2010 IMO Shortlist A3

Let $x_{1}, \ldots, x_{100}$ be nonnegative real numbers such that $x_{i}+x_{i+1}+x_{i+2} \leq 1$ for all $i=1, \ldots, 100$ (we put $x_{101}=x_{1}, x_{102}=x_{2}$ ). Find the maximal possible value of the sum $$S=\sum_{i=1}^{100} x_{i} x_{i+2}$$ (Russia) Answer. $\frac{25}{2}$.

令 $x_{1}, \ldots, x_{100}$ 为非负实数，使得 $x_{i}+x_{i+1}+x_{i+2} \leq 1$ 对于所有 $i=1, \ldots, 100$ (我们输入 $x_{101}=x_{1}, x_{102}=x_{2}$ )。求总和的最大可能值 $$S=\sum_{i=1}^{100} x_{i} x_{i+2}$$ (俄罗斯)答案。 $\frac{25}{2}$。