# Difference between revisions of "Cauchy-Schwarz Inequality"

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− | The Cauchy-Schwarz | + | 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. It has an elementary form, a complex form, and a general form. |

− | + | [https://en.wikipedia.org/wiki/Augustin-Louis_Cauchy|Augustin Louis Cauchy] wrote the first paper about the elementary form in 1821. The general form was discovered by [https://en.wikipedia.org/wiki/Viktor_Bunyakovsky|Viktor Bunyakovsky] in 1849 and independently by [https://en.wikipedia.org/wiki/Hermann_Amandus_Schwarz|Hermann Schwarz] in 1888. | |

− | + | == Elementary Form == | |

− | + | For any real numbers <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math>, | |

− | <math> | + | <cmath> |

− | + | \left( \sum_{i=1}^{n}a_ib_i \right)^2 \le \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right) | |

− | <math>\ | + | </cmath> |

− | + | with equality when there exists a nonzero constant <math>\mu</math> such that for all <math> 1 \le i \le n </math>, <math>\mu a_i = b_i </math>. | |

− | |||

− | This inequality is | + | === 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>(\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2</math> .The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2 = a^2b^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. | ||

+ | |||

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

+ | |||

+ | The inequality sometimes appears in the following form. | ||

+ | |||

+ | Let <math> a_1, \ldots, a_n </math> and <math> b_1, \ldots, b_n </math> be [[complex numbers]]. Then | ||

+ | <cmath> | ||

+ | \left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right) | ||

+ | </cmath> | ||

+ | This appears to be more powerful, but it follows from | ||

+ | <cmath> | ||

+ | \left| \sum_{i=1}^n a_ib_i \right| ^2 \le \left( \sum_{i=1}^n |a_i| \cdot |b_i| \right)^2 \le \left(\sum_{i=1}^n |a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right) | ||

+ | </cmath> | ||

+ | |||

+ | == Upper Bound on (Σa)(Σb) == | ||

+ | |||

+ | Let <math>a_1, a_2, \ldots, a_n</math> and <math>b_1, b_2, \ldots, b_n</math> be two sequences of positive real numbers with | ||

+ | <cmath> | ||

+ | 0 < m \le \frac{a_i}{b_i} \le M | ||

+ | </cmath> | ||

+ | for <math>1 \le i \le n</math>. Then | ||

+ | <cmath> | ||

+ | \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right) \le \frac{(M+m)^2}{4Mm} \left( \sum_{i=1}^{n}a_ib_i \right)^2, | ||

+ | </cmath> | ||

+ | with equality if and only if, for some ordering of the pairs <math>(a_i,b_i) \mapsto (a_{\sigma(i)},b_{\sigma(i)})</math>, some <math>0 \le j \le n</math> exists such that <math>a_{\sigma(i)}=mb_{\sigma(i)}</math> for <math>1 \le \sigma(i) \le j</math> and <math>a_{\sigma(i)}=Mb_{\sigma(i)}</math> for <math>j+1 \le \sigma(i) \le n</math>, and | ||

+ | <cmath> | ||

+ | m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2. | ||

+ | </cmath> | ||

+ | If we restrict that <math>m_1 \le a_i \le M_1</math> and <math>m_2 \le b_i \le M_2</math> for all <math>i</math>, then it's clear that for <math>a_i/b_i</math> to be <math>m=m_1/M_2</math> or <math>M=M_1/m_2</math> for all <math>i</math>, then <math>a_i=m_1 \Longleftrightarrow b_i=M_2</math> and <math>a_i=M_1 \Longleftrightarrow b_i=m_2</math>, so | ||

+ | <cmath> | ||

+ | m\sum_{\sigma(i)=1}^{j}b_{\sigma(i)}^2 = M\sum_{\sigma(i)=j+1}^{n}b_{\sigma(i)}^2 | ||

+ | </cmath> | ||

+ | is equivalent to | ||

+ | <cmath> | ||

+ | \begin{align*} | ||

+ | m(jM_2^2) = M((n-j)m_2^2) &\Longleftrightarrow m_1M_2j = M_1m_2(n-j)\\ | ||

+ | &\Longleftrightarrow j = \left(\frac{M_1m_2}{M_1m_2+m_1M_2}\right) n. | ||

+ | \end{align*} | ||

+ | </cmath> | ||

+ | (When this is not an integer, the maximum occurs when <math>j</math> is either the ceiling or floor of the right-hand side.) In the special case that <math>a_ib_i = k > 0</math> is constant for all <math>i</math>, we have <math>M_1=1/m_2</math> and <math>m_1=1/M_2</math>, so here <math>j</math> must be <math>n/2</math>. | ||

+ | |||

+ | === Proof === | ||

+ | |||

+ | Note that for all <math>i</math>, we have | ||

+ | <cmath> | ||

+ | 0 \le \left(\frac{a_i}{b_i}-m\right)\left(M-\frac{a_i}{b_i}\right) = \frac{1}{b_i^2}(a_ib_iM-a_i^2-b_i^2Mm+a_ib_im) | ||

