Proving an inequality without an integral: $\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}$

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 $$\frac {1}{x+1}\leq\frac{df}{dx}(\xi)\leq \frac {1}{x}$$

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.