View source for Cauchy–Schwarz inequality

{{short description|Mathematical inequality relating inner products and norms}}
The '''Cauchy–Schwarz inequality''' (also called '''Cauchy–Bunyakovsky–Schwarz inequality''')<ref>{{cite web|url=https://mathshistory.st-andrews.ac.uk/Biographies/Schwarz/|title=Hermann Amandus Schwarz|first1=J.J.|last1=O'Connor|first2=E.F.|last2=Robertson|website=[[Organization:University of St Andrews|University of St Andrews]], Scotland }}</ref><ref name="EncyclopediaOfMath">{{springer|id=b/b017770|title=Bunyakovskii inequality|first=V. I.|last=Bityutskov}}</ref><ref>{{cite web|url=http://www.wwu.edu/faculty/curgus/Courses/Math_pages/Math_504/Cauchy-Schwarz-Bunyakovsky.html|title=Cauchy-Bunyakovsky-Schwarz inequality|first=Branko|last=Ćurgus|website=Western Washington University|department=Department of Mathematics }}</ref><ref>{{cite web|url=https://mathcs.clarku.edu/~djoyce/ma130/cauchy.pdf |archive-url=https://ghostarchive.org/archive/20221009/https://mathcs.clarku.edu/~djoyce/ma130/cauchy.pdf |archive-date=2022-10-09 |url-status=live|title=Cauchy's inequality|first=David E.|last=Joyce|website=Clark University|department=Department of Mathematics and Computer Science }}</ref> is an upper bound on the [[Inner product space|inner product]] between two vectors in an inner product space in terms of the product of the vector [[Norm (mathematics)|norms]]. It is considered one of the most important and widely used [[Inequality (mathematics)|inequalities]] in mathematics.<ref name="Steele">{{cite book|last=Steele|first=J. Michael|year=2004|title=The Cauchy–Schwarz Master Class: an Introduction to the Art of Mathematical Inequalities|publisher=The Mathematical Association of America|isbn=978-0521546775|page=1|url=http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_index.html|quote=...there is no doubt that this is one of the most widely used and most important inequalities in all of mathematics.}}</ref>

The inequality for sums was published by {{harvs|first=Augustin-Louis|last=Cauchy|year=1821|txt=yes}}. The corresponding inequality for integrals was published by {{harvs|first=Viktor|last=Bunyakovsky|txt=yes|year=1859}}<ref name="EncyclopediaOfMath"/> and {{harvs|txt=yes|authorlink=Hermann Schwarz|first=Hermann|last=Schwarz|year=1888}}. Schwarz gave the modern proof of the integral version.<ref name="Steele" /> 

== Statement of the inequality ==
The Cauchy–Schwarz inequality states that for all vectors <math>\mathbf{u}</math> and <math>\mathbf{v}</math> of an [[Inner product space|inner product space]]

{{NumBlk|:|<math>\left |\langle \mathbf{u}, \mathbf{v} \rangle\right |^2 \leq \langle \mathbf{u}, \mathbf{u} \rangle \cdot \langle \mathbf{v}, \mathbf{v} \rangle,</math>|{{EquationRef|Cauchy–Schwarz inequality [written using only the inner product]}}}}
where <math>\langle \cdot, \cdot \rangle</math> is the inner product. Examples of inner products include the real and complex [[Dot product|dot product]]; see the [[Inner product space#Some examples|examples in inner product]]. Every inner product gives rise to a Euclidean <math>l_2</math> [[Norm (mathematics)|norm]], called the {{em|canonical}} or [[Inner product space#Norm|{{em|induced}} {{em|norm}}]], where the norm of a vector <math>\mathbf{u}</math> is denoted and defined by
<math display="block">\|\mathbf{u}\| := \sqrt{\langle \mathbf{u}, \mathbf{u} \rangle},</math> where <math>\langle \mathbf{u}, \mathbf{u} \rangle</math> is always a non-negative real number (even if the inner product is complex-valued). 
By taking the square root of both sides of the above inequality, the Cauchy–Schwarz inequality can be written in its more familiar form:<ref name="Strang5">{{cite book|last=Strang|first=Gilbert|date=19 July 2005|title=Linear Algebra and its Applications|edition=4th|chapter=3.2|publisher=Cengage Learning|location=Stamford, CT|isbn=978-0030105678|pages=154–155}}</ref><ref name=":0">{{cite book|last1=Hunter|first1=John K.|last2=Nachtergaele|first2=Bruno|year=2001|title=Applied Analysis|publisher=World Scientific|isbn=981-02-4191-7|url=https://books.google.com/books?id=oOYQVeHmNk4C}}</ref>

{{NumBlk|:|<math>|\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\|  \|\mathbf{v}\|.</math>|{{EquationRef|Cauchy–Schwarz inequality - written using norm and inner product}}}}

Moreover, the two sides are equal if and only if <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are [[Linear independence|linearly dependent]].<ref>{{cite book|last1=Bachmann|first1=George|last2=Narici|first2=Lawrence|last3=Beckenstein|first3=Edward|date=2012-12-06|title=Fourier and Wavelet Analysis|publisher=Springer Science & Business Media|isbn=9781461205050|page=14|url=https://books.google.com/books?id=PkHhBwAAQBAJ}}</ref><ref>{{cite book|last=Hassani|first=Sadri|year=1999|title=Mathematical Physics: A Modern Introduction to Its Foundations|publisher=Springer|isbn=0-387-98579-4|page=29|quote=Equality holds iff <c{{pipe}}c>=0 or {{pipe}}c>=0. From the definition of {{pipe}}c>, we conclude that {{pipe}}a> and {{pipe}}b> must be proportional.}}</ref><ref>{{cite book|last1=Axler|first1=Sheldon|date=2015|title=Linear Algebra Done Right, 3rd Ed.|publisher=Springer International Publishing|isbn=978-3-319-11079-0|page=172|url=https://books.google.com/books?id=CQWwoQEACAAJ|quote=This inequality is an equality if and only if one of u, v is a scalar multiple of the other.}}</ref>

