prove-that-n-1-a-n-b-n-p-1-p-n-1-a-n-p-1-p-n-1-b-n-p-1-p-

Question Number 9758 by richard last updated on 31/Dec/16

prove that

{(\underset{n = 1}{\sum^{\infty}} {(a_{n} + b_{n})}^{p})}^{1 / p} ⩽ {(\underset{n = 1}{\sum^{\infty}} a_{n}^{p})}^{1 / p} + {(\underset{n = 1}{\sum^{\infty}} b_{n}^{p})}^{1 / p}

Commented by FilupSmith last updated on 04/Jan/17

Attempting

∴ {(\underset{n = 1}{\sum^{\infty}} (\underset{v = 0}{\sum^{p}} (\begin{matrix} p \\ v \end{matrix}) a_{n}^{p - v} b_{n}^{p}))}^{1 / p} ⩽ {(\underset{n = 1}{\sum^{\infty}} a_{n}^{p})}^{1 / p} + {(\underset{n = 1}{\sum^{\infty}} b_{n}^{p})}^{1 / p}

let :

A = (\underset{n = 1}{\sum^{\infty}} (a_{n}^{p}))

B = (\underset{n = 1}{\sum^{\infty}} (b_{n}^{p}))

{(\underset{n = 1}{\sum^{\infty}} (\underset{v = 0}{\sum^{p}} (\begin{matrix} p \\ v \end{matrix}) a_{n}^{p - v} b_{n}^{p}))}^{1 / p} ⩽ \frac{1}{A^{p}} + \frac{1}{B^{p}}

{(\underset{n = 1}{\sum^{\infty}} (\underset{v = 0}{\sum^{p}} (\begin{matrix} p \\ v \end{matrix}) a_{n}^{p - v} b_{n}^{p}))}^{1 / p} ⩽ \frac{A^{p} + B^{p}}{{(A B)}^{p}}

{(\underset{n = 1}{\sum^{\infty}} (\underset{v = 0}{\sum^{p}} (\begin{matrix} p \\ v \end{matrix}) a_{n}^{p - v} b_{n}^{p}))}^{1 / p} ⩽ \frac{{(\underset{n = 1}{\sum^{\infty}} (a_{n}^{p}))}^{p} + {(\underset{n = 1}{\sum^{\infty}} (b_{n}^{p}))}^{p}}{{(\underset{n = 1}{\sum^{\infty}} (a_{n}^{p})) (\underset{n = 1}{\sum^{\infty}} (b_{n}^{p}))}^{p}}

Attempting