I'm an 8th grader. After browsing aops.com, a math contest website, I've seen a lot of problems solved by Cauchy Schwarz. I'm only in geometry (have not started learning trigonometry yet). So can anyone explain Cauchy Schwarz in layman's terms, as if you are explaining it to someone who has just started geo in 8th grade?

  • $\begingroup$ Do you know what an inner product/ the dot product is? $\endgroup$
    – user137731
    Nov 8, 2015 at 21:27
  • 5
    $\begingroup$ What is an inner product? $\endgroup$
    – Schneider
    Nov 8, 2015 at 21:28
  • $\begingroup$ The Cauchy Schwarz inequality is an inequality that has to do with norms (like absolute value) and inner products (like the dot product). If you've never heard of the dot product (or some other inner product) then I'm not sure that the CS inequality will be at all useful to you yet. $\endgroup$
    – user137731
    Nov 8, 2015 at 21:31
  • 3
    $\begingroup$ Sal's video is pretty easy to follow: khanacademy.org/math/linear-algebra/vectors_and_spaces/… $\endgroup$
    – Kyle Hale
    Nov 8, 2015 at 23:19
  • 3
    $\begingroup$ Cauchy Schwarz says that if $u$ and $v$ are vectors, the part of $v$ that is along $u$ is shorter than all of $v$, and vice versa. $\endgroup$ Nov 9, 2015 at 13:18

8 Answers 8


In geometry terms that you can understand, the Cauchy-Schwarz inequality says that:

Among all the parallelograms with sides a and b, the rectangle is the one with the largest area.

Usually you can use this inequality when you are looking for an upper (or lower) bound of an expression.

I wanted to give you an example, but my geometry studies are too far to remember an easy demonstration. Maybe you can ask somebody to give you a compass-and-straightedge demonstration of the equivalence between Cauchy-Schwarz and triangle inequalities.

  • 16
    $\begingroup$ +1 for the purely geometric interpretation that anyone can understand. Plus, I never independently saw this interpretation of Cauchy-Schwarz. $\endgroup$ Nov 9, 2015 at 16:43
  • 1
    $\begingroup$ Proof: the area of a parallelogram is base times height. Base is constant, but height is sin(theta) times the side length. The maximum value of sine is 1, where height is the side length, at theta = 90 degrees. If that angle is 90 degrees, the parallelogram is a rectangle. $\endgroup$
    – bjb568
    Nov 14, 2015 at 3:30
  • $\begingroup$ @N47 +1 nice answer $\endgroup$
    – Cloud JR K
    Jul 8, 2018 at 10:38

Since you said you browse AoPS, maybe you're interested in math contests. In math contests, the following forms of Cauchy-Schwarz (C-S) inequality are used:

For all $a_i,b_i\in\Bbb R$:

$$\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$$

Equality holds if and only if either $a_1=b_1k, a_2=b_2k,\ldots, a_n=b_nk$ for some $k\in\Bbb R$ or $a_1=a_2=\cdots=a_n=0$ or $b_1=b_2=\cdots=b_n=0$.

For all $a_i,b_i\in\Bbb R^+$:

$$\sqrt{a_1+a_2+\cdots+a_n}\sqrt{b_1+b_2+\cdots+b_n}\ge \sqrt{a_1b_1}+\sqrt{a_2b_2}+\cdots+\sqrt{a_nb_n}$$

Equality holds if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.

The following is also called Titu's lemma, or Engel's form of C-S inequality (to prove it, multiply both sides by $b_1+b_2+\cdots+b_n$ and apply the first form of C-S inequality):

For $b_i\in\Bbb R^+, a_i\in\Bbb R$:

$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$

with equality if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.

The more general is Hölder's inequality:

For all $a_i,b_i,c_i\in\Bbb R$:


$$\ge (a_1b_1c_1+a_2b_2c_2+\cdots+a_nb_nc_n)^3$$

The same holds for all $a_i,b_i,c_i,d_i\in\Bbb R$, etc. I.e., any number of arrays $a_i,b_i,c_i,d_i,\ldots$. In general, all these inequalities are called Hölder's inequalities.

