Difference between revisions of "Cauchy-Schwarz Inequality"

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The '''Cauchy-Schwarz Inequality''' (which is known by other names, including Cauchy's Inequality) states that, for two sets of real numbers <math>a_1,a_2,\ldots,a_n</math> and <math>b_1,b_2,\ldots,b_n</math>, the following [[inequality]] is always true:
 
The '''Cauchy-Schwarz Inequality''' (which is known by other names, including Cauchy's Inequality) states that, for two sets of real numbers <math>a_1,a_2,\ldots,a_n</math> and <math>b_1,b_2,\ldots,b_n</math>, the following [[inequality]] is always true:
 
  
 
<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>
 
<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>
  
 
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[[Equality condition|Equality]] holds if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}</math>.
[[Equality condition| equality]] holds if and only if <math>\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}</math>.
 
 
 
  
 
== Proof ==
 
== Proof ==
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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+ \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>
 
<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]],  
  
 
<math>{\left(\sum a_ib_i\right)}^2 \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right).</math>  
 
<math>{\left(\sum a_ib_i\right)}^2 \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right).</math>  
 
 
   
 
   
 
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>.
 
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>.
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== Contest Problem Solving ==
 
== Contest Problem Solving ==
 
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]].
 
 
  
 
== Other Resources ==
 
== Other Resources ==

Revision as of 20:12, 2 February 2007

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,

${\left(\sum a_ib_i\right)}^2 \leq \left(\sum a_i^2\right)\left(\sum b_i^2\right).$

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 $\frac{a_1}{b_1}=\frac{a_2}{b_2}= \cdots =\frac{a_n}{b_n}$.

Contest Problem Solving

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

Other Resources

Books

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