AM–GM inequality: Difference between revisions

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| publisher = Springer
| title = The Mathematical Gardner
| year = 1981}}</ref>]]| isbn = 978-1-4684-6688-1
}}</ref>]]
In [[mathematics]], the '''inequality of arithmetic and geometric means''', or more briefly the '''AM–GM inequality''', states that the [[arithmetic mean]] of a list of non-negative [[real number]]s is greater than or equal to the [[geometric mean]] of the same list; and further, that the two means are equal [[if and only if]] every number in the list is the same (in which case they are both that number).
 
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:<math>\exp \left( \frac{\ln {x_1} + \ln {x_2} + \cdots + \ln {x_n}}{n} \right).</math>
 
Note: This does not apply exclusively to the exp() function and natural logarithms. The base b of the exponentiation could be any positive real number if the logarithm is of base b.
 
== The inequality ==
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==Proofs of the AM–GM inequality==
The AM-GMAM–GM inequality is also known for the variety of methods that can be used to prove it.
 
===Proof using Jensen's inequality===
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:<math>
\frac{x_1 + x_2}{2} \ge> \sqrt{x_1 x_2}</math>
 
as desired.
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(in which case the first arithmetic mean and first geometric mean are both equal to&nbsp;{{math|''x''<sub>1</sub>}}, and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all {{math|2<sup>''k''</sup>}} numbers are equal, it is not possible for both inequalities to be equalities, so we know that:
 
:<math>\frac{x_1 + x_2 + \cdots + x_{2^k}}{2^k} \ge> \sqrt[2^k]{x_1 x_2 \cdots x_{2^k}}</math>
 
as desired.
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One stronger version of this, which also gives strengthened version of the unweighted version, is due to Aldaz. In particular,
There is a similar inequality for the [[weighted arithmetic mean]] and [[weighted geometric mean]]. Specifically, let the nonnegative numbers {{math|''x''<sub>1</sub>, ''x''<sub>2</sub>, . . . , ''x<sub>n</sub>''}} and the nonnegative weights {{math|''w''<sub>1</sub>, ''w''<sub>2</sub>, . . . , ''w<sub>n</sub>''}} be given. Assume further that the sum of the weights is 1. Then
:<math>\sum_{i=1}^n w_ix_i \geq \prod_{i=1}^n x_i^{w_i} + \sum_{i=1}^n w_i\left(x_i^{\frac{1}{2}} -\sum_{i=1}^n w_ix_i^{\frac{1}{2}} \right)^2 </math>.<ref>{{cite journal |last1=Aldaz |first1=J.M. |title=Self-Improvement of the Inequality Between Arithmetic and Geometric Means |journal=Journal of Mathematical Inequalities |date=2009 |volume=3 |issue=2 |pagepages=213-216213–216 |doi=10.7153/jmi-03-21 |url=http://jmi.ele-math.com/03-21/Self-improvement-of-the-inequality-between-arithmetic-and-geometric-means |access-date=11 January 2023|doi-access=free }}</ref>
 
====Proof using Jensen's inequality====
 
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{{notelist}}
{{reflist|group=note}}
 
==References==
{{Reflist}}