Difference between revisions of "Cauchy-Schwarz Inequality"
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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, | ||
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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> | <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> | ||
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or, in the more compact [[sigma notation]], | or, in the more compact [[sigma notation]], | ||
| − | <math>\left(\sum a_ib_i\right) | + | |
| + | <math>\left(\sum a_ib_i\right) \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 == | ||
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 01:38, 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 
and 
, the following inequality is always true:
Equality holds if and only if 
.
Proof
There are many ways to prove this; one of the more well-known is to consider the equation
.
Expanding, we find the equation to be of the form
where 
, 
, and 
. 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 
. Either way, the discriminant of the equation is nonpositive. Taking the discriminant, 
and substituting the above values of A, B, and C leaves us with the Cauchy-Schwarz Inequality,
or, in the more compact sigma notation,
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 .
Contest Problem Solving
This inequality is used very frequently to solve Olympiad-level Inequality problems, such as those on the USAMO and IMO.
