Question

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
\(1a_1^2+2a_2^2+\cdots+na_n^2?\)

Answer 1

have you tried lagrange multiplier method?

Answer 2

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 \( g(a_1,a_2,\cdots , a_n) = 1\)

Answer 3

The answer for \(n=5\) is \(\dfrac{1}{15}\) for all \(a_n\). Looks like AM-GM again.

Answer 4

@Callisto @ganeshie8 @Kainui
Any way without using generalised mean inequality?
\[
\sqrt{\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}}\geq\frac{2}{n(n+1)}\sum_{k=0}^nka_k=\frac{2}{n(n+1)}\quad\text{Generalised mean inequality}\\
\sqrt{\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}}\geq\frac{2}{n(n+1)}\sum_{k=0}^nka_k=\frac{2}{n(n+1)}\text{ iff }a_1=a_2=a_3=\dots=a_n\\
\begin{align*}
\sqrt{\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}}&=\frac{2}{n(n+1)}\\
\frac{2\sum_{k=1}^nka_k^2}{n(n+1)}&=\frac{4}{n^2(n+1)^2}\\
\sum_{k=1}^nka_k^2&=\frac{2}{n(n+1)}\\
\end{align*}
\]

Answer 5

@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.

Answer 6

Oh boy D: I dunno

Answer 7

@ganeshie8 We must use generalised mean inequality?