zmudz
  • zmudz
Assume that \( 1a_1+2a_2+\cdots+na_n=1, \) where the \(a_j\) are real numbers. As a function of \(n\), what is the minimum value of \(1a_1^2+2a_2^2+\cdots+na_n^2?\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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BAdhi
  • BAdhi
have you tried lagrange multiplier method?
BAdhi
  • BAdhi
take \[g(a_1,a_2,\cdots , a_n) = a_1+2a_2+\cdots +na_n\] and \[f(a_1,a_2,\cdots ,a_n) = a_1^2+2a_2^2+\cdots+na_n^2\] with \(g(a_1,a_2,\cdots , a_n) = 1\) \(g_{a_n} = n\) and \(f_{a_n} = 2na_n\) use \(\nabla g(a_1,a_2,\cdots , a_n) = \lambda \nabla f(a_1,a_2,\cdots ,a_n)\) with \( g(a_1,a_2,\cdots , a_n) = 1\)
thomas5267
  • thomas5267
The answer for \(n=5\) is \(\dfrac{1}{15}\) for all \(a_n\). Looks like AM-GM again.

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thomas5267
  • thomas5267
@Callisto @ganeshie8 @Kainui Any way without using generalised mean inequality? \[ \sqrt{\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}}\geq\frac{2}{n(n+1)}\sum_{k=0}^nka_k=\frac{2}{n(n+1)}\quad\text{Generalised mean inequality}\\ \sqrt{\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}}\geq\frac{2}{n(n+1)}\sum_{k=0}^nka_k=\frac{2}{n(n+1)}\text{ iff }a_1=a_2=a_3=\dots=a_n\\ \begin{align*} \sqrt{\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}}&=\frac{2}{n(n+1)}\\ \frac{2\sum_{k=1}^nka_k^2}{n(n+1)}&=\frac{4}{n^2(n+1)^2}\\ \sum_{k=1}^nka_k^2&=\frac{2}{n(n+1)}\\ \end{align*} \]
thomas5267
  • thomas5267
@zepdrix You seem to know this stuff. Can you help? He mentioned in one of his recent questions that he is studying pre-calc. I really don't want to invoke the generalised mean inequality since I don't think it is included in pre-calc. I didn't even know AM-GM inequality when I am studying it! The answer is correct though.
zepdrix
  • zepdrix
Oh boy D: I dunno
thomas5267
  • thomas5267
@ganeshie8 We must use generalised mean inequality?

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