<a name="docA001">
2009-06-08-19-10 start
This file http://freeman2.com/tute0008.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view 
point wrong, please ask a math expert near by.

<a name="docA002">,<a name="textbook">
This file is a note for reading inequality
book written by Professor J. Michael Steele
The Cauchy-Schwarz Master Class  　★★★★★
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">
Freeman study mechanical engineering. 
Engineering mathematics do not teach 
inequality. Above book is first inequality
book. First time read, it was very hard.
Although high school time learned
  a*a + b*b >= 2*a*b
But this little knowledge do not help.

<a name="docA004">
This file follow textbook chapter section
order, but not continuous, Freeman skip 
those uncertain sections/problems.

This file first function is to learn 
inequality. Second function is to learn
how to use html code to write math 
equations.

<a name="docA005">　Index begin　Index this file
On 2009-01-27-10-08 Freeman accessed
the next page
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html
save as sftw.umac.mo_text_math_eqn_good.htm
Above page is the main reference for html
math equation.

In order to let reader to build  html math
equation, previous file tute0007.htm page 
end has math symbol and internal code.

This file third function is to display how
to draw curves in web page. The main engine
is XYGraph v2.3 - Technical Figures

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">
2009-08-23-15-00 start
□　Exercise 1.14 solution

Exercise 1.14 problem (a) is solved by hint
Exercise 1.14 problem (b) is out of LiuHH's
reach. What is "cardinality of a set B⊂Z3" ?
Without clear understand the definition, 
LiuHH is unable to solve problem (b)
LiuHH major in mechanical engineering.
LiuHH is mathematics admirer and outsider.

<a name="docA007">
To solve problems in
The Cauchy-Schwarz Master Class
LiuHH's goal is to solve 50% of the exercise
problems. In reality, solve 40% is good enough.
Please do not be surprise that LiuHH skip some
text section, skip some problems.
2009-08-23-15-12 here

<a name="ch02d001">　Index begin　Index this file
2009-10-13-11-38 start
■　Exercise 2.13 problem statement
　　textbook page 36
（An AM-GM Inequality for Complex Numbers）

Consider a set S of n complex numbers
z1,z2,...,zn for which the polar forms
  zj=ρjeiθj ---eqn.AH001
satisfies the constraints
  0≦ρj＜∞ ---eqn.AH002
<a name="ch02d002">
and
  0≦|θj|＜ψ＜π/2 ---eqn.AH003
with
  1≦j≦n ---eqn.AH004
As one sees in Figure 2.5, the spread in
the arguments of the zj∈S is bounded
by 2ψ. 
<a name="ch02d003">
Show that for such numbers one has
the bound
  cos(ψ)|z1z2...zn|1/n≦
        |z1+z2+...+zn|/n ---eqn.2.35
Here one should note that if the
  zj j=1,2,3 ... n 
are all real numbers, then one can take
ψ=0, in which case the bound 2.35 
recaptures the usual AM-GM inequality.
2009-10-13-11-59 here

<a name="ch02d004">
2009-10-13-12-38 start
■　Exercise 2.13 hint
　　textbook page 235

In general one has
 |w|≧|Re w| ---eqn.AH005
and
 Re(w+z)=Re(w)+Re(z) ---eqn.AH006
so from
 Re(zj)=ρjcosθj ---eqn.AH007
<a name="ch02d005">
we find
  |z1+z2+...+zn|
 ≧|Re(z1+z2+...+zn)|
 =|z1|cosθ1+|z2|cosθ2+...+|zn|cosθn
 ≧(|z1|+|z2|+...+|zn|)cosψ
 ≧n(|z1|*|z2|*...*|zn|)1/n*cosψ
<a name="ch02d006">
where we first used the fact that cosine
is monotone decreasing on [0,π/2] and
then we applied the AM-GM inequality to
the non-negative real numbers zj j=1,2,...,n
This exercise is based on Wilf (1963)
Mitrinovic (1970) notes that versions
of this bound may be traced back at
least to Petrovitch (1917). There are
also informative generalizations given
by Diaz and Metcalf (1966)
2009-10-13-13-00 stop





<a name="ch02d007">　Index begin　Index this file
2009-10-13-13-20 start
■　Exercise 2.13 solution


For a set S of n complex numbers
z1,z2,...,zn for which the polar forms
  zj=ρjeiθj ---eqn.AH001
<a name="ch02d008">
problem statement require the following
constraints
  0≦ρj＜∞ ---eqn.AH002
eqn.AH002 is always true. Because the
ρj definition for complex a+b*i is
  ρj = sqrt(a*a+b*b) ---eqn.AH030

<a name="ch02d009">
Second constraint is
  0≦|θj|＜ψ＜π/2 ---eqn.AH003
with
  1≦j≦n ---eqn.AH004
θj definition for complex a+b*i is
  θj = atan2(b,a) ---eqn.AH031

eqn.AH003 is a special requirement. Not
general true for any complex number.

