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 November 30th, 2018, 03:38 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 711 Thanks: 96 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: 2,574 Thanks: 1421 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 $\dfrac{A}{\sqrt[n]{p}} \leq \sqrt[n]{\dfrac{\sum \limits_{k=1}^p~a_k^n}{p}} \leq A$ Thanks from idontknow Tags explanation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sadmath Geometry 8 September 13th, 2015 06:24 AM Vaki Calculus 7 April 7th, 2014 04:12 PM ranna Calculus 11 February 1st, 2011 09:34 PM Male_volence Calculus 2 February 13th, 2008 04:20 AM rain Real Analysis 3 December 31st, 1969 04:00 PM

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