Question 140
Given an integer $n \ge 3$. Let $\frac{n(n-1)}{2}$ non-negative real numbers $x_{i,j}$ ($1 \le i \lt j \le n$) satisfy: for any $1 \le i \lt j \lt k \le n$, we have $x_{i,j} + x_{j,k} \le x_{i,k}$. Prove that:
$$
\left\lfloor \frac{n^2}{4} \right\rfloor \cdot \sum_{1 \le i \lt j \le n} x_{i,j}^4 \ge \left( \sum_{1 \le i \lt j \le n} x_{i,j}^2 \right)^2 .
$$
$$
\left\lfloor \frac{n^2}{4} \right\rfloor \cdot \sum_{1 \le i \lt j \le n} x_{i,j}^4 \ge \left( \sum_{1 \le i \lt j \le n} x_{i,j}^2 \right)^2 .
$$