# How can I prove that $\frac {x-1}{x}\leq \log x\leq x-1$

How can I prove that $$\frac {x-1}{x}\leq \log x\leq x-1$$ for any $x>1$

The inequalities are tight for $x=1$.

Then for $x>1$, deriving,

$$\frac1{x^2}\le\frac1x\le1$$ holds.

I will do one inequality and you can do the other in a similar way.

Let $f(x)=\log x -\frac{x-1}{x}=\log x-1+\frac{1}{x}$. Note that $f(1)=0$ and $f'(x)=\frac{1}{x}-\frac{1}{x^2}$. So $f'(x)\gt 0$ if $x\gt 1$, the function $f$ is increasing.

$\log x = \int_{1}^{x} \frac{1}{x} dx$

$\frac{1}{x} < 1$ for all $x>1$

therefore, $\log x = \int_{1}^{x} \frac{1}{x} dx < \int_{1}^{x} 1 dx = x-1$

for the left side:

note that $\frac{1}{x^2} < \frac{1}{x}$ for all $x>1$

A sketch: $\log x$ is concave, so its graph lies below its tangent line at any point, including $(1,0)$. This gives the right-hand inequality. The left-hand inequality follows from the same fact, after a variable change $x \rightarrow {1 \over x}$.

For $\frac{x-1}{x}\leq \log x$, denote $f(x)=\log x-\frac{x-1}{x}$. You have $f(1)=0$. $$f'(x)=\frac{1}{x}-\frac{1}{x^2}\leq 0\Leftrightarrow x\in [1,\infty)$$ and so $f$ is increasing. For $$\log x\leq x-1$$ again if we denote $g(x)=x-1-\log x$ we have $g(1)=0$ and $g'(x)\ge 0$ for $x\ge 1$

For any $x > 1$, the mean value theorem gives $$\inf_{c \in (1,x)} \frac{1}{c} (x-1) \leq \log(x) - \log(1) \leq \sup_{c \in (1,x)} \frac{1}{c} (x-1)$$ $$\frac{x-1}{x} \leq \log(x) \leq x -1 .$$