Chứng tỏ rằng \(\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{70}} > \frac{4}{3}\).

Đặt \(A = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{70}}\)

Ta có: \(A = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}} + \frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}} + \frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}}\)

\(A = \left( {\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}}} \right) + \left( {\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}}} \right) + \left( {\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}}} \right)\)

Nhận thấy \(\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}} > \frac{1}{{30}} + \frac{1}{{30}} + .... + \frac{1}{{30}}\) hay \(\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}} > \frac{1}{{30}}.20 = \frac{2}{3}\)

                  \(\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}} > \frac{1}{{50}} + \frac{1}{{50}} + .... + \frac{1}{{50}}\) hay \(\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}} > \frac{1}{{50}}.20 = \frac{2}{5}\).

                  \(\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}} > \frac{1}{{70}} + \frac{1}{{70}} + .... + \frac{1}{{70}}\) hay \(\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}} > \frac{1}{{70}}.20 = \frac{2}{7}\).

Do đó, \(A > \frac{2}{3} + \frac{2}{5} + \frac{2}{7}\) hay \(A > \frac{{142}}{{105}} > \frac{{140}}{{105}} = \frac{4}{3}\).

Vậy \(A > \frac{4}{3}\) (đpcm)