3. If for all positive numbers $x, y$, we have $\sqrt{x}+\sqrt{y} \leq a \sqrt{x+y}$, then the minimum value of the real number $a$ is
Answer: $\sqrt{2}$
Solution: From $\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1$, we get $\frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leq \sqrt{2}$, with equality when $x=y$.