Lower Bound of Natural Logarithm

Theorem

 * $\forall x \in \R_{>0}: 1 - \dfrac 1 x \le \ln x$

where $\ln x$ denotes the natural logarithm of $x$.

Proof 1
Let $x > 0$.

Proof 2
Let $x > 0$.

Note that:


 * $1 - \dfrac 1 x \le \ln x$

is logically equivalent to:


 * $1 - \dfrac 1 x - \ln x \le 0$

Let $f \left({x}\right) = 1 - \dfrac 1 x - \ln x$.

Then:

Note that $f' \left({1}\right) = 0$.

Also, $f'' \left({1}\right) < 0$.

So by the Second Derivative Test, $x = 1$ is a local maximum.

On $\left ({0 \,.\,.\, 1} \right)$:
 * $f' \left({x}\right) > 0$

By Derivative of Monotone Function, $f$ is strictly increasing on that interval.

On $\left ({1 \,.\,.\, +\infty} \right)$:
 * $f'\left({x}\right) < 0$

By Derivative of Monotone Function, $f$ is strictly decreasing on that interval.

So $x = 1$ yields a global maximum, at which by Logarithm of 1 is 0:


 * $f \left({1}\right) = 1 - 1 - 0 = 0$

That is:
 * $\forall x > 0: f \left({x}\right) \le 0$

and so by definition of $f \left({x}\right)$:


 * $1 - \dfrac 1 x - \ln x \le 0$

Illustration

 * LowerBoundForLn.png

Also see

 * Bounds of Natural Logarithm