Mixing
Proofs of the Cauchy-Schwarz Inequality?
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How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?
real-analysis inequality big-list inner-product-space
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show 3 more comments
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How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?
real-analysis inequality big-list inner-product-space
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I've slightly edited the body your question in order to make it self-contained. I've also added thereal-analysistag.
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– t.b.
Feb 24 '11 at 12:58
14
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I would try the book "The Cauchy-Schwarz Masterclass".
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– user3533
Feb 24 '11 at 13:03
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@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
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– Vafa Khalighi
Feb 24 '11 at 13:30
2
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@Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/…
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– Qiaochu Yuan
Feb 24 '11 at 13:35
3
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Your first question is essentially unanswerable, except, maybe, by "many"...
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– Mariano Suárez-Álvarez
Feb 24 '11 at 22:52
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show 3 more comments
$begingroup$
How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?
real-analysis inequality big-list inner-product-space
$endgroup$
How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?
real-analysis inequality big-list inner-product-space
real-analysis inequality big-list inner-product-space
edited Jul 2 '12 at 9:55
community wiki
5 revs, 2 users 67%
Vafa Khalighi
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I've slightly edited the body your question in order to make it self-contained. I've also added thereal-analysistag.
$endgroup$
– t.b.
Feb 24 '11 at 12:58
14
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I would try the book "The Cauchy-Schwarz Masterclass".
$endgroup$
– user3533
Feb 24 '11 at 13:03
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@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
$endgroup$
– Vafa Khalighi
Feb 24 '11 at 13:30
2
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@Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/…
$endgroup$
– Qiaochu Yuan
Feb 24 '11 at 13:35
3
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Your first question is essentially unanswerable, except, maybe, by "many"...
$endgroup$
– Mariano Suárez-Álvarez
Feb 24 '11 at 22:52
|
show 3 more comments
$begingroup$
I've slightly edited the body your question in order to make it self-contained. I've also added thereal-analysistag.
$endgroup$
– t.b.
Feb 24 '11 at 12:58
14
$begingroup$
I would try the book "The Cauchy-Schwarz Masterclass".
$endgroup$
– user3533
Feb 24 '11 at 13:03
$begingroup$
@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
$endgroup$
– Vafa Khalighi
Feb 24 '11 at 13:30
2
$begingroup$
@Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/…
$endgroup$
– Qiaochu Yuan
Feb 24 '11 at 13:35
3
$begingroup$
Your first question is essentially unanswerable, except, maybe, by "many"...
$endgroup$
– Mariano Suárez-Álvarez
Feb 24 '11 at 22:52
$begingroup$
I've slightly edited the body your question in order to make it self-contained. I've also added the
real-analysis tag.$endgroup$
– t.b.
Feb 24 '11 at 12:58
$begingroup$
I've slightly edited the body your question in order to make it self-contained. I've also added the
real-analysis tag.$endgroup$
– t.b.
Feb 24 '11 at 12:58
14
14
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I would try the book "The Cauchy-Schwarz Masterclass".
$endgroup$
– user3533
Feb 24 '11 at 13:03
$begingroup$
I would try the book "The Cauchy-Schwarz Masterclass".
$endgroup$
– user3533
Feb 24 '11 at 13:03
$begingroup$
@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
$endgroup$
– Vafa Khalighi
Feb 24 '11 at 13:30
$begingroup$
@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
$endgroup$
– Vafa Khalighi
Feb 24 '11 at 13:30
2
2
$begingroup$
@Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/…
$endgroup$
– Qiaochu Yuan
Feb 24 '11 at 13:35
$begingroup$
@Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/…
$endgroup$
– Qiaochu Yuan
Feb 24 '11 at 13:35
3
3
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Your first question is essentially unanswerable, except, maybe, by "many"...
$endgroup$
– Mariano Suárez-Álvarez
Feb 24 '11 at 22:52
$begingroup$
Your first question is essentially unanswerable, except, maybe, by "many"...
$endgroup$
– Mariano Suárez-Álvarez
Feb 24 '11 at 22:52
|
show 3 more comments
6 Answers
6
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oldest
votes
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}newcommand{i}{mathrm{i}}newcommand{text}[1]{mathrm{#1}}newcommand{root}[2]{^{#2}sqrt[#1]} newcommand{derivative}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{abs}[1]{leftvert,{#1},rightvert}newcommand{x}[0]{times}newcommand{summ}[3]{sum^{#2}_{#1}#3}newcommand{s}[0]{space}newcommand{i}[0]{mathrm{i}}newcommand{kume}[1]{mathbb{#1}}newcommand{bold}[1]{textbf{#1}}newcommand{italic}[1]{textit{#1}}newcommand{kumedigerBETA}[1]{rm #1!#1}$
Here's a simple proof:
$|vec{x}cdotvec{y}| leq |vec{x}||vec{y}| $
Substitute $|vec{x}cdotvec{y}| = |vec{x}||vec{y}|cos theta$
$|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 $p(t)=||tvec{y}-vec{x}||^2$
As there's an absolute value, it must be equal to or bigger than $0$.
$p(t)=||tvec{y}-vec{x}||^2geq 0$
$p(t)=(tvec{y}-vec{x})(tvec{y}-vec{x})geq 0$
$p(t)=t^2(vec{y}cdot vec{y})-2t(vec{x}cdotvec{y})+vec{x}cdot vec{x}geq0$
Let's substitute some things.
