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

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

Given: \((a + b) \propto \sqrt{ab}\) \[ \Rightarrow (a + b) = k\sqrt{ab} \quad \text{[where \(k\) is a non-zero constant of proportionality]} \Rightarrow \frac{a + b}{\sqrt{ab}} = k \Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{k}{2} \quad \text{[dividing both sides by 2]} \] Now, applying the method of componendo and dividendo: \[ \frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{k + 2}{k - 2} \] Notice: \[ a + b + 2\sqrt{ab} = (\sqrt{a} + \sqrt{b})^2 \quad \text{and} \quad a + b - 2\sqrt{ab} = (\sqrt{a} - \sqrt{b})^2 \] So, \[ \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{k + 2}{k - 2} \Rightarrow \left(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}\right)^2 = \frac{k + 2}{k - 2} \Rightarrow \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{\frac{k + 2}{k - 2}} = \text{constant} \] \(\therefore (\sqrt{a} + \sqrt{b}) \propto (\sqrt{a} - \sqrt{b})\) — (Proved)