Prove that for a set of points $\left\{P_{1}, P_{2}, \cdots, P_{n}\right\}$ in the plane, let $a_{i}$ be the number of points $P_{i}$ that are a unit distance from $P_{i}$. Assume $a_{i} \geqslant 1$, then the number of pairs of points that are a unit distance apart is
$$A=\frac{a_{1}+a_{2}+\cdots+a_{n}}{2}$$

Let $C_{i}$ be the circle with center at point $P_{i}$ and radius 1.
Since each pair of circles has at most 2 intersection points, the total number of intersection points of all $C_{i}$ is at most
$$2 \mathrm{C}_{n}^{2}=n(n-1)$$

The point $P_{i}$ appears as an intersection point of $C_{j}$ $\mathrm{C}_{a_{j}}^{2}$ times, so
$$n(n-1) \geqslant \sum_{j=1}^{n} \mathrm{C}_{a_{j}}^{2}=\sum_{j=1}^{n} \frac{a_{j}\left(a_{j}-1\right)}{2} \geqslant \frac{1}{2} \sum_{j=1}^{n}\left(a_{j}-1\right)^{2} .$$

By the Cauchy-Schwarz inequality, we get

Thus
$$\left[\sum_{j=1}^{n}\left(a_{j}-1\right)\right]^{2} \leqslant n \cdot \sum_{j=1}^{n}\left(a_{j}-1\right)^{2} \leqslant n \cdot 2 n(n-1)<2 n^{3}$$

Therefore
$$\sum_{j=1}^{n}\left(a_{j}-1\right)<\sqrt{2} \cdot \sqrt{n^{3}}$$
$$A=\frac{\sum_{j=1}^{n} a_{j}}{2}<\frac{n+\sqrt{2 n^{3}}}{2}<2 \sqrt{n^{3}}$$

Hence the proposition is true.