A conjecture about power sum : $e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$ and $a+b+c=3$












4












$begingroup$


Hello I have this to propose :




Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$




For a generalization I have this conjecture :




Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$




In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .



Thanks in advance .






real-analysis inequality exponential-function jensen-inequality







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  • $begingroup$
    @MartinR Maybe OP is trying to create a contest-math problem.
    $endgroup$
    – yurnero
    Feb 1 at 13:42
















4












$begingroup$


Hello I have this to propose :




Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$




For a generalization I have this conjecture :




Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$




In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .



Thanks in advance .






real-analysis inequality exponential-function jensen-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    @MartinR Maybe OP is trying to create a contest-math problem.
    $endgroup$
    – yurnero
    Feb 1 at 13:42














4












4








4





$begingroup$


Hello I have this to propose :




Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$




For a generalization I have this conjecture :




Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$




In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .



Thanks in advance .






real-analysis inequality exponential-function jensen-inequality







share|cite|improve this question











$endgroup$




Hello I have this to propose :




Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$




For a generalization I have this conjecture :




Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$




In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .



Thanks in advance .






real-analysis inequality exponential-function jensen-inequality




real-analysis inequality exponential-function jensen-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 13:43









Martin R

30.8k33561




30.8k33561










asked Feb 1 at 13:21









FatsWallersFatsWallers

1447




1447












  • $begingroup$
    @MartinR Maybe OP is trying to create a contest-math problem.
    $endgroup$
    – yurnero
    Feb 1 at 13:42


















  • $begingroup$
    @MartinR Maybe OP is trying to create a contest-math problem.
    $endgroup$
    – yurnero
    Feb 1 at 13:42
















$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42




$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42










2 Answers
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$begingroup$

By Jensen
$$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
$$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
$$sum_{cyc}c^2(a-b)^2geq0.$$
The second inequality is wrong.



Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
    $$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
    where $f(x)=e^x$.
    Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
    $$
    e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
    $$
    For the second one, I am very confident that this approach should transfer without much hassle.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      3












      $begingroup$

      By Jensen
      $$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
      $$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
      $$sum_{cyc}c^2(a-b)^2geq0.$$
      The second inequality is wrong.



      Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        By Jensen
        $$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
        $$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
        $$sum_{cyc}c^2(a-b)^2geq0.$$
        The second inequality is wrong.



        Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          By Jensen
          $$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
          $$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
          $$sum_{cyc}c^2(a-b)^2geq0.$$
          The second inequality is wrong.



          Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$






          share|cite|improve this answer











          $endgroup$



          By Jensen
          $$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
          $$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
          $$sum_{cyc}c^2(a-b)^2geq0.$$
          The second inequality is wrong.



          Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 14:43

























          answered Feb 1 at 13:38









          Michael RozenbergMichael Rozenberg

          110k1896201




          110k1896201























              2












              $begingroup$

              For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
              $$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
              where $f(x)=e^x$.
              Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
              $$
              e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
              $$
              For the second one, I am very confident that this approach should transfer without much hassle.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
                $$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
                where $f(x)=e^x$.
                Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
                $$
                e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
                $$
                For the second one, I am very confident that this approach should transfer without much hassle.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
                  $$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
                  where $f(x)=e^x$.
                  Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
                  $$
                  e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
                  $$
                  For the second one, I am very confident that this approach should transfer without much hassle.






                  share|cite|improve this answer









                  $endgroup$



                  For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
                  $$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
                  where $f(x)=e^x$.
                  Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
                  $$
                  e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
                  $$
                  For the second one, I am very confident that this approach should transfer without much hassle.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 1 at 13:38









                  AaronAaron

                  2,015415




                  2,015415






























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