  • $\begingroup$ What does the 2 over the 1 on the a and b mean? $\endgroup$
    – Schneider
    Nov 8, 2015 at 21:46
  • $\begingroup$ @3.14 $a_1^2=a_1\cdot a_1$. We're applying the square function on $a_1$ (here $1$ is an index, denoting the $1$'st member of the sequence $a_1,a_2,\ldots,a_n$). $\endgroup$
    – user236182
    Nov 8, 2015 at 21:48
  • 4
    $\begingroup$ @3.14 : I'd like to apologize on behalf of all of us here for the rude comment by another user (which, I'm glad to see, the moderators have deleted). I like this question very much, and I hope you'll stick around despite that poor welcome. $\endgroup$
    – user21467
    Nov 9, 2015 at 15:04
  • $\begingroup$ Maybe it would clarify this for the asker if there were an example using two or three dimensional vectors using $x$, $y$, and maybe $z$ instead of indicies. $\endgroup$ Nov 9, 2015 at 16:47
  • $\begingroup$ @ToddWilcox You can fully use these inequalities in contests without knowing what a vector is. And you can prove them without using vectors as well. $\endgroup$
    – user236182
    Nov 9, 2015 at 19:17

Imagine that someone is pushing you. By doing so, a force is acting on you. That force has a direction (pointing from the person pushing you toward yourself) and a magnitude (measured by a non-negative real number, the amount of force that person is using to push you). Direction plus magnitude is what is known as a vector. So, somebody is applying a vector upon you. Now, suppose that you choose some direction, without magnitude. Just a direction. Since a direction plus magnitude is a vector, a direction without magnitude is vector with a fixed magnitude (if it's fixed, you can't control it, so effectively it is not there). So, when I say that you choose a direction, I mean you choose a vector whose magnitude is normalized, meaning it's magnitude is 1. Imagine simply you standing in place, someone is pushing you, and you are pointing at some direction. The question is: what is the component of the force acting on you in the direction you are pointing at? Asking the same question: How much will you move in the direction you are pointing at as a result of the person pushing you (in a direction not necessarily the same as the direction you are pointing at)? The Cauchy-Schwarz inequality says: the absolute value of the component of a vector at any given unit direction is bounded by the magnitude of the vector. In your case, the absolute displacement in physical space that you will experience in the direction you are pointing at as a result of someone pushing you is never larger than the total force that person used in order to push you.

  • 2
    $\begingroup$ This answer boils down to its statement: "the absolute value of the component of a vector at any given unit direction is bounded by the magnitude of the vector". This is spot on - but for an 8-th grader, a picture would complete the story and make this a really good answer. $\endgroup$
    – Keith
    Nov 9, 2015 at 0:20
  • 3
    $\begingroup$ I'm having trouble imagining an 8th grader following a sentence like, "Since a direction plus magnitude is a vector, a direction without magnitude is vector with a fixed magnitude (if it's fixed, you can't control it, so effectively it is not there)." $\endgroup$
    – djechlin
    Nov 9, 2015 at 5:43
  • $\begingroup$ @djechlin maybe. I wonder what the 8th grader thinks of this discussion. $\endgroup$ Nov 9, 2015 at 6:46
  • $\begingroup$ "A direction plus a magnitude is a vector, so a direction without magnitude is still a vector, only it has magnitude, which is 'fixed', which means it's not there." $\endgroup$
    – djechlin
    Nov 9, 2015 at 15:27
  • 1
    $\begingroup$ "A vector is a magnitude plus a direction, but if we want to talk about just directions, we can consider all of the vectors with a certain magnitude (for instance 1, since it's convenient), leaving only the direction to vary. For any direction, there's a unique vector with magnitude 1 in that direction." $\endgroup$
    – hobbs
    Nov 10, 2015 at 23:20

We denote by $x^{2}$ the number $x\cdot x$. We denote by $s\in S$ the fact that an element $s$ belongs to a set $S$. For example, $1\in\mathbb{R}$: $1$ belongs to the set of all real numbers.

