Let $\{X_i\}_{i=1}^{N}$ be a set of positive numbers and $\epsilon > 0$ such that $$ \sum_{i=1}^{N}{\frac{1}{X_i}} \geq \epsilon$$ Is it possible to find an upper bound such that $$ \sum_{i=1}^{N}{X_i} \leq f(\epsilon)$$ where $f$ is a smooth function.
Upper bound for the sum of positive numbers
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real-analysis
inequality
convex-analysis
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1No. If you choose $X_1 = 1/\epsilon$ then $X_2, X_3, ...$ can be arbitrarily large. – 2017-02-15
1 Answers
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No. For a given $\epsilon > 0$ you can choose $X_1 = 1/\epsilon$ and arbitrary positive $X_2, X_3, ...$. Then $\sum_{i=1}^{N}{\frac{1}{X_i}} \geq \epsilon$ is satisfied and $\sum_{i=1}^{N}{X_i}$ can be arbitrarily large.
Note that the inequality between the harmonic and the arithmetic mean provides an estimate in the opposite direction: $$ \sum_{i=1}^{N}{\frac{1}{X_i}} < \varepsilon \Longrightarrow \sum_{i=1}^{N}{X_i} > \frac{N^2}{\varepsilon} $$