5. 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) $\sqrt{6}$
5. D.
$$
\begin{array}{l}
\text { Let } y=\sqrt{x-3}+\sqrt{6-x}(3<x<6) \text {. Then } \\
y^{2} \leqslant 2[(x-3)+(6-x)]=6 \\
\Rightarrow 0<y \leqslant \sqrt{6} .
\end{array}
$$

Therefore, the maximum value of the real number $k$ is $\sqrt{6}$.