 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-1/5a/b/solution

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

∣ x \cdot y ∣ \leq ∥ x ∥∥ y ∥

. For

y \neq = 0

, positivity of

∥ x - t y ∥^{2}

t = (x \cdot y) /∥ y ∥^{2}

gives

∥ x ∥^{2} - (x \cdot y)^{2} /∥ y ∥^{2} \geq 0

;

y = 0

is immediate.

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