We will first define some concepts.

To stay simple, we will restrict to a euclidean space $\mathbb{R}^{n}$, that is $$\mathbb{R}^{n}:=\{(x_{1},x_{2},\dots,x_{n})\,\vert\,x_{1},x_{2},\dots,x_{n}\in\mathbb{R}\}$$ The elements $(x_{1},x_{2},\dots,x_{n})$ are called vectors or points of $\mathbb{R}^{n}$ and have the following properties:

  • $(x_{1},x_{2},\dots,x_{n})+(y_{1},y_{2},\dots,y_{n})=(x_{1}+y_{1},x_{2}+y_{2},\dots,x_{n}+y_{n})$
  • $\lambda(x_{1},x_{2},\dots,x_{n})=(\lambda x_{1},\lambda x_{2},\dots,\lambda x_{n})$ where $\lambda$ is some real number.

The terms $x_{1},x_{2},\dots,x_{n}$ are called the components of the vector $(x_{1},x_{2},\dots,x_{n})$.

Example: $\mathbb{R}^{2}$. It is often called "cartesian plane". You probably denote by $(x,y)$ an element of the cartesian plane. As $n$ (in $\mathbb{R}^{n}$) can be arbitrarly big, you may eventually not have enough letters in the alphabet to denote all the components. It is why we prefer the notation $(x_{1},x_{2})$ instead of $(x,y)$. Be aware that $x_{1}$ is the abscissa of $(x_{1},x_{2})$ and $x_{2}$ is the ordinate of $(x_{1},x_{2})$. For example, $(1,2)$ has $1$ for abscissa and $2$ for ordinate. We can represent it as follows:


Actually, $\mathbb{R}^{n}$ is called euclidean when it is equipped with a euclidean distance. Let's define it.

Let $X,Y$ be two points in $\mathbb{R}^{n}$, such that $X=(x_{1},x_{2},\dots,x_{n})$ and $Y=(y_{1},y_{2},\dots,y_{n})$. The euclidean distance between $X$ and $Y$ is defined as follows:

$$\begin{align}d_{E}(X,Y) &=d_{E}((x_{1},x_{2},\dots,x_{n})\,,\,(y_{1},y_{2},\dots,y_{n}))\\ &=\sqrt{(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+\dots+(x_{n}-y_{n})^{2}} \end{align}$$

This is the usual distance you can think about: "the distance as the crow flies between $X$ and $Y$".

Example: $\mathbb{R}^{2}$ equipped with $d_{E}((x_{1},x_{2})\,,\,(y_{1},y_{2}))=\sqrt{(x_{1}-y_{1})^{2}+(y_{1}-y_{2})^{2}}$ is a euclidean space. If we take $(x_{1},x_{2})=(1,2)$ and $(y_{1},y_{2})=(3,-1)$, the distance between $(1,2)$ and $(3,-1)$ is $$d_{E}((1,2)\,,\,(3,-1))=\sqrt{(1-3)^{2}+(2-(-1))^{2}}=\sqrt{(-2)^{2}+3^{2}}=\sqrt{13}$$ Visually, it gives:


We can define a very useful concept: the scalar product (also called "dot product" or "inner product"):

The scalar product between $X$ and $Y$ is defined as follows:

$$\begin{align}\langle X,Y\rangle &=\left\langle (x_{1},x_{2},\dots,x_{n})\,,\,(y_{1},y_{2},\dots,y_{n})\right\rangle\\ &=x_{1}\cdot y_{1}+x_{2}\cdot y_{2}+\dots+x_{n}\cdot y_{n} \end{align}$$

where $x_{i}\cdot y_{i}$ is the usual multiplication of the components $x_{i}$ and $y_{i}$ (for all $i=1,2,\dots,n$).

