When I came across the Cauchy-Schwarz inequality the other day, I found it really weird that this was its own thing, and it had lines upon lines of proof.

I've always thought the geometric definition of dot multiplication: $$|{\bf a }||{\bf b }|\cos \theta$$ is equivalent to the other, algebraic definition: $$a_1\cdot b_1+a_2\cdot b_2+\cdots+a_n\cdot b_n$$ And since the inequality is directly implied by the geometric definition (the fact that $\cos(\theta)$ is $1$ only when $\bf a$ and $\bf b$ are collinear), then shouldn't the Cauchy-Schwarz inequality be the world's most obvious and almost-no-proof-needed thing?

Can someone correct me on where my thought process went wrong?

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    $\begingroup$ The definition of $\cos \theta$ in multidimensional setting is sometimes that, however that does not prove it is in $[-1, 1]$ without Cauchy-Schwarz (or its equivalent steps). $\endgroup$
    – Macavity
    Mar 3, 2016 at 10:25
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    $\begingroup$ Cauchy-Schwarz inequality is valid in the much more general context of inner product spaces (possible infinite dimensional, such as function spaces $L^2$ or $\ell^2$). $\endgroup$
    – Bernard
    Mar 3, 2016 at 10:32
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    $\begingroup$ Cauchy-Swartz is not a geometric identity, it is a metric one. $\endgroup$
    – Masacroso
    Mar 3, 2016 at 10:46
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    $\begingroup$ Because someone named it? Anything can be named. $\endgroup$
    – PyRulez
    Mar 3, 2016 at 11:32
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    $\begingroup$ In addition to that: any result that just keeps popping up in different contexts deserves to have a name attached to it. $\endgroup$ Mar 3, 2016 at 12:25

6 Answers 6


Side note: it's actually the Cauchy-Schwarz-Bunyakovsky inequality, and don't let anyone tell you otherwise.

The problem with using the geometric definition is that you have to define what an angle is. Sure, in three dimensional space, you have pretty clear ideas about what an angle is, but what do you take as $\theta$ in your equation when $i$ and $j$ are $10$ dimensional vectors? Or infinitely-dimensional vectors? What if $i$ and $j$ are polynomials?

The Cauchy-Schwarz inequality tells you that anytime you have a vector space and an inner product defined on it, you can be sure that for any two vectors $u,v$ in your space, it is true that $\left|\langle u,v\rangle\right| \leq \|u\|\|v\|$.

Not all vector spaces are simple $\mathbb R^n$ businesses, either. You have the vector space of all continuous functions on $[0,1]$, for example. You can define the inner product as

$$\langle f,g\rangle=\int_0^1 f(x)g(x)dx$$

and use Cauchy-Schwarz to prove that for any pair $f,g$, you have $$\left|\int_{0}^1f(x)g(x)dx\right| \leq \sqrt{\int_0^1 f^2(x)dx\int_0^1g^2(x)dx}$$

which is not a trivial inequality.

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    $\begingroup$ I thought it was the Cauchy-Bunyakovsky-Schwarz Inequality...I appeared to have been mistaken. $\endgroup$
    – S.C.B.
    Mar 3, 2016 at 11:22
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    $\begingroup$ I would add that this inequality actually is the definition of an angle : the angle between $u$ and $v$ is the only $\theta$ such that $cos(\theta) = \frac{\langle u,v\rangle}{||u||\cdot ||v||}$, the existence being equivalent to the above inequality. $\endgroup$ Mar 3, 2016 at 11:32
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    $\begingroup$ Actually, now that I think of it, you can just say that $u$ and $v$ span a plane, and you can pick your favorite definition of (non oriented, of course) angle in this plane. $\endgroup$ Mar 3, 2016 at 12:27
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    $\begingroup$ Looking at Buniakovsky's paper, in hindsight at least, it does not look like he added anything that Cauchy didn't know. The limit of a discrete inequality, filled in the obvious way with values of a function on an interval, gives the analogue for integrals instead of sums. Maybe the addition of B's name to the inequality is a Russian thing that carried over to the Eastern European countries? $\endgroup$
    – zyx
    Mar 3, 2016 at 22:31
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    $\begingroup$ @MXY On the contrary - C-B-S is the correct, chronological sequence that corresponds to successive generalizations and more in line with its less abstract use when the Schwartz part is dropped. C-B-S is the name used in Russian literature when the S part is included. $\endgroup$
    – A.S.
    Mar 4, 2016 at 1:11
  1. The inequality is ubiquitous, so some name is needed.