<a name="ch02d010">
Simple words for eqn.AH003 is that
complex number real part must be
positive. It can not be zero. Because
for pure imaginary number (real=0) 
θj=+π/2 if imaginary>0
θj=-π/2 if imaginary<0
pure imaginary number violate eqn.AH003.

Next follow Exercise 2.13 hint and do
some explanation.

<a name="ch02d011">
In general one has
 |w|≧|Re w| ---eqn.AH005
Because |w| is an arbitrary 2D vector
composed by (real, imaginary)
|w| is sqrt(real*real + imag*imag)
|Re w| is |real|
vector length is greater than or equal to
its component. This is a simple |cosθ|≦1
reason.

<a name="ch02d012">　Index begin　Index this file
and
 Re(w+z)=Re(w)+Re(z) ---eqn.AH006
Re(w) is vector w component on real axis
Re(z) is vector z component on real axis
Both on real axis, their sum is simple
equality eqn.AH006.

<a name="ch02d013">
so from
 Re(zj)=ρjcosθj ---eqn.AH007
[ρj,θj] is a complex number polar form.
ρjcosθj is complex number real part.
ρjsinθj is complex number imag part.

<a name="ch02d014">
we find
  |z1+z2+...+zn| ---eqn.AH008
 // next line apply eqn.AH005
 ≧|Re(z1+z2+...+zn)| ---eqn.AH009
 // next line apply eqn.AH006
 =|z1|cosθ1+|z2|cosθ2+...+|zn|cosθn ---eqn.AH010
<a name="ch02d015">
 // next line apply eqn.AH003
 // Unique for Exercise 2.13
 // please note: 80º  >  60º
 // but      cos(80º) < cos(60º)
 // ψ ≧ θj , cos(ψ) ≦ cos(θj)
 // without AH003, AH011 is NOT true.
 ≧(|z1|+|z2|+...+|zn|)cosψ ---eqn.AH011
<a name="ch02d016">
 // next line is AM-GM inequality
 // AM-GM ineq. apply to REAL only
 // absolute value is REAL number
 ≧n(|z1|*|z2|*...*|zn|)1/n*cosψ ---eqn.AH012
where we first used the fact that cosine
is monotone decreasing on [0,π/2] and
then we applied the AM-GM inequality to 
the non-negative real numbers zj j=1,2,...,n

<a name="ch02d017">
Net result is
  |z1+z2+...+zn|
 ≧n(|z1|*|z2|*...*|zn|)1/n*cosψ ---eqn.AH013

eqn.AH013 and eqn.2.35 are the same.
Problem solved.

Follow Prof. J. Michael Steele steps,
everything become easy !!
2009-10-13-14-03 stop



<a name="ch02d018">　Index begin　Index this file
2009-10-13-16-57 start
■　Exercise 2.14 problem statement
　　textbook page 36
（A Leap-Forward Fall-Back Tour de Force）

One can use Cauchy's leap-forward fall-back
method of induction to prove that for all
non-negative x1,x2,...,xm
and for all integer power n=1,2,... one has
the bound

<a name="ch02d020">
2009-10-13-17-08 here
This is a special case of the power mean
inequality which we develop at length in
Chapter 8, but here the focus is on mastery
of technique. This exercise leads to one
of the more sustained application of
Cauchy's method that one is likely to meet.
2009-10-13-17-11 here




<a name="ch02d021">
2009-10-13-17-12 here
■　Exercise 2.14 hint
　　textbook page 235

Take x≧0 and y≧0 and consider 
the hypothesis
H(n), ((x+y)/2)n ≦ (xn + yn)/2 ---eqn.AH014
To prove H(n+1) we note by H(n) that

<a name="ch02d025">　Index begin　Index this file
2009-10-13-17-39 here
Induction then confirm the validity of
H(n) for all n≧1.
[[Alert !! CHANGE !!
Above power n vary, number of elements m=2
Below power n constant, but m vary. 
2009-10-17-11-38]]
Now by H(n) applied twice we find

<a name="ch02d028">
2009-10-13-18-02 here
and this argument can be repeated to show
that for each set of 2k nonnegative
numbers x1,x2,...,x2k we have

<a name="ch02d030">
[[Alert !! CHANGE !!
Above use H(n), parameter is n, n=power.
Below use Hnew(m), parameter is m, 
m=number of elements.
2009-10-17-11-44]]
2009-10-13-18-13 here
Cauchy's trick of padding a sequence of
length m with extra terms to get a sequence
of length 2k now runs into difficulty,
so a new twist is needed. One idea that 
works is to use a full backwards induction.

