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 $\map f x = 1 - \dfrac 1 x - \ln x$.

Then:

Note that $\map {f'} 1 = 0$.

Also, $\map {f''} 1 < 0$.

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

On $\openint 0 1$:
 * $\map {f'} x > 0$

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

On $\openint 1 \to$:
 * $\map {f'} x < 0$

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:


 * $\map f 1 = 1 - 1 - 0 = 0$

That is:
 * $\forall x > 0: \map f x \le 0$

and so by definition of $\map f x$:


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