cho a b khác 0 cm a^2 b^2 (ab 1/a b)>2

Lời giải.

Ta có:

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

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

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

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