Difference between revisions of "Cauchy-Schwarz Inequality"
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− | + | In [[algebra]], the '''Cauchy-Schwarz Inequality''', also known as the '''Cauchy–Bunyakovsky–Schwarz Inequality''' or informally as '''Cauchy-Schwarz''', is an [[inequality]] with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra. | |
− | = | + | Its elementary algebraic formulation is often referred to as '''Cauchy's Inequality''' and states that for any list of reals <math>a_1, a_2, \ldots, a_n</math> and <math>b_1, b_2, \ldots, b_n</math>, <cmath>(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,</cmath> with equality if and only if there exists a constant <math>t</math> such that <math>a_i = t b_i</math> for all <math>1 \leq i \leq n</math>, or if one list consists of only zeroes. Along with the [[AM-GM Inequality]], Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests. |
− | + | Its vector formulation states that for any vectors <math>\overrightarrow{v}</math> and <math>\overrightarrow{w}</math> in <math>\mathbb{R}^n</math>, where <math>\overrightarrow{v} \cdot \overrightarrow{w}</math> is the [[dot product]] of <math>\overrightarrow{v}</math> and <math>\overrightarrow{w}</math> and <math>\| \overrightarrow{v} \|</math> is the [[norm]] of <math>\overrightarrow{v}</math>, <cmath>\|\overrightarrow{v}\| \|\overrightarrow{w}\| \geq |\overrightarrow{v} \cdot \overrightarrow{w}|</cmath> with equality if and only if there exists a scalar <math>t</math> such that <math>\overrightarrow{v} = t \overrightarrow{w}</math>, or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems. | |
− | < | ||
− | <math> | ||
− | \ | ||
− | |||
− | </ | ||
− | with equality | ||
− | + | The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality. | |
− | 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> | + | == Proofs == |
+ | Here is a list of proofs of Cauchy-Schwarz. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | == Lemmas == | ||
=== Complex Form === | === Complex Form === | ||
− | |||
The inequality sometimes appears in the following 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 | 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) | |
− | \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> | |
− | This appears to be more powerful, but it follows | + | \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> |
− | |||
− | \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 ) | ||
− | </ | ||
− | |||
== General Form == | == General Form == | ||
− | Let <math>V </math> be a [[vector space]], and let <math> \langle \cdot, \cdot \rangle : V \times V \ | + | 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 , | |
− | \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>. | 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>. | ||
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Consider the polynomial of <math> t </math> | 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 . | |
− | \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. | 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. | ||
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We consider | 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 . | |
− | \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 | 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 . | |
− | \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 | 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} , | |
− | \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. | 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> | + | <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>, | + | <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> | |
− | <math> | ||
− | \ | ||
− | </math>. | ||
− | </ | ||
==Problems== | ==Problems== | ||
+ | |||
===Introductory=== | ===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]]) | *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=== | ===Intermediate=== | ||
− | *Let <math>ABC </math> be a triangle such that | + | |
− | < | + | *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 | + | \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. |
− | 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]]) | ([[2002 USAMO Problems/Problem 2|Source]]) | ||
+ | |||
===Olympiad=== | ===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 | *<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} | \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} | ||
− | </ | + | </cmath> |
− | |||
− | |||
is least. | is least. | ||
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− | [[Category: | + | [[Category:Algebra]] |
− | [[Category: | + | [[Category:Inequalities]] |
Revision as of 04:28, 18 May 2023
In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra.
Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals and , with equality if and only if there exists a constant such that for all , or if one list consists of only zeroes. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests.
Its vector formulation states that for any vectors and in , where is the dot product of and and is the norm of , with equality if and only if there exists a scalar such that , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems.
The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality.
Contents
Proofs
Here is a list of proofs of Cauchy-Schwarz.
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.
Lemmas
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
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: .
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.