# Lower Bound of Natural Logarithm/Proof 2

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## Theorem

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

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

$\blacksquare$