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$p(t)=color{blue}{a}t^2+color{red}{b}t+color{green}{c}geq0$
Its minimum value must be $large frac{-color{red}{b}}{2color{blue}{a}}$
Substituting $large t= frac{-color{red}{b}}{2color{blue}{a}}$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{-color{red}{b}}{2color{blue}{a}})^2+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{color{red}{b}^2}{4color{blue}{a}^2})+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Forget the $large p(t)$ function side (LHS)
$frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Multiply by $large 4color{blue}{a}$
$color{red}{b}^2-2color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$-color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$4color{blue}{a}color{green}{c}geq color{red}{b}^2$
De-substitute
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$4color{blue}{(vec{y}cdot vec{y})}color{green}{(vec{x}cdot vec{x})}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $large vec{v}cdotvec{v}=||vec{v}||^2$
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkume{R}$)
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(|-2vec{x}cdotvec{y}|)}^2$
As the both sides are not negative, you can square root both sides.
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|-2vec{x}cdotvec{y}|}$
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{2|vec{x}cdotvec{y}|}$
$largecolor{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|vec{x}cdotvec{y}|}$
This one was from KhanAcademy
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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
$$ begin{align}
0 leq |x + ty |^2 &= langle x + ty, x + tyrangle \
&= langle x,xrangle + langle x,t yrangle + langle yt, xrangle + langle ty,tyrangle \
&= langle x,xrangle + bar{t} langle x,yrangle + t overline{langle x,yrangle} + |t|^2 langle y,yrangle \
&= langle x,xrangle + 2 Re(t overline{langle x,yrangle}) + |t|^2 langle y,yrangle
end{align}$$
Now choose $t := -frac{langle x, y rangle}{langle y, y rangle}$. Then we get
$$ 0 leq langle x,xrangle + 2 Re(- frac{|langle x,yrangle|^2}{langle y, y rangle}) + frac{|langle x,yrangle|^2}{langle y, y rangle} = langle x, x rangle - frac{|langle x,yrangle|^2}{langle y, y rangle}$$
And hence $|langle x,y rangle| leq |x||y| $.
Note that if $y = lambda x$ for $lambda in mathbb C$ then equality holds:
$$ |lambda|^2 |langle x, x rangle| = |lambda|^2 |x||x| $$
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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).
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– t.b.
Jul 2 '12 at 9:55
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@t.b. Like this?
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– Rudy the Reindeer
Jul 2 '12 at 10:22
1
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This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
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– t.b.
Jul 2 '12 at 10:26
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In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
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– t.b.
Jul 2 '12 at 10:33
1
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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 $t$).
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– wildildildlife
Jul 2 '12 at 11:10
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show 4 more comments
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Here is a nice simple proof. Fix, $X,Yin mathbb{R}^n$ then we wish to show
$$
Xcdot Y leq |X||Y|
$$
the trick is to construct a suitable vector $Zin mathbb{R}^n$ and then use the property of the dot product $Zcdot Z geq 0$. Take
$$
Z = frac{X}{|X|}-frac{Y}{|Y|}
$$
then we compute $Zcdot Z$
begin{align}
Zcdot Z &= frac{Xcdot X}{|X|^2}-2frac{Xcdot Y}{|X||Y|}+frac{Ycdot Y}{|Y|^2}\
&=2 - 2frac{Xcdot Y}{|X||Y|}
end{align}
then we use $Zcdot Z geq 0$ to write
begin{align}
2-2frac{Xcdot Y}{|X||Y|}geq 0\
2geq 2frac{Xcdot Y}{|X||Y|}\
|X||Y|geq Xcdot Y
end{align}
and we are done.
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Without loss of generality, assume $|y|=1$. Write $x=left<x,yright>y+z$. Then $z$ is orthogonal to $y$, because
$$left<x,yright>=left<(left<x,yright>y+z),yright>=left<x,yright>left<y,yright>+left<z,yright>,$$
indeed yields $left<z,yright>=0$. Hence
$$|x|^2=left<x,xright>=|left<x,yright>|^2+left<z,zright>geq |left<x,yright>|^2,$$
with equality iff $z= 0$, i.e. $xinmathbb{F}y$.
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I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:
$langle x,yrangle=langle y,xrangle$
$langle ax+y,zrangle=alangle x,zrangle+langle y,zrangle$
$langle x,xranglegeq0$
Then
$0leqlangle lx+y,lx+yrangle=l^2langle x,xrangle+llangle x,yrangle+llangle y,xrangle+langle y,yrangle=l^2langle x,xrangle+2llangle x,yrangle+langle y,yrangle$
$Let:a=langle x,xrangle, b=langle x,yrangle,c=langle y,yrangle$, then the equation becomes
$al^2+bl+cgeq0$
This is a quadratic equation in $l$ with at most 1 real root. Therefore
$b^2-4acleq 0$
$implies4{langle x,yrangle}^2-4langle x,xranglelangle y,yrangleleq 0$
$implies{langle x,yrangle}^2leqlangle x,xranglelangle y,yrangle$
Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($
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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
$$ text{LHS} = ( sum_i x_i y_i )^2 = (sum_i x_i y_i)(sum_j x_j y_j) = sum_isum_j x_iy_ix_jy_j. $$
For the RHS write
$$
text{RHS}=
frac{1}{2}(sum_i x_i^2)(sum_j y_j^2)
+frac{1}{2} (sum_j x_j^2)(sum_i y_i^2)
- frac{1}{2} sum_i sum_j (x_iy_j -x_jy_i)^2
\ =
frac{1}{2}sum_isum_jleft(
x_i^2 y_j^2 + x_j^2y_i^2 - x_i^2 y_j^2 - x_j^2y_i^2 + 2 x_i y_i x_j y_j
right)
=
sum_isum_j x_iy_ix_jy_j .
$$
This shows that LHS$=$RHS and finishes the proof.
