 Solution

ID: past-exam-of-the-algebra-and-analysis-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/2/b/ii/solution

In part (i), set every

b_{i} = 1

. Its squared norm is

n

n i = 1 \sum n a_{i}^{2} \geq (i = 1 \sum n a_{i})^{2} .

The equality condition is

a_{1} = \dots = a_{n}

, because

a

(1, \dots, 1)

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