Difference between revisions of "Cauchy-Schwarz Inequality"

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<math> \displaystyle ({a_1}^2+{a_2}^2+...+{a_n}^2)({b_1}^2+{b_2}^2+...+{b_n}^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2</math>
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<math> \displaystyle ({a_1}^2+{a_2}^2+ \cdots +{a_n}^2)({b_1}^2+{b_2}^2+ \cdots +{b_n}^2) \geq (a_1b_1+a_2b_2+ \cdots +a_nb_n)^2</math>
  
  
Equality holds if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}</math>.
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Equality holds if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}</math>.
  
  
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There are many ways to prove this; one of the more well-known is to consider the equation
 
There are many ways to prove this; one of the more well-known is to consider the equation
  
<math> \displaystyle (a_1x+b_1)^2+(a_2x+b_2)^2+...+(a_nx+b_n)^2=0</math>.   
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<math> \displaystyle (a_1x+b_1)^2+(a_2x+b_2)^2+ \cdots +(a_nx+b_n)^2=0</math>.   
  
 
Expanding, we find the equation to be of the form  
 
Expanding, we find the equation to be of the form  
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where <math>A=\sum_{i=1}^n a_i^2</math>, <math>B=2\sum_{j=1}^n a_jb_j</math>, and <math>C=\sum_{k=1}^n b_k^2.</math>.  By the [[Trivial inequality | Trivial Inequality]], we know that the left-hand-side of the original equation is always at least 0, so either both roots are [[complex numbers]], or there is a double root at <math>x=0</math>.  Either way, the [[discriminant]] of the equation is nonpositive.  Taking the [[discriminant]], <math>B^2-4AC \leq 0</math> and substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality,  
 
where <math>A=\sum_{i=1}^n a_i^2</math>, <math>B=2\sum_{j=1}^n a_jb_j</math>, and <math>C=\sum_{k=1}^n b_k^2.</math>.  By the [[Trivial inequality | Trivial Inequality]], we know that the left-hand-side of the original equation is always at least 0, so either both roots are [[complex numbers]], or there is a double root at <math>x=0</math>.  Either way, the [[discriminant]] of the equation is nonpositive.  Taking the [[discriminant]], <math>B^2-4AC \leq 0</math> and substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality,  
  
<math> \displaystyle (a_1b_1+a_2b_2+...+a_nb_n)^2 \leq (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2),</math>
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<math> \displaystyle (a_1b_1+a_2b_2+ \cdots +a_nb_n)^2 \leq (a_1^2+a_2^2+ \cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2),</math>
  
 
or, in the more compact [[sigma notation]],  
 
or, in the more compact [[sigma notation]],  
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Note that this also gives us the equality case; equality holds if and only if the discriminant is equal to 0, which is true if and only if the equation has 0 as a double root, which is true if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}=...=\frac{a_n}{b_n}</math>.
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Note that this also gives us the equality case; equality holds if and only if the discriminant is equal to 0, which is true if and only if the equation has 0 as a double root, which is true if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}</math>.
  
 
This inequality is used very frequently to solve Olympiad-level Inequality problems, such as those on the [[United States of America Mathematics Olympiad | USAMO]] and [[International Mathematics Olympiad | IMO]].
 
This inequality is used very frequently to solve Olympiad-level Inequality problems, such as those on the [[United States of America Mathematics Olympiad | USAMO]] and [[International Mathematics Olympiad | IMO]].

Revision as of 01:36, 18 June 2006

The Cauchy-Schwarz Inequality (which is known by other names, including Cauchy's Inequality) states that, for two sets of real numbers $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$, the following inequality is always true:


$\displaystyle ({a_1}^2+{a_2}^2+ \cdots +{a_n}^2)({b_1}^2+{b_2}^2+ \cdots +{b_n}^2) \geq (a_1b_1+a_2b_2+ \cdots +a_nb_n)^2$


Equality holds if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}$.


Proof

There are many ways to prove this; one of the more well-known is to consider the equation

$\displaystyle (a_1x+b_1)^2+(a_2x+b_2)^2+ \cdots +(a_nx+b_n)^2=0$.

Expanding, we find the equation to be of the form

$\displaystyle Ax^2 + Bx + C,$

where $A=\sum_{i=1}^n a_i^2$, $B=2\sum_{j=1}^n a_jb_j$, and $C=\sum_{k=1}^n b_k^2.$. By the Trivial Inequality, we know that the left-hand-side of the original equation is always at least 0, so either both roots are complex numbers, or there is a double root at $x=0$. Either way, the discriminant of the equation is nonpositive. Taking the discriminant, $B^2-4AC \leq 0$ and substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality,

$\displaystyle (a_1b_1+a_2b_2+ \cdots +a_nb_n)^2 \leq (a_1^2+a_2^2+ \cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2),$

or, in the more compact sigma notation, $\left(\sum a_ib_i\right)$ $\leq \left(\sum a_i^2\right)\left(\sum b_i^2\right).$


Note that this also gives us the equality case; equality holds if and only if the discriminant is equal to 0, which is true if and only if the equation has 0 as a double root, which is true if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}$.

This inequality is used very frequently to solve Olympiad-level Inequality problems, such as those on the USAMO and IMO.

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