Sum of positive elements divided by their “weighted” product - inequality












2














I have following expression,



$$ frac{sum_{i=1}^n x_i}{prod_{i=1}^nx_i^{p_i}} $$



where $p_i$s satisfy $sum p_i = 1$ and $p_i in [0,1]$ and $x_igeq0$, $forall i in 1dots n$.



I think that this expression is always $geq 1$, however, I don't know how to prove it.



Is there anything I can conclude?



Thanks.






calculus inequality summation products







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  • 3




    The weighted AM-GM inequality yields $prod_i x_i^{p_i} leq left(sum_i p_i x_iright)^{sum_i p_i} = left(sum_i p_i x_iright)^1 = sum_i p_i x_i leq sum_i 1 x_i = sum_i x_i$.
    – darij grinberg
    4 hours ago










  • @darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.]
    – coffeemath
    4 hours ago










  • @coffeemath: Good idea.
    – darij grinberg
    3 hours ago
















2














I have following expression,



$$ frac{sum_{i=1}^n x_i}{prod_{i=1}^nx_i^{p_i}} $$



where $p_i$s satisfy $sum p_i = 1$ and $p_i in [0,1]$ and $x_igeq0$, $forall i in 1dots n$.



I think that this expression is always $geq 1$, however, I don't know how to prove it.



Is there anything I can conclude?



Thanks.






calculus inequality summation products







share|cite|improve this question




















  • 3




    The weighted AM-GM inequality yields $prod_i x_i^{p_i} leq left(sum_i p_i x_iright)^{sum_i p_i} = left(sum_i p_i x_iright)^1 = sum_i p_i x_i leq sum_i 1 x_i = sum_i x_i$.
    – darij grinberg
    4 hours ago










  • @darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.]
    – coffeemath
    4 hours ago










  • @coffeemath: Good idea.
    – darij grinberg
    3 hours ago














2












2








2


1





I have following expression,



$$ frac{sum_{i=1}^n x_i}{prod_{i=1}^nx_i^{p_i}} $$



where $p_i$s satisfy $sum p_i = 1$ and $p_i in [0,1]$ and $x_igeq0$, $forall i in 1dots n$.



I think that this expression is always $geq 1$, however, I don't know how to prove it.



Is there anything I can conclude?



Thanks.






calculus inequality summation products







share|cite|improve this question















I have following expression,



$$ frac{sum_{i=1}^n x_i}{prod_{i=1}^nx_i^{p_i}} $$



where $p_i$s satisfy $sum p_i = 1$ and $p_i in [0,1]$ and $x_igeq0$, $forall i in 1dots n$.



I think that this expression is always $geq 1$, however, I don't know how to prove it.



Is there anything I can conclude?



Thanks.






calculus inequality summation products




calculus inequality summation products






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









darij grinberg

10.4k33062




10.4k33062










asked 4 hours ago









Michael MarkMichael Mark

1119




1119








  • 3




    The weighted AM-GM inequality yields $prod_i x_i^{p_i} leq left(sum_i p_i x_iright)^{sum_i p_i} = left(sum_i p_i x_iright)^1 = sum_i p_i x_i leq sum_i 1 x_i = sum_i x_i$.
    – darij grinberg
    4 hours ago










  • @darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.]
    – coffeemath
    4 hours ago










  • @coffeemath: Good idea.
    – darij grinberg
    3 hours ago














  • 3




    The weighted AM-GM inequality yields $prod_i x_i^{p_i} leq left(sum_i p_i x_iright)^{sum_i p_i} = left(sum_i p_i x_iright)^1 = sum_i p_i x_i leq sum_i 1 x_i = sum_i x_i$.
    – darij grinberg
    4 hours ago










  • @darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.]
    – coffeemath
    4 hours ago










  • @coffeemath: Good idea.
    – darij grinberg
    3 hours ago








3




3




The weighted AM-GM inequality yields $prod_i x_i^{p_i} leq left(sum_i p_i x_iright)^{sum_i p_i} = left(sum_i p_i x_iright)^1 = sum_i p_i x_i leq sum_i 1 x_i = sum_i x_i$.
– darij grinberg
4 hours ago




