Suppose that $$\sum_{i=1}^n x_i \ge a$$ where $a>0$ and $x_i\in (0, b]$ for all $i$. Are there any bounding inequalities we can determine for $$\sum_{i=1}^n \frac{1}{x_i}?$$ I understand that $\sum_{i=1}^n \frac{1}{x_i} \ge \frac{n}{b}$, but I'm hoping to have some restriction that utilizes $a$. I have found this question that starts with the knowledge that $\sum_{i=1}^n x_i = a$. Following the details of the explanation in that question, I eventually conclude that $$\sum_{i=1}^n \frac{1}{x_i} \ge \frac{n^2}{\sum_{i=1}^n x_i},$$ but I'm not sure that this allows me to utilize the initial inequality $\sum_{i=1}^n x_i \ge a$. Am I missing something?
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1$\begingroup$ what is the range of $x_i$? $\endgroup$– NN2Commented Dec 4, 2023 at 21:02
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$\begingroup$ Are $x_i$ positive? $\endgroup$– userCommented Dec 4, 2023 at 21:05
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$\begingroup$ If $n \geq 2$ you can't say anything (why?). $\endgroup$– PhoemueXCommented Dec 4, 2023 at 21:06
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$\begingroup$ Note that in the question you link to, it is assumed that all summands are positive. That sort of detail is critical. Did you mean to add that assumption? $\endgroup$– luluCommented Dec 4, 2023 at 21:12
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1$\begingroup$ there is no lower bound as you can make $x_i$'s as large as you want. There is no upper bound since you can make at least one $x_i$ as small (positive) as you want. $\endgroup$– dezdichadoCommented Dec 4, 2023 at 22:56
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