== Special cases ==

=== Sedrakyan's lemma - Positive real numbers ===
[[Sedrakyan's inequality]], also called [[Biography:Harald Bergström|Bergström]]'s inequality, Engel's form, the T2 lemma, or Titu's lemma, states that for real numbers <math>u_1, u_2, \dots, u_n</math> and positive real numbers <math>v_1, v_2, \dots, v_n</math>:
<math display=block>\frac{\left(\sum_{i=1}^n u_i\right)^2 }{\sum_{i=1}^n v_i} \leq \sum_{i=1}^n \frac{u_i^2}{v_i} \quad \text{ or equivalently, } \quad  \frac{\left(u_1 + u_2 + \cdots + u_n\right)^2}{v_1 + v_2 + \cdots + v_n} \leq \frac{u^2_1}{v_1} + \frac{u^2_2}{v_2} + \cdots + \frac{u^2_n}{v_n} .</math>

It is a direct consequence of the Cauchy–Schwarz inequality, obtained by using the [[Dot product|dot product]] on <math>\R^n</math> upon substituting <math>u_i' = u_i / \sqrt{v_i}</math> and <math>v_i' = \sqrt{v_i}</math>. This form is especially helpful when the inequality involves fractions where the numerator is a [[Square number|perfect square]].

=== R<sup>2</sup> - The plane ===
[[File:Cauchy-Schwarz inequation in Euclidean plane.gif|thumb|300px|Cauchy-Schwarz inequality in a unit circle of the Euclidean plane]]
The real vector space <math>\R^2</math> denotes the 2-dimensional plane. It is also the 2-dimensional [[Euclidean space]] where the inner product is the [[Dot product|dot product]]. 
If <math>\mathbf{u} = \left(u_1, u_2\right)</math> and <math>\mathbf{v} = \left(v_1, v_2\right)</math> then the Cauchy–Schwarz inequality becomes:
<math display=block>\langle \mathbf{u}, \mathbf{v} \rangle^2 = (\|\mathbf{u}\| \|\mathbf{v}\| \cos \theta)^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,</math>
where <math>\theta</math> is the [[Angle|angle]] between <math>\mathbf{u}</math> and <math>\mathbf{v}</math>.

The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates <math>u_1</math>, <math>u_2</math>, <math>v_1</math>, and <math>v_2</math> as
<math display=block>\left(u_1 v_1 + u_2 v_2\right)^2 \leq \left(u_1^2 + u_2^2\right) \left(v_1^2 + v_2^2\right),</math>
where equality holds if and only if the vector <math>\left(u_1, u_2\right)</math> is in the same or opposite direction as the vector <math>\left(v_1, v_2\right)</math>, or if one of them is the zero vector.

{{anchor|Real Euclidean space}}

=== R<sup>''n''</sup> - ''n''-dimensional Euclidean space ===
{{anchor|real_number_proof}}In [[Euclidean space]] <math>\R^n</math> with the standard inner product, which is the [[Dot product|dot product]], the Cauchy–Schwarz inequality becomes:
<math display=block>\left(\sum_{i=1}^n u_i v_i\right)^2 \leq \left(\sum_{i=1}^n u_i^2\right) \left(\sum_{i=1}^n v_i^2\right).</math>

The Cauchy–Schwarz inequality can be proved using only elementary algebra in this case by observing that the difference of the right and the left hand side is 
<math> \frac{1}{2} \sum_{i,j=1,\ldots,n} (u_i v_j - u_j v_i)^2 \ge 0</math> 


or by considering the following quadratic polynomial in <math>x</math>
<math display=block> \left(u_1 x + v_1\right)^2 + \cdots + \left(u_n x + v_n\right)^2 = \left(\sum_i u_i^2\right) x^2 + 2 \left(\sum_i u_i v_i\right) x + \sum_i v_i^2.</math>

Since the latter polynomial is nonnegative, it has at most one real root, hence its [[Discriminant|discriminant]] is less than or equal to zero. That is,
<math display=block>\left(\sum_i u_i v_i\right)^2 - \left(\sum_i {u_i^2}\right)  \left(\sum_i {v_i^2}\right) \leq 0.</math>


=== C<sup>''n''</sup> - ''n''-dimensional Complex space===
If <math>\mathbf{u}, \mathbf{v} \in \Complex^n</math> with <math>\mathbf{u} = \left(u_1, \ldots, u_n\right)</math> and <math>\mathbf{v} = \left(v_1, \ldots, v_n\right)</math> (where <math>u_1, \ldots, u_n \in \Complex</math> and <math>v_1, \ldots, v_n \in \Complex</math>) and if the inner product on the vector space <math>\Complex^n</math> is the canonical complex inner product (defined by <math>\langle \mathbf{u}, \mathbf{v} \rangle := u_1 \overline{v_1} + \cdots + u_{n} \overline{v_n},</math> where the bar notation is used for [[Complex conjugate|complex conjugation]]), then the inequality may be restated more explicitly as follows: 
<math display=block>|\langle \mathbf{u}, \mathbf{v} \rangle|^2 
= \left|\sum_{k=1}^n u_k\bar{v}_k\right|^2 
\leq \langle \mathbf{u}, \mathbf{u} \rangle \langle \mathbf{v}, \mathbf{v} \rangle 
= \left(\sum_{k=1}^n u_k \bar{u}_k\right) \left(\sum_{k=1}^n v_k \bar{v}_k\right) 
= \sum_{j=1}^n \left|u_j\right|^2 \sum_{k=1}^n \left|v_k\right|^2.</math>

That is,
<math display=block>\left|u_1 \bar{v}_1 + \cdots + u_n \bar{v}_n\right|^2 \leq \left(\left|u_1\right|^2 + \cdots + \left|u_n\right|^2\right) \left(\left|v_1\right|^2 + \cdots + \left|v_n\right|^2\right).</math>

=== ''L''<sup>2</sup> ===

For the inner product space of square-integrable complex-valued [[Function (mathematics)|functions]], the following inequality:
<math display=block>\left|\int_{\R^n} f(x) \overline{g(x)}\,dx\right|^2  \leq  \int_{\R^n} |f(x)|^2\,dx \int_{\R^n} |g(x)|^2 \,dx.</math>

The Hölder inequality is a generalization of this.

