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?

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
$$A=1+2 a^2 \left(x-\frac{1}{a}\right)^2 \sinh ^2\left(\frac{\pi }{a}\right)+O\left(\left(x-\frac{1}{a}\right)^3\right)$$
$$f(x)=\cosh ^{-1}(A)=-2 \left(x-\frac{1}{a}\right) \left(a \sinh \left(\frac{\pi
   }{a}\right)\right)+O\left(\left(x-\frac{1}{a}\right)^3\right)$$
$$f(x)-x f'(x)=2 \sinh \left(\frac{\pi }{a}\right)+O\left(\left(x-\frac{1}{a}\right)^2\right)$$
