# Proofs of the Cauchy-Schwarz Inequality?

How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?

• I've slightly edited the body your question in order to make it self-contained. I've also added the real-analysis tag. – t.b. Feb 24 '11 at 12:58
• I would try the book "The Cauchy-Schwarz Masterclass". – user3533 Feb 24 '11 at 13:03
• @user3533: excellent. I got that book. It seems that this textbooks also talks about few other inequalities that I was going to study. Do you know any other references that basically is about different kind of inequalities and their proofs? Thanks – Vafa Khalighi Feb 24 '11 at 13:30
• @Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/… – Qiaochu Yuan Feb 24 '11 at 13:35
• Your first question is essentially unanswerable, except, maybe, by "many"... – Mariano Suárez-Álvarez Feb 24 '11 at 22:52

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## Here's a simple proof:

$|\vec{x}\cdot\vec{y}| \leq \|\vec{x}\|\|\vec{y}\|$

Substitute $|\vec{x}\cdot\vec{y}| = \|\vec{x}\|\|\vec{y}\|\cos \theta$

$\|\vec{x}\|\|\vec{y}\|\cos \theta \leq \|\vec{x}\|\|\vec{y}\|$

Divide both sides by $\|\vec{x}\|\|\vec{y}\|$

$\cos \theta \leq 1$

-Hey, I was looking for a "more serious" proof!

Then here you are!

## Here's another simple proof:

This is projecting a vector to another one (Click the gif if it doesn't load):

You drag its end in a line that is perpendicular to the other vector. Then multiply the length of the new vector with the old vector.

Do you know what the multiplication is equal to? The dot product of the vectors

When you project that vector, its norm (length) becomes lower - or stays the same if one of them is a scalar multiple of the other one.

^^ That was the proof. Think about it.

Source: $3$Blue$1$Brown

Wait, I look for a "really serious" proof!

Here you are.

## Another proof:

Let $p(t)=||t\vec{y}-\vec{x}||^2$

As there's an absolute value, it must be equal to or bigger than $0$.

$p(t)=||t\vec{y}-\vec{x}||^2\geq 0$

$p(t)=(t\vec{y}-\vec{x})(t\vec{y}-\vec{x})\geq 0$

$p(t)=t^2(\vec{y}\cdot \vec{y})-2t(\vec{x}\cdot\vec{y})+\vec{x}\cdot \vec{x}\geq0$

Let's substitute some things.

$p(t)=t^2\underbrace{(\vec{y}\cdot \vec{y})}_\color{blue}{\large a}+t\underbrace{(-2\vec{x}\cdot\vec{y})}_\color{red}{\large b}+\underbrace{(\vec{x}\cdot \vec{x})}_\color{green}{\large c}\geq0$

$p(t)=\color{blue}{a}t^2+\color{red}{b}t+\color{green}{c}\geq0$

Its minimum value must be $\large \frac{-\color{red}{b}}{2\color{blue}{a}}$

Substituting $\large t= \frac{-\color{red}{b}}{2\color{blue}{a}}$

$p(\frac{-\color{red}{b}}{2\color{blue}{a}})=\color{blue}{a}(\frac{-\color{red}{b}}{2\color{blue}{a}})^2+\color{red}{b}(\frac{-\color{red}{b}}{2\color{blue}{a}})+\color{green}{c}\geq0$

$p(\frac{-\color{red}{b}}{2\color{blue}{a}})=\color{blue}{a}(\frac{\color{red}{b}^2}{4\color{blue}{a}^2})+\color{red}{b}(\frac{-\color{red}{b}}{2\color{blue}{a}})+\color{green}{c}\geq0$

$p(\frac{-\color{red}{b}}{2\color{blue}{a}})=\frac{\color{red}{b}^2}{4\color{blue}{a}}+\frac{-\color{red}{b}^2}{2\color{blue}{a}}+\color{green}{c}\geq0$

Forget the $\large p(t)$ function side (LHS)

$\frac{\color{red}{b}^2}{4\color{blue}{a}}+\frac{-\color{red}{b}^2}{2\color{blue}{a}}+\color{green}{c}\geq0$

