# Difference between revisions of "Cauchy-Schwarz Inequality"

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with equality when there exist constants <math>\mu, \lambda </math> not both zero such that for all <math> 1 \le i \le n </math>, <math>\mu a_i = \lambda b_i </math>. | with equality when there exist constants <math>\mu, \lambda </math> not both zero such that for all <math> 1 \le i \le n </math>, <math>\mu a_i = \lambda b_i </math>. | ||

− | === | + | === Discussion === |

− | + | Consider the vectors <math> \mathbf{a} = \langle a_1, \ldots a_n \rangle </math> and <math> {} \mathbf{b} = \langle b_1, \ldots b_n \rangle </math>. If <math>\theta </math> is the [[angle]] formed by <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, then the left-hand side of the inequality is equal to the square of the [[dot product]] of <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, or <math> \left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cos\theta \right)^2 </math>. The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \right)^2 </math>. The inequality then follows from <math> |\cos\theta | \le 1 </math>, with equality when one of <math> \mathbf{a,b} </math> is a multiple of the other, as desired. Note that this is not actually a proof; rather, this result is used to establish that we can define the angles between two vectors in this way. Indeed, the proof that structures such as vectors exist in <math>\mathbb{R}_n</math> using [[linear algebra]] requires this very inequality, and so this would be circular logic. | |

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− | Consider the vectors <math> \mathbf{a} = \langle a_1, \ldots a_n \rangle </math> and <math> {} \mathbf{b} = \langle b_1, \ldots b_n \rangle </math>. If <math>\theta </math> is the [[angle]] formed by <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, then the left-hand side of the inequality is equal to the square of the [[dot product]] of <math> \mathbf{a} </math> and <math> \mathbf{b} </math>, or <math> \left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \cos\theta \right)^2 </math>. The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| \cdot ||\mathbf{b}|| \right)^2 </math>. The inequality then follows from <math> |\cos\theta | \le 1 </math>, with equality when one of <math> \mathbf{a,b} </math> is a multiple of the other, as desired. Note that this is not actually a proof; rather, this result is used to establish that we can define the angles between two vectors in this way. | ||

=== Complex Form === | === Complex Form === |

## Revision as of 11:12, 8 April 2008

The **Cauchy-Schwarz Inequality** (which is known by other names, including **Cauchy's Inequality**, **Schwarz's Inequality**, and the **Cauchy-Bunyakovsky-Schwarz Inequality**) is a well-known inequality with many elegant applications.

## Contents

## Elementary Form

For any real numbers and ,

,

with equality when there exist constants not both zero such that for all , .

### Discussion

Consider the vectors and . If is the angle formed by and , then the left-hand side of the inequality is equal to the square of the dot product of and , or . The right hand side of the inequality is equal to . The inequality then follows from , with equality when one of is a multiple of the other, as desired. Note that this is not actually a proof; rather, this result is used to establish that we can define the angles between two vectors in this way. Indeed, the proof that structures such as vectors exist in using linear algebra requires this very inequality, and so this would be circular logic.

### Complex Form

The inequality sometimes appears in the following form.

Let and be complex numbers. Then

.

This appears to be more powerful, but it follows immediately from

.

## General Form

Let be a vector space, and let be an inner product. Then for any ,

,

with equality if and only if there exist constants not both zero such that .

### Proof 1

Consider the polynomial of

.

This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., must be less than or equal to , with equality when or when there exists some scalar such that , as desired.

### Proof 2

We consider

.

Since this is always greater than or equal to zero, we have

.

Now, if either or is equal to , then . Otherwise, we may normalize so that , and we have

,

with equality when and may be scaled to each other, as desired.

### Examples

The elementary form of the Cauchy-Schwarz inequality is a special case of the general form, as is the **Cauchy-Schwarz Inequality for Integrals**: for integrable functions ,

,

with equality when there exist constants not both equal to zero such that for ,

.

## Problems

### Introductory

- Consider the function , where is a positive integer. Show that . (Source)

### Intermediate

- Let be a triangle such that

,

where and denote its semiperimeter and inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine those integers. (Source)

### Olympiad

- is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which

is least.

(Source)

## Other Resources

### Books

- The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
- Problem Solving Strategies by Arthur Engel contains significant material on inequalities.