Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions  $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?

We know that for $|x| < 1$, $$\log(1-x) = -x-\frac{1}{2}x^2 - \frac{1}{3}x^3 - \dotsb.$$
Since $$1 + \frac{1}{x-1} = \frac{x-1+1}{x-1} = \frac{x}{x-1},$$
we see that 
$$\log \left(1+\frac{1}{x-1}\right) = \log \left(\frac{x}{x-1}\right) = -\log\left(\frac{x-1}{x}\right) = -\log\left(1-\frac{1}{x}\right).$$
So if $x \ge 2$ then $0 < \frac{1}{x} < 1$, so we can apply the power series:
$$-\log\left(1-\frac{1}{x}\right) = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \dotsb > \frac{1}{x}.$$
