How do you prove that $\ln(\frac{x+1}{x}) \leq 1/x$? [duplicate]
HINT:
$$f(x)=\log x$$
The MVT gives
$$\log(x+1)-\log x=\frac{1}{c}$$
for $x<c<x+1$.
And $\frac{1}{c}<\frac{1}{x}$.
Can you finish?
you have $$\ln\left(\frac{x+1}x\right)= \ln(x+1) - \ln x = \int_x^{x+1} \frac{dt}t = \frac1c < \frac1 x, \text{ since } x < c < x+1.$$