IMC 1999 · Day 2 · P9

The inequality $$0 \le x^3 - \frac{3}{4} x + \frac{1}{4} = (x + 1) \left( x - \frac{1}{2} \right)^2$$ holds for $x \ge -1$.

Substituting $x_1, \dots, x_n$, we obtain $$0 \le \sum_{i=1}^{n} \left( x_i^3 - \frac{3}{4} x_i + \frac{1}{4} \right) = \sum_{i=1}^{n} x_i^3 - \frac{3}{4} \sum_{i=1}^{n} x_i + \frac{n}{4} = 0 - \frac{3}{4} \sum_{i=1}^{n} x_i + \frac{n}{4},$$ so $\sum\limits_{i=1}^{n} x_i \le \frac{n}{3}$.

Remark. Equailty holds only in the case when $n = 9k$, $k$ of the $x_1, \dots, x_n$ are $-1$, and $8k$ of them are $\frac{1}{2}$.