November 30th, 2018, 03:38 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 479 Thanks: 73  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 03:50 AM. 
November 30th, 2018, 08:28 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,404 Thanks: 1306 
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 $\dfrac{A}{\sqrt[n]{p}}$ 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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