== Applications ==

=== Analysis ===
In any [[Inner product space|inner product space]], the [[Triangle inequality|triangle inequality]] is a consequence of the Cauchy–Schwarz inequality, as is now shown: 
<math display="block">\begin{alignat}{4}
\|\mathbf{u} + \mathbf{v}\|^2 
&= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v} \rangle && \\
&= \|\mathbf{u}\|^2 + \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{u} \rangle + \|\mathbf{v}\|^2 ~ && ~ \text{ where } \langle \mathbf{v}, \mathbf{u} \rangle = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \\
&= \|\mathbf{u}\|^2 + 2 \operatorname{Re} \langle \mathbf{u}, \mathbf{v} \rangle + \|\mathbf{v}\|^2 && \\
&\leq \|\mathbf{u}\|^2 + 2|\langle \mathbf{u}, \mathbf{v} \rangle| + \|\mathbf{v}\|^2 && \\
&\leq \|\mathbf{u}\|^2 + 2\|\mathbf{u}\|\|\mathbf{v}\| + \|\mathbf{v}\|^2 ~ && ~ \text{ using CS}\\
&=(\|\mathbf{u}\| + \|\mathbf{v}\|)^2. && 
\end{alignat}</math>

Taking square roots gives the triangle inequality:
<math display=block>\|\mathbf{u} + \mathbf{v}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|.</math>

The Cauchy–Schwarz inequality is used to prove that the inner product is a [[Continuous function|continuous function]] with respect to the [[Topology|topology]] induced by the inner product itself.<ref>{{cite book|last1=Bachman|first1=George|last2=Narici|first2=Lawrence|date=2012-09-26|title=Functional Analysis|publisher=Courier Corporation|isbn=9780486136554|pages=141|url=https://books.google.com/books?id=_lTDAgAAQBAJ}}</ref><ref>{{cite book|last=Swartz|first=Charles|date=1994-02-21|title=Measure, Integration and Function Spaces|publisher=World Scientific|isbn=9789814502511|pages=236|url=https://books.google.com/books?id=SsbsCgAAQBAJ}}</ref>

=== Geometry ===
The Cauchy–Schwarz inequality allows one to extend the notion of "angle between two vectors" to any real inner-product space by defining:<ref>{{cite book|last=Ricardo|first=Henry|date=2009-10-21|title=A Modern Introduction to Linear Algebra|publisher=CRC Press|isbn=9781439894613|pages=18|url=https://books.google.com/books?id=s7bMBQAAQBAJ}}</ref><ref>{{cite book|last1=Banerjee|first1=Sudipto|last2=Roy|first2=Anindya|date=2014-06-06|title=Linear Algebra and Matrix Analysis for Statistics|publisher=CRC Press|isbn=9781482248241|pages=181|url=https://books.google.com/books?id=WDTcBQAAQBAJ}}</ref>
<math display=block>\cos\theta_{\mathbf{u} \mathbf{v}} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|}.</math>

The Cauchy–Schwarz inequality proves that this definition is sensible, by showing that the right-hand side lies in the interval {{math|[&minus;1,&nbsp;1]}} and justifies the notion that (real) [[Hilbert space]]s are simply generalizations of the [[Euclidean space]]. It can also be used to define an angle in complex inner-product spaces, by taking the absolute value or the real part of the right-hand side,<ref>{{cite book|last=Valenza|first=Robert J.|date=2012-12-06|title=Linear Algebra: An Introduction to Abstract Mathematics|publisher=Springer Science & Business Media|isbn=9781461209010|pages=146|url=https://books.google.com/books?id=7x8MCAAAQBAJ}}</ref><ref>{{cite book|last=Constantin|first=Adrian|date=2016-05-21|title=Fourier Analysis with Applications|publisher=Cambridge University Press|isbn=9781107044104|pages=74|url=https://books.google.com/books?id=JnMZDAAAQBAJ}}</ref> as is done when extracting a metric from [[Fidelity of quantum states|quantum fidelity]].

=== Probability theory ===
<!-- For the multivariate case,{{clarify|reason=define GE operator here|date=July 2011}}<ref>{{cite journal| last=Gautam|first=Tripathi|title=A matrix extension of the Cauchy-Schwarz inequality|journal=Economics Letters|date=4 December 1998|url=http://web2.uconn.edu/tripathi/published-papers/cs.pdf |archive-url=https://ghostarchive.org/archive/20221009/http://web2.uconn.edu/tripathi/published-papers/cs.pdf |archive-date=2022-10-09 |url-status=live|doi=10.1016/s0165-1765(99)00014-2}}</ref>
<math display=block>\operatorname{Var}(Y) \geq \operatorname{Cov} (Y, X) \operatorname{Var}^{-1}(X) \operatorname{Cov}(X, Y)</math>
This inequality means that the difference is semidefinite positive. -->

Let <math>X</math> and <math>Y</math> be [[Random variable|random variable]]s, then the covariance inequality<ref>{{cite book|last=Mukhopadhyay|first=Nitis|date=2000-03-22|title=Probability and Statistical Inference|publisher=CRC Press|isbn=9780824703790|pages=150|url=https://books.google.com/books?id=TMSnGkr_DxwC}}</ref><ref>{{cite book|last=Keener|first=Robert W.|date=2010-09-08|title=Theoretical Statistics: Topics for a Core Course|publisher=Springer Science & Business Media|isbn=9780387938394|pages=71|url=https://books.google.com/books?id=aVJmcega44cC}}</ref> is given by:
<math display=block>\operatorname{Var}(X) \geq \frac{\operatorname{Cov}(X, Y)^2}{\operatorname{Var}(Y)}.</math>

After defining an inner product on the set of random variables using the expectation of their product,
<math display=block>\langle X, Y \rangle := \operatorname{E}(X Y),</math>
the Cauchy–Schwarz inequality becomes
<math display=block>|\operatorname{E}(XY)|^2 \leq \operatorname{E}(X^2) \operatorname{E}(Y^2).</math>

To prove the covariance inequality using the Cauchy–Schwarz inequality, let <math>\mu = \operatorname{E}(X)</math> and <math>\nu = \operatorname{E}(Y),</math> then
<math display=block>\begin{align}
|\operatorname{Cov}(X, Y)|^2 
&= |\operatorname{E}((X - \mu)(Y - \nu))|^2 \\
&= |\langle X - \mu, Y - \nu \rangle |^2\\
&\leq \langle X - \mu, X - \mu \rangle \langle Y - \nu, Y - \nu \rangle \\
& = \operatorname{E}\left((X - \mu)^2\right) \operatorname{E}\left((Y - \nu)^2\right) \\
& = \operatorname{Var}(X) \operatorname{Var}(Y),
\end{align}</math>
where <math>\operatorname{Var}</math> denotes [[Variance|variance]] and <math>\operatorname{Cov}</math> denotes [[Covariance|covariance]].

