Exercise 6.A.22

Answers

Proof. Let $x = (1, \dots, 1)$ ( $n$ ones) and $a = (a_{1}, \dots, a_{n})$ . Then

\begin{aligned} {(\frac{a_{1} + \dots + a_{n}}{n})}^{2} & = \frac{| ⟨ x, a ⟩ |^{2}}{n^{2}} \\ \leq \frac{| | x | |^{2} | | a | |^{2}}{n^{2}} \\ = \frac{| | a | |^{2}}{n} \\ = \frac{a_{1}^{2} + \dots + a_{n}^{2}}{n} \end{aligned}

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

celiopassos

2017-10-06 00:00