Upper Bound of Natural Logarithm

Theorem
Let $\ln y$ be the natural logarithm of $y$ where $y \in \R, y > 0$.

Then:
 * $(1): \quad \ln y \le y - 1$
 * $(2): \quad \forall s \in \R: s > 0: \ln x \le \dfrac {x^s} s$

Proof
First, to show that $\ln y \le y - 1$:

From Logarithm is Strictly Increasing and Strictly Concave, $\ln$ is (strictly) concave.

From Mean Value of Concave Real Function:
 * $\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$

From Derivative of Natural Logarithm:
 * $D \ln 1 = \dfrac 1 1 = 1$

So:
 * $\ln y - \ln 1 \le \left({y - 1}\right)$

But from Logarithm of 1 is 0:
 * $\ln 1 = 0$

Hence the result.

Next, to show that $\ln x \le \dfrac {x^s} s$:

The result follows by dividing both sides by $s$.