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I'm stuck at proof of finding minimum of the expression

$\ \sum_{k=1}^{n}a_{k}^{2}+\left(\sum_{k=1}^n a_k\right)^2\\ \sum_{k=1}^{n}p_{k}a_{k}=1\\ $

So my first thought is to square

$1=\sum_{k=1}^{n}p_{k}a_{k}=\sum_{k=1}^{n}(p_{k}-\beta )a_{k}+\beta (\sum_{k=1}^{n}a_{k}) \\$

Divide it in three expressions and use cauchy-schwarz inequality:

$\ (\sum_{k=1}^{n}(p_{k}-\beta )a_{k})^{2} \leq\sum_{k=1}^{n}(p_{k}-\beta)^{2}\sum_{k=1}^{n}a_{k}^{2}\\ (\sum_{k=1}^{n}\beta a_{k})^{2}\leq \sum_{k=1}^{n}\beta^{2} \sum_{k=1}^{n}a_{k}^{2}\\ 2\beta \sum_{k=1}^{n}a_{k} \sum_{k=1}^{n}(p_{k}-\beta) \leq?$

Adding inequalities:

$\ 1 \leq \sum_{k=1}^{n}((p_{k}-\beta)^{2}+\beta^{2})\sum_{k=1}^{n}a_{k}^{2} + \sum_{k=1}^{n}a_{k} \sum_{k=1}^{n}(p_{k}-\beta) $

I don't know if it is correct or not, but I suppose I need to transform last inequality, because it's different in proof :

$\ 1=\sum_{k=1}^{n}p_{k}a_{k}=\sum_{k=1}^{n}(p_{k}-\beta )a_{k}+\beta (\sum_{k=1}^{n}a_{k})\leqslant (\sum_{k=1}^{n}(p_{k}-\beta )^{2}+\beta ^{2})((\sum_{k=1}^{n}a_{k}^{2})+(\sum_{k=1}^{n}a_{k})^{2})\Rightarrow \sum_{k=1}^{n}a_{k}^{2})+(\sum_{k=1}^{n}a_{k})^{2}\geqslant (\sum_{k=1}^{n}(p_{k}-\beta )^{2}+\beta ^{2})^{-1}$

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    I'm trying to repeat and understand proof http://math.stackexchange.com/questions/165760/inequality-using-lagrange-multipliers-and-cauchy-schwarz-inequality but I don't know how to extract $\ \left(\sum_{k=1}^n a_k\right)^2$.2012-12-01

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