# Intuition behind logarithm inequality: $1 – \frac1x \leq \log x \leq x-1$

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:
$$\log(1+x) = \sum_{i=1}^\infty (-1)^{n+1}\frac{x^n}{n} \leq (-1)^{1+1}\frac{x^1}{1} = x.$$

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.

#### Solutions Collecting From Web of "Intuition behind logarithm inequality: $1 – \frac1x \leq \log x \leq x-1$"

Take the upper bound:
$$\ln {x} \leq x-1$$
Apply it to $1/x$:
$$\ln \frac{1}{x} \leq \frac{1}{x} – 1$$
This is the same as
$$\ln x \geq 1 – \frac{1}{x}.$$

Starting from the fairly well-known,
$$1 – y \leq e^{-y}$$
Rearranging,
$$1 – e^{-y} \leq y$$
Substituting $y = \ln x$,
$$1 – \frac{1}{x} \leq \ln x$$