  2. As there is no cosine in the statement of the inequality, it cannot be called "cosine inequality" or anything like that.

  3. The geometric interpretation with cosines only works for finite-dimensional real Euclidean space, but the inequality holds and is used more generally than that. That is Schwarz' contribution.

  4. Schwarz founded the field of functional analysis (infinite-dimensional metrized linear algebra) with his proof of the inequality. That is important enough to warrant a name. In terms of consequences per line of proof it is one of the greatest arguments of all time.

  5. The Schwarz proof was part of the historical realization that Euclidean geometry, with its mysterious angle measure that seems to depend on notions of arc-length from calculus, is the theory of a vector space equipped with a quadratic form. That is a major shift in viewpoint.

  6. Stating the inequality in terms of cosines assumes that the inner product restricts to the standard Euclidean one on the 2 (or fewer) dimensional subspace spanned by the two vectors, and that you have proved the inequality for holds for standard Euclidean space of 2 dimensions or less. How do you know those things are correct without a much longer argument? That argument will, probably, include somewhere a proof of the Cauchy-Schwarz inequality, maybe written for 2-dimensions but working for the whole $n$-dimensional space, so it might as well be stated as a direct proof for $n$ dimensions. Which is what Cauchy and Schwarz did.

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    $\begingroup$ In my opinion, $1.$ is one of the most important aspects, which the other answers barely mention. After all, we have a name for the empty set, for the trivial topology, Euler's formula etc. $\endgroup$
    – Aloizio Macedo
    Mar 4, 2016 at 0:36
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    $\begingroup$ @AloizioMacedo "why is it ubiquitous" is an important question. e.g. (3), (4), (5). $\endgroup$
    – djechlin
    Mar 4, 2016 at 1:45
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    $\begingroup$ @goblin Yes, you can use CS to define an angle. But the point zyx is making is that you cannot start from a geometric interpretation to derive CS in general, since in a general setting you don't have a geometrical interpretation to start with. That comes only after you prove CS. $\endgroup$
    – bartgol
    Mar 4, 2016 at 6:36
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    $\begingroup$ @goblin: in my personal experience (20 years as a mathematician working in functional analysis) the CS inequality is a daily thing, while I have never used the notion of angle in a vector space. $\endgroup$ Mar 4, 2016 at 12:12
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    $\begingroup$ @A.S.: divorced from the concrete? Like in quantum mechanics, which is the most basic description of reality that humans have come up with? Like Sobolev spaces, which are a basic staple of applied analysis? Like the finite-method element (computational fluid dynamics)? Like partial differential equations (the language of classical physics)? $\endgroup$ Mar 4, 2016 at 20:57

Cauchy-Schwarz is not just that. The result that you stated is just a special case of Cauchy-Schwarz in Euclidean spaces. But it's still valid in any inner product space, equipped with any inner product. The proof is still easy though, but nobody said that the proof had to be long and difficult to give it a name. The fact is that Cauchy-Schwarz inequality is very useful in many applications, from geometry to probability theory, and that's why it's worth having its own name.


In short, it deserves a name, because it is important enough to devote a full book to this inequality: The Cauchy-Schwarz Master Class. An Introduction to the Art of Mathematical Inequalities, 2004, J. M. Steele.

First, and historically, the inequality progressively emerged in three bodies of works, one involving finite sums, the others with integral formulas, in one and two dimensions, where the notion of cosine might be less evident (back then). On page 10 of this book, a glimpse of the story:

Augustin-Louis Cauchy (1789–1857) published his famous inequality in 1821 in the second of two notes on the theory of inequalities that formed the final part of his book Cours d’Analyse Algébrique

This bound [in the form of integrals] first appeared in print in a Mémoire by Victor Yacovlevich Bunyakovsky which was published by the Imperial Academy of Sciences of St. Petersburg in 1859.

In particular, it does not seem to have been known in Göttingen in 1885 when Hermann Amandus Schwarz (1843–1921) was engaged in his fundamental work on the theory of minimal surfaces [with a] need for a two-dimensional integral analog of Cauchy’s inequality.

