Cho $0<b<a$. Chứng minh rằng:
    $\frac{a-b}{a}<\ln \frac{a}{b}<\frac{a-b}{b}$.
Chúng ta viết lại bất đẳng thức để làm xuất hiện hàm $F(x)$:
   $\displaystyle\frac{a-b}{a}<\ln \frac{a}{b}<\frac{a-b}{b}\Leftrightarrow (a-b)\tfrac{1}{a}<\ln a-\ln b<(a-b)\tfrac{1}{b}$
   $ \displaystyle \Leftrightarrow \frac{1}{a}< \frac{\ln a-\ln b}{a-b}<\frac{1}{b}$.
Xét hàm số $F(x)=\ln x $ khả vi và liên tục trên $[b,a]\subset (0,+\infty )$ theo định lí Lagrange luôn tồn tại $c\in(b,a)$ sao cho:
   $ \displaystyle F^'(c)=\frac{F(a)-F(b)}{a-b}\Leftrightarrow \frac{1}{c}=\frac{\ln a- \ln b}{a-b}$.
Ta có: $ \displaystyle 0<b<c<a\Leftrightarrow \frac{1}{a}<\frac{1}{c}<\frac{1}{b}\Leftrightarrow \frac{1}{a}< \frac{\ln a-\ln b}{a-b}<\frac{1}{b}$.

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