Cho biểu thức sau A = 2023/2022^2 + 1 + 2023/2022^2 + 2 + . . . + 2023/2022^2 + 2022 . Chứng minh rằng A > 1 .

Ta có: \(A = \frac{{2023}}{{{{2022}^2} + 1}} + \frac{{2023}}{{{{2022}^2} + 2}} + ... + \frac{{2023}}{{{{2022}^2} + 2022}}\)

\(A = 2023.\left( {\frac{1}{{{{2022}^2} + 1}} + \frac{1}{{{{2022}^2} + 2}} + ... + \frac{1}{{{{2022}^2} + 2022}}} \right)\)

Vì \(\frac{1}{{{{2022}^2} + 1}} > \frac{1}{{{{2022}^2} + 2}} > ... > \frac{1}{{{{2022}^2} + 2022}}\)

Nên \(A > 2023.\frac{1}{{{{2022}^2} + 2022}}.2022 = \frac{{2023.2022}}{{2022.(2022 + 1)}} = 1\)

Vậy \(A > 1\).