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












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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♦ Jan 5 at 21:26


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












    $begingroup$



    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











    $endgroup$



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


    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





      $begingroup$



      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











      $endgroup$





      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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      share|cite|improve this question













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      edited Jan 5 at 21:23









      Michael Rozenberg

      107k1894199




      107k1894199










      asked Jan 5 at 21:02









      Amihai ZivanAmihai Zivan

      1,87411625




      1,87411625




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


      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♦ Jan 5 at 21:26


      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
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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









                  $endgroup$



                  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 Jan 5 at 21:19









                  Michael RozenbergMichael Rozenberg

                  107k1894199




                  107k1894199























                      1












                      $begingroup$

                      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









                      $endgroup$


















                        1












                        $begingroup$

                        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









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          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









                          $endgroup$



                          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 Jan 5 at 21:21









                          Mostafa AyazMostafa Ayaz

                          15.7k3939




                          15.7k3939















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