Stronger than AM-GM and a conjectureQuestion about Rudin's Theorem 8.22 proof (Stirling's formula) - part...

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Stronger than AM-GM and a conjecture


Question about Rudin's Theorem 8.22 proof (Stirling's formula) - part (b)If $a_1a_2cdots a_n =1$, then $prod_{i=1}^n (1+a_i) geq 2^n$A conjecture similar to the Hardy inequalityA inequality about $x_1,x_2,ldots, x_n$Prove that for a family of $100n$ subsets of a $[n]$ such that each subset has size $m$, there exists two whose intersection is at least $m^2/10n$.Hard inequality with condition ($xyz=1$)Inequality using JensenRefinement of a strong inequalityPower sum inequalityA nice power sum inequalityA conjecture about power sum : $e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$ and $a+b+c=3$













1












$begingroup$



Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$




A possible way to generalize this is the following conjecture:




Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$




Any hints would be appreciable to solve this. Thanks in advance.






real-analysis inequality contest-math a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
    $endgroup$
    – Martin R
    Feb 28 at 7:47






  • 6




    $begingroup$
    The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
    $endgroup$
    – Martin R
    Feb 28 at 7:54










  • $begingroup$
    I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 8:48










  • $begingroup$
    @GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
    $endgroup$
    – Martin R
    Feb 28 at 10:13










  • $begingroup$
    @MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 10:25
















1












$begingroup$



Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$




A possible way to generalize this is the following conjecture:




Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$




Any hints would be appreciable to solve this. Thanks in advance.






real-analysis inequality contest-math a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
    $endgroup$
    – Martin R
    Feb 28 at 7:47






  • 6




    $begingroup$
    The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
    $endgroup$
    – Martin R
    Feb 28 at 7:54










  • $begingroup$
    I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 8:48










  • $begingroup$
    @GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
    $endgroup$
    – Martin R
    Feb 28 at 10:13










  • $begingroup$
    @MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 10:25














1












1








1





$begingroup$



Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$




A possible way to generalize this is the following conjecture:




Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$




Any hints would be appreciable to solve this. Thanks in advance.






real-analysis inequality contest-math a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$





Let $a_i>0$ be $n$ numbers such that $prod_{i=1}^{n}a_ileq 1$ then we have:
$$sum_{i=1}^{n}a_igeq n left(prod_{i=1}^{n}a_i right)^{largeleft({n+sum_{i=1}^{n}a_i-n(prod_{i=1}^{n}a_i)^{1/n}}right)^{-1}}.$$




A possible way to generalize this is the following conjecture:




Let $f(x)$ be a increasing and convex function and $a_i$ be $n$ real numbers such that $f(a_i)>0$ with $prod_{i=1}^{n}f(a_i)leq 1$ then we have:
$$sum_{i=1}^{n}f(a_i)geq n left( prod_{i=1}^{n}f(a_i) right)^{large left(n+sum_{i=1}^{n}f(a_i)-nfleft(sum_{i=1}^{n}a_i/nright)right)^{-1}}.$$




Any hints would be appreciable to solve this. Thanks in advance.






real-analysis inequality contest-math a.m.-g.m.-inequality




real-analysis inequality contest-math a.m.-g.m.-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 28 at 10:17









GNUSupporter 8964民主女神 地下教會

14k82650




14k82650










asked Feb 28 at 7:37









FatsWallersFatsWallers

1307




1307








  • 1




    $begingroup$
    Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
    $endgroup$
    – Martin R
    Feb 28 at 7:47






  • 6




    $begingroup$
    The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
    $endgroup$
    – Martin R
    Feb 28 at 7:54










  • $begingroup$
    I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 8:48










  • $begingroup$
    @GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
    $endgroup$
    – Martin R
    Feb 28 at 10:13










  • $begingroup$
    @MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 10:25














  • 1




    $begingroup$
    Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
    $endgroup$
    – Martin R
    Feb 28 at 7:47






  • 6




    $begingroup$
    The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
    $endgroup$
    – Martin R
    Feb 28 at 7:54










  • $begingroup$
    I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 8:48










  • $begingroup$
    @GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
    $endgroup$
    – Martin R
    Feb 28 at 10:13










  • $begingroup$
    @MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Feb 28 at 10:25








1




1




$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47




$begingroup$
Where do these inequalities come from? Why do you think they are true? – And to which mathematical contest is the question related?
$endgroup$
– Martin R
Feb 28 at 7:47




6




6




$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54




$begingroup$
The first inequality is formulated as a fact. Could you provide a reference for those who (like me) are not familiar with that stronger version of the AM-GM inequality?
$endgroup$
– Martin R
Feb 28 at 7:54












$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48




$begingroup$
I have tried to improve the readability of your question by enlarging the indices. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 8:48












$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13




$begingroup$
@GNUSupporter8964民主女神地下教會: It seems to be that your edit changed the expression in the second inequality. For example, it was $n+sum_{i=1}^{n}f(a_i)-nf(frac{sum_{i=1}^{n}a_i}{n})$ the exponent on the RHS.
$endgroup$
– Martin R
Feb 28 at 10:13