Example: If we take $(1,2)$ and $(3,-1)$ in $\mathbb{R}^{2}$, $\langle (1,2)\,,\,(3,-1)\rangle=1\cdot 3+2\cdot(-1)=3-2=1$. It admits a geometric interpretation but we will not develop it here. It has very interesting properties, but we will not develop them here.

We will call a norm of the vector $X$ and we will denote it $\Vert X\Vert=\Vert(x_{1},x_{2},\dots,x_{n})\Vert$ the square root of the scalar product between $X$ and itself:

$$\Vert X\Vert=\sqrt{\langle X\,,\,X\rangle}$$

Example: In $\mathbb{R}^{2}$, the norm of $(1,2)$ is $\Vert (1,2)\Vert\sqrt{\langle (1,2)\,,\,(1,2)\rangle}=\sqrt{1\cdot 1+2\cdot 2}=\sqrt{5}$. You can compute the norm of $(3,-1)$, which is $\Vert (3,-1)\Vert=\sqrt{10}$. Visually, the norm of $(x_{1},x_{2})$ is the euclidean distance between $(x,y)$ and $(0,0)$:


You have know all the required concepts to understand what Cauchy-Schwarz is. We denote by $\vert a\vert$ the absolute value of $a$.

If $X,Y$ are two vectors in a euclidean space $\mathbb{R}^{n}$, then we have the following relation, known as the "Cauchy-Schwarz inequality":

$$\vert\langle X\,,\,Y\rangle\vert\le\Vert X\Vert\cdot\Vert Y\Vert$$

Which is read "the scalar product between $X$ and $Y$ is less or equal than the product of the norm of $X$ by the norm of $Y$".

Example: In $\mathbb{R}^{2}$, we have shown that $\langle (1,2)\,,\,(3,-1)\rangle=1$ but $\Vert (1,2)\Vert=\sqrt{5}$ and $\Vert (3,-1)\Vert=\sqrt{10}$, so that $1\le\sqrt{5}\sqrt{10}=5\sqrt{2}$ and the Cauchy-Schwarz inequality is verified.

You can actually define the previous concepts in a more general context, but it requires more time (and more concepts.

  • 1
    $\begingroup$ Nice LAAARGE intelligently coloured drawings with displayed unit squares: perfect, +1. $\endgroup$ Nov 9, 2015 at 10:35
  • $\begingroup$ An excellent introduction. $\endgroup$
    – Titus
    Nov 11, 2015 at 8:19

Skip if you know what dot products are

Let's say you have two vectors and you want to know how much they line up with each other.

You want something that's really big if they point the same direction and really small if they point in the opposite direction.

Enter the dot product: $$ \vec{a} \cdot \vec{b} = \left(|\vec{a}|\right) \left(|\vec{b}|\right)\cos{\theta} $$

...where $\theta$ is the angle between the vectors and where $|\vec{v}|$ is the length of $\vec{v}$.

The dot product does other stuff too, but this is pretty much all you need to understand CS.

Stop skipping!

The Cauchy-Schwarz inequality concerns the absolute value of the dot product. Because a length must be positive, only the cosine can make the dot product negative. Therefore: $$ |\vec{a} \cdot \vec{b}| = \left(|\vec{a}|\right) \left(|\vec{b}|\right)|\cos{\theta}| $$

Because $0\leq|\cos{\theta}|\leq1$, it is pretty clear that: $$ \left(|\vec{a}|\right) \left(|\vec{b}|\right)|\cos{\theta}| \leq \left(|\vec{a}|\right) \left(|\vec{b}|\right) $$

Or equivalently, $$ |\vec{a} \cdot \vec{b}| \leq \left(|\vec{a}|\right) \left(|\vec{b}|\right) $$

And that's the Cauchy-Schwarz inequality.

To see it in action, search google for a dot product applet (or use this one). Make both vectors have length 1. 1*1 is 1, so CS tells us the maximum magnitude of the dot product is 1. If you swing one of the vectors around (without changing its length), does this seem legit?

So what?