==Proofs==

There are many different proofs<ref>{{cite journal|last1=Wu|first1=Hui-Hua|last2=Wu|first2=Shanhe|date=April 2009|title=Various proofs of the Cauchy-Schwarz inequality|journal=Octogon Mathematical Magazine|issn=1222-5657|isbn=978-973-88255-5-0|volume=17|issue=1|pages=221–229|url=http://www.uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_19.pdf |archive-url=https://ghostarchive.org/archive/20221009/http://www.uni-miskolc.hu/~matsefi/Octogon/volumes/volume1/article1_19.pdf |archive-date=2022-10-09 |url-status=live|access-date=18 May 2016}}</ref> of the Cauchy–Schwarz inequality other than those given below.<ref name="Steele" /><ref name=":0" /> 
When consulting other sources, there are often two sources of confusion. First, some authors define {{math|⟨⋅,⋅⟩}} to be linear in the [[Inner product space#Alternative definitions, notations and remarks|second argument]] rather than the first. 
Second, some proofs are only valid when the field is <math>\mathbb R</math> and not <math>\mathbb C.</math><ref>{{cite book|last1=Aliprantis|first1=Charalambos D.|last2=Border|first2=Kim C.|date=2007-05-02|title=Infinite Dimensional Analysis: A Hitchhiker's Guide|publisher=Springer Science & Business Media|isbn=9783540326960|url=https://books.google.com/books?id=4hIq6ExH7NoC}}</ref>

This section gives proofs of the following theorem:

{{math theorem|name=Cauchy–Schwarz inequality|note=|style=|math_statement= 
Let <math>\mathbf{u}</math> and <math>\mathbf{v}</math> be arbitrary vectors in an [[Inner product space|inner product space]] over the scalar field <math>\mathbb{F},</math> where <math>\mathbb F</math> is the field of real numbers <math>\R</math> or complex numbers <math>\Complex.</math> Then

{{NumBlk|:|<math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right| \leq \|\mathbf{u}\| \|\mathbf{v}\|</math>|{{EquationRef|Cauchy–Schwarz Inequality}}}}

with {{visible anchor|Characterization of Equality in Cauchy–Schwarz|text=equality holding}} in the {{EquationNote|Cauchy–Schwarz Inequality}} [[If and only if|if and only if]] <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are [[Linear independence|linearly dependent]]. 

Moreover, if <math>|\langle \mathbf{u}, \mathbf{v} \rangle| = \|\mathbf{u}\| \|\mathbf{v}\|</math> and <math>\mathbf{v} \neq \mathbf{0}</math> then <math>\mathbf{u} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2} \mathbf{v}.</math>
}} 


In all of the proofs given below, the proof in the trivial case where at least one of the vectors is zero (or equivalently, in the case where <math>\|\mathbf{u}\|\|\mathbf{v}\|= 0</math>) is the same. It is presented immediately below only once to reduce repetition. It also includes the easy part of the proof of the Equality Characterization given above; that is, it proves that if <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent then <math>\left|\langle \mathbf{u}, \mathbf{v} \rangle\right| = \|\mathbf{u}\| \|\mathbf{v}\|.</math> 

{{collapse top|title=Proof of the trivial parts: Case where a vector is <math>\mathbf{0}</math> and also one direction of the Equality Characterization|left=true}}
<!-- Note to Editors:
Anyone reading this part of the proof is very likely to be new to this subject, which is why this proof is much more detailed than the proofs given elsewhere in this article.
-->
By definition, <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent if and only if one is a scalar multiple of the other. 
If <math>\mathbf{u} = c \mathbf{v}</math> where <math>c</math> is some scalar then 
<math display=block>|\langle \mathbf{u}, \mathbf{v} \rangle| 
= |\langle c \mathbf{v}, \mathbf{v} \rangle| 
= |c \langle \mathbf{v}, \mathbf{v} \rangle| 
= |c|\|\mathbf{v}\| \|\mathbf{v}\|
=\|c \mathbf{v}\| \|\mathbf{v}\|
=\|\mathbf{u}\| \|\mathbf{v}\|</math>

which shows that equality holds in the {{EquationNote|Cauchy–Schwarz Inequality}}. 
The case where <math>\mathbf{v} = c \mathbf{u}</math> for some scalar <math>c</math> is very similar, with the main difference between the complex conjugation of <math>c:</math>
<math display=block>|\langle \mathbf{u}, \mathbf{v} \rangle| 
= |\langle \mathbf{u}, c \mathbf{u} \rangle| 
= \left|\overline{c} \langle \mathbf{u}, \mathbf{u} \rangle\right| 
= \left|\overline{c}\right| \|\mathbf{u}\| \|\mathbf{u}\|
= |c| \|\mathbf{u}\| \|\mathbf{u}\|
=\|\mathbf{u}\| \|c \mathbf{u}\|
=\|\mathbf{u}\| \|\mathbf{v}\|.</math> 

If at least one of <math>\mathbf{u}</math> and <math>\mathbf{v}</math> is the zero vector then <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are necessarily linearly dependent (just scalar multiply the non-zero vector by the number <math>0</math> to get the zero vector; for example, if <math>\mathbf{u} = \mathbf{0}</math> then let <math>c = 0</math> so that <math>\mathbf{u} = c \mathbf{v}</math>), which proves the converse of this characterization in this special case; that is, this shows that if at least one of <math>\mathbf{u}</math> and <math>\mathbf{v}</math> is <math>\mathbf{0}</math> then the Equality Characterization holds. 

If <math>\mathbf{u} = \mathbf{0},</math> which happens if and only if <math>\|\mathbf{u}\|= 0,</math> then <math>\|\mathbf{u}\| \|\mathbf{v}\| = 0</math> and <math>|\langle \mathbf{u}, \mathbf{v} \rangle| 
= |\langle \mathbf{0}, \mathbf{v} \rangle| 
= |0| = 0</math> so that in particular, the Cauchy–Schwarz inequality holds because both sides of it are <math>0.</math> 
The proof in the case of <math>\mathbf{v} = \mathbf{0}</math> is identical. 
{{collapse bottom}}

Consequently, the Cauchy–Schwarz inequality only needs to be proven only for non-zero vectors and also only the non-trivial direction of the Equality Characterization must be shown. 

