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