Lower Bound of Natural Logarithm/Proof 2

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Theorem

Bound of Natural Logarithm

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

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

$\blacksquare$


Illustration

LowerBoundForLn.png