Mathematical relationships
In mathematics , the QM-AM-GM-HM inequalities , also known as the mean inequality chain , state the relationship between the harmonic mean , geometric mean , arithmetic mean , and quadratic mean (also known as root mean square). Suppose that
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{\displaystyle x_{1},x_{2},\ldots ,x_{n}}
real numbers . Then
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{\displaystyle 0<{\frac {n}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+\cdots +{\frac {1}{x_{n}}}}}\leq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\leq {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\leq {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}{n}}}.}
[ 1] These inequalities often appear in mathematical competitions and have applications in many fields of science.
Proof
There are three inequalities between means to prove. There are various methods to prove the inequalities, including mathematical induction , the Cauchy–Schwarz inequality , Lagrange multipliers , and Jensen's inequality . For several proofs that GM ≤ AM, see Inequality of arithmetic and geometric means .
AM-QM inequality
From the Cauchy–Schwarz inequality on real numbers , setting one vector to (1, 1, ...) :
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{\displaystyle \left(\sum _{i=1}^{n}x_{i}\cdot 1\right)^{2}\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}1^{2}\right)=n\,\sum _{i=1}^{n}x_{i}^{2},}
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{\displaystyle \left({\frac {\sum _{i=1}^{n}x_{i}}{n}}\right)^{2}\leq {\frac {\sum _{i=1}^{n}x_{i}^{2}}{n}}}
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{\displaystyle x_{i}}
HM-GM inequality
The reciprocal of the harmonic mean is the arithmetic mean of the reciprocals
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{\displaystyle 1/x_{1},\dots ,1/x_{n}}
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{\displaystyle 1/{\sqrt[{n}]{x_{1}\dots x_{n}}}}
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{\displaystyle x_{i}>0}
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{\displaystyle {\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}\leq {\sqrt[{n}]{x_{1}\dots x_{n}}}.}
[ 2]
n = 2 caseThe semi-circle used to visualize the inequalities When n = 2, the inequalities become
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{\displaystyle {\frac {2x_{1}x_{2}}{x_{1}+x_{2}}}\leq {\sqrt {x_{1}x_{2}}}\leq {\frac {x_{1}+x_{2}}{2}}\leq {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}}{2}}}}
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{\displaystyle x_{1},x_{2}>0,}
[ 3] which can be visualized in a semi-circle whose diameter is [AB ] and center D .
Suppose AC = x 1 and BC = x 2 . Construct perpendiculars to [AB ] at D and C respectively. Join [CE ] and [DF ] and further construct a perpendicular [CG ] to [DF ] at G . Then the length of GF can be calculated to be the harmonic mean, CF to be the geometric mean, DE to be the arithmetic mean, and CE to be the quadratic mean. The inequalities then follow easily by the Pythagorean theorem .
Comparison of harmonic, geometric, arithmetic, quadratic and other mean values of two positive real numbers
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{\displaystyle x_{1}}
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{\displaystyle x_{2}}
Tests
To infer the correct order, the four expressions can be evaluated with two positive numbers.
For
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10
{\displaystyle x_{1}=10}
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40
{\displaystyle x_{2}=40}
16
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20
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34
{\displaystyle 16<20<25<5{\sqrt {34}}}
See also
References
^ Djukić, Dušan (2011). The IMO compendium: a collection of problems suggested for the International Mathematical Olympiads, 1959-2009 . Problem books in mathematics. International mathematical olympiad. New York: Springer. p. 7. ISBN 978-1-4419-9854-5 ^ Sedrakyan, Hayk; Sedrakyan, Nairi (2018), Sedrakyan, Hayk; Sedrakyan, Nairi (eds.), "The HM-GM-AM-QM Inequalities" , Algebraic Inequalities , Problem Books in Mathematics, Cham: Springer International Publishing, p. 23, doi :10.1007/978-3-319-77836-5_3 , ISBN 978-3-319-77836-5 , retrieved 2023-11-26 ^ Sedrakyan, Hayk; Sedrakyan, Nairi (2018), Sedrakyan, Hayk; Sedrakyan, Nairi (eds.), "The HM-GM-AM-QM Inequalities" , Algebraic Inequalities , Problem Books in Mathematics, Cham: Springer International Publishing, p. 21, doi :10.1007/978-3-319-77836-5_3 , ISBN 978-3-319-77836-5 , retrieved 2023-11-26
External links
![{\displaystyle 0<{\frac {n}{{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+\cdots +{\frac {1}{x_{n}}}}}\leq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\leq {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\leq {\sqrt {\frac {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}{n}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67057c5febf2a512872008a8a53dc58f9a4bfb03)
![{\displaystyle 1/{\sqrt[{n}]{x_{1}\dots x_{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9ae01f87b35eefae3c510c9f64cef1427e24ae2)
![{\displaystyle {\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}\leq {\sqrt[{n}]{x_{1}\dots x_{n}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f4b2233952c31eb61a26630ef67d8e2883b8c9b)
