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zmudz
 one year ago
Prove that if \(x_i > 0\) for all \(i\) then
\begin{align*}
&(x_1^{19} + x_2^{19} + \cdots + x_n^{19})(x_1^{93} + x_2^{93} + \cdots + x_n^{93}) \\
&\geq (x_1^{20} + x_2^{20} + \cdots + x_n^{20})(x_1^{92} + x_2^{92} + \cdots + x_n^{92}).
\end{align*}
Also, find when equality holds.
zmudz
 one year ago
Prove that if \(x_i > 0\) for all \(i\) then \begin{align*} &(x_1^{19} + x_2^{19} + \cdots + x_n^{19})(x_1^{93} + x_2^{93} + \cdots + x_n^{93}) \\ &\geq (x_1^{20} + x_2^{20} + \cdots + x_n^{20})(x_1^{92} + x_2^{92} + \cdots + x_n^{92}). \end{align*} Also, find when equality holds.

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Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well two simple cases where equality hold: \(x_i=1\) for all i or n=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 can you please help me with my question when you can this is urgent!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think we may use this inequality http://www.artofproblemsolving.com/wiki/index.php/Muirhead's_Inequality notice that \((93,19,0,\ldots) \succ (92,20,0,\ldots) \)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0thinking of what @Empty have said i'll add its equal whenever xi a constant for all i :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wiki says that muirhead is actually a generalization of Jensen's inequality https://en.wikipedia.org/wiki/Jensen%27s_inequality

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I was actually looking at the CaucySchwartz inequality but I don't know how to solve this problem yet I am trying to cook up some ideas though. https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! I know the CauchySchwarz inequality from class; however, we never covered the Jensen's inequality or Muirhead. But any help at all is much appreciated. Let me know how it goes. I'm still stuck.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1WOLG \(x_1\leq x_2\leq x_3\leq.....\leq x_n\) LHS:= \[x_1^{19}\left[\begin{matrix}x_1^{93}\\x_2^{93}\\::::\\x_n^{93}\end{matrix}\right]+x_2^{19}\left[\begin{matrix}x_1^{93}\\x_2^{93}\\::::\\x_n^{93}\end{matrix}\right]+::::+x_n^{19}\left[\begin{matrix}x_1^{93}\\x_2^{93}\\::::\\x_n^{93}\end{matrix}\right]\] RHS:= \[x_1^{20}\left[\begin{matrix}x_1^{92}\\x_2^{92}\\::::\\x_n^{92}\end{matrix}\right]+x_2^{20}\left[\begin{matrix}x_1^{92}\\x_2^{92}\\::::\\x_n^{92}\end{matrix}\right]+::::+x_n^{20}\left[\begin{matrix}x_1^{92}\\x_2^{92}\\::::\\x_n^{92}\end{matrix}\right]\] The first entry of the first matrix of the LHS is \(x_1^{19}*x_1^{93}=x_1^{112}\geq x_1^{20}*x_1^{92}=x_1^{112}\) og the RHS The second entry: LHS \(x_1^{19}*x_2^{93}=x_1^{19}*x_2^{92}x_2\geq x_1^{19}*x_2^{92}x_1=x_1^{20}*x_2^{92}=\text{the second entry of the RHS}\) Same as other entries and other matrix. That shows \(LHS\geq RHS\)
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