Prove that $ \ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x}$? How can one prove that 
$$\ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x} \, ?$$
 A: 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
$$
\log \left(\frac{x+1}{x}\right) > \frac{1}{x+1}.
$$
A: Hint:
1) Consider $f(x)=\ln\left(\dfrac{1+x}{x}\right)-\dfrac{1}{1+x}$, on its domain.
2) Prove that (when it's defined) $$f'(x)=-\frac{1}{x(1+x)^2}.$$
3) Deduce the variations of $f$.
4) Show that $$\lim_{x\to -\infty} f(x)=\lim_{x\to +\infty} f(x)=0.$$
5) Conclude.
A: Letting $u=1/x$ (with $x\gt0$ or $x\lt-1$, so that $u\gt-1$ but $u\not=0$), we have
$$\ln\left(1+{1\over x}\right)\gt{1\over1+x}\iff\ln(1+u)\gt{1\over1+1/u}
\iff\int_1^{1+u}{dt\over t}\gt {u\over1+u}$$
But the final inequality is guaranteed by the fact that $1/t$ is a decreasing function for $t\in(0,\infty)$. (For $u\gt0$ this is easy to see; for $-1\lt u\lt0$ it requires some care with minus signs.)
A: Suppose $x>0$. Noting that $f(t)=\frac1t$ is decreasing in $[x,x+1]$, one has
$$ \ln(x+1)-\ln x=\int_x^{x+1}\frac1tdt>\int_x^{x+1}\frac1{x+1}dt=\frac1{x+1}$$
or
$$ \ln\frac{x+1}{x}>\frac1{x+1}. $$
A: You can note that
$$
\frac{x}{1+x}=1-\frac{1}{1+x}
$$
Setting $t=1/(1+x)$, the inequality becomes
$$
-\ln(1-t)>t
$$
Consider $f(t)=t+\ln(1-t)$ defined for $t<1$. Then
$$
f'(t)=1-\frac{1}{1-t}=-\frac{t}{1-t}
$$
and so $f$ has a maximum at $0$. Since $f(0)=0$, the inequality $f(t)<0$ is satisfied as soon as $t\ne0$, which is true for our choice of $t=1/(1+x)$.
A: This is a heuristic:
Assuming $x>0,$ we have $$x\log\left(\frac{1+x}{x}\right)>\left(\frac{1+x}{x}\right)^{-1}.$$ Letting $A=1+1/x,$ we see that we need to show that $$xA\log A>1.$$ Now, since $A>1$ for all $x>0,$ 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 $x>0.$
