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?

hint :
$$f(x)=ln(1+\frac{1}{x-1})-\frac{1}{x}\\x≥2\\ f'<0$$f(x) is decreasing function ,but f(x) is above the x axis $$\\f(2)>0,f(\infty)>0\\f(x)>0\\so\\ln(1+\frac{1}{x-1})-\frac{1}{x}≥0\\ln(1+\frac{1}{x-1})≥\frac{1}{x}$$
