Lower Bound of Natural Logarithm

Theorem
Let $\ln y$ be the natural logarithm of $y$ where $y \in \R_{> 0}$.

Then $\ln$ satisfies the inequality:


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

Proof
Let $y > 0$.

Note that:


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

is logically equivalent to:


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

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

Then:

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

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

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

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

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

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

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

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


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

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

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


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

Also see

 * Upper Bound of Natural Logarithm