Often, objects are named afterward, as a recognition of the previous works.

I have discovered the book recently, and I believe it deserves attention, because of the many implications of this inequality, interesting tricks and subtle reasoning. For instance, the book offers an inductive proof in finite dimensions, which he deems novel. There are a few books on "inequalities", not so many on only one of them, especially when considered basic. Because this inequality is paradigmatic. The text:

is designed to coach readers toward mastery of the most fundamental mathematical inequalities.

Cauchy-Bunyakovsky-Schwarz is used in a systematic way to open to the geometry of squares, convexity, the power means ladder, majorization, Schur convexity, exponential sums, and the inequalities of Hölder, Hilbert, and Hardy...

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    $\begingroup$ Isn't this a bit circular? You're just shifting the question to "Why does the C-S inequality even have a book written about it?" $\endgroup$ Mar 3, 2016 at 16:46
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    $\begingroup$ There was a little bit of irony in my answer, indeed. But often, many objects are named afterward, as a recognition of the previous works. I have discovered the boook recently, and I believe it deserves attention, because of its many implications and subtle reasoning. There are a few books on "inequalities", not so many on only one of them, especially when considered basic. $\endgroup$ Mar 3, 2016 at 17:26
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    $\begingroup$ @NajibIdrissi noting the existence of a few hundred pages of answer and linking to it isn't circular, though. $\endgroup$
    – djechlin
    Mar 4, 2016 at 1:46
  • $\begingroup$ @djechlin If the answer requires a full book to be answered, then the question was too broad. If it doesn't require the full book, then key points (other than why the equality is named after these people...) could have least be mentioned in the answer; link-only answers are discouraged. And this answer is not phrased as you suggest: it literally says "It deserves a name, because it is important enough to devote a full book to [it]", not "read this book to understand why". $\endgroup$ Mar 4, 2016 at 12:22
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    $\begingroup$ @NajibIdrissi okay, just downvote the answer or something. $\endgroup$
    – djechlin
    Mar 4, 2016 at 18:14

The Cauchy-Schwarz inequality can be stated and proven as a more general algebraic result (i.e. independent of vector spaces) which can then be applied to the components of vectors in inner product spaces.

It says that given two finite sequences of $n$ numbers $(a_i)_{i=1, n}$ and $(b_i)_{i=1, n}$ then $|\sum_{i=1,n} a_i.b_i| \le (\sum_{i=1,n} |a_i|^2)^{1/2}.(\sum_{i=1,n} |b_i|^2)^{1/2}$

See here https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch13.pdf (page reference number.293) for a proof for real numbers which is very easily generalised to complex numbers.

  • $\begingroup$ ...If, of course, the sums in question exist. $\endgroup$
    – 5xum
    Mar 3, 2016 at 11:07
  • $\begingroup$ @5xum. Well I did specify finite sequences. But, yes you are correct in that for infinite sequences there are some convergence requirements. $\endgroup$ Mar 3, 2016 at 11:10
  • $\begingroup$ Whoops, missed the "finite" part. Sorry about that. $\endgroup$
    – 5xum
    Mar 3, 2016 at 11:10
  • $\begingroup$ @5xum. Quite OK: as you see I forgot about the infinite part. $\endgroup$ Mar 3, 2016 at 11:11
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    $\begingroup$ It should be the contrary : your inequality is a very special case of Cauchy-Schwarz for $L^2$ spaces. $\endgroup$ Mar 3, 2016 at 11:27

Flip your argument around. The Cauchy-Schwarz inequality makes thinking about angles seem outmodish, vestigial, obsolete. You don't need protractors to do geometric analysis anymore. You just need algebra: just one quadratic inequality. So much of what your proof using $\cos \theta$ has been doing only requires a simple algebraic inequality, and neither $\cos$ nor $\theta$.

Why algebra is important is out of scope to this answer.

  • $\begingroup$ You should back up your claims here by e.g. giving an example or two of how to use the CS inequality to do things that we once needed angles to do. $\endgroup$ Mar 4, 2016 at 2:51
  • $\begingroup$ Also, I think you should be more specific. For example, if you're saying that we should replace the notion "angle between $x$ and $y$" with the value $$\frac{\langle x,y \rangle}{\|x\| \|y\|},$$ then you should make this explicit. $\endgroup$ Mar 4, 2016 at 2:52

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