If \((\sqrt{a} + \sqrt{b}) \propto (\sqrt{a} - \sqrt{b})\), then show that \((a + b) \propto \sqrt{ab}\).

Q.If \((\sqrt{a} + \sqrt{b}) \propto (\sqrt{a} - \sqrt{b})\), then show that \((a + b) \propto \sqrt{ab}\).

Given: \((\sqrt{a} + \sqrt{b}) \propto (\sqrt{a} - \sqrt{b})\) ⇒ \((\sqrt{a} + \sqrt{b}) = k(\sqrt{a} - \sqrt{b})\), where \(k\) is a non-zero constant ⇒ \(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = k\) ⇒ Squaring both sides: \[ \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = k^2 \Rightarrow \frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = k^2 \] Now applying componendo and dividendo: \[ \frac{(a + b + 2\sqrt{ab}) + (a + b - 2\sqrt{ab})}{(a + b + 2\sqrt{ab}) - (a + b - 2\sqrt{ab})} = \frac{k^2 + 1}{k^2 - 1} \Rightarrow \frac{2(a + b)}{4\sqrt{ab}} = \frac{k^2 + 1}{k^2 - 1} \] Simplifying: \[ \frac{a + b}{\sqrt{ab}} = \frac{2(k^2 + 1)}{k^2 - 1} = \text{constant} \] Therefore, \((a + b) \propto \sqrt{ab}\) — Proven.