Chứng minh với ab ≥ 1 thì \(\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} \ge \frac{2}{{1 + ab}}\).

Ta có: \(\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} \ge \frac{2}{{1 + ab}}\)

\( \Leftrightarrow \frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} - \frac{2}{{1 + ab}} \ge 0\)

\( \Leftrightarrow \left( {\frac{1}{{1 + {a^2}}} - \frac{1}{{1 + ab}}} \right) + \left( {\frac{1}{{1 + {b^2}}} - \frac{1}{{1 + ab}}} \right) \ge 0\)

\( \Leftrightarrow \frac{{1 + ab - 1 - {a^2}}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)}} + \frac{{1 + ab - 1 - {b^2}}}{{\left( {1 + {b^2}} \right)\left( {1 + ab} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{ab - {a^2}}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)}} + \frac{{ab - {b^2}}}{{\left( {1 + {b^2}} \right)\left( {1 + ab} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{\left( {ab - {a^2}} \right)\left( {1 + {b^2}} \right) + \left( {1 + {a^2}} \right)\left( {ab - {b^2}} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{a\left( {b - a} \right)\left( {1 + {b^2}} \right) + \left( {1 + {a^2}} \right)b\left( {a - b} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{\left( {b - a} \right)\left[ {a\left( {1 + {b^2}} \right) - b\left( {1 + {a^2}} \right)} \right]}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{\left( {b - a} \right)\left[ {a + a{b^2} - b - {a^2}b} \right]}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{\left( {b - a} \right)\left[ {\left( {a - b} \right) - ab\left( {a - b} \right)} \right]}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\)

\( \Leftrightarrow \frac{{{{\left( {b - a} \right)}^2}\left( {ab - 1} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\)

Vì ab ≥ 1 nên ab – 1 ≥ 0

Mà (b – a)² ≥ 0

Suy ra (b – a)²(ab – 1) ≥ 0

Vì (a² + 1) > 0, (b² + 1) > 0, (ab + 1) > 0

Nên (a² + 1)(b² + 1)(ab + 1) > 0

Suy ra \(\frac{{{{\left( {b - a} \right)}^2}\left( {ab - 1} \right)}}{{\left( {1 + {a^2}} \right)\left( {1 + ab} \right)\left( {1 + {b^2}} \right)}} \ge 0\) với mọi a, b, ab ≥ 1

Vậy \(\frac{1}{{1 + {a^2}}} + \frac{1}{{1 + {b^2}}} \ge \frac{2}{{1 + ab}}\).