===Proof 1===

The special case of <math>\mathbf{v} = \mathbf{0}</math> was proven above so it is henceforth assumed that <math>\mathbf{v} \neq \mathbf{0}.</math> 
The Cauchy–Schwarz {{em|in}}equality (and the rest of the theorem) is an almost immediate corollary of the following {{em|equality}}:{{sfn|Halmos|1982|pp=2, 167}} 

{{NumBlk|:|<math>\frac{1}{\|\mathbf{v}\|^2} \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^2 ~=~ \|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - |\langle \mathbf{u}, \mathbf{v} \rangle|^2</math>|{{EquationRef|Eq. 1}}|LnSty=1px dashed black}} 

{{math proof|drop=hidden|title=Deducing {{EquationNote|Cauchy–Schwarz Inequality|Cauchy–Schwarz}} from {{EquationNote|Eq. 1}}|proof=
Because the left hand side of {{EquationNote|Eq. 1}} is non-negative, so is the right hand side, which proves that <math>|\langle \mathbf{u}, \mathbf{v} \rangle|^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,</math> from which the {{EquationNote|Cauchy-Schwarz Inequality}} follows (by taking the [[Square root|square root]] of both sides). 

If <math>|\langle \mathbf{u}, \mathbf{v} \rangle| = \|\mathbf{u}\| \|\mathbf{v}\|</math> then the right hand side (and thus also the left hand side) of {{EquationNote|Eq. 1}} is <math>0,</math> which is only possible if <math>\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v} = \mathbf{0}.</math><ref group=note>In fact, it follows immediately from {{EquationRef|Eq. 1}} that when <math>\mathbf{v} \neq \mathbf{0},</math> then <math>|\langle \mathbf{u}, \mathbf{v} \rangle| = \|\mathbf{u}\|\|\mathbf{v}\|</math> if and only if <math>\|\mathbf{v}\|^2 \mathbf{u} = \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}.</math></ref> 
Thus <math>\mathbf{u} = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2} \mathbf{v},</math> which shows that <math>\mathbf{u}</math> and <math>\mathbf{v}</math> are linearly dependent.{{sfn|Halmos|1982|pp=2, 167}} <math>\blacksquare</math>
}}

Equality {{EquationRef|Eq. 1}} is readily verified by elementarily expanding <math>\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^2</math> (via the definition of the norm) and then simplifying:

{{math proof|drop=hidden|title=Proof of {{EquationRef|Eq. 1}}|proof=
Let <math>V = \|\mathbf{v}\|^2</math> and <math>c = \langle \mathbf{u}, \mathbf{v} \rangle</math> so that <math>\bar{c} c = |c|^2 = |\langle \mathbf{u}, \mathbf{v} \rangle|^2</math> and <math>\bar{c} = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} = \langle \mathbf{v}, \mathbf{u} \rangle.</math> 
Then
<math display=block>\begin{alignat}{4}
\left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^2 
&= \|V \mathbf{u} - c \mathbf{v}\|^2 
= \langle V \mathbf{u} - c \mathbf{v}, V \mathbf{u} - c \mathbf{v} \rangle && ~\quad\text{ By definition of the norm } \\[0.5ex]
&= \langle V \mathbf{u}, V \mathbf{u} \rangle
 - \langle V \mathbf{u}, c \mathbf{v} \rangle
 - \langle c \mathbf{v}, V \mathbf{u} \rangle
 + \langle c \mathbf{v}, c \mathbf{v} \rangle && ~\quad\text{ Expand } \\[0.5ex]
&= V^2 \langle \mathbf{u}, \mathbf{u} \rangle
 - V \bar{c} \langle \mathbf{u}, \mathbf{v} \rangle
 - c V \langle \mathbf{v}, \mathbf{u} \rangle
 \,+ c \bar{c} \langle \mathbf{v}, \mathbf{v} \rangle && ~\quad\text{ Pull out scalars (note that } V := \|\mathbf{v}\|^2 \text{ is real) } \\[0.5ex]
&= V^2\|\mathbf{u}\|^2
 ~~- V \bar{c} c
 ~~~~~~~~- c V \bar{c}
 ~~~~~~~~+ c \bar{c} V && ~\quad\text{ Use definitions of } c := \langle \mathbf{u}, \mathbf{v} \rangle \text{ and } V \\[0.5ex]
&= V^2\|\mathbf{u}\|^2 ~~- V \bar{c} c 
~=~ V \left[V\|\mathbf{u}\|^2 - \bar{c} c\right] && ~\quad\text{ Simplify } \\[0.5ex]
&= \|\mathbf{v}\|^2 \left[\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - |\langle \mathbf{u}, \mathbf{v} \rangle|^2\right] && ~\quad\text{ Rewrite in terms of } \mathbf{u} \text{ and } \mathbf{v}. \\
\end{alignat}
</math>
Dividing by <math>\|\mathbf{v}\|^2 \neq 0</math> completes the proof. <math>\blacksquare</math>
}}

This expansion does not require <math>\mathbf{v}</math> to be non-zero; however, <math>\mathbf{v}</math> must be non-zero in order to divide both sides by <math>\|\mathbf{v}\|^2</math> and to deduce the Cauchy–Schwarz inequality from it. 
Swapping <math>\mathbf{u}</math> and <math>\mathbf{v}</math> gives rise to:
<math display=block>\left\|\|\mathbf{u}\|^2 \mathbf{v} - \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \mathbf{u}\right\|^2 ~=~ \|\mathbf{u}\|^2 \left[\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - |\langle \mathbf{u}, \mathbf{v} \rangle|^2\right]</math>
and thus
<math display=block>\begin{alignat}{4}
\|\mathbf{u}\|^2\|\mathbf{v}\|^2 \left[\|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - |\langle \mathbf{u}, \mathbf{v} \rangle|^2\right]
~&=~ \|\mathbf{u}\|^2 \left\|\|\mathbf{v}\|^2 \mathbf{u} - \langle \mathbf{u}, \mathbf{v} \rangle \mathbf{v}\right\|^2 \\
~&=~ \|\mathbf{v}\|^2 \left\|\|\mathbf{u}\|^2 \mathbf{v} - \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \mathbf{u}\right\|^2. \\
\end{alignat}
</math>

