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.