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Question Number 106315 by pticantor last updated on 04/Aug/20

s h o w t h a t

⊛ \forall (x_{1}, x_{2}, \dots . ., x_{n}) \in R^{n}

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

⊛ a, b > 0

p (x) = x^{n} + a x + b = 0

c o u l d n o t h a v e m o r e t h a n 3

r e a l s s o l u t i o n s

Answered by Dwaipayan Shikari last updated on 04/Aug/20

\underset{k = 1}{\sum^{n}} x_{k}^{2} . \underset{k = 1}{\sum^{n}} 1 ⩾ {(Σ (x_{k} . 1))}^{2} {C a u c h y s c h w a r z i n e q u a l i t y}

n \underset{k = 1}{\sum^{n}} x_{k}^{2} ⩾ {(\underset{k = 1}{\sum^{n}} x_{k})}^{2} {\sum^{n} (1) = n

I t m e a n s

n (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots .) ⩾ {(x_{1} + x_{2} + x_{3} + x_{4} + \dots . .)}^{2}

(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + \dots) ⩾ {(\frac{x_{1} + x_{2} + x_{3} + x_{4} + \dots \dots}{\sqrt{n}})}^{2}