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