show that ⊛∀ (x_1 ,x_2 ,.....,x_(n ) )∈R^n (Σ_(k=1) ^n x_k )^2 ≤nΣ_(k=1) ^n x_k ^2 ⊛a,b>0 p(x)=x^n +ax+b=0 could not have more than 3 reals solutions


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

$s h o w t h a t$ $⊛ \forall (x_{1}, x_{2}, . . . . ., x_{n}) \in R^{n}$ ${(\underset{k = 1}{\sum^{n}} x_{k})}^{2} ⩽ n \underset{k = 1}{\sum^{n}} x_{k}^{2}$ $⊛ a, b > 0$ $p (x) = x^{n} + a x + b = 0$ $c o u l d n o t h a v e m o r e t h a n 3$ $r e a l s s o l u t i o n s$
	Answered by Dwaipayan Shikari last updated on 04/Aug/20
	$Σ_(k=1) ^n x_k ^2 .Σ_(k=1) ^n 1≥(Σ(x_k .1))^2 {Cauchy schwarz inequality} nΣ_(k=1) ^n x_k ^2 ≥(Σ_(k=1) ^n x_k )^2 {Σ^n (1)=n It means n(x_1 ^2 +x_2 ^2 +x_3 ^2 +x_4 ^2 +....)≥(x_1 +x_2 +x_3 +x_4 +.....)^2 (x_1 ^2 +x_2 ^2 +x_3 ^2 +x_4 ^2 +...)≥(((x_1 +x_2 +x_3 +x_4 +......)/(√n)))^2$
	$\underset{k = 1}{\sum^{n}} x_{k}^{2} . \underset{k = 1}{\sum^{n}} 1 ⩾ {(Σ (x_{k} . 1))}^{2} {C a u c h y s c h w a r z i n e q u a l i t y}$ $n \underset{k = 1}{\sum^{n}} x_{k}^{2} ⩾ {(\underset{k = 1}{\sum^{n}} x_{k})}^{2} {\sum^{n} (1) = n$ $I t m e a n s$ $n (x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + . . . .) ⩾ {(x_{1} + x_{2} + x_{3} + x_{4} + . . . . .)}^{2}$ $(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + . . .) ⩾ {(\frac{x_{1} + x_{2} + x_{3} + x_{4} + . . . . . .}{\sqrt{n}})}^{2}$
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