Calculate infimum of $f(x): (0,\infty) \to \mathbb R$, $f(x) = \ln(e^x-1)+\frac2x-x$
I calculate derivative $$f'(x)= (\ln(e^x-1)+\frac2x-x)' = \frac1{e^x-1}e^x-\frac2{x^2}-1 = \frac{e^x}{e^x-1}-\frac{2-x^2}{x^2}=\\ =\frac{e^xx^2-2e^x+2-e^xx^2+x^2}{(e^x-1)x^2}= 0 <=> x^2+2-2e^x=0 $$
(1) The obvious solution is $x=0$, but how can I prove that there is no more solutions?
If $x=0$ is only solution, then there is no local extremes in $(0,\infty)$ So (if exist) infimum should be $\min{(\lim_{x\to0}f(x),\lim_{x\to\infty}f(x))}$ (there is no local extremas, so $f(x)$ is increasing or decreasing).
$$\lim_{x\to0^+}\ln(e^x-1)+\frac2x-x = +\infty$$
$$\lim_{x\to\infty}\ln(e^x-1)+\frac2x-x = 0$$
(2) So it's look like the result is $0$ is it correct?
(3) If (1) and (2) are true then whether what I wrote is enough?
So you want to show $x^2+2-2e^x \neq 0$ for $x>0$. The Maclaurin series tells you that $2e^x=2+x^2$ plus some more positive terms for positive $x$. So you have this result. So everything looks OK (though I would expect some more details on the limit calculations in a written up solution).