Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...)

share|cite|improve this question

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 $$ \sum_i\frac{a_i}{b_i}\geqslant\sum_i\frac{a_i}B=\sum_ia_i\cdot\frac1{\sum\limits_kb_k}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.