# Cauchy-Schwarz inequality - find minimum.

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:

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

$\ 1 \leq \sum_{k=1}^{n}((p_{k}-\beta)^{2}+\beta^{2})\sum_{k=1}^{n}a_{k}^{2} + \sum_{k=1}^{n}a_{k} \sum_{k=1}^{n}(p_{k}-\beta)$
$\ 1=\sum_{k=1}^{n}p_{k}a_{k}=\sum_{k=1}^{n}(p_{k}-\beta )a_{k}+\beta (\sum_{k=1}^{n}a_{k})\leqslant (\sum_{k=1}^{n}(p_{k}-\beta )^{2}+\beta ^{2})((\sum_{k=1}^{n}a_{k}^{2})+(\sum_{k=1}^{n}a_{k})^{2})\Rightarrow \sum_{k=1}^{n}a_{k}^{2})+(\sum_{k=1}^{n}a_{k})^{2}\geqslant (\sum_{k=1}^{n}(p_{k}-\beta )^{2}+\beta ^{2})^{-1}$
Are the $p_k$'s probabilities? That is, we can assume that $p_k > 0$ for all $k$, $1 \leq k \leq n$ and $\sum_k p_k = 1$?? Can some of the $p_k$'s be $0$ as long as the sum is $1$? – Dilip Sarwate Dec 1 '12 at 21:49
I'm not sure what the problem is. Are you trying to minimize the expression on the first line? If so, what does the second line with the $p_k$ have to do with it? What variable are you trying to minimize over? – Jonathan Christensen Dec 1 '12 at 21:49
I'm trying to repeat and understand proof math.stackexchange.com/questions/165760/… but I don't know how to extract $\ \left(\sum_{k=1}^n a_k\right)^2$. – DDT Dec 1 '12 at 22:13