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