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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?

BSplitter
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    what is the range of $x_i$? – NN2 Dec 04 '23 at 21:02
  • Are $x_i$ positive? – user Dec 04 '23 at 21:05
  • If $n \geq 2$ you can't say anything (why?). – PhoemueX Dec 04 '23 at 21:06
  • 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? – lulu Dec 04 '23 at 21:12
  • Yes, all the $x_i$ are positive. – BSplitter Dec 04 '23 at 22:54
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    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. – dezdichado Dec 04 '23 at 22:56
  • @dezdichado I've edited the question. After some thought, I've determined that $x_i \in (0,b] \quad \forall i$ for some constant $b$. – BSplitter Dec 04 '23 at 22:58
  • @BSplitter if $x_i \in (0,b]$ then the supremum is $+\infty$ and the maximum can be never reached (just take $x_1 \to 0$). The minimum is equal to $n/b$ and occurs if and only if $x_i = b$. If you want to " have some restriction that utilizes $a$" then you may impose that $x_i \in [c,d]$ for example – NN2 Dec 29 '23 at 02:44

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