<a name="ch02d031">
Specifically, we now let Hnew(m) denote
the hypothesis that

<a name="ch02d032">　Index begin　Index this file
2009-10-13-18-27 here
for any set of m nonnegative real numbers
x1,x2,...,xm. We already know that Hnew(m)
is valid when m is any power of two, so
to prove that Hnew(m) is valid for all
m=1,2,... we just need to show that for
m≧2, the hypothesis Hnew(m) implies
Hnew(m-1).

<a name="ch02d033">
Given m-1 nonnegative reals
 S={x1,x2,...,xm-1} ---eqn.AH020
we introduce a new variable y by setting
 y=(x1+x2+...+xm-1)/(m-1) ---eqn.AH021
Since y is equal to 
 (x1+x2+...+xm-1+y)/m ---eqn.AH022
we see that when we apply H(m) to the 
m-elements set S∪{y}, we obtain the bound

<a name="ch02d035">
and when we clear yn to the left side,
we find

<a name="ch02d036">
2009-10-13-18-48 here
This inequality is precisely what one 
needed to establish the validity of
Hnew(m-1), so the solution of the 
problem is complete. This solution is 
guided by the one given by Shklarsky, 
Chentzov, and Yaglom (1993, pp 391-392)
2009-10-13-18-52 stop




<a name="ch02d037">　Index begin　Index this file
2009-10-13-20-48 start
■　Exercise 2.14 solution


Exercise 2.14 eqn.2.36 is not Cauchy's
inequality (chapter one). 
Because Cauchy's inequality work with
two sequences, but eqn.2.36 has just
one sequence.

<a name="ch02d038">
eqn.2.36 is not AM-GM inequality 
(chapter two). Because eqn.2.36 do not
have geometric mean, we do not see
(a1*a2*...*an)1/n in eqn.2.36

eqn.2.36 is power mean. We must clearly
understand problem nature first!

<a name="ch02d039">
Second key point to observe is that number
of elements m and power n are different.
m and n are not related.

eqn.2.36 says take Arithmetic Mean for
sequence x1,x2,...,xm first, second step
rise AM to n th power. This value is
less than sequence elements individually
rise to n th power first, then take AM
for the n th power sequence.

<a name="ch02d040">
Exercise 2.14 hint start from simplest
sequence, two elements sequence {x,y}

hint said
[[
Take x≧0 and y≧0 and consider the hypothesis
H(n), ((x+y)/2)n ≦ (xn + yn)/2 ---eqn.AH014
]]
We assume chapter eight power mean is
correct. accept eqn.AH014 as true, 
prove it later at chapter eight.

H(n) is defined by eqn.AH014. 
     power n is parameter
Hnew(m) is defined by eqn.14.49.
     number of elements m is parameter.

<a name="ch02d041">
Based on eqn.AH014, move to eqn.AH015
[(x+y)/2]^(n+1) = [(x+y)/2]*[(x+y)/2]^n ---eqn.AH015 aux1
this is simple thing like a^3=a*a^2.

inequality in eqn.AH016 is the result 
of application of eqn.AH014.

equality in eqn.AH016 is multiply out
the left term.

<a name="ch02d042">　Index begin　Index this file
in eqn.AH016 right hand side add a zero
zero = xn+1+yn+1 -xn+1-yn+1 ---eqn.AH025
positive xn+1+yn+1 add to the same term.
negative -xn+1-yn+1 add to yxn+xyn
factor to eqn.AH017. In eqn.AH017 (x－y)
and (xn－yn) have same sign, then
[(x－y)(xn－yn)]/4 is never negative.
In eqn.AH017 drop a minus(non-negative)
term get the inequality in eqn.AH017.
(was 5-3 = 5-3, now drop '-3' get 5-3<5)