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add a comment |
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6 Answers
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6 Answers
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}newcommand{i}{mathrm{i}}newcommand{text}[1]{mathrm{#1}}newcommand{root}[2]{^{#2}sqrt[#1]} newcommand{derivative}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{abs}[1]{leftvert,{#1},rightvert}newcommand{x}[0]{times}newcommand{summ}[3]{sum^{#2}_{#1}#3}newcommand{s}[0]{space}newcommand{i}[0]{mathrm{i}}newcommand{kume}[1]{mathbb{#1}}newcommand{bold}[1]{textbf{#1}}newcommand{italic}[1]{textit{#1}}newcommand{kumedigerBETA}[1]{rm #1!#1}$
Here's a simple proof:
$|vec{x}cdotvec{y}| leq |vec{x}||vec{y}| $
Substitute $|vec{x}cdotvec{y}| = |vec{x}||vec{y}|cos theta$
$|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 $p(t)=||tvec{y}-vec{x}||^2$
As there's an absolute value, it must be equal to or bigger than $0$.
$p(t)=||tvec{y}-vec{x}||^2geq 0$
$p(t)=(tvec{y}-vec{x})(tvec{y}-vec{x})geq 0$
$p(t)=t^2(vec{y}cdot vec{y})-2t(vec{x}cdotvec{y})+vec{x}cdot vec{x}geq0$
Let's substitute some things.
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$p(t)=color{blue}{a}t^2+color{red}{b}t+color{green}{c}geq0$
Its minimum value must be $large frac{-color{red}{b}}{2color{blue}{a}}$
Substituting $large t= frac{-color{red}{b}}{2color{blue}{a}}$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{-color{red}{b}}{2color{blue}{a}})^2+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{color{red}{b}^2}{4color{blue}{a}^2})+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Forget the $large p(t)$ function side (LHS)
$frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Multiply by $large 4color{blue}{a}$
$color{red}{b}^2-2color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$-color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$4color{blue}{a}color{green}{c}geq color{red}{b}^2$
De-substitute
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$4color{blue}{(vec{y}cdot vec{y})}color{green}{(vec{x}cdot vec{x})}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $large vec{v}cdotvec{v}=||vec{v}||^2$
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkume{R}$)
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(|-2vec{x}cdotvec{y}|)}^2$
As the both sides are not negative, you can square root both sides.
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|-2vec{x}cdotvec{y}|}$
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{2|vec{x}cdotvec{y}|}$
$largecolor{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|vec{x}cdotvec{y}|}$
This one was from KhanAcademy
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}newcommand{i}{mathrm{i}}newcommand{text}[1]{mathrm{#1}}newcommand{root}[2]{^{#2}sqrt[#1]} newcommand{derivative}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{abs}[1]{leftvert,{#1},rightvert}newcommand{x}[0]{times}newcommand{summ}[3]{sum^{#2}_{#1}#3}newcommand{s}[0]{space}newcommand{i}[0]{mathrm{i}}newcommand{kume}[1]{mathbb{#1}}newcommand{bold}[1]{textbf{#1}}newcommand{italic}[1]{textit{#1}}newcommand{kumedigerBETA}[1]{rm #1!#1}$
Here's a simple proof:
$|vec{x}cdotvec{y}| leq |vec{x}||vec{y}| $
Substitute $|vec{x}cdotvec{y}| = |vec{x}||vec{y}|cos theta$
$|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 $p(t)=||tvec{y}-vec{x}||^2$
As there's an absolute value, it must be equal to or bigger than $0$.
$p(t)=||tvec{y}-vec{x}||^2geq 0$
$p(t)=(tvec{y}-vec{x})(tvec{y}-vec{x})geq 0$
$p(t)=t^2(vec{y}cdot vec{y})-2t(vec{x}cdotvec{y})+vec{x}cdot vec{x}geq0$
Let's substitute some things.
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$p(t)=color{blue}{a}t^2+color{red}{b}t+color{green}{c}geq0$
Its minimum value must be $large frac{-color{red}{b}}{2color{blue}{a}}$
Substituting $large t= frac{-color{red}{b}}{2color{blue}{a}}$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{-color{red}{b}}{2color{blue}{a}})^2+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{color{red}{b}^2}{4color{blue}{a}^2})+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Forget the $large p(t)$ function side (LHS)
$frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Multiply by $large 4color{blue}{a}$
$color{red}{b}^2-2color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$-color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$4color{blue}{a}color{green}{c}geq color{red}{b}^2$
De-substitute
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$4color{blue}{(vec{y}cdot vec{y})}color{green}{(vec{x}cdot vec{x})}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $large vec{v}cdotvec{v}=||vec{v}||^2$
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkume{R}$)
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(|-2vec{x}cdotvec{y}|)}^2$
As the both sides are not negative, you can square root both sides.
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|-2vec{x}cdotvec{y}|}$
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{2|vec{x}cdotvec{y}|}$
$largecolor{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|vec{x}cdotvec{y}|}$
This one was from KhanAcademy
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add a comment |
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}newcommand{i}{mathrm{i}}newcommand{text}[1]{mathrm{#1}}newcommand{root}[2]{^{#2}sqrt[#1]} newcommand{derivative}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{abs}[1]{leftvert,{#1},rightvert}newcommand{x}[0]{times}newcommand{summ}[3]{sum^{#2}_{#1}#3}newcommand{s}[0]{space}newcommand{i}[0]{mathrm{i}}newcommand{kume}[1]{mathbb{#1}}newcommand{bold}[1]{textbf{#1}}newcommand{italic}[1]{textit{#1}}newcommand{kumedigerBETA}[1]{rm #1!#1}$
Here's a simple proof:
$|vec{x}cdotvec{y}| leq |vec{x}||vec{y}| $
Substitute $|vec{x}cdotvec{y}| = |vec{x}||vec{y}|cos theta$
$|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 $p(t)=||tvec{y}-vec{x}||^2$
As there's an absolute value, it must be equal to or bigger than $0$.
