Cho \(A = \frac{1}{{31}} + \frac{1}{{32}} + \frac{1}{{33}} + ... + \frac{1}{{60}}.\) Chứng minh rằng: \(A > \frac{3}{5}.\)

Hướng dẫn giải:

Ta có: \(\frac{1}{{31}} > \frac{1}{{40}};\)

\(\frac{1}{{32}} > \frac{1}{{40}};\)

\(\frac{1}{{33}} > \frac{1}{{40}};\)

…

\(\frac{1}{{40}} = \frac{1}{{40}}.\)

Cộng vế theo vế ta được:

\[\frac{1}{{31}} + \frac{1}{{32}} + \frac{1}{{33}} + ..... + \frac{1}{{40}} > \underbrace {\frac{1}{{40}} + \frac{1}{{40}} + \frac{1}{{40}} + ..... + \frac{1}{{40}}}_{10\,\,so\,\,hang} = 10 \cdot \frac{1}{{40}} = \frac{1}{4}.\]

Tương tự ta được:

\[\frac{1}{{41}} + \frac{1}{{42}} + \frac{1}{{43}} + ..... + \frac{1}{{50}} > \underbrace {\frac{1}{{50}} + \frac{1}{{50}} + \frac{1}{{50}} + ..... + \frac{1}{{50}}}_{10\,\,so\,\,hang} = \frac{1}{5};\]

\[\frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + ..... + \frac{1}{{60}} > \underbrace {\frac{1}{{60}} + \frac{1}{{60}} + \frac{1}{{60}} + ..... + \frac{1}{{60}}}_{10\,\,so\,\,hang} = \frac{1}{6}.\]

Suy ra \(\frac{1}{{31}} + \frac{1}{{32}} + \frac{1}{{33}} + ..... + \frac{1}{{60}} > \frac{1}{4} + \frac{1}{5} + \frac{1}{6}.\)

Mà \(\frac{1}{4} + \frac{1}{6} = \frac{5}{{12}} > \frac{2}{5}\) nên \(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} > \frac{2}{5} + \frac{1}{5} = \frac{3}{5}.\)

Do đó \(\frac{1}{{31}} + \frac{1}{{32}} + \frac{1}{{33}} + ..... + \frac{1}{{60}} > \frac{3}{5}.\)

Vậy \(A > \frac{3}{5}.\)