The weighted AM-GM inequality yields $prod_i x_i^{p_i} leq left(sum_i p_i x_iright)^{sum_i p_i} = left(sum_i p_i x_iright)^1 = sum_i p_i x_i leq sum_i 1 x_i = sum_i x_i$.
– darij grinberg
4 hours ago












@darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.]
– coffeemath
4 hours ago




@darijgrinberg That comment could be made an answer. [Anyway I'd upvote it.]
– coffeemath
4 hours ago












@coffeemath: Good idea.
– darij grinberg
3 hours ago




@coffeemath: Good idea.
– darij grinberg
3 hours ago










2 Answers
2






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oldest

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3














Applying the weighted AM-GM inequality to the weights $p_1, p_2, ldots, p_n$, we obtain
begin{equation}
dfrac{p_1 x_1 + p_2 x_2 + cdots + p_n x_n}{1} geq sqrt[1]{x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}}
end{equation}
(since $p_1 + p_2 + cdots + p_n = sum_{i=1}^n p_i = 1$). This simplifies to
begin{equation}
p_1 x_1 + p_2 x_2 + cdots + p_n x_n geq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} .
end{equation}
Hence,
begin{align}
x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}
leq p_1 x_1 + p_2 x_2 + cdots + p_n x_n
= sum_{i=1}^n underbrace{p_i}_{substack{leq 1 \ text{(since $p_i in left[0,1right]$)}}} x_i leq sum_{i=1}^n x_i ,
end{align}
so that
begin{align}
sum_{i=1}^n x_i leq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} = prod_{i=1}^n x_i^{p_i} .
end{align}






share|cite|improve this answer





























    4














    Concavity of $log$ gives
    $$logleft( prod_{i=1}^nx_i^{p_i}right) = sum_{i=1}^n p_ilog x_i stackrel{mbox{concavity}}{leq} logleft( sum_{i=1}^np_i x_iright) stackrel{0leq p_ileq 1}{leq} logleft( sum_{i=1}^nx_iright)$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      Applying the weighted AM-GM inequality to the weights $p_1, p_2, ldots, p_n$, we obtain
      begin{equation}
      dfrac{p_1 x_1 + p_2 x_2 + cdots + p_n x_n}{1} geq sqrt[1]{x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}}
      end{equation}
      (since $p_1 + p_2 + cdots + p_n = sum_{i=1}^n p_i = 1$). This simplifies to
      begin{equation}
      p_1 x_1 + p_2 x_2 + cdots + p_n x_n geq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} .
      end{equation}
      Hence,
      begin{align}
      x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}
      leq p_1 x_1 + p_2 x_2 + cdots + p_n x_n
      = sum_{i=1}^n underbrace{p_i}_{substack{leq 1 \ text{(since $p_i in left[0,1right]$)}}} x_i leq sum_{i=1}^n x_i ,
      end{align}
      so that
      begin{align}
      sum_{i=1}^n x_i leq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} = prod_{i=1}^n x_i^{p_i} .
      end{align}






      share|cite|improve this answer


























        3














        Applying the weighted AM-GM inequality to the weights $p_1, p_2, ldots, p_n$, we obtain
        begin{equation}
        dfrac{p_1 x_1 + p_2 x_2 + cdots + p_n x_n}{1} geq sqrt[1]{x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}}
        end{equation}
        (since $p_1 + p_2 + cdots + p_n = sum_{i=1}^n p_i = 1$). This simplifies to
        begin{equation}
        p_1 x_1 + p_2 x_2 + cdots + p_n x_n geq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} .
        end{equation}
        Hence,
        begin{align}
        x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}
        leq p_1 x_1 + p_2 x_2 + cdots + p_n x_n
        = sum_{i=1}^n underbrace{p_i}_{substack{leq 1 \ text{(since $p_i in left[0,1right]$)}}} x_i leq sum_{i=1}^n x_i ,
        end{align}
        so that
        begin{align}
        sum_{i=1}^n x_i leq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} = prod_{i=1}^n x_i^{p_i} .
        end{align}






