Upper Bound of Natural Logarithm/Proof 1

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Theorem

Let $\ln x$ be the natural logarithm of $x$ where $x \in \R_{>0}$.


Then:

$\ln x \le x - 1$


Proof

From Logarithm is Strictly Concave:

$\ln$ is (strictly) concave.

From Mean Value of Concave Real Function:

$\ln x - \ln 1 \le \paren {\dfrac \d {\d x} \ln 1} \paren {x - 1}$

From Derivative of Natural Logarithm:

$\dfrac \d {\d x} \ln 1 = \dfrac 1 1 = 1$

So:

$\ln x - \ln 1 \le \paren {x - 1}$

But from Logarithm of 1 is 0:

$\ln 1 = 0$

Hence the result.

$\blacksquare$


Sources