Simple proof of $\ln x \leq x-1$

Question

Is this simple proof for $\ln x \leq x-1$ valid?

Proof:

Since $\ln$ is concave, let $y, x \in \mathbb{R}_{++}.$ We have that $\ln(y) \leq \ln(x) + \frac{d\ln x}{dx}(y-x)$. Since this is valid $\forall x, y \in \mathbb{R}_{++}$, it must be valid for an arbitrary $y$ and $x=1$. Then, we get that $\ln(y) \leq \frac{1}{x}(y-x) \Rightarrow \ln(y) \leq y-1$.

Thanks

Answer 1

A simply way is to consider $f(x)=x-\ln x -1$ for $x>0$ and note that

• $f'(x)=1-\frac1x=0 \implies x=1 \quad f(1)=0$
• $f''(x)=\frac1{x^2}>0$ then $f(1)$ is a minimum

Answer 2

The proof given in the OP is acceptable given $(i)$ $\log(1)=0$, $(ii)$ $\frac{d\log(x)}{dx}=\frac1x$ and $(iii)$ the logarithm is concave.

Another way forward it to note that since your tacit definition of $\log(x)$ is $\log(x)=\int_1^x \frac1t\,dt$, $x>0$, then we have from elementary analysis of the integral

$$\frac{x-1}{x}=\int_1^x \frac1x \,dt\le \log(x)\le \int_1^x \frac11\,dt=x-1$$