Với mọi a, b, chứng minh: a^2 + b^2 ≥ ( a + b )^2 /2

a) Xét hiệu \[\frac{{{a^2} + {b^2}}}{2} - {\left( {\frac{{a + b}}{2}} \right)^2} = \frac{{{a^2} + {b^2}}}{2} - \frac{{{a^2} + 2ab + {b^2}}}{4}\]

                                                 \[ = \frac{{2{a^2} + 2{b^2} + {a^2} - 2ab + {b^2}}}{4} = \frac{{{{\left( {a - b} \right)}^2}}}{4} \ge 0\].

Do đó, \[\frac{{{a^2} + {b^2}}}{2} - {\left( {\frac{{a + b}}{2}} \right)^2}\] ≥ 0 .

Vậy \[\frac{{{a^2} + {b^2}}}{2} \ge {\left( {\frac{{a + b}}{2}} \right)^2}\] (đpcm).

b) Xét hiệu a² + b² − \[\frac{{{{\left( {a + b} \right)}^2}}}{2}\] = \[\frac{{2{a^2} + 2{b^2} - {{\left( {a + b} \right)}^2}}}{2}\]

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

Suy ra a² + b² − \[\frac{{{{\left( {a + b} \right)}^2}}}{2}\] ≥ 0.

Vậy \[{a^2} + {b^2} \ge \frac{{{{\left( {a + b} \right)}^2}}}{2}\] (đpcm).