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?

Expand $\ln(1+\frac{1}{x-1})=\ln(\frac{x}{x-1})$ using Taylor series. You will get:
$$\ln(\frac{x}{x-1})=\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2+O(x^3)$$
Now you have to show that:
$$\frac{1}{x-1}-\frac12\left(\frac{1}{x-1}\right)^2\geq\frac1x$$
$$\frac{2x-3}{2\cdot(x-1)^2}\geq\frac1x$$
Which is equivalent to:
$$x>1$$