Multiply by $\large 4\color{blue}{a}$

$\color{red}{b}^2-2\color{red}{b}^2+4\color{blue}{a}\color{green}{c}\geq0$

$-\color{red}{b}^2+4\color{blue}{a}\color{green}{c}\geq0$

$4\color{blue}{a}\color{green}{c}\geq \color{red}{b}^2$

De-substitute

$p(t)=t^2\underbrace{(\vec{y}\cdot \vec{y})}_\color{blue}{\large a}+t\underbrace{(-2\vec{x}\cdot\vec{y})}_\color{red}{\large b}+\underbrace{(\vec{x}\cdot \vec{x})}_\color{green}{\large c}\geq0$

$4\color{blue}{(\vec{y}\cdot \vec{y})}\color{green}{(\vec{x}\cdot \vec{x})}\geq \color{red}{(-2\vec{x}\cdot\vec{y})}^2$

Using the identity $\large \vec{v}\cdot\vec{v}=||\vec{v}||^2$

$4\color{blue}{||\vec{y}||^2}\color{green}{||\vec{x}||^2}\geq \color{red}{(-2\vec{x}\cdot\vec{y})}^2$

Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)\in\kume{R}$)

$4\color{blue}{||\vec{y}||^2}\color{green}{||\vec{x}||^2}\geq \color{red}{(|-2\vec{x}\cdot\vec{y}|)}^2$

As the both sides are not negative, you can square root both sides.

$2\color{blue}{||\vec{y}||}\color{green}{||\vec{x}||}\geq \color{red}{|-2\vec{x}\cdot\vec{y}|}$

$2\color{blue}{||\vec{y}||}\color{green}{||\vec{x}||}\geq \color{red}{2|\vec{x}\cdot\vec{y}|}$

$\large\color{blue}{||\vec{y}||}\color{green}{||\vec{x}||}\geq \color{red}{|\vec{x}\cdot\vec{y}|}$

This one was from KhanAcademy

Here is one:

Claim: $|\langle x,y \rangle| \leq \|x\|\|y\|$

Proof: If one of the two vectors is zero then both sides are zero so we may assume that both $x,y$ are non-zero. Let $t \in \mathbb C$. Then

\begin{align} 0 \leq \|x + ty \|^2 &= \langle x + ty, x + ty\rangle \\ &= \langle x,x\rangle + \langle x,t y\rangle + \langle yt, x\rangle + \langle ty,ty\rangle \\ &= \langle x,x\rangle + \bar{t} \langle x,y\rangle + t \overline{\langle x,y\rangle} + |t|^2 \langle y,y\rangle \\ &= \langle x,x\rangle + 2 \Re(t \overline{\langle x,y\rangle}) + |t|^2 \langle y,y\rangle \end{align}

Now choose $t := -\frac{\langle x, y \rangle}{\langle y, y \rangle}$. Then we get $$0 \leq \langle x,x\rangle + 2 \Re(- \frac{|\langle x,y\rangle|^2}{\langle y, y \rangle}) + \frac{|\langle x,y\rangle|^2}{\langle y, y \rangle} = \langle x, x \rangle - \frac{|\langle x,y\rangle|^2}{\langle y, y \rangle}$$

And hence $|\langle x,y \rangle| \leq \|x\|\|y\|$.

Note that if $y = \lambda x$ for $\lambda \in \mathbb C$ then equality holds: $$|\lambda|^2 |\langle x, x \rangle| = |\lambda|^2 \|x\|\|x\|$$

• I think that a proof of the Cauchy-Schwarz inequality should also include a discussion of the equality case (which is also straightforward from this argument). – t.b. Jul 2 '12 at 9:55
• @t.b. Like this? – Rudy the Reindeer Jul 2 '12 at 10:22
• This is half of what I had in mind. More interesting is the fact that if equality $\lvert\langle x,y\rangle\rvert = \lVert x\rVert \lVert y \rVert$ holds then $y = \lambda x$ or $x = 0$. – t.b. Jul 2 '12 at 10:26
• In other words: equality holds if and only if $x$ and $y$ are linearly dependent. – t.b. Jul 2 '12 at 10:33
• It doesn't help to use the angle, since that is defined in terms of the inner product. It's just that $\|x+ty\|^2=0$ iff $x+ty=0$ iff $x=-ty$ (with your choice of $t$). – wildildildlife Jul 2 '12 at 11:10