<a name="ch02d043">
eqn.AH015 begin at [(x+y)/2]^(n+1)
eqn.AH017  end  at ≦ [x^(n+1) + y^(n+1)]/2
combine get
  [(x+y)/2]^(n+1)
  ≦ [x^(n+1) + y^(n+1)]/2 ---eqn.AH026
through induction
H(n), ((x+y)/2)n ≦ (xn + yn)/2 ---eqn.AH014
imply 
H(n+1) eqn.AH026
<a name="ch02d044">
in eqn.AH014 change n to n+1 get eqn.AH026
which is true.
Induction then confirm the validity of
H(n) for all n≧1.
2009-10-13-21-48 here

<a name="ch02d045">
Our problem eqn.2.36 has two parameters
m and n.
[[
Second key point to observe is that number
of elements m and power n are different.
m and n are not related.
]]

<a name="ch02d046">
Induction eqn.AH026 solved parameter n.
Induction eqn.AH026 lock m at 2 elements.
We need extend m to more than two.

eqn.AH018 start extend m from 2 to 4 to 8
etc.

<a name="ch02d047">　Index begin　Index this file
re-write eqn.AH018 left hand side as
[(x1+x2)/2 + (x3+x4)/2]/2 ---eqn.AH027
here (x1+x2)/2 is one element
 and (x3+x4)/2 is second element.
apply eqn.AH014 to these two element get
eqn.AH018 . From eqn.AH018 to eqn.AH019
left term is 
apply eqn.AH014 to x1 and x2 two elements
and
apply eqn.AH014 to x3 and x4 two elements
apply twice, get eqn.AH019 left term.
Equality in eqn.AH019 is simple work
of re-arrangement.

<a name="ch02d048">
Compare beginning of eqn.AH018 and ending
of eqn.AH019. m=2 terms extend to m=4
terms. Follow the same reason, m=2k
k=1,2,3... any integer are true.
m=2,4,8,16,32 ... etc are true.
m=3,5,6,7,9, ...  etc are not covered.

<a name="ch02d049">
Exercise 2.14 hint start at ch02d030
solve this problem: to extend m to
any integer, not just m=2k.

Hint use Hnew(m) denote eqn.14.49. 
eqn.AH018 and eqn.AH019 proved Hnew(m)
is true for m=2k. Now assume m=2k.
Start from Hnew(m) m=2k if we can
prove Hnew(m-1) is also true, then
we break the m=2k limit, and m can be
any integer. Next prove 
Hnew(m) is true → Hnew(m-1) is true

H(n) is defined by eqn.AH014. 
     power n is parameter
Hnew(m) is defined by eqn.14.49.
     number of elements m is parameter.

<a name="ch02d050">
Now m=2k. We build m-1 nonnegative 
real sequence
 S={x1,x2,...,xm-1} ---eqn.AH020
(here m-1 is 2k-1, for example 3,7,15
 etc.)
To satisfy 2k requirement we need add 
one more term to S (eqn.AH020)

<a name="ch02d051">
Hint said
we introduce a new variable y by setting
 y=(x1+x2+...+xm-1)/(m-1) ---eqn.AH021

eqn.AH021 is easy for us, follower. But
It is hard work for first developer !!

<a name="ch02d052">　Index begin　Index this file
Hint said
Since y is equal to 
 (x1+x2+...+xm-1+y)/m ---eqn.AH022

Is that ("y is equal to") true?
eqn.AH021 is defined, no question asked.
Start from eqn.AH021, verify eqn.AH022.

<a name="ch02d053">
re-write eqn.AH021 as
 (m-1)*y=(x1+x2+...+xm-1) ---eqn.AH028
that is
 m*y-y=(x1+x2+...+xm-1)
that is
 m*y=(x1+x2+...+xm-1+y)
that is
 y=(x1+x2+...+xm-1+y)/m ---eqn.AH029
TRUE !!

<a name="ch02d054">
Now S (eqn.AH020) has m-1 elements, plus
one more element y (eqn.AH022 or eqn.AH029)
m-1+1= m = 2k, we can apply eqn.14.48

<a name="ch02d055">
2009-10-13-22-50 here
Now substitute eqn.AH029 to eqn.AH023 aux1
left side, get eqn.AH023.
In eqn.AH023, move m to left hand side,
cancel one yn, then move (m-1)
back to eqn.AH023 right denominator. We
get eqn.AH024.

<a name="ch02d056">
eqn.AH024 is for m-1 terms.

