x f'(x)=2 \sinh \left(\frac{\pi }{a}\right)+O\left(\left(x-\frac{1}{a}\right)^2\right)$$
A: Partial answer
Remarks: It appears that $h''<0$ on $(0, \frac{1}{a})$, if $a \ge 1$ (I think that proof is not very difficult, according to the expression of $h''$ below).
The difficult case is $a < 1$ in which $h''$ can be positive.
Let us prove that if $a \ge 2$, then $f(x) - xf'(x) > 0$ on $(0, \frac{1}{a})$.
We have $f = \ln (1 + h^2 + h\sqrt{h^2 + 2})$ where
$$h(x) = \frac{(1 - a^2x^2)(\mathrm{e}^{\pi x} - \mathrm{e}^{-\pi x})}{(a^2x^2 + 1)\sqrt{2}}.$$
Note: $h(x) > 0$ for each $x$ in $(0, \frac{1}{a})$.
Let $F(x) = f(x) - xf'(x)$. We have $F(0) = 0$.
It suffices to prove that $F'(x) > 0$ on $(0, \frac{1}{a})$.
We have
$$F'(x) = -xf''(x) = \frac{x}{(h^2+2)^{3/2}}[h(h')^2 - (h^2+2)h''].$$
It suffices to prove that $h'' < 0$ on $(0, \frac{1}{a})$.
We have
$$h'' = -\frac{\mathrm{e}^{2\pi x} - 1}{(a^2x^2+1)^3\mathrm{e}^{\pi x}\sqrt{2}}
\left(P \frac{\mathrm{e}^{2\pi x} + 1}{\mathrm{e}^{2\pi x} - 1} + C
\right)$$
where
$$P = 8\pi a^4 x^3 + 8\pi a^2 x,$$
$$C = \pi^2 a^6 x^6+\pi^2 a^4 x^4-\pi^2 a^2 x^2-12 a^4 x^2-\pi^2+4 a^2.$$
Since $\frac{\mathrm{e}^{2\pi x} + 1}{\mathrm{e}^{2\pi x} - 1}
\ge \frac{\mathrm{e}^{2\pi/a} + 1}{\mathrm{e}^{2\pi/a} - 1} \ge \frac{a}{2\pi} + \frac{1}{2}$
for $0 < x < \frac{1}{a}$ (easy to prove), it suffices to prove that
$P (\frac{a}{2\pi} + \frac{1}{2}) + C > 0$ or
\begin{align}
&\pi^2 a^6 x^6 + \pi^2 a^4 x^4  - \pi^2 a^2 x^2 - 12 a^4 x^2  - \pi^2 + 4 a^2\\
&\qquad\qquad + (4 \pi a^4 x^3 + 4 \pi a^2 x) + (4 a^5 x^3 + 4 a^3 x)> 0.
\end{align}
Since $4 \pi a^4 x^3 + 4 \pi a^2 x \ge 8\pi a^3 x^2$ and $4 a^5 x^3 + 4 a^3 x \ge 8a^4 x^2$ by AM-GM, it suffices to prove that
$$\pi^2 a^6 x^6 + \pi^2 a^4 x^4  - \pi^2 a^2 x^2 - 12 a^4 x^2  - \pi^2 + 4 a^2 + 8\pi a^3 x^2 + 8a^4 x^2
> 0$$
or
$$[\pi a^4 x^4+2 \pi a^2 x^2+\pi+ 2 a (1 - a^2x^2)] (\pi a^2 x^2-\pi+2 a) > 0$$
which is clearly true.
