Is $f(x)= \frac{1}{x} - \frac{1}{e^x-1}$ monotonic?

Question

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!

Answer 1

Let $f(x)$ be the function

$$f(x)=\frac{1}{x}-\frac{1}{e^x-1}$$

Then, we have

$$\begin{align} f'(x)&=-\frac1{x^2}+\frac{e^x}{(e^x-1)^2}\\\\ &=\frac{-(e^x-1)^2+x^2e^x}{x^2(e^x-1)^2}\\\\ &=\left(\frac{e^x}{x^2(e^x-1)^2}\right)\left(x^2-4\sinh^2(x/2)\right)\\\\ &\ge 0 \end{align}$$

since the hyperbolic sine satisfies the inequality

$$|\sinh(x)|\ge |x|$$

Answer 2

A function is monotone if its first derivative doesnt change sign. So here $1/x$ is continuously decreasing also $e^x$ is increasing so $\frac{1}{e^x-1}$ is decreasing function hence the function $f(x)$ is nonincreasing montonic function.

Answer 3

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 $y \ge 0$.