$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$ strict inequality when $p_i neq p_j$ for $i neq j$











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Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:



$$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$



(I used Jensen's inequality for this)



Now the next question is why we do we have the strict inequality if all $x_i$ are different:



$$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.



Can somebody answer this? Thanks in advance!






real-analysis inequality







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    up vote
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    favorite
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    Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:



    $$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$



    (I used Jensen's inequality for this)



    Now the next question is why we do we have the strict inequality if all $x_i$ are different:



    $$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.



    Can somebody answer this? Thanks in advance!






    real-analysis inequality







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      Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:



      $$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$



      (I used Jensen's inequality for this)



      Now the next question is why we do we have the strict inequality if all $x_i$ are different:



      $$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.



      Can somebody answer this? Thanks in advance!






      real-analysis inequality







      share|cite|improve this question













      Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:



      $$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$



      (I used Jensen's inequality for this)



      Now the next question is why we do we have the strict inequality if all $x_i$ are different:



      $$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.



      Can somebody answer this? Thanks in advance!






      real-analysis inequality




      real-analysis inequality






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      user610431

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          You mentioned that you used Jensen's inequality, which tells you that
          $$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$



          since $x mapsto mathrm{exp}(x)$ is convex.



          But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.



          So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.






          share|cite|improve this answer





















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            up vote
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            down vote



            accepted










            You mentioned that you used Jensen's inequality, which tells you that
            $$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$



            since $x mapsto mathrm{exp}(x)$ is convex.



            But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.



            So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              You mentioned that you used Jensen's inequality, which tells you that
              $$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$



              since $x mapsto mathrm{exp}(x)$ is convex.



              But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.



              So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                You mentioned that you used Jensen's inequality, which tells you that
                $$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$



                since $x mapsto mathrm{exp}(x)$ is convex.



                But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.



                So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.






                share|cite|improve this answer












                You mentioned that you used Jensen's inequality, which tells you that
                $$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$



                since $x mapsto mathrm{exp}(x)$ is convex.



                But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.



                So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Clive Newstead

                49.2k472132




                49.2k472132






























                     

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