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| == content Timberland Factory Outlet Melbourne ==
| | In [[mathematics]], '''Mahler's inequality''', named after [[Kurt Mahler]], states that the [[geometric mean]] of the term-by-term sum of two finite sequences of positive numbers is greater than or equal to the sum of their two separate geometric means: |
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| | :<math>\prod_{k=1}^n (x_k + y_k)^{1/n} \ge \prod_{k=1}^n x_k^{1/n} + \prod_{k=1}^n y_k^{1/n}</math> |
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| | when ''x''<sub>''k''</sub>, ''y''<sub>''k''</sub> > 0 for all ''k''. |
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| <li>[http://114.255.231.28/discuz/viewthread.php?tid=12964861&extra= http://114.255.231.28/discuz/viewthread.php?tid=12964861&extra=]</li>
| | == Proof == |
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| | By the [[inequality of arithmetic and geometric means]], we have: |
| <li>[http://www.sunplay.net/forum.php?mod=viewthread&tid=126516&fromuid=24881 http://www.sunplay.net/forum.php?mod=viewthread&tid=126516&fromuid=24881]</li>
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| | :<math>\prod_{k=1}^n \left({x_k \over x_k + y_k}\right)^{1/n} \le {1 \over n} \sum_{k=1}^n {x_k \over x_k + y_k},</math> |
| <li>[http://enseignement-lsf.com/spip.php?article64#forum25676111 http://enseignement-lsf.com/spip.php?article64#forum25676111]</li>
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| | : <math>\prod_{k=1}^n \left({y_k \over x_k + y_k}\right)^{1/n} \le {1 \over n} \sum_{k=1}^n {y_k \over x_k + y_k}.</math> |
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| | Hence, |
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| | :<math>\prod_{k=1}^n \left({x_k \over x_k + y_k}\right)^{1/n} + \prod_{k=1}^n \left({y_k \over x_k + y_k}\right)^{1/n} \le {1 \over n} n = 1.</math> |
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| | Clearing denominators then gives the desired result. |
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| | == See also == |
| | *[[Minkowski's inequality]] |
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| | == References == |
| | *http://eom.springer.de/M/m064060.htm |
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| | [[Category:Inequalities]] |
| | [[Category:Articles containing proofs]] |
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| | {{mathanalysis-stub}} |
In mathematics, Mahler's inequality, named after Kurt Mahler, states that the geometric mean of the term-by-term sum of two finite sequences of positive numbers is greater than or equal to the sum of their two separate geometric means:
when xk, yk > 0 for all k.
Proof
By the inequality of arithmetic and geometric means, we have:
and
Hence,
Clearing denominators then gives the desired result.
See also
References
Template:Mathanalysis-stub