# Lower Bound of Natural Logarithm/Proof 2

## Proof

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:

 $\displaystyle f \left({x}\right)$ $=$ $\displaystyle 1 - \dfrac 1 x - \ln x$ $\displaystyle \implies \ \$ $\displaystyle f' \left({x}\right)$ $=$ $\displaystyle \frac 1 {x^2} - \frac 1 x$ Derivative of Constant, Power Rule for Derivatives, Derivative of Natural Logarithm Function $\displaystyle$ $=$ $\displaystyle \frac {1 - x} {x^2}$ $\displaystyle \implies \ \$ $\displaystyle f'' \left({x}\right)$ $=$ $\displaystyle - \frac 2 {x^3} + \frac 1 {x^2}$ Power Rule for Derivatives

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 $f \left({x}\right)$:

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

$\blacksquare$