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. |
− | + | 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. | |
− | |||
− | |||
− | |||
− | |||
− | == | + | == 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. | 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. | ||
Line 28: | Line 26: | ||
</cmath> | </cmath> | ||
− | == | + | === A Useful Inequality === |
+ | Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality: | ||
− | + | For any real numbers <math>(a_1,a_2,...,a_n)</math> and <math>(b_1,b_2,...,b_n)</math> where <math>(b_i>0, i\in \{1,2,..,n\})</math> the following is true: <cmath> \frac{a_1^{2}}{b_1}+\frac{a_2^{2}}{b_2}+\ldots+\frac{a_n^{2}}{b_n}\geq\frac{\left(a_1+a_2+...+a_n\right)^{2}}{b_1+b_2+...+b_n}.</cmath> | |
+ | |||
+ | == Real Vector Spaces == | ||
+ | |||
+ | 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> | <cmath> | ||
− | + | \langle \mathbf{a,b} \rangle^2 \le \langle \mathbf{a,a} \rangle \langle \mathbf{b,b} \rangle , | |
</cmath> | </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>. The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers. | |
+ | |||
+ | === Proof 1 === | ||
+ | |||
+ | Consider the polynomial of <math> t </math> | ||
<cmath> | <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> | </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> | <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> | </cmath> | ||
− | + | Since this is always greater than or equal to zero, we have | |
<cmath> | <cmath> | ||
− | + | \langle \mathbf{a,b} \rangle \le \frac{1}{2} \langle \mathbf{a,a} \rangle + \frac{1}{2} \langle \mathbf{b,b} \rangle . | |
</cmath> | </cmath> | ||
− | is | + | 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> | <cmath> | ||
− | \ | + | \langle \mathbf{a,b} \rangle \le 1 = \langle \mathbf{a,a} \rangle^{1/2} \langle \mathbf{b,b} \rangle^{1/2} , |
− | |||
− | |||
− | \ | ||
</cmath> | </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> | ||
+ | |||
+ | == Complex Vector Spaces == | ||
+ | |||
+ | For any two vectors <math>\mathbf{a}, \mathbf{b}</math> in the complex vector space <math>W</math>, the following holds: | ||
+ | <cmath>|\langle \mathbf{a}, \mathbf{b}\rangle| \leq ||\mathbf{a}||||\mathbf{b}||</cmath> with equality holding only when <math>\mathbf{a}, \mathbf{b}</math> are linearly dependent. | ||
=== Proof === | === Proof === | ||
+ | The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021. | ||
− | + | Define the unit vectors <math>\mathbf{u}</math>, <math>\mathbf{v}</math> as <math>\mathbf{u} = \frac{\mathbf{a}}{||\mathbf{a}||}</math> and <math>\mathbf{v} = \frac{\mathbf{b}}{||\mathbf{b}||}</math>. Put <math>\gamma = \frac{\langle \mathbf{u},\mathbf{v}\rangle}{|\langle \mathbf{u},\mathbf{v}\rangle|}</math>. In other words, <math>\gamma</math> is the complex argument of <math>\langle \mathbf{u},\mathbf{v}\rangle</math> and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let <math>\mathbf{p} = \frac{1}{2}(\mathbf{u}+\gamma \mathbf{v})</math> and <math>\mathbf{q} = \frac{1}{2}(\mathbf{u}-\gamma \mathbf{v})</math>. Importantly, we have | |
+ | <cmath> | ||
+ | \langle \mathbf{p},\mathbf{q}\rangle = \frac{1}{4}(||\mathbf{u}||^2 -\langle \mathbf{u},\gamma \mathbf{v}\rangle + \langle \gamma \mathbf{v},\mathbf{u}\rangle - \gamma \bar{\gamma}||\mathbf{v}||^2) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \frac{1}{4}(1 -\langle \mathbf{u},\gamma \mathbf{v}\rangle + \langle \gamma \mathbf{v},\mathbf{u}\rangle - 1) | ||
+ | </cmath> | ||
<cmath> | <cmath> | ||
− | + | = \frac{1}{4}(\langle \gamma \mathbf{v},\mathbf{u}\rangle - \overline{\langle \gamma \mathbf{v},\mathbf{u}\rangle}) | |
</cmath> | </cmath> | ||
− | |||
<cmath> | <cmath> | ||
− | + | = \frac{1}{2}\operatorname{Im}(\langle \gamma \mathbf{v},\mathbf{u}\rangle) | |
</cmath> | </cmath> | ||
− | |||
<cmath> | <cmath> | ||
− | \ | + | = \frac{1}{2}\operatorname{Im}(\gamma\overline{\langle \mathbf{u},\mathbf{v}\rangle}) |
− | |||
− | |||
− | |||
</cmath> | </cmath> | ||
− | |||
<cmath> | <cmath> | ||
− | \ | + | = 0. |
− | \ | + | </cmath> |
