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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: drininotes''
by Koro on 2002-05-23 05:54:45 |
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| You could add the general means inequality:
If 0\neq x\in R, the x-mean of the nonnegative
numbers a_1,...,a_n is
M_x = [(1/n)\sum_{k=1}^n a_k^x]^{1/x}
Given real numbers x,y such that xy\neq 0
and x<y, we have
M_x \leq M_y (*)
The equality holds if and only if a_1 = ... = a_n.
Additionally, if we define M_0 to be the
geometric mean (a_1a_2...a_n)^(1/n), we have
that (*) holds for arbitrary real numbers x<y.
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