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I can't find a reference to the fairly simple idea that:

$\Sigma (a_i) \Sigma(b_i) \geq \Sigma(a_ib_i)$ for $a_i,b_i \geq 0$

This is obviously equal for the case where $i$ goes only to 1 and for the n=2 case, the left hand side expands out to include every term on the right hand side plus a few cross terms. I just can't find its name or an elegant proof other than expanding the left hand side. It looks maybe related to the arithmetic-geometric mean inequality? Thanks

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$\sum{a_i}\sum{b_i} - \sum{a_i}{b_i} = \sum_{i\neq j}a_ib_j\geq0$

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You don’t have to expand the lefthand side completely:

$$\begin{align*} \left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^nb_j\right)&=\sum_{i=1}^n\sum_{j=1}^na_ib_j\\ &=\sum_{1\le i,j\le n}a_ib_j\\ &=\sum_{1\le i=j\le n}a_ib_j+\sum_{{1\le i,j\le n}\atop{i\ne j}}a_ib_j\\ &\ge\sum_{1\le i=j\le n}a_ib_j\\ &=\sum_{i=1}^na_ib_i\;. \end{align*}$$

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If everything is nonnegative, then each $a_i$ is less than or equal to the sum of all the $a$'s. So $$ a_i\le\sum_k a_k\tag{*} $$ and so $$\sum_i a_ib_i\stackrel{(1)}\le \sum_i \left(\sum_k a_k\right) b_i \stackrel{(2)}= \sum_k a_k \sum_i b_i $$ In (1) we use (*) and the fact that all the $b$'s are nonnegative. In (2) the quantity $\sum_ka_k$ is constant, and so can be pulled out of the sum.

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Also probabilistic interpretation.

WLOG assume $0<\sum b_i<\infty$ (otherwise inequality is trivial). Then

$B_i= b_i/ \sum (b_i)$ is a probability measure, and therefore

$$ \sum a_i B_i \le \max_{i} a_i \le \sum a_i.$$

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