MathLabs

第2题

In a competition, there 是 a a contestants 且 b b judges, 其中 b≥3 b\ge3 是an odd 整数. Each judge rates 每个contestant as either “pass” 或“fail”. 设 k k 是a 数 使得, 对于any two judges, their ratings coincide 对于在most k k contestants. 证明 ka≥b−12b\frac{k}{a}\ge\frac{b-1}{2b}.
第 4/4 步：结论 inequality
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通俗地说

Simplify double-counting inequality.

ka≥b−12b\frac{k}{a}\ge\frac{b-1}{2b}
详细分析

Beca使用 b≥3 b\ge3, divide combined inequality 由positive quantity ab(b−1)/2 a b(b-1)/2. Th是gives exactly k/a≥(b−1)/(2b) k/a\ge(b-1)/(2b).