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This question already has an answer here:

Let $\tfrac{a_1}{b_1},\dots,\tfrac{a_n}{b_n}$ where $a_i,b_i>0$. How can one prove that $$\frac{\sum_i a_i}{\sum_i b_i}\leq \max_j \tfrac{a_j}{b_j}$$?

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marked as duplicate by Martin R, Shubham Johri, Community Jan 5 at 21:26

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Let for all $k$ we have $$\frac{a_j}{b_j}\geq\frac{a_k}{b_k}$$ or $$a_jb_k\geq a_kb_j.$$ Thus, $$a_j\sum_{k=1}^nb_k\geq\sum_{k=1}^na_kb_j=b_j\sum_{k=1}^na_k.$$

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Let $${a_k\over b_k}=\max_i{a_i\over b_i}$$therefore $${a_i\over b_i}\le {a_k\over b_k}$$for any $i$ which is equivalent to $$a_ib_k\le a_kb_i$$by a summation we finally have $$b_k\sum_i a_i\le a_k\sum_i b_i$$or $${\sum_i a_i\over \sum_i b_i}\le {a_k\over b_k}=\max_i{a_i\over b_i}$$

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