A curious inequality concerning binomial coefficients












8












$begingroup$


Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.



Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
$$
sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
$$
The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.






inequalities binomial-coefficients







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    8












    $begingroup$


    Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.



    Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
    $$
    sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
    $$
    The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.






    inequalities binomial-coefficients







    share|cite|improve this question









    $endgroup$















      8












      8








      8


      1



      $begingroup$


      Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.



      Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
      $$
      sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
      $$
      The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.






      inequalities binomial-coefficients







      share|cite|improve this question









      $endgroup$




      Has anyone seen an inequality of this form before? It seems to be true (based on extensive testing), but I am not able to prove it.



      Let $a_1,a_2,ldots,a_k$ be non-negative integers such that $sum_i a_i = A$. Then, for any non-negative integer $B le A$:
      $$
      sum_{(b_1,ldots,b_k): sum_i b_i = B} prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}} ge {binom{A}{B}}^{2-k}.
      $$
      The sum on the left is over all tuples $(b_1,b_2,ldots,b_k)$ of non-negative integers, with $b_i le a_i$ for all $i$, whose sum is equal to $B$.






      inequalities binomial-coefficients




      inequalities binomial-coefficients






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 13 hours ago









      Navin K.Navin K.

      682




      682






















          1 Answer
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          12












          $begingroup$

          By Cauchy–Bunyakovsky–Schwarz inequality we have
          $$
          left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
          left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
          .$$



          Thus it suffices to prove that
          $$
          sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
          $$
          But RHS is just the sum of the same guys without the restriction $sum b_i=B$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Brilliant! Thank you.
            $endgroup$
            – Navin K.
            5 hours ago











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          12












          $begingroup$

          By Cauchy–Bunyakovsky–Schwarz inequality we have
          $$
          left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
          left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
          .$$



          Thus it suffices to prove that
          $$
          sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
          $$
          But RHS is just the sum of the same guys without the restriction $sum b_i=B$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Brilliant! Thank you.
            $endgroup$
            – Navin K.
            5 hours ago
















          12












          $begingroup$

          By Cauchy–Bunyakovsky–Schwarz inequality we have
          $$
          left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
          left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
          .$$



          Thus it suffices to prove that
          $$
          sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
          $$
          But RHS is just the sum of the same guys without the restriction $sum b_i=B$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Brilliant! Thank you.
            $endgroup$
            – Navin K.
            5 hours ago














          12












          12








          12





          $begingroup$

          By Cauchy–Bunyakovsky–Schwarz inequality we have
          $$
          left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
          left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
          .$$



          Thus it suffices to prove that
          $$
          sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
          $$
          But RHS is just the sum of the same guys without the restriction $sum b_i=B$.






          share|cite|improve this answer









          $endgroup$



          By Cauchy–Bunyakovsky–Schwarz inequality we have
          $$
          left(sum prod_i frac{binom{a_i}{b_i}}{binom{A-a_i}{B-b_i}}right)left(sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}right)geqslant
          left(sum prod_ibinom{a_i}{b_i}right)^2=binom{A}{B}^2
          .$$



          Thus it suffices to prove that
          $$
          sumprod_i binom{a_i}{b_i}binom{A-a_i}{B-b_i}leqslant binom{A}B^k.
          $$
          But RHS is just the sum of the same guys without the restriction $sum b_i=B$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 11 hours ago









          Fedor PetrovFedor Petrov

          50.8k6117233




          50.8k6117233












          • $begingroup$
            Brilliant! Thank you.
            $endgroup$
            – Navin K.
            5 hours ago


















          • $begingroup$
            Brilliant! Thank you.
            $endgroup$
            – Navin K.
            5 hours ago
















          $begingroup$
          Brilliant! Thank you.
          $endgroup$
          – Navin K.
          5 hours ago




          $begingroup$
          Brilliant! Thank you.
          $endgroup$
          – Navin K.
          5 hours ago


















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