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

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

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