1 The maximum value of the real number $k$ for which the inequality $\sqrt{x-3}+\sqrt{6-x} \geqslant k$ has solutions for $x$ is ( ).
(A) $\sqrt{6}-\sqrt{3}$
(B) $\sqrt{3}$
(C) $\sqrt{6}+\sqrt{3}$
(D) $\sqrt{6}$
Let $y=\sqrt{x-3}+\sqrt{6-x}, 3 \leqslant x \leqslant 6$, then
$$
y^{2}=(\sqrt{x-3}+\sqrt{6-x})^{2} \leqslant 2[(x-3)+(6-x)]=6 \text {, }
$$

Therefore, $0<y \leqslant \sqrt{6}$, when $x=\frac{9}{2}$, $y=\sqrt{6}$, so the maximum value of $y$ is $\sqrt{6}$, and the maximum value of the real number $k$ is $\sqrt{6}$. Hence, the correct choice is D.