1. **Assume one of the numbers $c$ or $d$ lies between $a$ and $b$:**
- Without loss of generality, assume $a < c < b$. This implies $a < c$ and $c < b$.

2. **Consider the inequality $\sqrt{(a+b)(c+d)} > \sqrt{ab} + \sqrt{cd}$:**
- We need to show that $\sqrt{(a+b)(c+d)} > \sqrt{ab} + \sqrt{cd}$.

3. Expand and simplify the left-hand side:
$$\sqrt{(a+b)(c+d)} = \sqrt{ac + ad + bc + bd}$$

4. Expand and simplify the right-hand side:
$$\sqrt{ab} + \sqrt{cd}$$

5. Compare the terms:
- We need to show that:
$$\sqrt{ac + ad + bc + bd} > \sqrt{ab} + \sqrt{cd}$$

6. Square both sides to eliminate the square roots:
$$(ac + ad + bc + bd) > (\sqrt{ab} + \sqrt{cd})^2$$
$$ac + ad + bc + bd > ab + cd + 2\sqrt{abcd}$$

7. Rearrange the inequality:
$$ac + ad + bc + bd - ab - cd > 2\sqrt{abcd}$$
$$(ac - ab) + (ad - cd) + (bc - cd) > 2\sqrt{abcd}$$

8. Simplify the terms:
$$a(c - b) + d(a - c) + b(c - d) > 2\sqrt{abcd}$$

9. **Since $a < c < b$, we have $c - b < 0$, $a - c < 0$, and $c - d < 0$:
- This implies that the left-hand side is negative, which contradicts the right-hand side being positive.

10. Therefore, the inequality holds if one of the numbers $c$ or $d$ lies between $a$ and $b$:**
$$\sqrt{(a+b)(c+d)} > \sqrt{ab} + \sqrt{cd}$$

11. **Consider the case where none of the numbers $c$ or $d$ lies between $a$ and $b$:**
- For example, let $a = 1$, $b = 2$, $c = 3$, $d = 4$:
$$\sqrt{(a+b)(c+d)} = \sqrt{(1+2)(3+4)} = \sqrt{21}$$
$$\sqrt{ab} + \sqrt{cd} = \sqrt{1 \cdot 2} + \sqrt{3 \cdot 4} = \sqrt{2} + \sqrt{12}$$
$$\sqrt{21} - (\sqrt{2} + \sqrt{12}) < 0$$

12. **Thus, the inequality can be false if none of the numbers $c$ or $d$ lies between $a$ and $b$:**
- This shows that the inequality does not hold in this case.

The final answer is $\boxed{ \sqrt{(a+b)(c+d)} > \sqrt{ab} + \sqrt{cd} }$ if one of the numbers $c$ or $d$ lies between $a$ and $b$, or one of $a$ or $b$ lies between $c$ and $d$. Otherwise, the inequality can be false.