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Question Number 71508 by Cmr 237 last updated on 16/Oct/19

montrer que : \forall a, b \in R ona

2 ∣ ab ∣⩽ a^{2} + b^{2}

Endeduire que \forall x_{1}, \dots, x_{n} \in R on a :

{(\underset{i = 1}{\sum^{n}} ∣ x_{i} ∣)}^{2} ⩽ n \underset{i = 1}{\sum^{n}} x_{i}^{2}

please i need help

Commented by Cmr 237 last updated on 16/Oct/19

please can you prove it ?

Commented by Prithwish sen last updated on 16/Oct/19

let

k = \frac{x_{1} + x_{2} + \dots \dots . x_{n}}{n} \dots . . (i)

now \frac{x_{1}}{k}, \frac{x_{2}}{k}, \dots \dots . . \frac{x_{n}}{k} all are positive and not all

of them equal to 1

{[\frac{x_{i}}{k}]}^{2} - 1 ⩾ 0 equality holds only \frac{x_{i}}{k} = 1 (i \in N)

now {[\frac{x_{1}}{k}]}^{2} + {[\frac{x_{2}}{k}]}^{2} + \dots . . {[\frac{x_{n}}{k}]}^{2} - n ⩾ 2 [\frac{x_{1}}{k} + \frac{x_{2}}{k} + \dots . . + \frac{x_{n}}{k} - n] (from given condition)

or, \frac{x_{1}^{2} + x_{2}^{2} + \dots . + x_{n}^{2}}{k^{2}} - n ⩾ 2 (n - n) from (i)

or \frac{x_{1}^{2} + x_{2}^{2} + \dots \dots . + x_{n}^{2}}{n} ⩾ k^{2}

\Rightarrow x_{1}^{2} + x_{2}^{2} + \dots \dots + x_{n}^{2} ⩾ n {(\frac{x_{1} + x_{2} + \dots . . + x_{n}}{n})}^{2}

Double subscripts: use braces to clarify

Answered by turbo msup by abdo last updated on 17/Oct/19

w e h s v e \sum_{i = 1}^{n} a_{i} b_{i} ⩽ {(\sum_{i = 1}^{n} a_{i}^{2})}^{\frac{1}{2}} \times {(\sum_{i = 1}^{n} b_{i}^{2})}^{\frac{1}{2}} (h o l d e r)

f o r a l l n u m b e r s p o s i t i f s (a_{i}) s n d (b_{i})

l e t a_{i} = 1 s n d b_{i} =∣ x_{i} ∣ \Rightarrow

\sum_{i = 1}^{n} ∣ x_{i} ∣⩽ \sqrt{n} \times {(\sum_{i = 1}^{n} x_{i}^{2})}^{\frac{1}{2}} \Rightarrow

{(\sum_{i = 1}^{n} ∣ x_{i} ∣)}^{2} ⩽ n \times (\sum_{i = 1}^{n} x_{i}^{2})

Commented by Prithwish sen last updated on 17/Oct/19

Thank you sir .

Commented by mathmax by abdo last updated on 18/Oct/19

y o u a r e w e l c o m e .