+ | </cmath> | ||

+ | or | ||

+ | <cmath> | ||

+ | (M+m)a_ib_i \ge a_i^2+(Mm)b_i^2, | ||

+ | </cmath> | ||

+ | with equality if and only if <math>a_i=mb_i</math> or <math>a_i=Mb_i</math>. Summing up these inequalities over <math>1 \le i \le n</math>, we obtain from AM-GM that | ||

+ | <cmath> | ||

+ | \begin{align*} | ||

+ | (M+m)\sum_{i=1}^{n}a_ib_i &\ge \sum_{i=1}^{n}a_i^2 + (Mm)\sum_{i=1}^{n}b_i^2\\ | ||

+ | &\ge 2\sqrt{Mm \left(\sum_{i=1}^{n}a_i^2 \right) \left(\sum_{i=1}^{n}b_i^2 \right)}, | ||

+ | \end{align*} | ||

+ | </cmath> | ||

+ | and squaring gives us the desired bound. For equality to occur, we must have <math>a_i=mb_i</math> or <math>a_i=Mb_i</math> for all <math>i</math>. If, without loss of generality, <math>a_i=mb_i</math> for <math>1 \le i \le j</math> and <math>a_i=Mb_i</math> for <math>j+1 \le i \le n</math> for some <math>0 \le j \le n</math>, then for the AM-GM to reach equality we must have (assume <math>M>m</math> since <math>M=m</math> is trivial) | ||

+ | <cmath> | ||

+ | \begin{align*} | ||

+ | \sum_{i=1}^{n}a_i^2 &= Mm\sum_{i=1}^{n}b_i^2\\ | ||

+ | m^2\sum_{i=1}^{j}b_i^2 + M^2\sum_{i=j+1}^{n}b_i^2 &= Mm\sum_{i=1}^{j}b_i^2 + Mm\sum_{i=j+1}^{n}b_i^2\\ | ||

+ | (m-M)m\sum_{i=1}^{j}b_i^2 &= (m-M)M\sum_{i=j+1}^{n}b_i^2\\ | ||

+ | m\sum_{i=1}^{j}b_i^2 &= M\sum_{i=j+1}^{n}b_i^2. | ||

+ | \end{align*} | ||

+ | </cmath> | ||

+ | |||

+ | == General Form == | ||

+ | |||

+ | Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \to \mathbb{R} </math> be an [[inner product]]. Then for any <math> \mathbf{a,b} \in V </math>, | ||

+ | <cmath> | ||

+ | \langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle , | ||

+ | </cmath> | ||

+ | with equality if and only if there exist constants <math>\mu, \lambda </math> not both zero such that <math> \mu\mathbf{a} = \lambda\mathbf{b} </math>. | ||

+ | |||

+ | === Proof 1 === | ||

+ | |||

+ | Consider the polynomial of <math> t </math> | ||

+ | <cmath> | ||

+ | \langle t\mathbf{a + b}, t\mathbf{a + b} \rangle = t^2\langle \mathbf{a,a} \rangle + 2t\langle \mathbf{a,b} \rangle + \langle \mathbf{b,b} \rangle . | ||

+ | </cmath> | ||

+ | This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., <math> \langle \mathbf{a,b} \rangle^2 </math> must be less than or equal to <math> \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle </math>, with equality when <math> \mathbf{a = 0} </math> or when there exists some scalar <math>-t </math> such that <math> -t\mathbf{a} = \mathbf{b} </math>, as desired. | ||

+ | |||

+ | === Proof 2 === | ||

+ | |||

+ | We consider | ||

+ | <cmath> | ||

+ | \langle \mathbf{a-b, a-b} \rangle = \langle \mathbf{a,a} \rangle + \langle \mathbf{b,b} \rangle - 2 \langle \mathbf{a,b} \rangle . | ||

+ | </cmath> | ||

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

+ | <cmath> | ||

+ | \langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle . | ||

+ | </cmath> | ||

+ | Now, if either <math> \mathbf{a} </math> or <math> \mathbf{b} </math> is equal to <math> \mathbf{0} </math>, then <math> \langle \mathbf{a,b} \rangle^2 = \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle = 0 </math>. Otherwise, we may [[normalize]] so that <math> \langle \mathbf {a,a} \rangle = \langle \mathbf{b,b} \rangle = 1 </math>, and we have | ||

+ | <cmath> | ||

+ | \langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} , | ||

+ | </cmath> | ||

+ | with equality when <math>\mathbf{a} </math> and <math> \mathbf{b} </math> may be scaled to each other, as desired. | ||

+ | |||

+ | === Proof 3 === | ||

+ | |||

+ | Consider <math>a-\lambda b</math> for some scalar <math>\lambda</math>. Then: | ||

+ | <math>0\le||a-\lambda b||^2</math> (by the Trivial Inequality) | ||

+ | <math>=\langle a-\lambda b,a-\lambda b\rangle</math> | ||

+ | <math>=\langle a,a\rangle-2\lambda\langle a,b\rangle+\lambda^2\langle y,y\rangle</math> | ||

