3. If for all positive numbers $x, y$, we have $\sqrt{x}+\sqrt{y} \leqslant a \sqrt{x+y}$, then the minimum value of the real number $a$ is $\qquad$
By the Cauchy inequality, we have $(\sqrt{x}+\sqrt{y})^{2} \leqslant(1+1)(x+y) \Rightarrow \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2} \Rightarrow a \geqslant \sqrt{2}$, equality holds when $x=y$. Therefore, the minimum value of the real number $a$ is $\sqrt{2}$.