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I have two vectors $a$ and $b$. I have the two following quantities, $\sum_i a_i \frac{1}{b_i}$ and $\sum_i a_i \frac{1}{\sum_j b_j}$. I know that for every $i$, $0\leq a_i \leq b_i \leq 1$. Which inequality holds between the two sums?

I know that, calling $c_i = 1 / b_i$, the inverse holder inequality holds, that means, $\sum_i a_i \frac{1}{b_i} =|| a c||_1 \leq ||a||_2 ||c||_{-1}$, and I also know that $|| a||_2 \leq || a||_1$, by inclusion of the Lp spaces. Is it possible to show that $$\sum_i a_i \frac{1}{b_i} \leq \sum_i a_i \frac{1}{\sum b_i}$$ or that $$\sum_i a_i \frac{1}{b_i} \geq \sum_i a_i \frac{1}{\sum b_i} ?$$ (I am not sure wether one of those is true...)

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1 Answer 1

For every $i$, $b_i\leqslant B$ with $B=\sum\limits_kb_k$ since every $b_k$ is positive, hence $\frac1{b_i}\geqslant\frac1B$. Summing these and using the nonnegativity of every $a_i$, one gets $$ \sum_i\frac{a_i}{b_i}\geqslant\sum_i\frac{a_i}B=\sum_ia_i\cdot\frac1{\sum\limits_kb_k}.$$

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