# Assume that $1a_1+2a_2+\cdots+na_n=1$, where the $a_j$ are real numbers.

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
$$1a_1^2+2a_2^2+\cdots+na_n^2?$$

#### Solutions Collecting From Web of "Assume that $1a_1+2a_2+\cdots+na_n=1$, where the $a_j$ are real numbers."

hint: $1 = \left(1a_1+\sqrt{2}(\sqrt{2}a_2)+\cdots + \sqrt{n}(\sqrt{n}a_n)\right)^2\leq (1+2+\cdots + n)(1a_1^2+2a_2^2+\cdots + na_n^2)$

From Jensen’s inequality, the weighted average of the squares is greater than or equal to the square of the weighted average.


Lets
$\ds{\mathbf{r} \equiv\pars{\substack{\ds{a_{1}} \\ \ds{\vdots} \\[1mm] \ds{a_{n}}}}}$,

$\ds{\mathsf{A}}$ is a diagonal matrix with eigenvalues $\ds{\braces{i}}$ $\ds{\pars{i = 1,2,\ldots,n}}$ and
$\ds{\mathbf{u} \equiv \pars{\substack{\ds{1} \\[2mm] \ds{2} \\[2mm] \ds{\vdots} \\[2mm] \ds{n}}}}$.

The problem is reduced to find the minimum of $\ds{{\Large\varepsilon} \equiv \mathbf{r}^{\mrm{T}}\mathsf{A}\mathbf{r}}$ under the restriction $\ds{\mathbf{r}^{\mrm{T}}\mathbf{u} = 1}$. The Lagrange Multiplier Method yields

\begin{align}
\pars{~\lambda:\ Lagrange\ Multiplier~}
\\
\implies &
\mathbf{r} = \lambda\mathsf{A}^{-1}\mathbf{u} =
\lambda
\pars{\substack{\ds{1} \\[2mm] \ds{1} \\[2mm] \ds{\vdots} \\[2mm] \ds{1}}}
\implies
1 = \lambda \pars{1 + 2 + \cdots + n} \implies \lambda = {2 \over n\pars{n + 1}}
\\[5mm] \implies &
\mathbf{r} =
{2 \over n\pars{n + 1}}\,\pars{\substack{\ds{1} \\[2mm] \ds{1} \\[2mm] \ds{\vdots} \\[2mm] \ds{1}}}
\\[5mm] \implies &
{\Large\varepsilon} = {2 \over n\pars{n + 1}}\,\pars{1 + 2 + \cdots + n}\,
{2 \over n\pars{n + 1}} = \bbx{2 \over n\pars{n + 1}}
\end{align}