China Mathematical Competition (Jiangxi)

Let $k$ be a real number such that the inequality $\sqrt{x-3}+\sqrt{6-x}\ge k$ has a solution. The maximum value of $k$ is ( ). (A) $\sqrt{6}-\sqrt{3}$ (B) $\sqrt{3}$ (C) $\sqrt{6}+\sqrt{3}$ (D) $\sqrt{6}$