3. If for all positive numbers $x, y$, we have
$$
\sqrt{x}+\sqrt{y} \leqslant a \sqrt{x+y} \text {, }
$$

then the minimum value of the real number $a$ is $\qquad$.
3. $\sqrt{2}$.
$$
\begin{array}{l}
\text { Given }\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1 \\
\Rightarrow \frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2} .
\end{array}
$$

When $x=y$, the equality holds.