Cho biểu thức P = (1/2^2 - 1)(1/3^2 - 1)(1/4^2 - 1)...(1/2021^2 - 1). So sánh P với 1/2

\[P = \left( {\frac{1}{{{2^2}}} - 1} \right)\left( {\frac{1}{{{3^2}}} - 1} \right)\left( {\frac{1}{{{4^2}}} - 1} \right) \cdot \cdot \cdot \left( {\frac{1}{{{{2021}^2}}} - 1} \right)\]

\[ = \frac{{1 - {2^2}}}{{{2^2}}} \cdot \frac{{1 - {3^2}}}{{{3^2}}} \cdot \frac{{1 - {4^2}}}{{{4^2}}} \cdot \cdot \cdot \frac{{1 - {{2021}^2}}}{{{{2021}^2}}}\]

\[ = \frac{{\left( {1 - 2} \right)\left( {1 + 2} \right)}}{{{2^2}}} \cdot \frac{{\left( {1 - 3} \right)\left( {1 + 3} \right)}}{{{3^2}}} \cdot \cdot \cdot \frac{{\left( {1 - 2021} \right)\left( {1 + 2021} \right)}}{{{{2021}^2}}}\]

\[ = \frac{{\left( {1 \cdot 2 \cdot 3 \cdot \cdot \cdot 2020} \right) \cdot \left( {3 \cdot 4 \cdot 5 \cdot \cdot \cdot 2022} \right)}}{{\left( {2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021} \right)\left( {2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021} \right)}}\]

\[ = \frac{{1 \cdot 2 \cdot 3 \cdot \cdot \cdot 2020}}{{2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021}} \cdot \frac{{3 \cdot 4 \cdot 5 \cdot \cdot \cdot 2022}}{{2 \cdot 3 \cdot 4 \cdot \cdot \cdot 2021}}\]

\[ = \frac{1}{{2021}} \cdot \frac{{2022}}{2} = \frac{{1011}}{{2021}} > \frac{{1011}}{{2022}} = \frac{1}{2}\]

\[P > \frac{1}{2}\]