 In an aquarium there are nine small fish. The aquarium is cube shaped with a side length of two meters and is completely filled with water. Show that it is always possible to find two small fish with a distance of less than $\sqrt3$ meters.
1. **Understanding the Problem:**
   We need to show that in a cube-shaped aquarium with side length 2 meters, containing 9 small fish, there are always at least two fish that are less than $\sqrt{3}$ meters apart.

2. **Cube Geometry:**
   The cube has side length 2 meters. The maximum distance between any two points in the cube is the space diagonal, which is $\sqrt{3} \times 2 = 2\sqrt{3}$ meters. However, we are interested in distances less than $\sqrt{3}$ meters.

3. **Pigeonhole Principle:**
   The Pigeonhole Principle states that if $n$ items are put into $m$ containers, with $n > m$, then at least one container must contain more than one item. Here, we will use this principle to show that at least two fish must be within a smaller region of the cube.

4. **Dividing the Cube:**
   Divide the cube into 8 smaller cubes, each with side length 1 meter. Each smaller cube has a space diagonal of $\sqrt{3}$ meters. Therefore, the maximum distance between any two points within a smaller cube is $\sqrt{3}$ meters.

5. **Placing the Fish:**
   Since there are 9 fish and only 8 smaller cubes, by the Pigeonhole Principle, at least one of these smaller cubes must contain at least 2 fish.

6. **Distance Calculation:**
   The maximum distance between any two points within a smaller cube is $\sqrt{3}$ meters. Therefore, if at least two fish are in the same smaller cube, the distance between them must be less than or equal to $\sqrt{3}$ meters.

7. **Conclusion:**
   Since at least one smaller cube must contain at least two fish, and the maximum distance within a smaller cube is $\sqrt{3}$ meters, it follows that there are always at least two fish with a distance of less than $\sqrt{3}$ meters.

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