$p(t)=||tvec{y}-vec{x}||^2geq 0$
$p(t)=(tvec{y}-vec{x})(tvec{y}-vec{x})geq 0$
$p(t)=t^2(vec{y}cdot vec{y})-2t(vec{x}cdotvec{y})+vec{x}cdot vec{x}geq0$
Let's substitute some things.
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$p(t)=color{blue}{a}t^2+color{red}{b}t+color{green}{c}geq0$
Its minimum value must be $large frac{-color{red}{b}}{2color{blue}{a}}$
Substituting $large t= frac{-color{red}{b}}{2color{blue}{a}}$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{-color{red}{b}}{2color{blue}{a}})^2+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{color{red}{b}^2}{4color{blue}{a}^2})+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Forget the $large p(t)$ function side (LHS)
$frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Multiply by $large 4color{blue}{a}$
$color{red}{b}^2-2color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$-color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$4color{blue}{a}color{green}{c}geq color{red}{b}^2$
De-substitute
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$4color{blue}{(vec{y}cdot vec{y})}color{green}{(vec{x}cdot vec{x})}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $large vec{v}cdotvec{v}=||vec{v}||^2$
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkume{R}$)
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(|-2vec{x}cdotvec{y}|)}^2$
As the both sides are not negative, you can square root both sides.
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|-2vec{x}cdotvec{y}|}$
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{2|vec{x}cdotvec{y}|}$
$largecolor{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|vec{x}cdotvec{y}|}$
This one was from KhanAcademy
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$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}newcommand{i}{mathrm{i}}newcommand{text}[1]{mathrm{#1}}newcommand{root}[2]{^{#2}sqrt[#1]} newcommand{derivative}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{abs}[1]{leftvert,{#1},rightvert}newcommand{x}[0]{times}newcommand{summ}[3]{sum^{#2}_{#1}#3}newcommand{s}[0]{space}newcommand{i}[0]{mathrm{i}}newcommand{kume}[1]{mathbb{#1}}newcommand{bold}[1]{textbf{#1}}newcommand{italic}[1]{textit{#1}}newcommand{kumedigerBETA}[1]{rm #1!#1}$
Here's a simple proof:
$|vec{x}cdotvec{y}| leq |vec{x}||vec{y}| $
Substitute $|vec{x}cdotvec{y}| = |vec{x}||vec{y}|cos theta$
$|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 $p(t)=||tvec{y}-vec{x}||^2$
As there's an absolute value, it must be equal to or bigger than $0$.
$p(t)=||tvec{y}-vec{x}||^2geq 0$
$p(t)=(tvec{y}-vec{x})(tvec{y}-vec{x})geq 0$
$p(t)=t^2(vec{y}cdot vec{y})-2t(vec{x}cdotvec{y})+vec{x}cdot vec{x}geq0$
Let's substitute some things.
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$p(t)=color{blue}{a}t^2+color{red}{b}t+color{green}{c}geq0$
Its minimum value must be $large frac{-color{red}{b}}{2color{blue}{a}}$
Substituting $large t= frac{-color{red}{b}}{2color{blue}{a}}$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{-color{red}{b}}{2color{blue}{a}})^2+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=color{blue}{a}(frac{color{red}{b}^2}{4color{blue}{a}^2})+color{red}{b}(frac{-color{red}{b}}{2color{blue}{a}})+color{green}{c}geq0$
$p(frac{-color{red}{b}}{2color{blue}{a}})=frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Forget the $large p(t)$ function side (LHS)
$frac{color{red}{b}^2}{4color{blue}{a}}+frac{-color{red}{b}^2}{2color{blue}{a}}+color{green}{c}geq0$
Multiply by $large 4color{blue}{a}$
$color{red}{b}^2-2color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$-color{red}{b}^2+4color{blue}{a}color{green}{c}geq0$
$4color{blue}{a}color{green}{c}geq color{red}{b}^2$
De-substitute
$p(t)=t^2underbrace{(vec{y}cdot vec{y})}_color{blue}{large a}+tunderbrace{(-2vec{x}cdotvec{y})}_color{red}{large b}+underbrace{(vec{x}cdot vec{x})}_color{green}{large c}geq0$
$4color{blue}{(vec{y}cdot vec{y})}color{green}{(vec{x}cdot vec{x})}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $large vec{v}cdotvec{v}=||vec{v}||^2$
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(-2vec{x}cdotvec{y})}^2$
Using the identity $(f(x))^2=(|f(x)|)^2$ (where $f(x)inkume{R}$)
$4color{blue}{||vec{y}||^2}color{green}{||vec{x}||^2}geq color{red}{(|-2vec{x}cdotvec{y}|)}^2$
As the both sides are not negative, you can square root both sides.
