Proving an inequality without an integral: $\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}$ I would like to prove the following inequality without integration; could you help? 
$$\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}, \quad x > 0. $$
I can however differentiate this.
Thanks in advance. 
 A: Lemma:  For all real $x, e^x \ge 1 + x$.
Proof of lemma:  Consider $f(x)=e^x-x-1.$  Its first derivative ($e^x-1$) is $0$ when and only when $x=0$, and its second derivative ($e^x$) is positive for all real $x, $ so $e^x-x-1$ has a global minimum when $x=0\; (f(0)=0$) and $e^x-x-1\ge 0$ for all real $x.$
From the lemma, $\exp\left({\frac 1x}\right)\ge 1+ \frac 1 x,$
which implies (take logarithm of both sides, noting that preserves order) $ \frac 1 x \ge \ln\left(1 + \frac 1 x\right)=\ln\left(\frac{x+1}x\right)=\ln(x+1)-\ln(x),$
which is one part of the desired inequality.
Also from the lemma, $\exp\left(\frac {-1}{1+x}\right)\ge 1-\frac 1 {1+x}=\frac x {1+x}$ implies (taking reciprocal of both sides, reversing the order) $\exp\left(\frac {1}{1+x}\right)\le \frac {1+x} {x}$ so (again, taking logarithm of both sides) $ \frac 1 {1+x} \le \ln \frac{1+x}{x}=\ln (1+x) - \ln(x),$  the other part of the desired inequality.
A: Let $f(x):=\ln(x)$, then the Mean Value Theorem (Differentiation) says that there exists some $\xi\in (x,x+1)$
$$\ln (1+x)- \ln (x) = \frac{\ln (1+x)- \ln (x)}{(1+x)-x}= \frac{f (1+x)- f (x)}{(1+x)-x} =\frac{df}{dx}(\xi) $$
As $\frac{df}{dx}(x)= 1/x$ and as the function  $1/x$ is monotonic we know that
$$\frac {1}{x+1}\leq\frac{df}{dx}(\xi)\leq \frac {1}{x}$$
A: If we write $t=\frac1x$ the inequalities become $${t\over t+1}\leq\ln(1+t)\leq t, t>0$$  Observe that the three expressions are equal when $t=0,$ and then show, by differentiating, that $t-\ln(1+t)$ and $\ln(1+t)-{t\over1+t}$ are increasing functions.
A: The right inequality.
We need to prove that $f(x)\geq0,$ where  $$f(x)=\frac{1}{x}-\ln(1+x)+\ln{x}.$$
Indeed, $$f'(x)=-\frac{1}{x^2}-\frac{1}{1+x}+\frac{1}{x}=-\frac{1}{x^2(1+x)}<0,$$ which says
$$f(x)\geq\lim_{x\rightarrow+\infty}f(x)=\lim_{x\rightarrow+\infty}\left(\frac{1}{x}+\ln\frac{x}{1+x}\right)=0.$$
The left inequality.
We need to prove that $g(x)\geq0,$ where
$$g(x)=\ln(1+x)-\ln{x}-\frac{1}{1+x}.$$
We have
$$g'(x)=\frac{1}{1+x}-\frac{1}{x}+\frac{1}{(1+x)^2}=-\frac{1}{x(1+x)^2}<0,$$
which says $$g(x)\geq\lim_{x\rightarrow+\infty}g(x)=0$$ and we are done! 
A: This is trivial using the definition of $\ln(x)$ as the area below the hyperbola $1/x$ between 1 and $x$.
$\ln(x+1)-\ln(x)$ is the hyperbola area between $x$ and $x+1$, bounded by the rectangles with area $1/(x+1)$ and $1/x$.
