1. Let $L$ and $R$ be the left-hand side (LHS) and right-hand side (RHS) of the inequality, respectively. We have:
$$L = \left( \sum_{i,j=1}^{n} |x_i - x_j| \right)^2$$
and
$$R = \frac{2(n^2-1)}{3} \sum_{i,j=1}^{n} (x_i - x_j)^2.$$

2. Notice that $(x_i - x_j)^2 = (x_j - x_i)^2$. Therefore, we can rewrite $R$ as:
$$R = \frac{4(n^2-1)}{3} \sum_{i > j \ge 1}^{n} (x_i - x_j)^2.$$

3. Let $J = \sum_{i > j \ge 1}^{n} (x_i - x_j)^2$ and $K = \sum_{i=1}^{n} \sum_{j=1}^{i} j^2$. By the Cauchy-Schwarz inequality, we have:
$$JK \ge \left( \sum_{i > j \ge 1}^{n} (x_i - x_j) \right)^2.$$

4. We need to find the specific value of $K$. Using the known summation formulas:
$$\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2,$$
$$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6},$$
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}.$$

5. Therefore, we have:
$$K = \sum_{i=1}^{n-1} \frac{i(i+1)(2i+1)}{6} = \sum_{i=1}^{n} \frac{2i^3 + 3i^2 + i}{6}.$$

6. Simplifying $K$:
$$K = \frac{\frac{(n-1)^2 n^2}{2} + \frac{(n-1) n (2n-1)}{2} + \frac{(n-1) n}{2}}{6} = \frac{(n-1) n^2 (n+1)}{12}.$$

7. Now, we have:
$$R = \frac{4(n^2-1)}{3K} JK \ge \frac{4(n-1)(n+1)}{\frac{3(n-1)n^2(n+1)}{12}} I^2 = \frac{16}{n^2} I^2.$$

8. Finally, it suffices to prove $L = \frac{16}{n^2} I^2$. Note that:
$$L = 4 \left( \sum_{i \ge 1 \ge 1} |x_i - x_j| \right)^2.$$

9. Let $d_i = 2i - (n+1)$. For specific $x_p$, notice that if $i > p$, then $|x_i - x_p|$ will result in a negative $x_p$ term. If $i < p$, then $|x_p - x_i|$ will result in a positive $x_p$ term, and otherwise nothing. We now see that the coefficient of $x_p$ will be:
$$\left( (p-1) - 1 + 1 \right) - \left( n - (p+1) + 1 \right) = 2p - n - 1.$$

10. However, $\frac{nd_i}{2} = pn - \frac{n^2}{2} - \frac{n}{2} = \binom{n}{2} - n(n-p) = xc_i$. This means:
$$L = 4 \left( \frac{2}{n} \right)^2 I^2 = \frac{16}{n^2} I^2,$$
which proves the inequality.

11. Finally, we see that equality holds when $\frac{|x_i - x_j|}{|i - j|}$ is a constant, which gives equality in the Cauchy-Schwarz inequality. Since $i < j \implies x_i < x_j$, this means $x_1, x_2, \ldots, x_n$ must be an arithmetic sequence. $\blacksquare$