===Proof 2===

The special case of <math>\mathbf{v} = \mathbf{0}</math> was proven above so it is henceforth assumed that <math>\mathbf{v} \neq \mathbf{0}.</math> 
Let
<math display=block>\mathbf{z} := \mathbf{u} - \frac {\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}.</math>

It follows from the linearity of the inner product in its first argument that:
<math display=block>\langle \mathbf{z}, \mathbf{v} \rangle 
= \left\langle \mathbf{u} - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}, \mathbf{v} \right\rangle 
= \langle \mathbf{u}, \mathbf{v} \rangle - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{v}, \mathbf{v} \rangle 
= 0.</math>

Therefore, <math>\mathbf{z}</math> is a vector orthogonal to the vector <math>\mathbf{v}</math> (Indeed, <math>\mathbf{z}</math> is the [[Vector projection|projection]] of <math>\mathbf{u}</math> onto the plane orthogonal to <math>\mathbf{v}.</math>) We can thus apply the [[Pythagorean theorem#Inner product spaces|Pythagorean theorem]] to
<math display=block>\mathbf{u}= \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} + \mathbf{z}</math>
which gives
<math display=block>\|\mathbf{u}\|^2 
= \left|\frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}\right|^2 \|\mathbf{v}\|^2 + \|\mathbf{z}\|^2 
= \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{(\|\mathbf{v}\|^2 )^2} \,\|\mathbf{v}\|^2 + \|\mathbf{z}\|^2 
= \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{\|\mathbf{v}\|^2} + \|\mathbf{z}\|^2 \geq \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{\|\mathbf{v}\|^2}.</math>

The Cauchy–Schwarz inequality follows by multiplying by <math>\|\mathbf{v}\|^2</math> and then taking the square root. 
Moreover, if the relation <math>\geq</math> in the above expression is actually an equality, then <math>\|\mathbf{z}\|^2 = 0</math> and hence <math>\mathbf{z} = \mathbf{0};</math> the definition of <math>\mathbf{z}</math> then establishes a relation of linear dependence between <math>\mathbf{u}</math> and <math>\mathbf{v}.</math> The converse was proved at the beginning of this section, so the proof is complete. <math>\blacksquare</math>

===Proof for real inner products===

Let <math>(V, \langle \cdot, \cdot \rangle)</math> be a real inner product space. Consider an arbitrary pair <math>\mathbf{u}, \mathbf{v} \in V</math> and the function <math>p : \R \to \R</math> defined by <math>p(t) = \langle t\mathbf{u} + \mathbf{v}, t\mathbf{u} + \mathbf{v}\rangle.</math> 
Since the inner product is positive-definite, <math>p(t)</math> only takes non-negative values. On the other hand, <math>p(t)</math> can be expanded using the bilinearity of the inner product and using the fact that <math>\langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v}, \mathbf{u} \rangle</math> for real inner products:
<math display=block>p(t) = \Vert \mathbf{u} \Vert^2 t^2 + t \left[\langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{u} \rangle\right] + \Vert \mathbf{v} \Vert^2 = \Vert \mathbf{u} \Vert^2 t^2 + 2t \langle \mathbf{u}, \mathbf{v} \rangle + \Vert \mathbf{v} \Vert^2.</math> 
Thus, <math>p</math> is a polynomial of degree <math>2</math> (unless <math>u = 0,</math> which is a case that can be independently verified). Since the sign of <math>p</math> does not change, the discriminant of this polynomial must be non-positive:
<math display=block>\Delta = 4 \left(\langle \mathbf{u}, \mathbf{v} \rangle ^2 - \Vert \mathbf{u} \Vert^2 \Vert \mathbf{v} \Vert^2\right) \leqslant 0.</math>
The conclusion follows.

For the equality case, notice that <math>\Delta = 0</math> happens if and only if <math>p(t) = (t\Vert \mathbf{u} \Vert + \Vert \mathbf{v} \Vert)^2.</math> If <math>t_0 = -\Vert \mathbf{v} \Vert / \Vert \mathbf{u} \Vert,</math> then <math>p(t_0) = \langle t_0\mathbf{u} + \mathbf{v},t_0\mathbf{u} + \mathbf{v}\rangle = 0,</math> and hence <math>\mathbf{v} = -t_0\mathbf{u}.</math>

===Proof for the dot product===

The Cauchy–Schwarz inequality in the case where the inner product is the [[Dot product|dot product]] on <math>\R^n</math> is now proven.
The Cauchy–Schwarz inequality may be rewritten as <math>\left|\mathbf{a} \cdot \mathbf{b}\right|^2 \leq \left\|\mathbf{a}\right\|^2 \, \left\|\mathbf{b}\right\|^2</math> or equivalently, <math>\left(\mathbf{a} \cdot \mathbf{b}\right)^2 \leq \left(\mathbf{a} \cdot \mathbf{a}\right) \, \left(\mathbf{b} \cdot \mathbf{b}\right)</math> for <math>\mathbf{a} := \left(a_1, \ldots, a_n\right), \mathbf{b} := \left(b_1, \ldots, b_n\right) \in \R^n,</math> which expands to:
<math display=block>\left(a_1^2 + a_2^2 + \cdots + a_n^2\right) \left(b_1^2 + b_2^2 + \cdots + b_n^2\right) \geq \left(a_1b_1 + a_2b_2 + \cdots + a_nb_n\right)^2.</math> 

To simplify, let 
<math display=block>\begin{align} A &= a_1^2 + a_2^2 + \cdots + a_n^2, \\ B &= b_1^2 + b_2^2 + \cdots + b_n^2 \\ D &= a_1 b_1 + a_2 b_2 + \cdots + a_n b_n \\\end{align}</math>
so that the statement that remains to be to proven can be written as <math>A B \geq D^2,</math> which can be rearranged to <math>D^2 - A B \leq 0.</math> The [[Discriminant|discriminant]] of the [[Quadratic equation|quadratic equation]] <math>A x^2 + 2 D x + B</math> is <math>4 D^2 - 4 A B.</math> 

