8. (2005 National High School Mathematics League Jiangxi Province Preliminary) If for all positive numbers $x, y$, the inequality $\sqrt{x}+\sqrt{y} \leqslant a \sqrt{x+y}$ always holds, then the minimum value of $a$ is
8. Solution: From $\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1$. Therefore, $\frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2}$. Equality holds when $x=y$, hence the minimum value of $a$ is $\sqrt{2}$.