Problem 1532

1. **Base Case: $n = 2$**

For $n = 2$, we need to show that:
$$\sum_{1 \le k < j \le 2} kx_kx_j \le \frac{2-1}{3} \sum_{k=1}^2 kx_k$$
This simplifies to:
$$x_1 x_2 \le \frac{1}{3} (x_1 + 2x_2)$$
Since $x_1, x_2 \in [0, 1]$, we have:
$$x_1 x_2 \le \frac{1}{3} (x_1 + 2x_2)$$
This inequality holds because $x_1 x_2 \le x_1$ and $x_1 x_2 \le x_2$, and thus:
$$x_1 x_2 \le \frac{1}{3} (x_1 + 2x_2)$$

2. Inductive Step:

Assume the statement is true for $n = m$, i.e.,
$$\sum_{1 \le k < j \le m} kx_kx_j \le \frac{m-1}{3} \sum_{k=1}^m kx_k$$
We need to show it holds for $n = m+1$.

Consider:
$$\sum_{1 \le k < j \le m+1} kx_kx_j$$
This can be split as:
$$\sum_{1 \le k < j \le m+1} kx_kx_j = \sum_{1 \le k < j \le m} kx_kx_j + \sum_{k=1}^m kx_kx_{m+1}$$

By the inductive hypothesis:
$$\sum_{1 \le k < j \le m} kx_kx_j \le \frac{m-1}{3} \sum_{k=1}^m kx_k$$

Now consider the second term:
$$\sum_{k=1}^m kx_kx_{m+1}$$
Since $x_{m+1} \in [0, 1]$, we have:
$$\sum_{k=1}^m kx_kx_{m+1} \le x_{m+1} \sum_{k=1}^m kx_k$$

Combining these, we get:
$$\sum_{1 \le k < j \le m+1} kx_kx_j \le \frac{m-1}{3} \sum_{k=1}^m kx_k + x_{m+1} \sum_{k=1}^m kx_k$$

Factor out $\sum_{k=1}^m kx_k$:
$$\sum_{1 \le k < j \le m+1} kx_kx_j \le \left( \frac{m-1}{3} + x_{m+1} \right) \sum_{k=1}^m kx_k$$

Since $x_{m+1} \le 1$, we have:
$$\frac{m-1}{3} + x_{m+1} \le \frac{m-1}{3} + 1 = \frac{m+2}{3}$$

Therefore:
$$\sum_{1 \le k < j \le m+1} kx_kx_j \le \frac{m+2}{3} \sum_{k=1}^m kx_k$$

Notice that:
$$\sum_{k=1}^{m+1} kx_k = \sum_{k=1}^m kx_k + (m+1)x_{m+1}$$

Thus:
$$\sum_{1 \le k < j \le m+1} kx_kx_j \le \frac{m+2}{3} \left( \sum_{k=1}^m kx_k + (m+1)x_{m+1} \right)$$

Simplifying, we get:
$$\sum_{1 \le k < j \le m+1} kx_kx_j \le \frac{m+2}{3} \sum_{k=1}^{m+1} kx_k$$

This completes the inductive step.

$\blacksquare$