this follows from Holder's inequality,and in my proof,for the sake of simplicity,i'll use it

Minkowski's inequality states that

\(\displaystyle\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ \frac { 1 }{ p } }\le (\displaystyle\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } ) } ^{ \frac { 1 }{ p } }+(\displaystyle\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } )^{ \frac { 1 }{ p } }\)

for \(p>1\)

and \({x}_{n},{y}_{n}\ge0\)

**Proof**:

we know

\(\displaystyle\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p }=\displaystyle\sum _{ n=1 }^{ k }{ { x }_{ n }({ x }_{ n }+{ y }_{ n })^{ p-1 } } } +\displaystyle\sum _{ n=1 }^{ k }{ { y }_{ n }({ x }_{ n }+{ y }_{ n } } )^{ p-1 }\)

let's define \(a\) as \(a=\frac { p }{ p-1 } \) now by Holder's inequality we have

\(\displaystyle \sum_{n \mathop = 1}^k x_n \left({x_n + y_n}\right)^{p-1} + \sum_{n \mathop = 1}^k y_n \left({x_n + y_n}\right)^{p-1}\le((\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } )^{ \frac { 1 }{ p } }+(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } )^{ \frac { 1 }{ p } })((\sum _{ n=1 }^{ k }{ ({ x }_{ k }+{ y }_{ k })^{ p })^{ \frac { 1 }{ a } } } )=\\((\displaystyle\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } )^{ \frac { 1 }{ p } }+(\displaystyle\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } )^{ \frac { 1 }{ p } })(\displaystyle\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p } } )^{ \frac { 1 }{ a } }\)

dividing by \(\displaystyle\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p } } )^{ \frac { 1 }{ a } }\)

we get

\(\displaystyle\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ \frac { 1 }{ p } }\le (\displaystyle\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } ) } ^{ \frac { 1 }{ p } }+(\displaystyle\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } )^{ \frac { 1 }{ p } }\)

Hence proved

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## Comments

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TopNewestvery good job dude :O

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thanks :)

i tried 2 proofs,one was just too long,so i went with the simpler one :)

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Great proof!!!!

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thanks!

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You should post your findings in the wiki page, hölder's inequality.

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Can someone clearly prove for what conditions do we have the equality?

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Give introduction to minkowski's inequality

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