I am trying to get an good estimate of the function $f(x) = \frac{\cosh(ax)}{\sinh^2(bx)}$ when $x>0$ and $a, b$ are positive constant with $a\geq b.$ I formulated $f(x)\leq \frac{a}{(bx)^2}$. I have tried graphically and have seen that not for every $a,b$ satisfying $a\geq b$, the bound hold. Supposing that $a> > b$, then the bound is false. Any suggestion how to get the estimate or any improved estimate of the function $f(x)$. Any help will be appreciated.
Update: I have tried forming the function $g(x)= bx - \sinh(bx)(\frac{cosh(ax)}{a})^{-1/2}$ and tried to show the function is decreasing and therefore $g(x)<g(0)$ and hence the result. But showing the function is decreasing with some condition on $a$ and $b$ is getting tuffer.
If you use series expansion followed by long division, you should get close to $x=0$
$$f(x) = \frac{\cosh(ax)}{\sinh^2(bx)}$$ $$f(x)=\frac{1}{b^2 x^2}+\left(\frac{a^2}{2 b^2}-\frac{1}{3}\right)+x^2 \left(\frac{a^4}{24 b^2}-\frac{a^2}{6}+\frac{b^2}{15}\right)+O\left(x^4\right)$$ but, stricto sensu, this is not a Taylor series.