<a name="ch02d057">　Index begin　Index this file
2009-10-13-23-04 here
y is defined in eqn.AH021 or eqn.AH029.
eqn.AH024 aux1 left hand side is yn
So eqn.AH024 aux1 and eqn.AH024 are 
same thing.
eqn.AH024 aux1 show complete equation,
show inequality is true for m-1 which
break 2k limitation.

<a name="ch02d058">
Up to here, we showed both power n and
number of elements m are flexible.
eqn.2.36 is verified, problem solved.

"Exercise 2.14 solution" is basically
a study/thinking record.

When I have question, write it down,
find answer, write it down. for example
at ch02d046
<a name="ch02d059">
I ask myself, what is the difference 
above this point and below this point?
Think and get answer, then write 
[[
Induction eqn.AH026 solved parameter n.
Induction eqn.AH026 lock m at 2 elements.
We need extend m to more than two.
]]
This kind thinking record will help 
reader save time.
2009-10-13-23-25 stop

2009-10-14-16-03 done proofread
2009-10-14-16-20 done spelling check

■■Chapter 02: The AM-GM Inequality
　　end here
Thank you for visiting Freeman's site.




<a name="ch03a001">　Index begin　Index this file
2009-10-16-12-53 start
■■Chapter 03: Lagrange's Identity
　　 and Minkowski's Conjecture.

■　Identity brief
Mathematics equation has three categories
1. equality equation
2. inequality equation
3. identity equation

<a name="ch03a002">
Difference between equality equation and
inequality equation is simple.
equation two sides have same value, that 
is equality equation.

equation two sides have different value, 
that is inequality equation.

<a name="ch03a003">
identity equation is equality equation,
Difference between identity and equality
is that 
equality equation is valid at few points,
identity equation is valid at ALL points.

For example
  (a+b)^2 = a*a + 2*a*b + b*b ---eqn.AH032
and
  [sin(x)]^2 + [cos(x)]^2 = 1 ---eqn.AH033
both are identity equations.

<a name="ch03a004">
eqn.AH032 is identity, because right hand 
side is expansion of left hand side. It is
symbolic equation, put any value to a or b
eqn.AH032 is always true.

eqn.AH033 is identity, because the definition
of sin(x) and cos(x) satisfy Pythagoras' 
theorem, which is true for any x value.

<a name="ch03a005">
Non-identity example is
 1+x = exp(x) ---eqn.AH034
Please click Draw 603
Red solid curve is f(x)=exp(x) ---eqn.AH035
Red  dash line  is g(x)=1+x    ---eqn.AH036
Two curves have one common point at x=0.
eqn.AH034 is true for x=0 only, 
eqn.AH034 is not true for all x.
So eqn.AH034 is equality equation,
eqn.AH034 is not identity equation.
One more step reveal this point better.
<a name="ch03a006">
Taylor series expansion for exp(x) is
Peirce Table Item 759
exp(x) = 1 + x + x^2/2! + x^3/3! + ... ---eqn.AH037
eqn.AH037 is an identity, it is true 
for any x.
eqn.AH034 says exp(x) = 1 + x that is
1 + x = 1 + x + x^2/2! + x^3/3! + ... ---eqn.AH038
cancel common term 1 + x get
 0 = x^2/2! + x^3/3! + ... ---eqn.AH039
eqn.AH039 is true only if x=0 !!
From eqn.AH038 we know eqn.AH034 can 
not be an identity.

<a name="ch03a007">
We draw curve for equality equation.
Drawing for identity equation is rare
and no fun. How many of us draw
  [sin(x)]^2 + [cos(x)]^2 = 1 ---eqn.AH033
?

<a name="ch03a008">　Index begin　Index this file
■　Identity change to Inequality

We know
 (a+b)^2 = a*a + 2*a*b + b*b ---eqn.AH032
is an equality and identity. If both
a and b are positive, and if we drop
2*a*b, then eqn.AH032 is not an identity
any more, the remaining terms become an
inequality
 (a+b)^2 ＞ a*a + b*b ---eqn.AH040

<a name="ch03a009">
Let us create another identity
 (a+b)^2 = a*a + 2*a*b + b*b ---eqn.AH032
 (a-b)^2 = a*a - 2*a*b + b*b ---eqn.AH041
Add two together, get
 (a+b)^2 + (a-b)^2 = 2*a*a + 2*b*b ---eqn.AH042
eqn.AH042 is true for any a and any b.
Now let us drop (a-b)^2 from eqn.AH042
get
 (a+b)^2 ≦ 2*a*a + 2*b*b ---eqn.AH043

<a name="ch03a010">
Start from eqn.AH042 get eqn.AH043 is easy.
But start from eqn.AH043, it is hard for
us to figure out the missing term (a-b)^2 

The missing term (a-b)^2 is important for
measuring the inequality eqn.AH043 error
magnitude. When we face an inequality,
like eqn.AH043, it would be wonderful if
we can measure its error, the difference
of two sides.
2009-10-16-13-49 here

<a name="ch03a011">　Index begin　Index this file
■　Cauchy Inequality and AM-GM Inequality

Cauchy Inequality use two sequences,
a1,a2,...,an and b1,b2,...,bn
Treat two sequences as components of
two vectors. Cauchy Inequality says
two vectors dot product value is less
than or equal to two vectors length 
product.

<a name="ch03a012">
If we normalize two vectors, it is 
easy to see that
Cauchy Inequality is basically a
statement of
 -1 ≦ cos(theta) ≦ +1 ---eqn.AH044

On the other hand, AM-GM Inequality 
need just one sequence (not two like
Cauchy) 
AM-GM Inequality need all sequence
elements value be non-negative.
(Cauchy allow negative value)

<a name="ch03a013">
AM-GM Inequality use two different
mean procedure arithmetic mean and
geometric mean to compare two mean
values. Assume simplest two elements
sequence [a1,a2]
AM = (a1+a2)/2   ---eqn.AH045
GM = (a1*a2)^(1/2) ---eqn.AH046
eqn.AH046 is two element product then
take root of two. 

<a name="ch03a014">
If we have n elements in a sequence,
we need n-element product then take 
root of n for GM evaluation.

Here we want to relate
Cauchy Inequality with a possible 
identity. and relate
AM-GM Inequality with a possible 
identity.

<a name="ch03a015">
Cauchy Inequality do not have "root
of n" operation.
AM-GM Inequality use sum divide by 
total number n (AM) and use product
then take root of n (GM).

Whether there is an identity which 
involve sum, division, product and 
roots of n? Although we can not say
no at this point, but it is very hard
to find such identity.

<a name="ch03a016">
Textbook introduce Lagrange's Identity 
which reduce to Cauchy's Inequality 
easily. When read textbook, LiuHH 
asked whether there is an identity 
and reduce to AM-GM Inequality? 
Hope someday find this answer.
2009-10-16-14-20 stop

2009-10-16-16-04 start
<a name="ch03a017">　Index begin　Index this file
■　Two terms two sequence identity

Two terms two sequence Cauchy's
Inequality is
(a1*b1+a2*b2)2≦(a12+a22)(b12+b22) ---eqn.AH047
Can we find an identity from eqn.AH047?
Try, carry out multiplication of right
side of eqn.AH047 get
<a name="ch03a018">
  (a12+a22)(b12+b22)
=a12*b12+a12*b22+a22*b12+a22*b22
 // re-group
=(a12*b12+a22*b22) + (a12*b22+a22*b12)
 // add one zero , red terms
=(a12*b12+a22*b22) + (a12*b22+a22*b12)
 +2*a1*b1*a2*b2 -2*a1*b1*a2*b2
=(a12*b12 +2*a1*b1*a2*b2 +a22*b22)
+(a12*b22 -2*a1*b1*a2*b2 +a22*b12)
 // re-group, always identity
=(a1*b1+a2*b2)2+(a1*b2-a2*b1)2

<a name="ch03a019">
Now write begin term and end term only
 (a12+a22)(b12+b22) ---eqn.3.1 text page 37
=(a1*b1+a2*b2)2+(a1*b2-a2*b1)2
eqn.3.1 is an identity.
Red terms are Cauchy's Inequality.
Black term (a1*b2-a2*b1)2 measure
the error of Cauchy's Inequality.
<a name="ch03a020">
Black term is dropped from Cauchy's 
Inequality.
Black term tell us important information
that is 
when Black term become zero, under that
condition, Cauchy's Inequality become 
equality.
<a name="ch03a021">
Black term become zero is when
 a1*b2-a2*b1 = 0 ---eqn.AH048
eqn.AH048 is same as
 a1/b1=a2/b2 = λ ---eqn.AH049
eqn.AH049 tell us that sequence [a1,a2]
and sequence [b1,b2] are in proportion.
Proportional constant is λ. 
<a name="ch03a022">
We can write
 a1=λ*b1 ---eqn.AH050
 a2=λ*b2 ---eqn.AH051

View from geometry, eqn.AH049 says 
vector a and vector b point to same 
direction. The only difference is 
two vector length may be different.