        share|cite|improve this answer
























          3












          3








          3






          Applying the weighted AM-GM inequality to the weights $p_1, p_2, ldots, p_n$, we obtain
          begin{equation}
          dfrac{p_1 x_1 + p_2 x_2 + cdots + p_n x_n}{1} geq sqrt[1]{x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}}
          end{equation}
          (since $p_1 + p_2 + cdots + p_n = sum_{i=1}^n p_i = 1$). This simplifies to
          begin{equation}
          p_1 x_1 + p_2 x_2 + cdots + p_n x_n geq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} .
          end{equation}
          Hence,
          begin{align}
          x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}
          leq p_1 x_1 + p_2 x_2 + cdots + p_n x_n
          = sum_{i=1}^n underbrace{p_i}_{substack{leq 1 \ text{(since $p_i in left[0,1right]$)}}} x_i leq sum_{i=1}^n x_i ,
          end{align}
          so that
          begin{align}
          sum_{i=1}^n x_i leq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} = prod_{i=1}^n x_i^{p_i} .
          end{align}






          share|cite|improve this answer












          Applying the weighted AM-GM inequality to the weights $p_1, p_2, ldots, p_n$, we obtain
          begin{equation}
          dfrac{p_1 x_1 + p_2 x_2 + cdots + p_n x_n}{1} geq sqrt[1]{x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}}
          end{equation}
          (since $p_1 + p_2 + cdots + p_n = sum_{i=1}^n p_i = 1$). This simplifies to
          begin{equation}
          p_1 x_1 + p_2 x_2 + cdots + p_n x_n geq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} .
          end{equation}
          Hence,
          begin{align}
          x_1^{p_1} x_2^{p_2} cdots x_n^{p_n}
          leq p_1 x_1 + p_2 x_2 + cdots + p_n x_n
          = sum_{i=1}^n underbrace{p_i}_{substack{leq 1 \ text{(since $p_i in left[0,1right]$)}}} x_i leq sum_{i=1}^n x_i ,
          end{align}
          so that
          begin{align}
          sum_{i=1}^n x_i leq x_1^{p_1} x_2^{p_2} cdots x_n^{p_n} = prod_{i=1}^n x_i^{p_i} .
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          darij grinbergdarij grinberg

          10.4k33062




          10.4k33062























              4














              Concavity of $log$ gives
              $$logleft( prod_{i=1}^nx_i^{p_i}right) = sum_{i=1}^n p_ilog x_i stackrel{mbox{concavity}}{leq} logleft( sum_{i=1}^np_i x_iright) stackrel{0leq p_ileq 1}{leq} logleft( sum_{i=1}^nx_iright)$$






              share|cite|improve this answer


























                4














                Concavity of $log$ gives
                $$logleft( prod_{i=1}^nx_i^{p_i}right) = sum_{i=1}^n p_ilog x_i stackrel{mbox{concavity}}{leq} logleft( sum_{i=1}^np_i x_iright) stackrel{0leq p_ileq 1}{leq} logleft( sum_{i=1}^nx_iright)$$






                share|cite|improve this answer
























                  4












                  4








                  4






                  Concavity of $log$ gives
                  $$logleft( prod_{i=1}^nx_i^{p_i}right) = sum_{i=1}^n p_ilog x_i stackrel{mbox{concavity}}{leq} logleft( sum_{i=1}^np_i x_iright) stackrel{0leq p_ileq 1}{leq} logleft( sum_{i=1}^nx_iright)$$






                  share|cite|improve this answer












                  Concavity of $log$ gives
                  $$logleft( prod_{i=1}^nx_i^{p_i}right) = sum_{i=1}^n p_ilog x_i stackrel{mbox{concavity}}{leq} logleft( sum_{i=1}^np_i x_iright) stackrel{0leq p_ileq 1}{leq} logleft( sum_{i=1}^nx_iright)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  trancelocationtrancelocation

                  9,5601622




                  9,5601622






























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