TITLE: Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$
QUESTION [55 upvotes]: One of fundamental inequalities on logarithm is:
$$ 1 - \frac1x \leq \log x \leq x-1 \quad\text{for all $x > 0$},$$
which you may prefer write in the form of
$$ \frac{x}{1+x} \leq \log{(1+x)} \leq x \quad\text{for all $x > -1$}.$$
The upper bound is very intuitive -- it's easy to derive from Taylor series as follows:
$$ \log(1+x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n} \leq (-1)^{1+1}\frac{x^1}{1} = x.$$
My question is: "what is the intuition behind the lower bound?" I know how to prove the lower bound of $\log (1+x)$ (maybe by checking the derivative of the function $f(x) = \frac{x}{1+x}-\log(1+x)$ and showing it's decreasing) but I'm curious how one can obtain this kind of lower bound. My ultimate goal is to come up with a new lower bound on some logarithm-related function, and I'd like to apply the intuition behind the standard logarithm lower-bound to my setting.

REPLY [25 votes]: If you don't already know that $\log(x)=\int_1^x\frac1t\,dt$, one way to define the logarithm function is: $$\text{$\log(x)$ is the area under the curve $y=\frac1t$ from $t=1$ to $t=x$.}$$ The picture below shows that this area is sandwiched between two rectangles, each of width $x-1$. The smaller rectangle has height $1/x$, while the larger one has height $1$. In other words, we have the following inequalities:
$$(x-1)\cdot \frac1x\le\log(x)\le(x-1)\cdot 1$$