Does $alpha_j-(Pi alpha_i)^{1/d}=beta_j-(Pi beta_i)^{1/d} $ force $alpha_i=beta_i$?












0












$begingroup$


Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:




If
$alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.




If we allow all of the differences to be zero, then uniqueness fails: set
$alpha_i=a,beta_i=b$ for $a neq b$.



I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.






systems-of-equations means a.m.-g.m.-inequality







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    0












    $begingroup$


    Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:




    If
    $alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.




    If we allow all of the differences to be zero, then uniqueness fails: set
    $alpha_i=a,beta_i=b$ for $a neq b$.



    I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.






    systems-of-equations means a.m.-g.m.-inequality







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:




      If
      $alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.




      If we allow all of the differences to be zero, then uniqueness fails: set
      $alpha_i=a,beta_i=b$ for $a neq b$.



      I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.






      systems-of-equations means a.m.-g.m.-inequality







      share|cite|improve this question









      $endgroup$




      Let $d>1$, and let $alpha_1,dots,alpha_d,beta_1,dots,beta_d$ be positive real numbers. Is there an easy proof of the following claim:




      If
      $alpha_j-(Pi_{i=1}^d alpha_i)^{1/d}=beta_j-(Pi_{i=1}^d beta_i)^{1/d} $ for every $1le jle d$ and not all of these differences are zero, then $alpha_i=beta_i$ for every $i$.




      If we allow all of the differences to be zero, then uniqueness fails: set
      $alpha_i=a,beta_i=b$ for $a neq b$.



      I think I have a proof for this uniqueness, but it is very cumbersome. I am interested in seeing other approaches.






      systems-of-equations means a.m.-g.m.-inequality




      systems-of-equations means a.m.-g.m.-inequality






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 11 '18 at 10:08









      Asaf ShacharAsaf Shachar

      5,64631141




      5,64631141






















          1 Answer
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          $begingroup$

          Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.



          The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives



          $$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
          with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
            $endgroup$
            – Asaf Shachar
            Dec 11 '18 at 15:02










          • $begingroup$
            Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
            $endgroup$
            – Dap
            Dec 11 '18 at 18:55










          • $begingroup$
            Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
            $endgroup$
            – Asaf Shachar
            Dec 12 '18 at 14:29













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          1 Answer
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          $begingroup$

          Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.



          The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives



          $$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
          with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
            $endgroup$
            – Asaf Shachar
            Dec 11 '18 at 15:02










          • $begingroup$
            Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
            $endgroup$
            – Dap
            Dec 11 '18 at 18:55










          • $begingroup$
            Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
            $endgroup$
            – Asaf Shachar
            Dec 12 '18 at 14:29


















          1












          $begingroup$

          Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.



          The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives



          $$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
          with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
            $endgroup$
            – Asaf Shachar
            Dec 11 '18 at 15:02










          • $begingroup$
            Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
            $endgroup$
            – Dap
            Dec 11 '18 at 18:55










          • $begingroup$
            Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
            $endgroup$
            – Asaf Shachar
            Dec 12 '18 at 14:29
















          1












          1








          1





          $begingroup$

          Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.



          The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives



          $$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
          with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$






          share|cite|improve this answer











          $endgroup$



          Assume that that condition holds. Then $beta_i=c+alpha_i$ for all $i,$ where $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$ If $c=0$ we're done, so assume not. By swapping $alpha$ and $beta$ we can assume $c> 0$ without loss of generality.



          The $d$-function $L^d$ Hölder inequality applied to the functions $f_i$ defined by $f_i(0)=c^{1/d}$ and $f_i(1)=alpha_i^{1/d},$ on the domain ${0,1}$ with counting measure, gives



          $$c+prod_{i=1}^d alpha_i^{1/d}leq prod_{i=1}^d |f_i|_d=prod_{i=1}^d (c+alpha_i)^{1/d}=prod_{i=1}^d beta_i^{1/d}$$
          with strict inequality unless all the $alpha_i$ are the same. But strict inequality would contradict $c=prod_{i=1}^dbeta_i^{1/d}-prod_{i=1}^dalpha_i^{1/d}.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 12:40

























          answered Dec 11 '18 at 10:51









          DapDap

          17.3k840




          17.3k840












          • $begingroup$
            Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
            $endgroup$
            – Asaf Shachar
            Dec 11 '18 at 15:02










          • $begingroup$
            Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
            $endgroup$
            – Dap
            Dec 11 '18 at 18:55










          • $begingroup$
            Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
            $endgroup$
            – Asaf Shachar
            Dec 12 '18 at 14:29




















          • $begingroup$
            Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
            $endgroup$
            – Asaf Shachar
            Dec 11 '18 at 15:02










          • $begingroup$
            Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
            $endgroup$
            – Dap
            Dec 11 '18 at 18:55










          • $begingroup$
            Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
            $endgroup$
            – Asaf Shachar
            Dec 12 '18 at 14:29


















          $begingroup$
          Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
          $endgroup$
          – Asaf Shachar
          Dec 11 '18 at 15:02




          $begingroup$
          Wow! How on earth did you think about this use of Holder's inequality? This seems like a very non-trivial approach...Thank you.
          $endgroup$
          – Asaf Shachar
          Dec 11 '18 at 15:02












          $begingroup$
          Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
          $endgroup$
          – Dap
          Dec 11 '18 at 18:55




          $begingroup$
          Hölder's inequality for finite sums is a standard tool in "problem solving"/"contest math" style inequalities. Though I had happened to have just read another question where its use was quite surprising to me: math.stackexchange.com/questions/3031422/…
          $endgroup$
          – Dap
          Dec 11 '18 at 18:55












          $begingroup$
          Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
          $endgroup$
          – Asaf Shachar
          Dec 12 '18 at 14:29






          $begingroup$
          Thank you. I still find this use of Hölder's inequality somewhat miraculous... so I asked another question about it here: math.stackexchange.com/questions/3036749/…. Thank you again for this very nice solution.
          $endgroup$
          – Asaf Shachar
          Dec 12 '18 at 14:29




















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