MathLabs

Problem 2

In a competition, there are a a contestants and b b judges, where b≥3 b\ge3 is an odd integer. Each judge rates each contestant as either “pass” or “fail”. Suppose k k is a number such that, for any two judges, their ratings coincide for at most k k contestants. Prove that ka≥b−12b\frac{k}{a}\ge\frac{b-1}{2b}.
Step 4 of 4: Finish the inequality
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In plain words

Simplify the double-counting inequality.

ka≥b−12b\frac{k}{a}\ge\frac{b-1}{2b}
Detailed analysis

Because b≥3 b\ge3, divide the combined inequality by the positive quantity ab(b−1)/2 a b(b-1)/2. This gives exactly k/a≥(b−1)/(2b) k/a\ge(b-1)/(2b).