 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/5a/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

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\frac{1}{2} \sum_{i, j} (x_{i} y_{j} - x_{j} y_{i})^{2} = \sum_{i, j} x_{i}^{2} y_{j}^{2} - \sum_{i, j} x_{i} x_{j} y_{i} y_{j} = ∣ x ∣^{2} ∣ y ∣^{2} - (x \cdot y)^{2} .

The left-hand side is nonnegative, so

∣ x ∣^{2} ∣ y ∣^{2} \geq ∣ x \cdot y ∣^{2} .

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