On its own, the CS inequality tells you that two vectors can't line up more than two parallel vectors. Eventually this will seem pretty elementary if it doesn't already.

However, it is commonly used to prove a lot of other, less easy-to-show things. Take a look at this wikipedia article for more.

  • $\begingroup$ I believe you have an error sir. The second equation after you say "stop skipping" You include the cosine in the equation. The purpose of the inequality is to extend the concept of the dot product into n dimensions and the cosine proofs of CS are careful not to reference the angle in n dimensions. The expression may be defeating the purpose. If your proof was correct it would be splashed onto every text book in multivariable calculus instead of the more difficult proofs I am seeing, you may wish to revisit your proof or show me where I am not correct? Thank you $\endgroup$
    – Sedumjoy
    May 14, 2017 at 22:26
  • $\begingroup$ You're right. This isn't a proof, but I hope it still provides some intuition $\endgroup$ May 15, 2017 at 12:46

This answer uses a tiny bit of trigonometry, in the form of the cosine function. All you need to know about the cosine function is that $\cos \phi$ increases from $-1$ to $1$ as $\phi$ goes from $180^\circ$ to $0^\circ$, hitting zero when $\phi$ is $\pm 90^\circ$.

Pick a point in space and call it "the origin." A straight arrow starting at the origin is called a vector. Lots of other things can be called vectors too, but arrows in space are the examples that motivate the whole business.

Two fundamental ideas in geometry are the ideas of length and angle. When talking about arrows, geometers like to package length and angle into a single measurement, called the inner product. The inner product $\langle v, w \rangle$ of two arrows $v$ and $w$ can be defined as $$\|v\|\|w\| \cos \theta_{v,w},$$ where $\|v\|$ is the length of $v$, $\|w\|$ is the length of $w$, and $\theta_{v,w}$ is the angle between $v$ and $w$. The Cauchy-Schwarz inequality is the fact that $$\langle v, w \rangle \le \|v\| \|w\|.$$

The Cauchy-Schwarz inequality is totally obvious from the way we defined the inner product, so why does it deserve a name? For three reasons:

  • The Cauchy-Schwarz inequality is equivalent to the triangle inequality, a fundamental fact about how distances work.
  • There are other ways to define the inner product between arrows in space, and most of them make the Cauchy-Schwarz inequality much less obvious.
  • Things other than arrows in space can be called vectors, and the Cauchy-Schwarz inequality is true for them too. For example, sound waves can be called vectors. The "length" of a sound wave is its total energy, and the "angle" between two sound waves is a measurement of how they interfere. The triangle inequality is the fact that when two sounds interfere, the energy of the combined sound can't be greater than the total energy of the two original sounds.
  • 1
    $\begingroup$ Note the asker does clarify that they haven't learned Trigonometry yet. $\endgroup$ Nov 9, 2015 at 16:44
  • $\begingroup$ @ToddWilcox Nonetheless, this might be motivation to learn some. To have C-S without trigonometry would be unusual, I think. Geometric interpretations lead quickly to trigonometry, and if this is out of contests then learning some trigonometry as an adjunct to geometry will rapidly become essential. $\endgroup$ Nov 10, 2015 at 21:12
  • 1
    $\begingroup$ @ToddWilcox: Whoops, I missed that! I'll edit a bit. $\endgroup$
    – Vectornaut
    Nov 10, 2015 at 23:57

Cauchy's inequality is (the square of) the triangle inequality for vectors (in $n$-dimensional space) with Pythagorean distance, $|x| = \sqrt \sum x_i^2$. Schwarz contributed a nice proof that generalizes well.


The most easy example I know, even useful in everyday life:

2 same rectangles each made out of edge length of a and b

are smaller or equal to

2 squares, each made of a and b separately.

Or in ever day terms: Squares make the best use of edge length to gain area.

General: |⟨X,Y⟩|≤∥X∥⋅∥Y∥

in a more concrete form (case):

|ab| ≤ |a||b|

... more concrete (rectangle vs square - case):

2 *(a * b) ≤ a * a + b * b


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .