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zmudz
 one year ago
Assume that
\(
1a_1+2a_2+\cdots+20a_{20}=1,
\)
where the \(a_j\) are real numbers and that these values minimize
\(1a_1^2+2a_2^2+\cdots+20a_{20}^2\)
Find \(a_{12}\).
zmudz
 one year ago
Assume that \( 1a_1+2a_2+\cdots+20a_{20}=1, \) where the \(a_j\) are real numbers and that these values minimize \(1a_1^2+2a_2^2+\cdots+20a_{20}^2\) Find \(a_{12}\).

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Recall Cauchy–Schwarz inequality : \[ \langle x,x\rangle\cdot \langle y,y \rangle \ge \langle x,y\rangle^2\] Let \(x=(1,\sqrt{2},\ldots,\sqrt{20})\) \(y=(1a_1,\sqrt{2}a_2,\ldots,\sqrt{20}a_{20})\) \((1+2+\cdots +20)\cdot (1a_1^2+2a_2^2+\cdots+20a_{20}^2)\ge(1a_1+2a_2+\cdots + 20a_{20})^2=1 \) \(\implies 1a_1^2+2a_2^2+\cdots+20a_{20}^2\ge \dfrac{2}{20*21} \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6The equality is achieved when \(x\) and \(y\) are parallel : \(y=kx\implies a_1=a_2=\cdots=a_{20}=a\) \(\implies a(1+2+\cdots +20)=1 \implies a=\dfrac{2}{20*21}\) so \(a_{12}=\dfrac{1}{210}\)
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