Chứng minh rằng: 

- Hướng dẫn giải

$C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{99}+\frac{1}{100}>1$$\iff C=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\mathrm{...}+\frac{1}{30}\right)+\mathrm{...}+\left(\frac{1}{91}+\frac{1}{92}+\mathrm{...}+\frac{1}{100}\right)$$C>\frac{1}{10}+\frac{10}{20}+\frac{10}{30}+\mathrm{...}+\frac{10}{100}>\frac{10}{20}+\frac{10}{30}+\frac{10}{60}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$