<s>[INST] 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. 
 [/INST] 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.
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