Difference between revisions of "Homogenization"

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If <math>a,b,c>0</math> and <math>a+b+c=1</math>, prove that <math>a^2+b^2+c^2+1\ge 4(ab+bc+ca)</math>.
 
If <math>a,b,c>0</math> and <math>a+b+c=1</math>, prove that <math>a^2+b^2+c^2+1\ge 4(ab+bc+ca)</math>.
 
==Solution==
 
==Solution==
  So all the terms except for the <math>1</math> are of the second degree. We substituting <math>a+b+c</math> for the <math>1</math>  
+
  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>  
  
 
. The inequality still gives a non-homogeneous inequality. So instead we square the condition to make  
 
. The inequality still gives a non-homogeneous inequality. So instead we square the condition to make  

Revision as of 20:54, 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 however 

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

to solve problems.

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