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|-2vec{x}cdotvec{y}|}$
$2color{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{2|vec{x}cdotvec{y}|}$
$largecolor{blue}{||vec{y}||}color{green}{||vec{x}||}geq color{red}{|vec{x}cdotvec{y}|}$
This one was from KhanAcademy
edited Sep 4 '17 at 16:31
community wiki
2 revs
MCCCS
add a comment |
add a comment |
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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
$$ begin{align}
0 leq |x + ty |^2 &= langle x + ty, x + tyrangle \
&= langle x,xrangle + langle x,t yrangle + langle yt, xrangle + langle ty,tyrangle \
&= langle x,xrangle + bar{t} langle x,yrangle + t overline{langle x,yrangle} + |t|^2 langle y,yrangle \
&= langle x,xrangle + 2 Re(t overline{langle x,yrangle}) + |t|^2 langle y,yrangle
end{align}$$
Now choose $t := -frac{langle x, y rangle}{langle y, y rangle}$. Then we get
$$ 0 leq langle x,xrangle + 2 Re(- frac{|langle x,yrangle|^2}{langle y, y rangle}) + frac{|langle x,yrangle|^2}{langle y, y rangle} = langle x, x rangle - frac{|langle x,yrangle|^2}{langle y, y rangle}$$
And hence $|langle x,y rangle| leq |x||y| $.
Note that if $y = lambda x$ for $lambda in mathbb C$ then equality holds:
$$ |lambda|^2 |langle x, x rangle| = |lambda|^2 |x||x| $$
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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).
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– t.b.
Jul 2 '12 at 9:55
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@t.b. Like this?
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– Rudy the Reindeer
Jul 2 '12 at 10:22
1
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This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
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– t.b.
Jul 2 '12 at 10:26
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In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
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– t.b.
Jul 2 '12 at 10:33
1
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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 $t$).
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– wildildildlife
Jul 2 '12 at 11:10
|
show 4 more comments
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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
$$ begin{align}
0 leq |x + ty |^2 &= langle x + ty, x + tyrangle \
&= langle x,xrangle + langle x,t yrangle + langle yt, xrangle + langle ty,tyrangle \
&= langle x,xrangle + bar{t} langle x,yrangle + t overline{langle x,yrangle} + |t|^2 langle y,yrangle \
&= langle x,xrangle + 2 Re(t overline{langle x,yrangle}) + |t|^2 langle y,yrangle
end{align}$$
Now choose $t := -frac{langle x, y rangle}{langle y, y rangle}$. Then we get
$$ 0 leq langle x,xrangle + 2 Re(- frac{|langle x,yrangle|^2}{langle y, y rangle}) + frac{|langle x,yrangle|^2}{langle y, y rangle} = langle x, x rangle - frac{|langle x,yrangle|^2}{langle y, y rangle}$$
And hence $|langle x,y rangle| leq |x||y| $.
Note that if $y = lambda x$ for $lambda in mathbb C$ then equality holds:
$$ |lambda|^2 |langle x, x rangle| = |lambda|^2 |x||x| $$
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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).
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– t.b.
Jul 2 '12 at 9:55
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@t.b. Like this?
$endgroup$
– Rudy the Reindeer
Jul 2 '12 at 10:22
1
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This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
$endgroup$
– t.b.
Jul 2 '12 at 10:26
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In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
$endgroup$
– t.b.
Jul 2 '12 at 10:33
1
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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 $t$).
$endgroup$
– wildildildlife
Jul 2 '12 at 11:10
|
show 4 more comments
$begingroup$
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
$$ begin{align}
0 leq |x + ty |^2 &= langle x + ty, x + tyrangle \
&= langle x,xrangle + langle x,t yrangle + langle yt, xrangle + langle ty,tyrangle \
&= langle x,xrangle + bar{t} langle x,yrangle + t overline{langle x,yrangle} + |t|^2 langle y,yrangle \
&= langle x,xrangle + 2 Re(t overline{langle x,yrangle}) + |t|^2 langle y,yrangle
end{align}$$
Now choose $t := -frac{langle x, y rangle}{langle y, y rangle}$. Then we get
$$ 0 leq langle x,xrangle + 2 Re(- frac{|langle x,yrangle|^2}{langle y, y rangle}) + frac{|langle x,yrangle|^2}{langle y, y rangle} = langle x, x rangle - frac{|langle x,yrangle|^2}{langle y, y rangle}$$
And hence $|langle x,y rangle| leq |x||y| $.
Note that if $y = lambda x$ for $lambda in mathbb C$ then equality holds:
$$ |lambda|^2 |langle x, x rangle| = |lambda|^2 |x||x| $$
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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
$$ begin{align}
0 leq |x + ty |^2 &= langle x + ty, x + tyrangle \
&= langle x,xrangle + langle x,t yrangle + langle yt, xrangle + langle ty,tyrangle \
&= langle x,xrangle + bar{t} langle x,yrangle + t overline{langle x,yrangle} + |t|^2 langle y,yrangle \
&= langle x,xrangle + 2 Re(t overline{langle x,yrangle}) + |t|^2 langle y,yrangle
end{align}$$
Now choose $t := -frac{langle x, y rangle}{langle y, y rangle}$. Then we get
$$ 0 leq langle x,xrangle + 2 Re(- frac{|langle x,yrangle|^2}{langle y, y rangle}) + frac{|langle x,yrangle|^2}{langle y, y rangle} = langle x, x rangle - frac{|langle x,yrangle|^2}{langle y, y rangle}$$
And hence $|langle x,y rangle| leq |x||y| $.
Note that if $y = lambda x$ for $lambda in mathbb C$ then equality holds:
$$ |lambda|^2 |langle x, x rangle| = |lambda|^2 |x||x| $$
edited Jul 2 '12 at 10:22
community wiki
2 revs
Matt N.
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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).
$endgroup$
– t.b.
Jul 2 '12 at 9:55
$begingroup$
@t.b. Like this?
$endgroup$
– Rudy the Reindeer
Jul 2 '12 at 10:22
1
$begingroup$
This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
$endgroup$
– t.b.
