Show $1-1/x <\ln(x) 1$?

Question

Show $1-1/x <\ln(x) <x-1$ for all $x>1$?

My attempt:

(1) $\ln (x)<x-1$

Suppose $h: \mathbb{R}^+ \to \mathbb{R}, t\mapsto t-1-\ln(t)$, then $h'(t)=0$ if $t=1$. Furthermore $h'(t)>0$ for all $t\in \mathbb{R}^+$.

(2) $1-1/x<\ln(x)$

Suppose $f: \mathbb{R}^+\to \mathbb{R}, t\mapsto \ln(t)-1+\frac{1}{t}$, then $f'(t)=0$ for $t=1$ and $f'(t)>0$ for all $t\in \mathbb{R}^+$.

Therefore the inequality is true.

Answer 1

$\forall t> 1, \,\, \dfrac1{t^2}<\dfrac1t<1$ thus for $x>1$ we have : $$\,\,\int_1^x\dfrac{dt}{t^2}<\int_1^x\dfrac{dt}{t}<\int_1^xdt$$

Answer 2

Let $$f(x)=x-1-\ln(x)$$ then $$f'(x)=1-\frac{1}{x}=\frac{x-1}{x}>0$$ for $x>1$ and $$f(1)=0$$ so $f(x)$ is strictly monotonously increasing .