Therefore, to complete the proof it is sufficient to prove that this quadratic either has no real roots or exactly one real root, because this will imply: 
<math display=block>4 \left(D^2 - A B\right) \leq 0.</math>

Substituting the values of <math>A, B, D</math> into <math>A x^2 + 2 D x + B</math> gives:
<math display=block>\begin{alignat}{4}
A x^2 + 2 D x + B 
&= \left(a_1^2 + a_2^2 + \cdots + a_n^2\right) x^2 + 2 \left(a_1 b_1 + a_2 b_2 + \cdots + a_n b_n\right) x + \left(b_1^2 + b_2^2 + \cdots + b_n^2\right) \\
&= \left(a_1^2 x^2 + 2a_1 b_1 x + b_1^2\right) + \left(a_2^2 x^2 + 2a_2 b_2 x + b_2^2\right) + \cdots + \left(a_n^2 x^2 + 2a_n b_n x + b_n^2\right) \\
&= \left(a_1x + b_1\right)^2 + \left(a_2 x + b_2\right)^2 + \cdots + \left(a_n x + b_n\right)^2 \\
&\geq 0
\end{alignat}</math>
which is a sum of terms that are each <math>\,\geq 0\,</math> by the [https://artofproblemsolving.com/wiki/index.php/Trivial_Inequality trivial inequality:] <math>r^2 \geq 0</math> for all <math>r \in \R.</math> 
This proves the inequality and so to finish the proof, it remains to show that equality is achievable.
The equality <math>a_i x = - b_i</math> is the equality case for Cauchy–Schwarz after inspecting 
<math display=block>\left(a_1 x + b_1\right)^2 + \left(a_2 x + b_2\right)^2 + \cdots + \left(a_n x + b_n\right)^2 \geq 0,</math>
which proves that equality is achievable. <math>\blacksquare</math>

== Generalizations ==
Various generalizations of the Cauchy–Schwarz inequality exist. [[Hölder's inequality]] generalizes it to <math>L^p</math> norms. More generally, it can be interpreted as a special case of the definition of the norm of a linear operator on a [[Banach space]] (Namely, when the space is a [[Hilbert space]]).  Further generalizations are in the context of [[Operator theory|operator theory]], e.g. for operator-convex functions and [[Operator algebra|operator algebra]]s, where the domain and/or range are replaced by a [[C*-algebra]] or W*-algebra.

An inner product can be used to define a [[Positive linear functional|positive linear functional]]. For example, given a Hilbert space <math>L^2(m), m</math> being a finite measure, the standard inner product gives rise to a positive functional <math>\varphi</math> by <math>\varphi (g) = \langle g, 1 \rangle.</math>  Conversely, every positive linear functional <math>\varphi</math> on <math>L^2(m)</math> can be used to define an  inner product <math>\langle f, g \rangle _\varphi := \varphi\left(g^* f\right),</math> where <math>g^*</math> is the [[Pointwise product|pointwise]] [[Complex conjugate|complex conjugate]] of <math>g.</math> In this language, the Cauchy–Schwarz inequality becomes<ref>{{cite book|last1=Faria|first1=Edson de|last2=Melo|first2=Welington de|date=2010-08-12|title=Mathematical Aspects of Quantum Field Theory|publisher=Cambridge University Press|isbn=9781139489805|pages=273|url=https://books.google.com/books?id=u9M9PFLNpMMC}}</ref>
<math display=block>\left|\varphi\left(g^* f\right)\right|^2 \leq \varphi\left(f^* f\right) \varphi\left(g^* g\right),</math>

which extends verbatim to positive functionals on C*-algebras:

{{math theorem|name=Cauchy–Schwarz inequality for  positive functionals on C*-algebras<ref>{{cite book|last=Lin|first=Huaxin|date=2001-01-01|title=An Introduction to the Classification of Amenable C*-algebras|publisher=World Scientific|isbn=9789812799883|pages=27|url=https://books.google.com/books?id=2qru8d7BCAAC}}</ref><ref>{{cite book|last=Arveson|first=W.|date=2012-12-06|title=An Invitation to C*-Algebras|publisher=Springer Science & Business Media|isbn=9781461263715|pages=28|url=https://books.google.com/books?id=d5TqBwAAQBAJ}}</ref>|note=|style=|math_statement= 
If <math>\varphi</math> is a positive linear functional on a C*-algebra <math>A,</math> then for all <math>a, b \in A,</math> <math>\left|\varphi\left(b^*a\right)\right|^2 \leq \varphi\left(b^*b\right) \varphi\left(a^*a\right).</math>
}}

The next two theorems are further examples in operator algebra.

{{math theorem|name=Kadison–Schwarz inequality<ref>{{cite book|last=Størmer|first=Erling|date=2012-12-13|title=Positive Linear Maps of Operator Algebras|series=Springer Monographs in Mathematics|publisher=Springer Science & Business Media|isbn=9783642343698|url=https://books.google.com/books?id=lQtKAIONqwIC}}</ref><ref>{{cite journal|last=Kadison|first=Richard V.|date=1952-01-01|title=A Generalized Schwarz Inequality and Algebraic Invariants for Operator Algebras|jstor=1969657|journal=Annals of Mathematics|doi=10.2307/1969657|volume=56|number=3|pages=494–503}}</ref>|note=Named after Richard Kadison|style=|math_statement= 
If <math>\varphi</math> is a unital positive map, then for every [[Normal operator|normal element]] <math>a</math> in its domain, we have <math>\varphi(a^*a) \geq \varphi\left(a^*\right) \varphi(a)</math> and <math>\varphi\left(a^*a\right) \geq \varphi(a) \varphi\left(a^*\right).</math>
}}

This extends the fact <math>\varphi\left(a^*a\right) \cdot 1 \geq \varphi(a)^* \varphi(a) = |\varphi(a)|^2,</math> when <math>\varphi</math> is a linear functional. The case when <math>a</math> is self-adjoint, that is, <math>a = a^*,</math> is sometimes known as '''Kadison's inequality'''.

