` a \ge b \to \sqrt{a} \ge \sqrt{b}`
`( \sqrt(a) - \sqrt(b))^2 = a + b - 2\sqrt(ab)`
` (\sqrt(a-b))^2 = a -b`
Ta có
` ( \sqrt(a) - \sqrt(b))^2 - (\sqrt(a-b))^2 = ( a + b - 2 \sqrt(ab)) - ( a-b)`
` = 2b - 2\sqrt(ab)`
` = 2( b - \sqrt(ab))`
Ta có ` b = \sqrt{b.b} \le \sqrt{a.b}`
` \to b - \sqrt(ab) \le 0`
`\to 2( b - \sqrt(ab)) \le 0`
`\to ( \sqrt(a) - \sqrt(b))^2 - (\sqrt(a-b))^2 \le 0`
`\to \sqrt(a) - \sqrt(b) \le \sqrt(a-b)`
Dấu `=` xảy ra khi ` b = \sqrt{b.b} = \sqrt{a.b} \to a = b`
