Sum of numerators divided by sum of denominators $leq$ the maximum fraction [duplicate]












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  • shorter proof of generalized mediant inequality?

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






inequality summation







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marked as duplicate by Martin R, Shubham Johri, Community♦ 2 days ago


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    -2















    This question already has an answer here:




    • shorter proof of generalized mediant inequality?

      2 answers




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






    inequality summation







    share|cite|improve this question















    marked as duplicate by Martin R, Shubham Johri, Community♦ 2 days ago


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















      -2












      -2








      -2








      This question already has an answer here:




      • shorter proof of generalized mediant inequality?

        2 answers




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






      inequality summation







      share|cite|improve this question
















      This question already has an answer here:




      • shorter proof of generalized mediant inequality?

        2 answers




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





      This question already has an answer here:




      • shorter proof of generalized mediant inequality?

        2 answers







      inequality summation




      inequality summation






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      edited 2 days ago









      Michael Rozenberg

      97.4k1589188




      97.4k1589188










      asked 2 days ago









      Amihai ZivanAmihai Zivan

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      1,87411624




      marked as duplicate by Martin R, Shubham Johri, Community♦ 2 days ago


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Martin R, Shubham Johri, Community♦ 2 days ago


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          2 Answers
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          Let for all $k$ we have
          $$frac{a_j}{b_j}geqfrac{a_k}{b_k}$$ or
          $$a_jb_kgeq a_kb_j.$$
          Thus, $$a_jsum_{k=1}^nb_kgeqsum_{k=1}^na_kb_j=b_jsum_{k=1}^na_k.$$






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            Let $${a_kover b_k}=max_i{a_iover b_i}$$therefore $${a_iover b_i}le {a_kover b_k}$$for any $i$ which is equivalent to $$a_ib_kle a_kb_i$$by a summation we finally have $$b_ksum_i a_ile a_ksum_i b_i$$or $${sum_i a_iover sum_i b_i}le {a_kover b_k}=max_i{a_iover b_i}$$






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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              Let for all $k$ we have
              $$frac{a_j}{b_j}geqfrac{a_k}{b_k}$$ or
              $$a_jb_kgeq a_kb_j.$$
              Thus, $$a_jsum_{k=1}^nb_kgeqsum_{k=1}^na_kb_j=b_jsum_{k=1}^na_k.$$






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                2














                Let for all $k$ we have
                $$frac{a_j}{b_j}geqfrac{a_k}{b_k}$$ or
                $$a_jb_kgeq a_kb_j.$$
                Thus, $$a_jsum_{k=1}^nb_kgeqsum_{k=1}^na_kb_j=b_jsum_{k=1}^na_k.$$






                share|cite|improve this answer
























                  2












                  2








                  2






                  Let for all $k$ we have
                  $$frac{a_j}{b_j}geqfrac{a_k}{b_k}$$ or
                  $$a_jb_kgeq a_kb_j.$$
                  Thus, $$a_jsum_{k=1}^nb_kgeqsum_{k=1}^na_kb_j=b_jsum_{k=1}^na_k.$$






                  share|cite|improve this answer












                  Let for all $k$ we have
                  $$frac{a_j}{b_j}geqfrac{a_k}{b_k}$$ or
                  $$a_jb_kgeq a_kb_j.$$
                  Thus, $$a_jsum_{k=1}^nb_kgeqsum_{k=1}^na_kb_j=b_jsum_{k=1}^na_k.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Michael RozenbergMichael Rozenberg

                  97.4k1589188




                  97.4k1589188























                      1














                      Let $${a_kover b_k}=max_i{a_iover b_i}$$therefore $${a_iover b_i}le {a_kover b_k}$$for any $i$ which is equivalent to $$a_ib_kle a_kb_i$$by a summation we finally have $$b_ksum_i a_ile a_ksum_i b_i$$or $${sum_i a_iover sum_i b_i}le {a_kover b_k}=max_i{a_iover b_i}$$






                      share|cite|improve this answer


























                        1














                        Let $${a_kover b_k}=max_i{a_iover b_i}$$therefore $${a_iover b_i}le {a_kover b_k}$$for any $i$ which is equivalent to $$a_ib_kle a_kb_i$$by a summation we finally have $$b_ksum_i a_ile a_ksum_i b_i$$or $${sum_i a_iover sum_i b_i}le {a_kover b_k}=max_i{a_iover b_i}$$






                        share|cite|improve this answer
























                          1












                          1








                          1






                          Let $${a_kover b_k}=max_i{a_iover b_i}$$therefore $${a_iover b_i}le {a_kover b_k}$$for any $i$ which is equivalent to $$a_ib_kle a_kb_i$$by a summation we finally have $$b_ksum_i a_ile a_ksum_i b_i$$or $${sum_i a_iover sum_i b_i}le {a_kover b_k}=max_i{a_iover b_i}$$






                          share|cite|improve this answer












                          Let $${a_kover b_k}=max_i{a_iover b_i}$$therefore $${a_iover b_i}le {a_kover b_k}$$for any $i$ which is equivalent to $$a_ib_kle a_kb_i$$by a summation we finally have $$b_ksum_i a_ile a_ksum_i b_i$$or $${sum_i a_iover sum_i b_i}le {a_kover b_k}=max_i{a_iover b_i}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 days ago









                          Mostafa AyazMostafa Ayaz

                          14.2k3937




                          14.2k3937















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