Difference between revisions of "Muirhead's Inequality"
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'''Muirhead's Inequality''' states that if a sequence <math>A</math> [[Majorization|majorizes]] a sequence <math>B</math>, then given a set of positive reals <math>x_1,x_2,\cdots,x_n</math>: | '''Muirhead's Inequality''' states that if a sequence <math>A</math> [[Majorization|majorizes]] a sequence <math>B</math>, then given a set of positive reals <math>x_1,x_2,\cdots,x_n</math>: | ||
| − | + | <center><math>\sum_{\text{sym}} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\geq \sum_{\text{sym}} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}</math></center> | |
| − | <math>\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}</math> | ||
== Example == | == Example == | ||
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The inequality is easier to understand given an example. Since the sequence <math>(5,1)</math> majorizes <math>(4,2)</math>, Muirhead's inequality states that for any positive <math>x,y</math>, | The inequality is easier to understand given an example. Since the sequence <math>(5,1)</math> majorizes <math>(4,2)</math>, Muirhead's inequality states that for any positive <math>x,y</math>, | ||
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== Usage on Olympiad Problems == | == Usage on Olympiad Problems == | ||
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A common [[Brute forcing|bruteforce]] technique with inequalities is to clear denominators, multiply everything out, and apply Muirhead's or [[Schur's Inequality|Schur's]]. However, it is worth noting that any inequality that can be proved directly with Muirhead can also be proved using the [[Arithmetic mean-geometric mean | Arithmetic Mean-Geometric Mean]] inequality. In fact, [[International Mathematics Olympiad | IMO]] gold medalist [[Thomas Mildorf]] says it is unwise to use Muirhead in an Olympiad solution; one should use an application of AM-GM instead. Thus, it is suggested that Muirhead be used only to verify that an inequality ''can'' be proved with AM-GM before demonstrating the full AM-GM proof. | A common [[Brute forcing|bruteforce]] technique with inequalities is to clear denominators, multiply everything out, and apply Muirhead's or [[Schur's Inequality|Schur's]]. However, it is worth noting that any inequality that can be proved directly with Muirhead can also be proved using the [[Arithmetic mean-geometric mean | Arithmetic Mean-Geometric Mean]] inequality. In fact, [[International Mathematics Olympiad | IMO]] gold medalist [[Thomas Mildorf]] says it is unwise to use Muirhead in an Olympiad solution; one should use an application of AM-GM instead. Thus, it is suggested that Muirhead be used only to verify that an inequality ''can'' be proved with AM-GM before demonstrating the full AM-GM proof. | ||
Revision as of 23:49, 22 April 2008
Muirhead's Inequality states that if a sequence 
majorizes a sequence 
, then given a set of positive reals 
:
Example
The inequality is easier to understand given an example. Since the sequence 
majorizes 
, Muirhead's inequality states that for any positive 
,
.
Usage on Olympiad Problems
A common bruteforce technique with inequalities is to clear denominators, multiply everything out, and apply Muirhead's or Schur's. However, it is worth noting that any inequality that can be proved directly with Muirhead can also be proved using the Arithmetic Mean-Geometric Mean inequality. In fact, IMO gold medalist Thomas Mildorf says it is unwise to use Muirhead in an Olympiad solution; one should use an application of AM-GM instead. Thus, it is suggested that Muirhead be used only to verify that an inequality can be proved with AM-GM before demonstrating the full AM-GM proof.
As an example, the above inequality can be proved using AM-GM as follows:
![$\frac{x^4+x^4+x^4+y^4}{4}\geq\sqrt[4]{x^{12}y^4}=x^3y$](http://latex.artofproblemsolving.com/4/3/e/43e7046bb7e98924c632961732191380a0c2e9c6.png)
![$\frac{x^4+y^4+y^4+y^4}{4}\geq\sqrt[4]{x^4y^{12}}=xy^3$](http://latex.artofproblemsolving.com/3/d/a/3da9a78e244a01a097b79713a99fce41ba63026f.png)
Adding these,
.
Multiplying both sides by 
(as both 
and 
are positive),
as desired.