<a name="ch03a023">
Over all conclusion is In 2-D space
when vector a and vector b point to 
same direction, Cauchy's Inequality 
become equality. Angle between 
vectors a and b are zero. (a, b are 
parallel). cosine(zero) = 1
inequality become equality. This is 
consistent with previous statement
at ch03a012.
2009-10-16-17-00 stop

2009-10-16-18-20 start
<a name="ch03a024">　Index begin　Index this file
■　Lagrange's Identity for n dimension

We reviewed two elements sequence.
But Cauchy's Inequality is unlimited 
in dimension. Assume we have two
sequences, each has n elements
sequence A: a1,a2,...,an
sequence B: b1,b2,...,bn
<a name="ch03a025">
Cauchy's Inequality in n dimension is
(a1*b1+a2*b2+...+an*bn)2 ---eqn.AH052
≦(a12+a22+...+an2)(b12+b22+...+bn2)
eqn.AH052 error term is the difference
of two side. Let Qn represent error
Qn = ---eqn.AH053
(a12+a22+...+an2)(b12+b22+...+bn2)
 - (a1*b1+a2*b2+...+an*bn)2

<a name="ch03a026">
To simplify eqn.AH053, in its positive
term (red)
let ai represent a1 ... an
let bj represent b1 ... bn
Red terms become
 SUM[i=1,i=n]{ai2}*SUM[j=1,j=n]{bj2}

<a name="ch03a027">
To simplify eqn.AH053, in its negative
term (blue)
Because blue terms are square, square 
represent product of two identical sets.
let aibi represent first  set a1b1 ... anbn
let ajbj represent second set a1b1 ... anbn
Blue terms become
 SUM[i=1,i=n]{aibi}*SUM[j=1,j=n]{ajbj}
Write in better equation as following

<a name="ch03a029">
2009-10-16-18-55 here
Can we find an identity with Qn
(eqn.3.2) as error term?
It is easier to work with a symmetry 
equation. a symmetry equation is if 
exchange i and j the result is same
as before.

<a name="ch03a030">
Negative term in eqn.3.2 is
aibiajbj
exchange i and j the result is
ajbjaibi
same as before !!

<a name="ch03a031">
Positive term in eqn.3.2 is
ai2bj2
exchange i and j the result is
aj2bi2
NOT same as before !! because [a] and
[b] are two different sequences.

<a name="ch03a032">　Index begin　Index this file
We must know i and j both sum whole 
range from 1 to n. Both are dummy
variables. That means if we replace
i or j with k (for example) result 
is the same. Based on this under-
standing, re-write eqn.3.2 as below.

<a name="ch03a034">
2009-10-16-19-12 here
Now in eqn.3.3 positive term is
symmetry.

In eqn.3.3, double sum is common, summed
terms can be put together, result is

<a name="ch03a036">
2009-10-16-19-30 here
Qn started from eqn.AH053 or eqn.3.2
Qn end at eqn.AH054
So eqn.3.2 = eqn.AH054 = Qn
The whole derivation is an identity.
Because it is symbolic equation 
calculation, result is valid for any
numerical value.

<a name="ch03a038">
2009-10-16-20-03 here
In eqn.3.4, the right most term with
coefficient 1/2 is the error term for
Cauchy's Inequality.

In eqn.3.4, the left term and the middle
term are Cauchy's Inequality.

Cauchy's Inequality is a corollary of
Lagrange's Identity.
2009-10-16-20-11 stop

2009-10-16-21-18 start
<a name="ch03a039">　Index begin　Index this file
■　n-dimensional equality condition

eqn.3.4 is a n-dimensional equation.
right end term is Qn, see eqn.AH054
Qn measure the magnitude of error
term in Cauchy's Inequality. When
Cauchy's Inequality become equality, 
then Qn must be zero which
require
 aibj－ajbi = 0 ---eqn.AH055
<a name="ch03a040">
re-write as
 ai/bi=aj/bj = λ ---eqn.AH056
index i and j are free, no constraint,
eqn.AH056 is valid for all i=1 to i=n
This is a condition for two sequences
a1,a2,...,an and b1,b2,...,bn
be proportional, or point to same 
direction in vector space.

