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I encountered this inequality when reading some paper, however, why it holds does not seem very obvious to me:

Consider nonnegative real numbers $a_1, ..., a_n, b_1, ..., b_n$, and $b_i \leq \frac{a_i}{n}$. Let $\beta := (\prod_{i=1}^n a_i)^{1/n}$, show that $\sum_i \frac{b_i}{a_i} \leq \sum_i \frac{b_i}{\beta}$.

Update: I noticed that $\beta$ is actually the geometric mean of $a_i$'s, and I'm still googling about properties of geometric mean.

Update2: I have solved this problem myself. I was overlooking the assumption that $b_i \leq \frac{a_i}{n}$, with which the problem is simply an application of the inequality of arithmetic mean and geometric mean. Thanks for the comments!

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    $\begingroup$ @Macavity Hi, I updated the question 10 mins ago, and I added that $a_i \geq n b_i$. Sorry for the confusion! $\endgroup$ – Yang Feb 17 at 4:18

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