Jul 2 '12 at 10:26
$begingroup$
In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
$endgroup$
– t.b.
Jul 2 '12 at 10:33
1
$begingroup$
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 $t$).
$endgroup$
– wildildildlife
Jul 2 '12 at 11:10
|
show 4 more comments
$begingroup$
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).
$endgroup$
– t.b.
Jul 2 '12 at 9:55
$begingroup$
@t.b. Like this?
$endgroup$
– Rudy the Reindeer
Jul 2 '12 at 10:22
1
$begingroup$
This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
$endgroup$
– t.b.
Jul 2 '12 at 10:26
$begingroup$
In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
$endgroup$
– t.b.
Jul 2 '12 at 10:33
1
$begingroup$
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 $t$).
$endgroup$
– wildildildlife
Jul 2 '12 at 11:10
$begingroup$
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).
$endgroup$
– t.b.
Jul 2 '12 at 9:55
$begingroup$
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).
$endgroup$
– t.b.
Jul 2 '12 at 9:55
$begingroup$
@t.b. Like this?
$endgroup$
– Rudy the Reindeer
Jul 2 '12 at 10:22
$begingroup$
@t.b. Like this?
$endgroup$
– Rudy the Reindeer
Jul 2 '12 at 10:22
1
1
$begingroup$
This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
$endgroup$
– t.b.
Jul 2 '12 at 10:26
$begingroup$
This is half of what I had in mind. More interesting is the fact that if equality $lvertlangle x,yranglervert = lVert xrVert lVert y rVert$ holds then $y = lambda x$ or $x = 0$.
$endgroup$
– t.b.
Jul 2 '12 at 10:26
$begingroup$
In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
$endgroup$
– t.b.
Jul 2 '12 at 10:33
$begingroup$
In other words: equality holds if and only if $x$ and $y$ are linearly dependent.
$endgroup$
– t.b.
Jul 2 '12 at 10:33
1
1
$begingroup$
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 $t$).
$endgroup$
– wildildildlife
Jul 2 '12 at 11:10
$begingroup$
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 $t$).
$endgroup$
– wildildildlife
Jul 2 '12 at 11:10
|
show 4 more comments
$begingroup$
Here is a nice simple proof. Fix, $X,Yin mathbb{R}^n$ then we wish to show
$$
Xcdot Y leq |X||Y|
$$
the trick is to construct a suitable vector $Zin mathbb{R}^n$ and then use the property of the dot product $Zcdot Z geq 0$. Take
$$
Z = frac{X}{|X|}-frac{Y}{|Y|}
$$
then we compute $Zcdot Z$
begin{align}
Zcdot Z &= frac{Xcdot X}{|X|^2}-2frac{Xcdot Y}{|X||Y|}+frac{Ycdot Y}{|Y|^2}\
&=2 - 2frac{Xcdot Y}{|X||Y|}
end{align}
then we use $Zcdot Z geq 0$ to write
begin{align}
2-2frac{Xcdot Y}{|X||Y|}geq 0\
2geq 2frac{Xcdot Y}{|X||Y|}\
|X||Y|geq Xcdot Y
end{align}
and we are done.
$endgroup$
add a comment |
$begingroup$
Here is a nice simple proof. Fix, $X,Yin mathbb{R}^n$ then we wish to show
$$
Xcdot Y leq |X||Y|
$$
the trick is to construct a suitable vector $Zin mathbb{R}^n$ and then use the property of the dot product $Zcdot Z geq 0$. Take
$$
Z = frac{X}{|X|}-frac{Y}{|Y|}
$$
then we compute $Zcdot Z$
begin{align}
Zcdot Z &= frac{Xcdot X}{|X|^2}-2frac{Xcdot Y}{|X||Y|}+frac{Ycdot Y}{|Y|^2}\
&=2 - 2frac{Xcdot Y}{|X||Y|}
end{align}
then we use $Zcdot Z geq 0$ to write
begin{align}
2-2frac{Xcdot Y}{|X||Y|}geq 0\
2geq 2frac{Xcdot Y}{|X||Y|}\
|X||Y|geq Xcdot Y
end{align}
and we are done.
$endgroup$
add a comment |
$begingroup$
Here is a nice simple proof. Fix, $X,Yin mathbb{R}^n$ then we wish to show
$$
Xcdot Y leq |X||Y|
$$
the trick is to construct a suitable vector $Zin mathbb{R}^n$ and then use the property of the dot product $Zcdot Z geq 0$. Take
$$
Z = frac{X}{|X|}-frac{Y}{|Y|}
$$
then we compute $Zcdot Z$
begin{align}
Zcdot Z &= frac{Xcdot X}{|X|^2}-2frac{Xcdot Y}{|X||Y|}+frac{Ycdot Y}{|Y|^2}\
&=2 - 2frac{Xcdot Y}{|X||Y|}
end{align}
then we use $Zcdot Z geq 0$ to write
begin{align}
2-2frac{Xcdot Y}{|X||Y|}geq 0\
2geq 2frac{Xcdot Y}{|X||Y|}\
|X||Y|geq Xcdot Y
end{align}
and we are done.