{{math theorem|name=Cauchy–Schwarz inequality|note=Modified Schwarz inequality for 2-positive maps<ref>{{cite book|last=Paulsen|first=Vern|year=2002|title=Completely Bounded Maps and Operator Algebras|series=Cambridge Studies in Advanced Mathematics|volume=78|publisher=Cambridge University Press|isbn=9780521816694|page=40|url=https://books.google.com/books?id=VtSFHDABxMIC&pg=PA40}}</ref>|style=|math_statement= 
For a 2-positive map <math>\varphi</math> between C*-algebras, for all <math>a, b</math> in its domain,
<math display=block>\varphi(a)^*\varphi(a) \leq \Vert\varphi(1)\Vert\varphi\left(a^*a\right), \text{ and }</math>
<math display=block>\Vert\varphi\left(a^* b\right)\Vert^2 \leq \Vert\varphi\left(a^*a\right)\Vert \cdot \Vert\varphi\left(b^*b\right)\Vert.</math>
}}

Another generalization is a refinement obtained by interpolating between both sides of the Cauchy–Schwarz inequality:  

{{math theorem|name=Callebaut's Inequality<ref>{{cite journal|last1=Callebaut|first1=D.K.|date=1965|title=Generalization of the Cauchy–Schwarz inequality|journal=J. Math. Anal. Appl.|volume=12|issue=3|pages=491–494|doi=10.1016/0022-247X(65)90016-8|doi-access=free}}</ref>|note=|style=|math_statement=  
For reals <math>0 \leqslant s \leqslant t \leqslant 1,</math> 
<math display=block>\left(\sum_{i=1}^n a_i b_i\right)^2 
~\leqslant~ \left(\sum_{i=1}^n a_i^{1+s} b_i^{1-s}\right) \left(\sum_{i=1}^n a_i^{1-s} b_i^{1+s}\right)
~\leqslant~ \left(\sum_{i=1}^n a_i^{1+t} b_i^{1-t}\right) \left(\sum_{i=1}^n a_i^{1-t} b_i^{1+t}\right)
~\leqslant~ \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right).</math>
}}

This theorem can be deduced from [[Hölder's inequality]].<ref>{{cite book|title=Callebaut's inequality|publisher=Entry in the AoPS Wiki|url=https://artofproblemsolving.com/wiki/index.php?title=Callebaut%27s_Inequality}}</ref> There are also non-commutative versions for operators and tensor products of matrices.<ref>{{cite journal|last1=Moslehian|first1=M.S.|last2=Matharu|first2=J.S.|last3=Aujla|first3=J.S.|date=2011|title=Non-commutative Callebaut inequality|journal=Linear Algebra and Its Applications|volume=436|issue=9|pages=3347–3353|doi=10.1016/j.laa.2011.11.024|arxiv=1112.3003|s2cid=119592971}}</ref>

Several matrix versions of the Cauchy–Schwarz inequality and [[Kantorovich inequality]] are applied to linear regression models.<ref>
{{cite journal |author1=Liu, Shuangzhe|author2=Neudecker, Heinz 
|year=1999
|title= A survey of Cauchy-Schwarz and Kantorovich-type matrix inequalities
|journal= Statistical Papers |volume=40 |pages=55–73 
|doi=10.1007/BF02927110 
|s2cid=122719088 
}}
</ref>
<ref>{{Cite journal| last1=Liu|first1=Shuangzhe|
last2= Trenkler|first2=Götz| last3=Kollo|first3=Tõnu| 
last4=von Rosen|first4=Dietrich| 
last5=Baksalary|first5=Oskar Maria| 
date= 2023|
title= Professor Heinz Neudecker and matrix differential calculus| 
journal= Statistical Papers| language=en|
doi= 10.1007/s00362-023-01499-w|s2cid=263661094 }}</ref>

== See also ==

* {{annotated link|Bessel's inequality}}
* {{annotated link|Hölder's inequality}}
* {{annotated link|Jensen's inequality}}
* {{annotated link|Kantorovich inequality}}
* {{annotated link|Kunita–Watanabe inequality}}
* {{annotated link|Minkowski inequality}}
* {{annotated link|Paley–Zygmund inequality}}

== Notes ==

{{reflist|group=note}}
{{reflist|group=proof}}

== Citations ==
{{reflist|30em}}

== References ==

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* {{citation|first=Viktor|last=Bunyakovsky|title=Sur quelques inegalités concernant les intégrales aux différences finies|journal=Mem. Acad. Sci. St. Petersbourg|volume=7|issue=1|year=1859|pages=6|url=http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/bunyakovsky.pdf |archive-url=https://ghostarchive.org/archive/20221009/http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/bunyakovsky.pdf |archive-date=2022-10-09 |url-status=live}}
* {{citation|first=A.-L.|last=Cauchy|title=Sur les formules qui résultent de l'emploie du signe et sur > ou <, et sur les moyennes entre plusieurs quantités|journal=Cours d'Analyse, 1er Partie: Analyse Algébrique 1821; OEuvres Ser.2 III 373-377|year=1821}}
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* {{Halmos A Hilbert Space Problem Book 1982}} <!-- {{sfn|Halmos|1982|pp=}} -->
* {{citation|first=R. V.|last=Kadison|title=A generalized Schwarz inequality and algebraic invariants for operator algebras|journal=Annals of Mathematics|volume=56|year=1952|doi=10.2307/1969657|pages=494–503|jstor=1969657|issue=3}}.
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* {{citation|first=V.|last=Paulsen|title=Completely Bounded Maps and Operator Algebras|publisher=Cambridge University Press|year=2003}}.
* {{citation|first=H. A.|last=Schwarz|year=1888|pages=318|journal=Acta Societatis Scientiarum Fennicae|volume=XV|title=Über ein Flächen kleinsten Flächeninhalts betreffendes Problem der Variationsrechnung|url=http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/Schwarz.pdf |archive-url=https://ghostarchive.org/archive/20221009/http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/Schwarz.pdf |archive-date=2022-10-09 |url-status=live}}
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{{refend}}

== External links ==

* [http://jeff560.tripod.com/c.html Earliest Uses: The entry on the Cauchy–Schwarz inequality has some historical information.]
* [http://people.revoledu.com/kardi/tutorial/LinearAlgebra/LinearlyIndependent.html#LinearlyIndependentVectors Example of application of Cauchy–Schwarz inequality to determine Linearly Independent Vectors] Tutorial and Interactive program.

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