Lower Bound of Natural Logarithm/Proof 2
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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 $\map f x$:
- $1 - \dfrac 1 x - \ln x \le 0$
$\blacksquare$