Vì \(a ≥ 0\) nên \(\sqrt a \) xác định, \(b ≥ 0\) nên \(\sqrt b \) xác định.
Ta có:
\({\left( {\sqrt a - \sqrt b } \right)^2} \ge 0 \)\( \Leftrightarrow a - 2\sqrt {ab} + b \ge 0\)
\(\Leftrightarrow a + b \ge 2\sqrt {ab} \)
\( \Leftrightarrow a + b + a + b \ge a + b + 2\sqrt {ab} \)
\( \Leftrightarrow 2(a + b) \ge {\left( {\sqrt a } \right)^2} + 2\sqrt {ab} + {\left( {\sqrt b } \right)^2}\)
\( \Leftrightarrow 2(a + b) \ge {\left( {\sqrt a + \sqrt b } \right)^2} \)
\(\displaystyle \Leftrightarrow {{a + b} \over 2} \ge {{{{\left( {\sqrt a + \sqrt b } \right)}^2}} \over 4} \)
\(\displaystyle \Leftrightarrow \sqrt {{{a + b} \over 2}} \ge \sqrt {{{{{\left( {\sqrt a + \sqrt b } \right)}^2}} \over 4}} \)
\(\displaystyle \Leftrightarrow \sqrt {{{a + b} \over 2}} \ge {{\sqrt a + \sqrt b } \over 2} \)
