Lower Bound of Natural Logarithm/Proof 2

Theorem

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

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

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:

\(\ds \map f x\)

\(=\)

\(\ds 1 - \dfrac 1 x - \ln x\)

\(\ds \leadsto \ \ \)

\(\ds \map {f'} x\)

\(=\)

\(\ds \frac 1 {x^2} - \frac 1 x\)

Derivative of Constant, Power Rule for Derivatives, Derivative of Natural Logarithm Function

\(\ds \)

\(=\)

\(\ds \frac {1 - x} {x^2}\)

\(\ds \leadsto \ \ \)

\(\ds \map {f} x\)

\(=\)

\(\ds - \frac 2 {x^3} + \frac 1 {x^2}\)

Power Rule for Derivatives

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 $\map f x$:

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

$\blacksquare$

Illustration