$endgroup$
Here is a nice simple proof. Fix, $X,Yin mathbb{R}^n$ then we wish to show
$$
Xcdot Y leq |X||Y|
$$
the trick is to construct a suitable vector $Zin mathbb{R}^n$ and then use the property of the dot product $Zcdot Z geq 0$. Take
$$
Z = frac{X}{|X|}-frac{Y}{|Y|}
$$
then we compute $Zcdot Z$
begin{align}
Zcdot Z &= frac{Xcdot X}{|X|^2}-2frac{Xcdot Y}{|X||Y|}+frac{Ycdot Y}{|Y|^2}\
&=2 - 2frac{Xcdot Y}{|X||Y|}
end{align}
then we use $Zcdot Z geq 0$ to write
begin{align}
2-2frac{Xcdot Y}{|X||Y|}geq 0\
2geq 2frac{Xcdot Y}{|X||Y|}\
|X||Y|geq Xcdot Y
end{align}
and we are done.
answered May 8 '18 at 2:52
community wiki
Eli Fonseca
add a comment |
add a comment |
$begingroup$
Without loss of generality, assume $|y|=1$. Write $x=left<x,yright>y+z$. Then $z$ is orthogonal to $y$, because
$$left<x,yright>=left<(left<x,yright>y+z),yright>=left<x,yright>left<y,yright>+left<z,yright>,$$
indeed yields $left<z,yright>=0$. Hence
$$|x|^2=left<x,xright>=|left<x,yright>|^2+left<z,zright>geq |left<x,yright>|^2,$$
with equality iff $z= 0$, i.e. $xinmathbb{F}y$.
$endgroup$
add a comment |
$begingroup$
Without loss of generality, assume $|y|=1$. Write $x=left<x,yright>y+z$. Then $z$ is orthogonal to $y$, because
$$left<x,yright>=left<(left<x,yright>y+z),yright>=left<x,yright>left<y,yright>+left<z,yright>,$$
indeed yields $left<z,yright>=0$. Hence
$$|x|^2=left<x,xright>=|left<x,yright>|^2+left<z,zright>geq |left<x,yright>|^2,$$
with equality iff $z= 0$, i.e. $xinmathbb{F}y$.
$endgroup$
add a comment |
$begingroup$
Without loss of generality, assume $|y|=1$. Write $x=left<x,yright>y+z$. Then $z$ is orthogonal to $y$, because
$$left<x,yright>=left<(left<x,yright>y+z),yright>=left<x,yright>left<y,yright>+left<z,yright>,$$
indeed yields $left<z,yright>=0$. Hence
$$|x|^2=left<x,xright>=|left<x,yright>|^2+left<z,zright>geq |left<x,yright>|^2,$$
with equality iff $z= 0$, i.e. $xinmathbb{F}y$.
$endgroup$
Without loss of generality, assume $|y|=1$. Write $x=left<x,yright>y+z$. Then $z$ is orthogonal to $y$, because
$$left<x,yright>=left<(left<x,yright>y+z),yright>=left<x,yright>left<y,yright>+left<z,yright>,$$
indeed yields $left<z,yright>=0$. Hence
$$|x|^2=left<x,xright>=|left<x,yright>|^2+left<z,zright>geq |left<x,yright>|^2,$$
with equality iff $z= 0$, i.e. $xinmathbb{F}y$.
answered Jul 2 '12 at 10:51
community wiki
wildildildlife
add a comment |
add a comment |
$begingroup$
I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:
$langle x,yrangle=langle y,xrangle$
$langle ax+y,zrangle=alangle x,zrangle+langle y,zrangle$
$langle x,xranglegeq0$
Then
$0leqlangle lx+y,lx+yrangle=l^2langle x,xrangle+llangle x,yrangle+llangle y,xrangle+langle y,yrangle=l^2langle x,xrangle+2llangle x,yrangle+langle y,yrangle$
$Let:a=langle x,xrangle, b=langle x,yrangle,c=langle y,yrangle$, then the equation becomes
$al^2+bl+cgeq0$
This is a quadratic equation in $l$ with at most 1 real root. Therefore
$b^2-4acleq 0$
$implies4{langle x,yrangle}^2-4langle x,xranglelangle y,yrangleleq 0$
$implies{langle x,yrangle}^2leqlangle x,xranglelangle y,yrangle$
Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($
$endgroup$
add a comment |
$begingroup$
I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:
$langle x,yrangle=langle y,xrangle$
$langle ax+y,zrangle=alangle x,zrangle+langle y,zrangle$
$langle x,xranglegeq0$
Then
$0leqlangle lx+y,lx+yrangle=l^2langle x,xrangle+llangle x,yrangle+llangle y,xrangle+langle y,yrangle=l^2langle x,xrangle+2llangle x,yrangle+langle y,yrangle$
$Let:a=langle x,xrangle, b=langle x,yrangle,c=langle y,yrangle$, then the equation becomes
$al^2+bl+cgeq0$
This is a quadratic equation in $l$ with at most 1 real root. Therefore
$b^2-4acleq 0$
$implies4{langle x,yrangle}^2-4langle x,xranglelangle y,yrangleleq 0$
$implies{langle x,yrangle}^2leqlangle x,xranglelangle y,yrangle$
Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($
$endgroup$
add a comment |
$begingroup$
I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:
$langle x,yrangle=langle y,xrangle$
$langle ax+y,zrangle=alangle x,zrangle+langle y,zrangle$
$langle x,xranglegeq0$
Then
$0leqlangle lx+y,lx+yrangle=l^2langle x,xrangle+llangle x,yrangle+llangle y,xrangle+langle y,yrangle=l^2langle x,xrangle+2llangle x,yrangle+langle y,yrangle$
