Ta có:
\(\displaystyle\eqalign{ & {\left( {a - b} \right)^2} \ge 0 \cr & \Leftrightarrow {a^2} + {b^2} - 2ab \ge 0 \cr & \Leftrightarrow {a^2} + {b^2} - 2ab + 2ab \ge 2ab \cr & \Leftrightarrow {a^2} + {b^2} \ge 2ab \cr} \)
Vì \(a > 0, b > 0\) nên \(ab > 0 \displaystyle \Rightarrow {1 \over {ab}} > 0\)
\(\displaystyle\eqalign{ &\Rightarrow \left( {{a^2} + {b^2}} \right).{1 \over {ab}} \ge 2ab.{1 \over {ab}} \cr & \Leftrightarrow {a \over b} + {b \over a} \ge 2 \cr & \Leftrightarrow 2 + {a \over b} + {b \over a} \ge 2 + 2 \cr & \Leftrightarrow 2 + {a \over b} + {b \over a} \ge 4 \cr & \Leftrightarrow 1 + 1 + {a \over b} + {b \over a} \ge 4 \cr & \Leftrightarrow {a \over a} + {a \over b} + {b \over b} + {b \over a} \ge 4\cr& \Leftrightarrow a\left( {{1 \over a} + {1 \over b}} \right) + b\left( {{1 \over a} + {1 \over b}} \right) \ge 4 \cr & \Leftrightarrow \left( {a + b} \right)\left( {{1 \over a} + {1 \over b}} \right) \ge 4 \cr} \)
