Prove that $ \ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x}$? How can one prove that 
$$\ln\left( \frac{1+x}{x}\right) &gt;\frac{1}{1+x} \, ?$$
Considering $f(t) = \log t$ and applying Lagrange's theorem in the interval $[x,x+1], x &gt; 0$ 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 $\frac{1}{\xi_x} &gt; \frac{1}{x+1}$ you can conclude that

$$
\log \left(\frac{x+1}{x}\right) &gt; \frac{1}{x+1}.
$$
This is a heuristic:

Assuming $x&gt;0,$ we have $$x\log\left(\frac{1+x}{x}\right)&gt;\left(\frac{1+x}{x}\right)^{-1}.$$ Letting $A=1+1/x,$ we see that we need to show that $$xA\log A&gt;1.$$ Now, since $A&gt;1$ for all $x&gt;0,$ but so that $A\to 1$ as $x\to \infty,$ and that $\log A$ therefore goes to $0$ in this limit, we need only show that $x\log A$ stays more than $1$ as $x\to \infty.$ That is, the limiting value must not fall below $1.$ But $$x\log A=\log\left(1+1/x\right)^x,$$ which indeed goes to $1$ near infinity, so that the inequality always holds for $x&gt;0.$