Đáp án:
$\begin{array}{l}
\sqrt a \left( {\sqrt {\dfrac{a}{b}} - 1} \right) - \sqrt b \left( {1 - \sqrt {\dfrac{b}{a}} } \right)\\
= \sqrt a .\dfrac{{\sqrt a - \sqrt b }}{{\sqrt b }} - \sqrt b .\dfrac{{\sqrt a - \sqrt b }}{{\sqrt a }}\\
= \left( {\sqrt a - \sqrt b } \right).\left( {\dfrac{{\sqrt a }}{{\sqrt b }} - \dfrac{{\sqrt b }}{{\sqrt a }}} \right)\\
= \left( {\sqrt a - \sqrt b } \right).\dfrac{{a - b}}{{\sqrt {ab} }}\\
= \left( {\sqrt a - \sqrt b } \right).\dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt {ab} }}\\
= \dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}.\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt {ab} }}\\
Do:\left\{ \begin{array}{l}
{\left( {\sqrt a - \sqrt b } \right)^2} \ge 0;\\
\left( {\sqrt a + \sqrt b } \right) > 0\\
\sqrt {ab} > 0
\end{array} \right.\left( {khi:a > 0;b > 0} \right)\\
\Rightarrow \dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}.\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt {ab} }} \ge 0\\
Vay\,\sqrt a \left( {\sqrt {\dfrac{a}{b}} - 1} \right) \ge \sqrt b \left( {1 - \sqrt {\dfrac{b}{a}} } \right)
\end{array}$
