1. The maximum value of the real number $k$ that makes the inequality $\sqrt{x-3}+\sqrt{6-x} \geqslant k$ solvable for $x$ is
A. $\sqrt{6}-\sqrt{3}$
B. $\sqrt{3}$
C. $\sqrt{6}+\sqrt{3}$
D. $\sqrt{6}$
By the Cauchy-Schwarz inequality, we have $(\sqrt{x-3}+\sqrt{6-x})^{2} \leqslant 2(x-3+6-x)=6$ $\Rightarrow \sqrt{x-3}+\sqrt{6-x} \leqslant \sqrt{6}$, equality holds when $x-3=6-x \Rightarrow x=\frac{9}{2}$. Therefore, $k \leqslant \sqrt{6}$. So the answer is $D$.