Cho a > 0, b > 0. Chứng minh rằng \(\frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }} \ge \sqrt a + \sqrt b \)
Ta có
\(\begin{array}{l}
\frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }} \ge \sqrt a + \sqrt b \\
\Leftrightarrow a\sqrt a + b\sqrt b \ge \sqrt a \sqrt b \left( {\sqrt a + \sqrt b } \right)\\
\Leftrightarrow \left( {\sqrt a + \sqrt b } \right)\left( {a + b - \sqrt a \sqrt b } \right) \ge \sqrt a \sqrt b \left( {\sqrt a + \sqrt b } \right)\\
\Leftrightarrow a + b - \sqrt a \sqrt b \ge \sqrt a \sqrt b \\
\Leftrightarrow \left( {{{\sqrt a }^2}} \right) + {\left( {\sqrt b } \right)^2} - 2\sqrt a \sqrt b \ge 0\\
\Leftrightarrow {\left( {\sqrt a - \sqrt b } \right)^2} \ge 0
\end{array}\)
(đúng với mọi a > 0, b > 0)
Do đó \(\frac{a}{{\sqrt b }} + \frac{b}{{\sqrt a }} \ge \sqrt a + \sqrt b \) (đpcm)
-- Mod Toán 10