Cho a > 0 và b > 0, chứng tỏ rằng
\(\left( {a + b} \right)\left( {{1 \over a} + {1 \over b}} \right) \ge 4\)
Ta có:
\(\eqalign{ & {\left( {a – b} \right)^2} \ge 0 \cr & \Leftrightarrow {a^2} + {b^2} – 2ab \ge 0 \cr & \Leftrightarrow {a^2} + {b^2} – 2ab + 2ab \ge 2ab \cr & \Leftrightarrow {a^2} + {b^2} \ge 2ab \cr} \)
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Vì a > 0, b > 0 nên ab ≥ 0 \( \Rightarrow {1 \over {ab}} > 0\)
\(\eqalign{ & \left( {{a^2} + {b^2}} \right).{1 \over {ab}} \ge 2ab.{1 \over {ab}} \cr & \Leftrightarrow {a \over b} + {b \over a} \ge 2 \cr & \Leftrightarrow 2 + {a \over b} + {b \over a} \ge 2 + 2 \cr & \Leftrightarrow 2 + {a \over b} + {b \over a} \ge 4 \cr & \Leftrightarrow 1 + 1 + {a \over b} + {b \over a} \ge 4 \cr & \Leftrightarrow a\left( {{1 \over a} + {1 \over b}} \right) + b\left( {{1 \over a} + {1 \over b}} \right) \ge 4 \cr & \Leftrightarrow \left( {a + b} \right)\left( {{1 \over a} + {1 \over b}} \right) \ge 4 \cr} \)