Prove that $ \ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x}$? How can one prove that 
$$\ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x} \, ?$$

Considering $f(t) = \log t$ and applying Lagrange's theorem in the interval $[x,x+1], x > 0$ you get
$$
f(x+1) - f(x) = f'(\xi_x) (x+1-x) = \frac{1}{\xi_x}, \quad \xi_x \in ]x,x+1[
$$
But, since $f(x+1)-f(x)= \log (x+1)-\log x = \log \left(\frac{x+1}{x}\right)$ and $\frac{1}{\xi_x} > \frac{1}{x+1}$ you can conclude that
$$
\log \left(\frac{x+1}{x}\right) > \frac{1}{x+1}.
$$