<a name="ch03a041">
Same as 2-D case, in n-D space
when vector a and vector b point to 
same direction, Cauchy's Inequality 
become equality.
2009-10-16-21-38 stop

2009-10-17-07-57 done proofread
2009-10-17-11-28 done spelling check




<a name="docB001">　Index begin　Index this file
2009-10-14-12-15 start　fig2.5
This file 
http://freeman2.com/tute0014.htm
has Graph Fig2.5 AM-GM Inequality for
Complex Numbers. Explain as following.
[[
Graph area size, W:  H:  [Plot3]　 figure down below 
x min:  , x max:  ; y min:  , y max:  ; 
Graph title: 
]]
<a name="docB002">
Drawing board size W: and H: can be 
specified. Larger size avoid points 
crowded together. 
Default W: 300 and H: 300
Click [Plot3] draw figure display at below.
<a name="docB003">　fig2.5
[[
x min:  , x max:  ; y min:  , y max:  ; 
]]
is NOT used in Fig2.5 Program auto detect
x, y range.
<a name="autoscale">　fig2.5
2009-10-14-21-57 changed code, add 
auto scale, user scale options. 
Suggest let auto scale run first, then
use user scale for fine tuning to get
better x/y min/max.

<a name="docB004">
Box 1 accept complex number in the form
a+i*b. When click [Plot3], program convert
a+i*b to rho,theta format. For example
1+2i  is  2.23606797749979,1.1071487177940904
Polar form is rho=sqrt(a*a+b*b)
      and     theta=atan2(b,a)
Polar form data is stored in Box 2.
<a name="docB005">　fig2.5
Box 1 accept data with different points
separator. Newline or ';' or ',' or tab.
Program use the separator to split data.

If you specify wrong separator, program
can not work.
"Box 1 complex separator" do not apply to 
Box 2.
Program draw Box 2 data, NOT Box 1 data.

<a name="docB006">
Box 2 accept data ONLY one line one point.
If your data is rho,theta polar format,
make sure they are one line one point.
Paste polar format to box 2 and delete
box 1 , make box 1 blank. Then click
[Plot3]

<a name="docB007">　fig2.5
Below Box 2, it has random number generator.
Click [random#] to get random number in
both Box 1 and Box 2. 
Modify Box 1 data, it become effective.
Modify Box 2 data, it does not work.
Because program look for box 1 first.

<a name="docB008">
[equal#] button generate [10^] box number
[fill] box number times and store to Box 1.
[equal#] do not generate complex number.
[equal#] button has no use here.

<a name="docB009">　fig2.5
Check box  +/0 , +/0/-  control complex
number real part. Imaginary part +/- can
not be controlled.
Exercise 2.13 you should click +/0 and
should NOT click +/0/-. See
Special point to Exercise 2.13 Sp1 Sp2

<a name="docB010">
If you click +/0/-, program still draw
points. But violate Exercise 2.13 basic
assumption.

Box 3 message is for debug output.
2009-10-14-12-53 stop

<a name="docB011">　fig2.5
2009-10-16-10-35 start
On 2009-10-16 add [Point Label: ] box,
allow user to re-define point label. 
If let [Point Label: ] box be empty, 
program use default point label 'A',
  'B','C', ...
If let [Point Label: ] box begin with
',', program delete point label, draw
point only.
<a name="docB012">
If let [Point Label: ] box be string
and use ',' as separator, for example
 [α,β,γ,δ,ε]
program use user defined point label.

[Point Label: ] box right end has [A],
[B], [C], [d1] four buttons.
Click [A], fill [,], draw point only.
Click [B], use Greek alphabet as label.
Click [C], use Chinese character as label.
Click [d1], empty box, use default value.
You can create your own string, use comma
as separator and paste string to
[Point Label: ] box.
2009-10-16-10-43 stop

<a name="docB013">
2009-11-05-20-34 start
Update 2009-11-05 change auto scale
code, see source code at time stamp
"9811052018"
2009-11-05-20-36 stop

<a name=20091217>
2009-12-17-11-09 start
Update 2009-12-17 change all tute*.htm
(from tute0007.htm to tute0023.htm)
first:
Correct 'Limit' link from '#docA06'
to '#docA006'
second:
Change Javascript index to read from
jslist1e.js so that update jslist1e.js
then update ALL tute*.htm.
2009-12-17-11-23 stop

<a name=20100815>
2010-08-15-21-52 start
Update 2010-08-15 add alert about
complex number separator error
Store input data at Box1
If "Box 1 complex separator:"
selection not present in data
then issue warning.
Long time not use tute0014.htm
even LiuHH wonder a while when
write study notes at 
tute0051.htm#ch14c194
At the same time add five link
at alert spot.
2010-08-15-21-58 stop

<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56