 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014-resit/11/ii/solution

Expanding the squared Euclidean norm and applying the Cauchy-Schwarz inequality gives

∣ a + b ∣^{2} = ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} \leq ∣ a ∣^{2} + 2∣ a ∣ ∣ b ∣ + ∣ b ∣^{2} = (∣ a ∣ + ∣ b ∣)^{2} .

Both sides are nonnegative, so taking square roots proves the triangle inequality,

∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣

. For nonzero vectors, equality requires a positive real proportionality factor.

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