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Difference between revisions of "Homogenization"

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==Solution==
 
==Solution==
  
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  So all the terms except for the <math>1</math> are of the second degree. We substituting <math>a+b+c</math> for <math>1</math>
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  So all the terms except for the <math>1</math> are of the second degree. We substituting  
  
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. The inequality still gives a non-homogeneous inequality. So instead we square the condition to make
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<math>a+b+c</math> for <math>1</math>. The inequality still gives a non-homogeneous inequality.  
  
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it second degree and get <math>a^2+b^2+c^2+2(ab+bc+ca)=1</math>. Now plugging this for <math>1</math> in the  
+
So instead we square the condition to make it second degree and get <cmath>a^2+b^2+c^2+2(ab+bc+ca)=1</cmath> Now plugging this for <math>1</math> in the  
  
 
inequality and simplifying gives <math>a^2+b^2+c^2\ge ab+bc+ca</math>, which is well-known by  
 
inequality and simplifying gives <math>a^2+b^2+c^2\ge ab+bc+ca</math>, which is well-known by  

Revision as of 21:57, 22 January 2014

Homogenizing is a useful technique to solve certain multivariate inequalities. Given an inequality of the form $P(a_1,a_2, \ldots, a_n) \ge 0$, where $P$ is a homogenous polynomial (that is, the degree of any term in the polynomial is the same), then we can arbitrarily impose a restraint of one order.

Example

If $a,b,c>0$ and $a+b+c=1$, prove that $a^2+b^2+c^2+1\ge 4(ab+bc+ca)$.

Solution

So all the terms except for the $1$ are of the second degree. We substituting

$a+b+c$ for $1$. The inequality still gives a non-homogeneous inequality.

So instead we square the condition to make it second degree and get \[a^2+b^2+c^2+2(ab+bc+ca)=1\] Now plugging this for $1$ in the

inequality and simplifying gives $a^2+b^2+c^2\ge ab+bc+ca$, which is well-known by Cauchy-Schwarz Inequality. 


We can use homogenization to help us solve these types of problems, especially inequalities.

It's use is not limited. After making something homogenous we can often apply well known inequalities 

to solve problems.

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