Question

Cho A=\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{20}\). Chứng minh A>\(\dfrac{27}{20}\)

Answer 1

Ta có : $\\$ $\dfrac{1}{4} > \dfrac{1}{20} \\ \dfrac{1}{5} > \dfrac{1}{20} \\ \dfrac{1}{6} > \dfrac{1}{20} \\ ... \\ \dfrac{1}{20}=\dfrac{1}{20} \\ \Rightarrow \dfrac{1}{4} + \dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{20} > \dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20} \\  \Rightarrow \dfrac{1}{4} + \dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{20} >  \dfrac{17}{20} > \dfrac{27}{20} \\ \Rightarrow đpcm$