<s>[INST] Is $f(x)= \frac{1}{x} - \frac{1}{e^x-1}$ monotonic? We have $\displaystyle f(x)= \frac{1}{x} - \frac{1}{e^x-1}$, additionally $f(0)=\frac{1}{2}$. Determine whether $f(x)$ is monotonic.
I tried to do this by checking if $f'(x)<0$, however it does not look very helpful. I don't know if it's tricky, or I am just blind to something obvious here.
Thanks for any hints in advance!
 [/INST] Observe that 
$$f(x) = \frac{1}{x} - \frac{1}{2} \coth \frac{x}{2} + \frac{1}{2}.$$ 
Hence 
$$f'(x) = -\frac{1}{x^2} + \frac{1}{4} \mathrm{cosech}^2 \frac{x}{2}.$$
So it is sufficient to prove that $4\sinh^2 \frac{x}{2} \ge x^2$ for $x \ge 0$. This follows from $\sinh y \ge y$ for $y \ge 0$.
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