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November 30th, 2018, 04:38 AM   #1
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Need explanation

Let $\displaystyle A$ denote the largest of the numbers $\displaystyle a_1 , a_2 , ... , a_p$ for $\displaystyle p,n \in N $
Need explanation for $\displaystyle \frac{A}{\sqrt[n]{p}}\leq \sqrt[\displaystyle n]{\frac{a^{n}_1+a^{n}_2 +...+a^{n}_p } {p}}\leq A$ how to show it is true (how to derive?)

Last edited by idontknow; November 30th, 2018 at 04:50 AM.
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November 30th, 2018, 09:28 AM   #2
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the only restriction on $a$ is that $\max{(a_k)} = A$

so if we choose $s$ as $a_1=A,~a_k = 0,~k=2,p$

the expression on the right becomes


Now consider the sequence $a_k = A$

This should produce the largest value of that expression

$\sqrt[n]{\dfrac{ p A^n}{p}} = A \sqrt[n]{\dfrac{p}{p}} = A$

and thus we see

$\dfrac{A}{\sqrt[n]{p}} \leq \sqrt[n]{\dfrac{\sum \limits_{k=1}^p~a_k^n}{p}} \leq A$
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