10. If for all positive numbers $x, y$, the inequality $\sqrt{x}+\sqrt{y}$ $\leqslant a \sqrt{x+y}$ holds, then the minimum value of $a$ is $\qquad$
10. $\sqrt{2}$.

From $\left(\frac{\sqrt{x}}{\sqrt{x+y}}\right)^{2}+\left(\frac{\sqrt{y}}{\sqrt{x+y}}\right)^{2}=1$, we have $\frac{\sqrt{x}}{\sqrt{x+y}}+\frac{\sqrt{y}}{\sqrt{x+y}} \leqslant \sqrt{2}$.
When $x=y$, the equality holds, hence the minimum value of $a$ is $\sqrt{2}$.