## zmudz one year ago 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 $$1a_1^2+2a_2^2+\cdots+na_n^2?$$

have you tried lagrange multiplier method?

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

3. thomas5267

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

4. 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*}

5. 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.

6. zepdrix

Oh boy D: I dunno

7. thomas5267

@ganeshie8 We must use generalised mean inequality?