$Let:a=langle x,xrangle, b=langle x,yrangle,c=langle y,yrangle$, then the equation becomes
$al^2+bl+cgeq0$
This is a quadratic equation in $l$ with at most 1 real root. Therefore
$b^2-4acleq 0$
$implies4{langle x,yrangle}^2-4langle x,xranglelangle y,yrangleleq 0$
$implies{langle x,yrangle}^2leqlangle x,xranglelangle y,yrangle$
Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($
$endgroup$
I like this proof for real vectors a lot. Recall that an inner product for real vectors has the following properties:
$langle x,yrangle=langle y,xrangle$
$langle ax+y,zrangle=alangle x,zrangle+langle y,zrangle$
$langle x,xranglegeq0$
Then
$0leqlangle lx+y,lx+yrangle=l^2langle x,xrangle+llangle x,yrangle+llangle y,xrangle+langle y,yrangle=l^2langle x,xrangle+2llangle x,yrangle+langle y,yrangle$
$Let:a=langle x,xrangle, b=langle x,yrangle,c=langle y,yrangle$, then the equation becomes
$al^2+bl+cgeq0$
This is a quadratic equation in $l$ with at most 1 real root. Therefore
$b^2-4acleq 0$
$implies4{langle x,yrangle}^2-4langle x,xranglelangle y,yrangleleq 0$
$implies{langle x,yrangle}^2leqlangle x,xranglelangle y,yrangle$
Not bad huh? Sadly it doesn't work out so nicely with complex vectors $:($
answered Jul 27 '14 at 20:42
community wiki
Pauly B
add a comment |
add a comment |
$begingroup$
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
$$ text{LHS} = ( sum_i x_i y_i )^2 = (sum_i x_i y_i)(sum_j x_j y_j) = sum_isum_j x_iy_ix_jy_j. $$
For the RHS write
$$
text{RHS}=
frac{1}{2}(sum_i x_i^2)(sum_j y_j^2)
+frac{1}{2} (sum_j x_j^2)(sum_i y_i^2)
- frac{1}{2} sum_i sum_j (x_iy_j -x_jy_i)^2
\ =
frac{1}{2}sum_isum_jleft(
x_i^2 y_j^2 + x_j^2y_i^2 - x_i^2 y_j^2 - x_j^2y_i^2 + 2 x_i y_i x_j y_j
right)
=
sum_isum_j x_iy_ix_jy_j .
$$
This shows that LHS$=$RHS and finishes the proof.
$endgroup$
add a comment |
$begingroup$
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
$$ text{LHS} = ( sum_i x_i y_i )^2 = (sum_i x_i y_i)(sum_j x_j y_j) = sum_isum_j x_iy_ix_jy_j. $$
For the RHS write
$$
text{RHS}=
frac{1}{2}(sum_i x_i^2)(sum_j y_j^2)
+frac{1}{2} (sum_j x_j^2)(sum_i y_i^2)
- frac{1}{2} sum_i sum_j (x_iy_j -x_jy_i)^2
\ =
frac{1}{2}sum_isum_jleft(
x_i^2 y_j^2 + x_j^2y_i^2 - x_i^2 y_j^2 - x_j^2y_i^2 + 2 x_i y_i x_j y_j
right)
=
sum_isum_j x_iy_ix_jy_j .
$$
This shows that LHS$=$RHS and finishes the proof.
$endgroup$
add a comment |
$begingroup$
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
$$ text{LHS} = ( sum_i x_i y_i )^2 = (sum_i x_i y_i)(sum_j x_j y_j) = sum_isum_j x_iy_ix_jy_j. $$
For the RHS write
$$
text{RHS}=
frac{1}{2}(sum_i x_i^2)(sum_j y_j^2)
+frac{1}{2} (sum_j x_j^2)(sum_i y_i^2)
- frac{1}{2} sum_i sum_j (x_iy_j -x_jy_i)^2
\ =
frac{1}{2}sum_isum_jleft(
x_i^2 y_j^2 + x_j^2y_i^2 - x_i^2 y_j^2 - x_j^2y_i^2 + 2 x_i y_i x_j y_j
right)
=
sum_isum_j x_iy_ix_jy_j .
$$
This shows that LHS$=$RHS and finishes the proof.
$endgroup$
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
$$ text{LHS} = ( sum_i x_i y_i )^2 = (sum_i x_i y_i)(sum_j x_j y_j) = sum_isum_j x_iy_ix_jy_j. $$
For the RHS write
$$
text{RHS}=
frac{1}{2}(sum_i x_i^2)(sum_j y_j^2)
+frac{1}{2} (sum_j x_j^2)(sum_i y_i^2)
- frac{1}{2} sum_i sum_j (x_iy_j -x_jy_i)^2
\ =
frac{1}{2}sum_isum_jleft(
x_i^2 y_j^2 + x_j^2y_i^2 - x_i^2 y_j^2 - x_j^2y_i^2 + 2 x_i y_i x_j y_j
right)
=
sum_isum_j x_iy_ix_jy_j .
$$
This shows that LHS$=$RHS and finishes the proof.
answered Jan 20 at 19:43
community wiki
Hashimoto
add a comment |
add a comment |
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$begingroup$
I've slightly edited the body your question in order to make it self-contained. I've also added the
real-analysistag.$endgroup$
– t.b.
Feb 24 '11 at 12:58
14
$begingroup$
I would try the book "The Cauchy-Schwarz Masterclass".
$endgroup$
– user3533
Feb 24 '11 at 13:03
$begingroup$
@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
$endgroup$
– Vafa Khalighi
Feb 24 '11 at 13:30
2
$begingroup$
@Vafa: many Olympiad preparation books include sections on inequalities. There is also Kedlaya's notes: artofproblemsolving.com/Resources/Papers/…
$endgroup$
– Qiaochu Yuan
Feb 24 '11 at 13:35
3
$begingroup$
Your first question is essentially unanswerable, except, maybe, by "many"...
$endgroup$
– Mariano Suárez-Álvarez
Feb 24 '11 at 22:52