Exercise 6.A.15

Answers

Proof. Let $a = (\sqrt{1} a_{1}, \dots, \sqrt{n} a_{n})$ and $b = (\frac{b_{1}}{\sqrt{1}}, \dots, \frac{b_{n}}{\sqrt{n}})$ . Then

\begin{aligned} (\sum_{j = 1}^{n} a_{j} b_{j}) & = {(⟨ a, b ⟩)}^{2} \\ \leq {(| | a | | | | b | |)}^{2} \\ = | | a | |^{2} | | b | |^{2} \\ = (\sum_{j = 1}^{n} j a_{j}^{2}) (\sum_{j = 1}^{n} \frac{b_{j}^{2}}{j}) \end{aligned}

where the second line follows from the Cauchy-Schwarz Inequality. □

celiopassos

2017-10-06 00:00