− | + | Since <math>\mathbf{u}=\mathbf{p}+\mathbf{q}</math> and <math>\mathbf{v} = \bar{\gamma}(\mathbf{p}-\mathbf{q})</math>, this calculation shows that <math>\mathbf{p}</math> and <math>\mathbf{q}</math> form an orthogonal basis of the linear subspace spanned by <math>\mathbf{u}</math> and <math>\mathbf{v}</math>. Thus we can think of <math>\mathbf{u}</math> and <math>\mathbf{v}</math> as lying on the unit sphere in this subspace, which is isomorphic to <math>\mathbb{C}^2</math>. Another thing to note is that | |
− | + | <cmath> | |
− | + | ||\mathbf{p}||^2 + ||\mathbf{q}||^2 = \langle \mathbf{p},\mathbf{p} \rangle + \langle \mathbf{q},\mathbf{q} \rangle | |
− | + | </cmath> | |
+ | <cmath> | ||
+ | = \frac{1}{4}(\langle \mathbf{u}+\gamma \mathbf{v},\mathbf{u}+\gamma \mathbf{v}\rangle + \langle \mathbf{u}-\gamma \mathbf{v},\mathbf{u}-\gamma \mathbf{v}\rangle) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \frac{1}{4}(2\langle \mathbf{u},\mathbf{u} \rangle + 2\gamma\bar{\gamma}\langle \mathbf{v},\mathbf{v} \rangle) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \frac{1}{4}(2||\mathbf{u}||^2 + 2||\mathbf{v}||^2) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = 1. | ||
</cmath> | </cmath> | ||
− | + | The previous two calculations established that <math>\mathbf{p}</math> and <math>\mathbf{q}</math> are orthogonal, and that the sum of their squared norms is <math>1</math>. Now we have | |
− | + | <cmath> | |
− | + | |\langle \mathbf{u},\mathbf{v}\rangle| = |\langle \mathbf{p}+\mathbf{q},\bar{\gamma}(\mathbf{p}-\mathbf{q})\rangle| | |
+ | </cmath> | ||
<cmath> | <cmath> | ||
− | \langle \mathbf{ | + | = |\gamma||\langle \mathbf{p}+\mathbf{q},\mathbf{p}-\mathbf{q}\rangle| |
</cmath> | </cmath> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
<cmath> | <cmath> | ||
− | \langle | + | = |\langle \mathbf{p}+\mathbf{q},\mathbf{p}-\mathbf{q}\rangle| |
</cmath> | </cmath> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
<cmath> | <cmath> | ||
− | + | = |||\mathbf{p}||^2 +\langle \mathbf{q},\mathbf{p}\rangle - \langle \mathbf{p},\mathbf{q}\rangle - ||\mathbf{q}||^2| | |
</cmath> | </cmath> | ||
− | |||
<cmath> | <cmath> | ||
− | + | = |||\mathbf{p}||^2 - ||\mathbf{q}||^2| | |
</cmath> | </cmath> | ||
− | |||
<cmath> | <cmath> | ||
− | + | = |||\mathbf{p}||^2 + ||\mathbf{q}||^2 - 2||\mathbf{q}||^2| | |
</cmath> | </cmath> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
<cmath> | <cmath> | ||
− | + | = |1 - 2||\mathbf{q}||^2| | |
</cmath> | </cmath> | ||
− | |||
<cmath> | <cmath> | ||
− | \ | + | \leq 1. |
</cmath> | </cmath> | ||
+ | Equality holds when either <math>||\mathbf{q}||=0</math> or <math>||\mathbf{q}||=1</math>, or equivalently when <math>\mathbf{u}=\pm \mathbf{v}</math> and <math>\mathbf{a} = \lambda \mathbf{b}</math>. Lastly, multiplying each side by <math>||\mathbf{a}||||\mathbf{b}||</math>, we have <cmath>|\langle \mathbf{a},\mathbf{b}\rangle| \leq ||\mathbf{a}||||\mathbf{b}||.</cmath> | ||
==Problems== | ==Problems== | ||
Line 161: | Line 181: | ||
[[Category:Algebra]] | [[Category:Algebra]] | ||
− | [[Category: | + | [[Category:Inequalities]] |
− |
Latest revision as of 15:28, 22 February 2024
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
A Useful Inequality
Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality:
For any real numbers and
where
the following is true:
Real Vector Spaces
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
. The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers.
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:
.
Complex Vector Spaces
For any two vectors in the complex vector space
, the following holds:
with equality holding only when
are linearly dependent.
Proof
The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021.
Define the unit vectors ,
as
and
. Put
. In other words,
is the complex argument of
and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let
and
. Importantly, we have
Since
and
, this calculation shows that
and
form an orthogonal basis of the linear subspace spanned by
and
. Thus we can think of
and
as lying on the unit sphere in this subspace, which is isomorphic to
. Another thing to note is that
The previous two calculations established that and
are orthogonal, and that the sum of their squared norms is
. Now we have
Equality holds when either
or
, or equivalently when
and
. Lastly, multiplying each side by
, 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.