Strange Mean Inequality [duplicate]

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.

Solution 1:

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.