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Difference between revisions of "Muirhead's Inequality"
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Before '''Muirhead's Inequality''' can be defined, it is first necessary to define another term.
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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 integers <math>\displaystyle x_1,x_2,\cdots,x_n</math>:
  
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Given two sequences of real numbers <math>\displaystyle A=a_1,a_2,\cdots,a_n</math> and <math>\displaystyle B=b_1,b_2,\cdots,b_n</math>, <math>A</math> is said to [[Majorization | majorize]] <math>B</math> if and only if all of the following are true:
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<math>\displaystyle \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>
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<math>\displaystyle a_1\geq b_1</math>
 
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<math>\displaystyle a_1+a_2\geq b_1+b_2</math>  
 
  
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<math>\displaystyle \vdots</math>
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=== Example ===
  
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<math>\displaystyle a_1+a_2+\cdots+a_{n-1}\geq b_1+b_2+\cdots+b_{n-1}</math>
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The inequality is easier to understand given an example.  Since the sequence <math>\displaystyle (5,1)</math> majorizes <math>\displaystyle (4,2)</math>, Muirhead's inequality states that for any positive <math>\displaystyle x,y</math>,
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<math>\displaystyle a_1+a_2+\cdots+a_n=b_1+b_2+\cdots+b_n</math>
 
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With this out of the way, it is now possible to define Muirhead's Inequality.
 
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Let <math>\displaystyle x_1,x_2,\cdots,x_n</math> be a set of positive integers, and let <math>\displaystyle A=a_1,a_2,\cdots,a_n</math>, and <math>\displaystyle B=b_1,b_2,\cdots,b_n</math> be sets of positive real numbers (note that they all contain the same number of terms) such that <math>\displaystyle A</math> majorizes <math>\displaystyle B</math>.  Then, using [[sigma notation]], it is possible to say that
 
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<math>\displaystyle \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>
 
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A concrete example will probably help to understand the inequality.  Since the sequence <math>\displaystyle 5,1</math> majorizes <math>\displaystyle 4,2</math>, Muirhead's inequality tells us that, for any positive <math>\displaystyle x,y</math>, we have
 
  
 
<math>\displaystyle x^5y^1+y^5x^1\geq x^4y^2+y^4x^2</math>
 
<math>\displaystyle x^5y^1+y^5x^1\geq x^4y^2+y^4x^2</math>
  
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=== 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.
  
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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; you should use an application of AM-GM instead.
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As an example, the above inequality can be proved using AM-GM as follows:
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As an example, here's how the above inequality can be proved using AM-GM:
 
  
 
<math>\displaystyle \frac{x^4+x^4+x^4+y^4}{4}\geq\sqrt[4]{x^{12}y^4}=x^3y</math>
 
<math>\displaystyle \frac{x^4+x^4+x^4+y^4}{4}\geq\sqrt[4]{x^{12}y^4}=x^3y</math>
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<math>\displaystyle \frac{x^4+y^4+y^4+y^4}{4}\geq\sqrt[4]{x^4y^{12}}=xy^3</math>
 
<math>\displaystyle \frac{x^4+y^4+y^4+y^4}{4}\geq\sqrt[4]{x^4y^{12}}=xy^3</math>
  
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Adding these, we get
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Adding these,
  
 
<math>\displaystyle x^4+y^4\geq x^3y+xy^3</math>
 
<math>\displaystyle x^4+y^4\geq x^3y+xy^3</math>
  
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Multiplying both sides of this by <math>\displaystyle xy</math> (we can do this because both <math>\displaystyle x</math> and <math>\displaystyle y</math> are positive), we get
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Multiplying both sides by <math>\displaystyle xy</math> (as both <math>\displaystyle x</math> and <math>\displaystyle y</math> are positive),
  
 
<math>\displaystyle x^5y+xy^5\geq x^4y^2+x^2y^4</math>
 
<math>\displaystyle x^5y+xy^5\geq x^4y^2+x^2y^4</math>
  
 
as desired.
 
as desired.
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Muirhead's inequality is just one of a good many well-known [[Inequality | inequalities]] that problem-solving students should learn.
 

Revision as of 22:38, 25 June 2006

Muirhead's Inequality states that if a sequence $A$ majorizes a sequence $B$, then given a set of positive integers $\displaystyle x_1,x_2,\cdots,x_n$:

$\displaystyle \sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}$

Example

The inequality is easier to understand given an example. Since the sequence $\displaystyle (5,1)$ majorizes $\displaystyle (4,2)$, Muirhead's inequality states that for any positive $\displaystyle x,y$,

$\displaystyle x^5y^1+y^5x^1\geq x^4y^2+y^4x^2$

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:

$\displaystyle \frac{x^4+x^4+x^4+y^4}{4}\geq\sqrt[4]{x^{12}y^4}=x^3y$

$\displaystyle \frac{x^4+y^4+y^4+y^4}{4}\geq\sqrt[4]{x^4y^{12}}=xy^3$

Adding these,

$\displaystyle x^4+y^4\geq x^3y+xy^3$

Multiplying both sides by $\displaystyle xy$ (as both $\displaystyle x$ and $\displaystyle y$ are positive),

$\displaystyle x^5y+xy^5\geq x^4y^2+x^2y^4$

as desired.

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