Proof of Minkowski's inequality

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

Minkowski's inequality states that

n=1k(xn+yn)1p(n=1kxnp)1p+(n=1kynp)1p\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>1p>1

and xn,yn0{x}_{n},{y}_{n}\ge0


Proof:

we know

n=1k(xn+yn)p=n=1kxn(xn+yn)p1+n=1kyn(xn+yn)p1\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 aa as a=pp1a=\frac { p }{ p-1 } now by Holder's inequality we have

n=1kxn(xn+yn)p1+n=1kyn(xn+yn)p1((n=1kxnp)1p+(n=1kynp)1p)((n=1k(xk+yk)p)1a)=((n=1kxnp)1p+(n=1kynp)1p)(n=1k(xn+yn)p)1a\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 n=1k(xn+yn)p)1a\displaystyle\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p } } )^{ \frac { 1 }{ a } }

we get

n=1k(xn+yn)1p(n=1kxnp)1p+(n=1kynp)1p\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

Note by Hamza A
3 years, 8 months ago

No vote yet
1 vote

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Comments

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

Jakub Bober - 3 years, 1 month ago

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very good job dude :O

Natanael Flores - 3 years, 8 months ago

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thanks :)

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

Hamza A - 3 years, 8 months ago

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Great proof!!!!

shivam mishra - 3 years, 8 months ago

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thanks!

Hamza A - 3 years, 8 months ago

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

Pi Han Goh - 3 years, 8 months ago

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Give introduction to minkowski's inequality

Supreetha Gowda - 1 year ago

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