9. The minimum distance from a point on the curve $C: \sqrt{x}+\sqrt{y}=1$ to the origin is $\qquad$
9. Solution Since $\sqrt{x}+\sqrt{y}=1$, then $x, y \in[0,1]$,
$$
1=(\sqrt{x}+\sqrt{y})^{2}=x+y+2 \sqrt{x y} \leqslant 2(x+y) \text {. }
$$

Also, $(x+y)^{2}=x^{2}+y^{2}+2 x y \leqslant 2\left(x^{2}+y^{2}\right)$,
so $x^{2}+y^{2} \geqslant \frac{1}{2}(x+y)^{2} \geqslant \frac{1}{8}, \sqrt{x^{2}+y^{2}} \geqslant \frac{\sqrt{2}}{4}$.
Equality holds if and only if $x=y=\frac{1}{4}$, hence the minimum distance to the origin is $\frac{\sqrt{2}}{4}$.