+ | <math>=||a||^2-2\lambda\langle a,b\rangle+\lambda^2||b||^2</math>. | ||

+ | Now, let <math>\lambda=\frac{\langle a,b\rangle}{||b||^2}</math>. Then, we have: | ||

+ | <math>0\le||a||^2-\frac{\langle a,b\rangle|^2}{||b||^2}</math> | ||

+ | <math>\implies\langle a,b\rangle|^2\le||a||^2||b||^2=\langle a,a\rangle\cdot\langle b,b\rangle</math>. <math>\square</math> | ||

+ | |||

+ | === 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 <math> f,g : [a,b] \mapsto \mathbb{R} </math>, | ||

+ | <cmath> | ||

+ | \biggl( \int_{a}^b f(x)g(x)dx \biggr)^2 \le \int_{a}^b \bigl[ f(x) \bigr]^2dx \cdot \int_a^b \bigl[ g(x) \bigr]^2 dx | ||

+ | </cmath> | ||

+ | with equality when there exist constants <math> \mu, \lambda </math> not both equal to zero such that for <math> t \in [a,b] </math>, | ||

+ | <cmath> | ||

+ | \mu \int_a^t f(x)dx = \lambda \int_a^t g(x)dx . | ||

+ | </cmath> | ||

+ | |||

+ | ==Problems== | ||

+ | |||

+ | ===Introductory=== | ||

+ | |||

+ | *Consider the function <math>f(x)=\frac{(x+k)^2}{x^2+1},x\in (-\infty,\infty)</math>, where <math>k</math> is a positive integer. Show that <math>f(x)\le k^2+1</math>. ([[User:Temperal/The_Problem_Solver's Resource Competition|Source]]) | ||

+ | * [http://www.mathlinks.ro/Forum/viewtopic.php?t=78687 (APMO 1991 #3)] Let <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_n</math>, <math>b_1</math>, <math>b_2</math>, <math>\cdots</math>, <math>b_n</math> be positive real numbers such that <math>a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n</math>. Show that | ||

+ | <cmath>\frac {a_1^2}{a_1 + b_1} + \frac {a_2^2}{a_2 + b_2} + \cdots + \frac {a_n^2}{a_n + b_n} \geq \frac {a_1 + a_2 + \cdots + a_n}{2}</cmath> | ||

+ | |||

+ | ===Intermediate=== | ||

+ | |||

+ | *Let <math>ABC</math> be a triangle such that | ||

+ | <cmath> | ||

+ | \left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2 , | ||

+ | </cmath> | ||

+ | where <math>s</math> and <math>r</math> denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle <math>ABC </math> is similar to a triangle <math>T </math> whose side lengths are all positive integers with no common divisor and determine those integers. | ||

+ | ([[2002 USAMO Problems/Problem 2|Source]]) | ||

+ | |||

+ | ===Olympiad=== | ||

+ | |||

+ | *<math>P</math> is a point inside a given triangle <math>ABC</math>. <math>D, E, F</math> are the feet of the perpendiculars from <math>P</math> to the lines <math>BC, CA, AB</math>, respectively. Find all <math>P</math> for which | ||

+ | <cmath> | ||

+ | \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} | ||

+ | </cmath> | ||

+ | is least. | ||

+ | |||

+ | ([[1981 IMO Problems/Problem 1|Source]]) | ||

+ | |||

+ | == Other Resources == | ||

+ | * [http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality Wikipedia entry] | ||

+ | |||

+ | ===Books=== | ||

+ | * [http://www.amazon.com/exec/obidos/ASIN/052154677X/artofproblems-20 The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities] by J. Michael Steele. | ||

+ | * [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequalities. | ||

+ | |||

+ | |||

+ | [[Category:Algebra]] | ||

+ | [[Category:Inequality]] | ||

+ | [[Category:Theorems]] |

## Latest revision as of 14:57, 17 January 2021

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. It has an elementary form, a complex form, and a general form.

Louis Cauchy wrote the first paper about the elementary form in 1821. The general form was discovered by Bunyakovsky in 1849 and independently by Schwarz in 1888.

## Contents

## Elementary Form

For any real numbers and , with equality when there exists a nonzero constant 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.

### 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 from

## Upper Bound on (Σa)(Σb)

Let and be two sequences of positive real numbers with for . Then with equality if and only if, for some ordering of the pairs , some exists such that for and for , and If we restrict that and for all , then it's clear that for to be or for all , then and , so is equivalent to (When this is not an integer, the maximum occurs when is either the ceiling or floor of the right-hand side.) In the special case that is constant for all , we have and , so here must be .

### Proof

Note that for all , we have or with equality if and only if or . Summing up these inequalities over , we obtain from AM-GM that and squaring gives us the desired bound. For equality to occur, we must have or for all . If, without loss of generality, for and for for some , then for the AM-GM to reach equality we must have (assume since is trivial)

## 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.

### Proof 3

Consider for some scalar . Then: (by the Trivial Inequality) . Now, let . Then, we have: .

### 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)
- (APMO 1991 #3) Let , , , , , , , be positive real numbers such that . Show that

### 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.