The maximum value of $k$ such that the enequality $\sqrt{x-3}+\sqrt{6-x}\geq k$ has a real solution is
$\text{(A)}\sqrt6-\sqrt3\qquad\text{(B)}\sqrt3\qquad\text{(C)}\sqrt3+\sqrt6\qquad\text{(D)}\sqrt6$
1. **Apply the Cauchy-Schwarz Inequality:**
   The Cauchy-Schwarz Inequality states that for any real numbers \(a_1, a_2, \ldots, a_n\) and \(b_1, b_2, \ldots, b_n\),
   \[
   (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2) \geq (a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2.
   \]
   In this problem, let \(a_1 = \sqrt{x-3}\) and \(a_2 = \sqrt{6-x}\), and \(b_1 = b_2 = 1\). Then,
   \[
   (\sqrt{x-3}^2 + \sqrt{6-x}^2)(1^2 + 1^2) \geq (\sqrt{x-3} \cdot 1 + \sqrt{6-x} \cdot 1)^2.
   \]

2. **Simplify the Inequality:**
   \[
   ((x-3) + (6-x))(2) \geq (\sqrt{x-3} + \sqrt{6-x})^2.
   \]
   Simplifying the left-hand side,
   \[
   (x-3 + 6-x)(2) = 6 \cdot 2 = 12.
   \]
   Therefore,
   \[
   12 \geq (\sqrt{x-3} + \sqrt{6-x})^2.
   \]

3. **Find the Maximum Value:**
   Taking the square root of both sides,
   \[
   \sqrt{12} \geq \sqrt{x-3} + \sqrt{6-x}.
   \]
   Since \(\sqrt{12} = 2\sqrt{3}\), we have
   \[
   2\sqrt{3} \geq \sqrt{x-3} + \sqrt{6-x}.
   \]
   Thus, the maximum value of \(\sqrt{x-3} + \sqrt{6-x}\) is \(2\sqrt{3}\).

4. **Verify the Equality Condition:**
   To achieve equality, we need
   \[
   2\sqrt{3} = \sqrt{x-3} + \sqrt{6-x}.
   \]
   Squaring both sides,
   \[
   12 = (\sqrt{x-3} + \sqrt{6-x})^2.
   \]
   This equality holds when \(x = \frac{3}{2}\), as shown in the solution.

Therefore, the maximum value of \(k\) such that the inequality \(\sqrt{x-3} + \sqrt{6-x} \geq k\) has a real solution is \(\sqrt{6}\).

The final answer is \( \boxed{ \sqrt{6} } \).