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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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up vote 15 down vote accepted

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 $$ \ln x \geq 1 - \frac{1}{x}. $$

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


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