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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.

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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 $y = \ln x$,

$$1 – \frac{1}{x} \leq \ln x$$

TADA!

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