x-i-1-i-n-n-real-number-positifs-wish-verfy-i-1-i-n-x-i-1-prove-that-1-i-n-x-i-2-1-n-

Question Number 26244 by abdo imad last updated on 22/Dec/17

{(x_{i})}_{1 ⩽ i ⩽ n} n r e a l n u m b e r p o s i t i f s w i s h v e r f y \sum_{i = 1}^{i = n} x_{i} = 1

p r o v e t h a t \sum_{1 ⩽ i ⩽ n} x_{i}^{2} ⩾ \frac{1}{n} .

Commented by abdo imad last updated on 28/Dec/17

f o r a l l s e c o n s e s o f r e a l n u m b e r s {(a_{i})}_{1 ⩽ i ⩽ n} a n d {(b_{i})}_{1 ⩽ i ⩽ n} p o s i t i f s

\sum_{i = 1}^{i = n} a_{i} b_{i} ⩽ {(\sum_{i = 1}^{i = n} a_{i}^{2})}^{\frac{1}{2}} . {(\sum_{i = 1}^{i = n} b_{i}^{2})}^{\frac{1}{2}} (h o l d e r i n e q u a l i t y)

l e t t a k e b_{i} = 1 \Rightarrow \sum_{i = 1}^{i = n} a_{i} ⩽ \sqrt{n} {(\sum_{i = 1}^{i = n} a_{i}^{2})}^{\frac{1}{2}} a n d f o r a_{i} x_{i}

w e o b t a i n {(\sum_{i =}^{i = n} x_{i})}^{2} ⩽ n (\sum_{i = 1}^{i = n} x_{i}^{2})

\Leftrightarrow 1 ⩽ n (\sum_{i = 1}^{i = n} x_{i}^{2}) \Rightarrow \sum_{i = 1}^{i = n} x_{i}^{2} ⩾ \frac{1}{n} .