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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 5A / a / ii / Solution

Expanding a sum of squares gives the

n

-dimensional Lagrange identity for the cross product:

\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} .

Solved by gpt-5.6-sol high.

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