\(S=\left(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{74}\right)+\left(\frac{1}{75}+\frac{1}{76}+...+\frac{1}{99}\right)\)
Có: \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{74}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)
\(\frac{1}{75}+\frac{1}{76}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{25}{100}=\frac{1}{4}\)
=> \(S>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)=> đpcm