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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

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

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**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.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}

\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}

\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}

\newcommand{\dd}{\mathrm{d}}

\newcommand{\ds}[1]{\displaystyle{#1}}

\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}

\newcommand{\ic}{\mathrm{i}}

\newcommand{\mc}[1]{\mathcal{#1}}

\newcommand{\mrm}[1]{\mathrm{#1}}

\newcommand{\pars}[1]{\left(\,{#1}\,\right)}

\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}

\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}

\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}

\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

&\mathsf{A}\mathbf{r} – \lambda\mathbf{u} = 0\,,\qquad

\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}

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