Power Means Inequality (the geometric part)

AI Thread Summary
The discussion centers on understanding the power means inequality, particularly how the geometric mean is defined as the limit of the power mean as n approaches zero. The formula for the power means is provided, highlighting that for r > s, the inequality holds true. There is confusion about why the limit of P(r) as r approaches 0 results in P(0), which is clarified by referencing the indeterminate form 1^∞ and suggesting the use of logarithms and L'Hopital's Rule to resolve it. Additionally, there is a side query about formatting fractions to appear larger. Overall, the conversation aims to clarify the mathematical principles behind the power means inequality.
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Hi everyone! So I'm trying to learn more about inequalities and the one I'm starting with is the power means inequality. But it all seems pretty intuitive except how they define the n=0 power mean (i.e. the geometric mean). I read that it's actually the limit as n->0, but I don't see why that's true. Any help? Thanks!
 
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Could you perhaps quote the 'power means' inequality? I may have seen it before under a different name, or under no name.
 
You're perhaps asking about
(a^x+b^x)^{1/x}
 
So the equality is that given r > s and x_1, x_2, \ldots x_n \epsilon \; \Re where s \geq 0 if any of the x_1, x_2, \ldots x_n = 0. Then:
P(r) \geq P(s)
where P(r) = \left \{ {\begin{array}{*{20}c}<br /> {(\frac{x_1^r + x_2^r \ldots + x_n^r}{n})^{1/r},} &amp; {r \neq 0} \\<br /> {\sqrt[n]{x_1 x_2 \ldots x_n},} &amp; {r = 0} \\<br /> \end{array}} \right.<br />
And supposedly if we take the limit of P(r) to 0 we get P(0) but I don't see how that's true.

On another note, how do I make my fraction look bigger. It looks really small. Thanks!
 
In the limit, you have an indeterminate form like 1^\infty. The standard way to deal with that kind of indeterminate form is to take the log and then use L'Hopital's Rule. If you'd like me to work it out for you, let me know, but I think you can probably get it yourself from there.
 
Success! Thanks!
 
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