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<record version="3" id="2934">
 <title>general means inequality</title>
 <name>GeneralMeansInequality</name>
 <created>2002-05-23 22:18:12</created>
 <modified>2002-05-23 23:42:27</modified>
 <type>Theorem</type>
 <creator id="3" name="drini"/>
 <author id="3" name="drini"/>
 <classification>
	<category scheme="msc" code="26D15"/>
 </classification>
 <synonyms>
	<synonym concept="general means inequality" alias="power means inequality"/>
 </synonyms>
 <related>
	<object name="ArithmeticGeometricMeansInequality"/>
	<object name="ArithmeticMean"/>
	<object name="GeometricMean"/>
	<object name="HarmonicMean"/>
	<object name="PowerMean"/>
	<object name="ProofOfArithmeticGeometricHarmonicMeansI"/>
	<object name="RootMeanSquare3"/>
	<object name="DerivationOfZerothWeightedPowerMean"/>
	<object name="ProofOfArithmeticGeometricHarmonicMeansInequality"/>
	<object name="ComparisonOfPythagoreanMeans"/>
 </related>
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 <content>The power means inequality is a generalization of arithmetic-geometric means inequality.

If $0\neq r\in\R$, the $r$-mean (or $r$-th power mean) of the nonnegative
numbers $a_1,\ldots,a_n$ is defined as
$$M^r(a_1,a_2,\ldots,a_n)= \left(\frac{1}{n}\displaystyle{\sum_{k=1}^n a_k^r}\right)^{1/r}$$

Given real numbers $x,y$ such that $xy\neq 0$
and $x&lt;y$, we have
$$M^x \leq M^y$$
and 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 the inequality above holds for arbitrary real numbers $x&lt;y$.

The mentioned inequality is a special case of this one, since $M^1$ is the arithmetic mean, $M^0$ is the geometric mean and $M^{-1}$ is the harmonic mean.

This inequality can be further generalized using weighted power means.</content>
</record>
