\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}=\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)\)
\(\ge\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{15+12+10}{60}=\frac{37}{60}>\frac{30}{60}=\frac{1}{2}\)
Ta có :
\(\frac{1}{31}>\frac{1}{60}\)
\(\frac{1}{32}>\frac{1}{60}\)
\(\frac{1}{33}>\frac{1}{60}\)
\(...\)
\(\frac{1}{59}>\frac{1}{60}\)
\(\frac{1}{60}=\frac{1}{60}\)
\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}.30=\frac{1}{2}\left(ĐPCM\right)\)
