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This problem was inspired by this question.

$\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } \ge \frac { a+\sqrt { ab } +\sqrt [ 3 ]{ abc } }{ 3 } $

The above can be proved using Hölder's inequality.

$\sqrt [ 3 ]{ a(\frac { a+b }{ 2 } )(\frac { a+b+c }{ 3 } ) } =\sqrt [ 3 ]{ (\frac { a }{ 3 } +\frac { a }{ 3 } +\frac { a }{ 3 } )(\frac { a }{ 3 } +\frac { a+b }{ 6 } +\frac { b }{ 3 } )(\frac { a+b+c }{ 3 } ) } \ge \sqrt [ 3 ]{ (\frac { a }{ 3 } +\frac { a }{ 3 } +\frac { a }{ 3 } )(\frac { a }{ 3 } +\frac { \sqrt { ab } }{ 3 } +\frac { b }{ 3 } )(\frac { a }{ 3 } +\frac { b }{ 3 } +\frac { c }{ 3 } ) } (\because \text{AM-GM})\\ \ge \frac { a+\sqrt { ab } +\sqrt [ 3 ]{ abc } }{ 3 } (\because \text{Holder's inequality)}$

However, I had trouble generalizing this inequality to

$\sqrt [ n ]{ \prod _{ i=1 }^{ n }{ { A }_{ i } } } \ge \frac { \sum _{ i=1 }^{ n }{ { G }_{ i } } }{ n } $

when ${ A }_{ i }=\frac { \sum _{ j=1 }^{ i }{ { a }_{ i } } }{ i } $

and ${ G }_{ i }=\sqrt [ i ]{ \prod _{ j=1 }^{ i }{ { a }_{ i } } } $ as I could not split the fractions as I did above.

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  • $\begingroup$ Do you have reason to believe that the generalization is true? $\endgroup$
    – David
    Jan 3, 2016 at 4:51
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    $\begingroup$ Yes, I do. I did manage to formulate a proof for n=4,5,6. However, the way I proved them is the same as n=3, and I believed that it was not necesarry to post them. However, I had trouble repeating the process for n>6. So it could be wrong. If you do find a counter example, please inform me. $\endgroup$
    – Chad Shin
    Jan 3, 2016 at 4:58
  • $\begingroup$ Well, the generalization of this problem is true. But the proof is difficult. :) $\endgroup$ Jan 3, 2016 at 5:31
  • $\begingroup$ @貓貓吃狗狗 I do not mean to annoy you, but am I correct in my assumption that your comment indicated your intent to give a solution/hint to the problem? $\endgroup$
    – Chad Shin
    Jan 5, 2016 at 6:28
  • $\begingroup$ This is a result of K. Kedlaya, "K. KEDLAYA, Proof of a mixed arithmetic-mean, geometric-mean inequality, Amer. Math. Monthly, 101 (1994), 355–357." A generalization could be found in the article : emis.de/journals/JIPAM/images/165_06_JIPAM/165_06.pdf $\endgroup$ Jan 9, 2016 at 19:43

1 Answer 1

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This result was conjectured by professor Finbarr Holand, and then it was proved by K. Kedlaya in an article in the American Mathematical Monthly that could be found here, and then it was generalized by Professor Holand in an article that could be found here.

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