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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            Fil:Tokke komm.svg

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            Skildring Skildring Tokke komm.svg Norsk nynorsk: Våpen for Tokke kommune i Telemark fylke, godkjent ved kongeleg resolusjon 6. februar 1987. Motiv: På gull grunn ein gåande svart bjørn. [1] Tokke ligg i eit gamalt bjørneområde og bjørnen er eit viktig innslag i fleire lokale historier. Bjørnen kan òg symbolisere naturen og dyrelivet i kommunen. English: Coat of arms of the Norwegian municipality of Tokke, Telemark. Dato 6. februar 1987 Kjelde Original by Stein Davidsen, based upon image from caplex.no. Opphavsperson Eige arbeid Vektorgrafikken vart laga med Inkscape. Løyve (Gjenbruk av denne fila) Public domain Public domain false false This official Norwegian coat of arms is ineligible for copyright, but its usage is restricted according to Norwegian penal law, independent of the copyright status of the depiction shown here. Dette biletet er sett saman av eit flagg, eit våpen, eit sigill eller eit anna offisielt insignium . Bruken av slike symbol er avgrensa i fleire land. Desse r
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