Here is a nice simple proof. Fix, $X,Y\in \mathbb{R}^n$ then we wish to show $$X\cdot Y \leq \|X\|\|Y\|$$ the trick is to construct a suitable vector $Z\in \mathbb{R}^n$ and then use the property of the dot product $Z\cdot Z \geq 0$. Take $$Z = \frac{X}{\|X\|}-\frac{Y}{\|Y\|}$$ then we compute $Z\cdot Z$ \begin{align} Z\cdot Z &= \frac{X\cdot X}{\|X\|^2}-2\frac{X\cdot Y}{\|X\|\|Y\|}+\frac{Y\cdot Y}{\|Y\|^2}\\ &=2 - 2\frac{X\cdot Y}{\|X\|\|Y\|} \end{align} then we use $Z\cdot Z \geq 0$ to write \begin{align} 2-2\frac{X\cdot Y}{\|X\|\|Y\|}\geq 0\\ 2\geq 2\frac{X\cdot Y}{\|X\|\|Y\|}\\ \|X\|\|Y\|\geq X\cdot Y \end{align} and we are done.

Here is the proof from Introductory Real Analysis'', Kolmogorov & Fomin, Silverman Translation. Assume all sums are from $$1$$ to $$n$$.

Lemma: $$( \sum_i x_i y_i )^2 = (\sum_i x_i^2)(\sum_i y_i^2) - \frac{1}{2} \sum_i \sum_j (x_iy_j -x_jy_i)^2$$

Proof of Cauchy-Schwarz: The third term in the Lemma is always non-positive, so clearly $$( \sum_i x_i y_i )^2 \leq (\sum_i x_i^2)(\sum_i y_i^2)$$ .

Proof of Lemma: The left hand side (LHS), and the right hand side (RHS) should be shown to be equal. For the LHS write $$\text{LHS} = ( \sum_i x_i y_i )^2 = (\sum_i x_i y_i)(\sum_j x_j y_j) = \sum_i\sum_j x_iy_ix_jy_j.$$ For the RHS write $$\text{RHS}= \frac{1}{2}(\sum_i x_i^2)(\sum_j y_j^2) +\frac{1}{2} (\sum_j x_j^2)(\sum_i y_i^2) - \frac{1}{2} \sum_i \sum_j (x_iy_j -x_jy_i)^2 \\ = \frac{1}{2}\sum_i\sum_j\left( x_i^2 y_j^2 + x_j^2y_i^2 - x_i^2 y_j^2 - x_j^2y_i^2 + 2 x_i y_i x_j y_j \right) = \sum_i\sum_j x_iy_ix_jy_j .$$ This shows that LHS$$=$$RHS and finishes the proof.

Without loss of generality, assume $\|y\|=1$. Write $x=\left<x,y\right>y+z$. Then $z$ is orthogonal to $y$, because $$\left<x,y\right>=\left<(\left<x,y\right>y+z),y\right>=\left<x,y\right>\left<y,y\right>+\left<z,y\right>,$$ indeed yields $\left<z,y\right>=0$. Hence $$\|x\|^2=\left<x,x\right>=|\left<x,y\right>|^2+\left<z,z\right>\geq |\left<x,y\right>|^2,$$ with equality iff $z= 0$, i.e. $x\in\mathbb{F}y$.

I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:

$\langle x,y\rangle=\langle y,x\rangle$

$\langle ax+y,z\rangle=a\langle x,z\rangle+\langle y,z\rangle$

$\langle x,x\rangle\geq0$

Then $0\leq\langle lx+y,lx+y\rangle=l^2\langle x,x\rangle+l\langle x,y\rangle+l\langle y,x\rangle+\langle y,y\rangle=l^2\langle x,x\rangle+2l\langle x,y\rangle+\langle y,y\rangle$

$Let\:a=\langle x,x\rangle, b=\langle x,y\rangle,c=\langle y,y\rangle$, then the equation becomes

$al^2+bl+c\geq0$

This is a quadratic equation in $l$ with at most 1 real root. Therefore

$b^2-4ac\leq 0$

$\implies4{\langle x,y\rangle}^2-4\langle x,x\rangle\langle y,y\rangle\leq 0$

$\implies{\langle x,y\rangle}^2\leq\langle x,x\rangle\langle y,y\rangle$

Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($