$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25




$begingroup$
@MartinR Nice catch! Nonetheless, this shows the importance of avoiding fractions in subscripts/superscripts. The letters were so small that I mistakenly thought that they were the same. There were old and experienced Math.SE users who can't read such small letters.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Feb 28 at 10:25










1 Answer
1






active

oldest

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4





+50







$begingroup$

For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$



The generalization is false, if my calculations are correct. Take
begin{align*}
n&=4\
a_1&=tfrac12\
a_i&=1qquad(2leq ileq 4)\
f(x)&=begin{cases}x/e^2& (xleq 7/8)\
7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
end{cases}\
f(a_1)&=1/2e^2\
f(a_i)&=1/eqquad(2leq ileq 4)\
A&=tfrac1nsum f(a_i)approx0.29283\
G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
end{align*}
Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$






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    1 Answer
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    active

    oldest

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    4





    +50







    $begingroup$

    For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
    But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
    where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$



    The generalization is false, if my calculations are correct. Take
    begin{align*}
    n&=4\
    a_1&=tfrac12\
    a_i&=1qquad(2leq ileq 4)\
    f(x)&=begin{cases}x/e^2& (xleq 7/8)\
    7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
    end{cases}\
    f(a_1)&=1/2e^2\
    f(a_i)&=1/eqquad(2leq ileq 4)\
    A&=tfrac1nsum f(a_i)approx0.29283\
    G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
    F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
    end{align*}
    Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$






    share|cite|improve this answer











    $endgroup$


















      4





      +50







      $begingroup$

      For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
      But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
      where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$



      The generalization is false, if my calculations are correct. Take
      begin{align*}
      n&=4\
      a_1&=tfrac12\
      a_i&=1qquad(2leq ileq 4)\
      f(x)&=begin{cases}x/e^2& (xleq 7/8)\
      7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
      end{cases}\
      f(a_1)&=1/2e^2\
      f(a_i)&=1/eqquad(2leq ileq 4)\
      A&=tfrac1nsum f(a_i)approx0.29283\
      G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
      F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
      end{align*}
      Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$






      share|cite|improve this answer











      $endgroup$
















        4





        +50







        4





        +50



        4




        +50



        $begingroup$

        For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
        But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
        where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$



        The generalization is false, if my calculations are correct. Take
        begin{align*}
        n&=4\
        a_1&=tfrac12\
        a_i&=1qquad(2leq ileq 4)\
        f(x)&=begin{cases}x/e^2& (xleq 7/8)\
        7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
        end{cases}\
        f(a_1)&=1/2e^2\
        f(a_i)&=1/eqquad(2leq ileq 4)\
        A&=tfrac1nsum f(a_i)approx0.29283\
        G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
        F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
        end{align*}
        Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$






        share|cite|improve this answer











        $endgroup$



        For the first inequality, write $A=tfrac1n sum a_i$ and $G=prod a_i^{1/n}.$ We need to prove $A^{1+A-G}geq G.$
        But $$A^{A-G}=(A^A)^{(A-G)/A}geq e^{-(A-G)/A}geq 1-(A-G)/A=G/A$$
        where the middle inequality uses $A^Ageq e^{-e^{-1}}geq e^{-1}.$



        The generalization is false, if my calculations are correct. Take
        begin{align*}
        n&=4\
        a_1&=tfrac12\
        a_i&=1qquad(2leq ileq 4)\
        f(x)&=begin{cases}x/e^2& (xleq 7/8)\
        7/8e^2 + 8(1/e-7/8e^2)(x-7/8)& (xgeq 7/8)
        end{cases}\
        f(a_1)&=1/2e^2\
        f(a_i)&=1/eqquad(2leq ileq 4)\
        A&=tfrac1nsum f(a_i)approx0.29283\
        G&=prod f(a_i)^{1/n}=(1/2e)^{1/4}/eapprox0.24092\
        F&=f(sum a_i/n)=f(7/8)=7/8e^2approx 0.11842.
        end{align*}
        Then $0.80718approx A^{A-F}<G/Aapprox 0.82274,$ but the desired inequality is equivalent to $A^{A-F}geq G/A.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 15 at 10:27

























        answered Mar 15 at 10:16









        DapDap

        19k842




        19k842






























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            Nidaros erkebispedøme

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            For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bj...
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            Birsay

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            Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser d...
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            Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...

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            Do VLANs within a subnet need to have their own subnet for router on a stick? What are the differences between the usage of 'it' and 'they'? Dragon forelimb placement How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? What typically incentivizes a professor to change jobs to a lower ranking university? How to write a macro that is braces sensitive? Modeling an IPv4 Address Writing rule stating superpower from different root cause is bad writing Can I ask the recruiters in my resume to put the reason why I am rejected? Approximately how much travel t...
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