Do có $2020$ cặp số nên
$\begin{array}{l} \left( {\dfrac{1}{{{2^2}}} - 1} \right)\left( {\dfrac{1}{{{3^2}}} - 1} \right)...\left( {\dfrac{1}{{{{2021}^2}}} - 1} \right)\\ A = \left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{{2021}^2}}}} \right)\\ A = \left( {1 - \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{2}} \right)\left( {1 - \dfrac{1}{3}} \right)\left( {1 + \dfrac{1}{3}} \right)...\left( {1 - \dfrac{1}{{2021}}} \right)\left( {1 + \dfrac{1}{{2021}}} \right)\\ A = \dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{{2020}}{{2021}}.\dfrac{{2022}}{{2021}}\\ A = \dfrac{{1.2.3.4.2020}}{{2.3.4....2021}}.\dfrac{{3.4.5...2022}}{{2.3.4...2021}}\\ A = \dfrac{1}{{2021}}.\dfrac{{2022}}{2} = \dfrac{{2022}}{{4042}} > \dfrac{{2022}}{{4044}} = \dfrac{1}{2}\\ \Rightarrow A > \dfrac{1}{2} \end{array}$
