Inequality Study 44th file   Update 2010-06-14
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The Cauchy-Schwarz Master Class   J. Michael Steele   ★★★★★
This file is personal home work. No one
proofread. Cannot promise correctness.
If you suspect any view point wrong,
please ask a math expert near by.
Freeman 2009-06-19-10-46
Please use MSIE browser to read this file.
Did not test other browser. This file is
written under MSIE 6.0 




<a name="docA001">　Index begin　Index this file
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 stop








<a name="ch12c001">　Index begin　Index this file
2010-05-25-18-02 start
■　Problem 12.2 (A Symmetric 
　　Appetizer)
Show that for nonnegative x, y, 
and z one has the bound
<a name="ch12c002">
  x2y3+x2z3+y2x3+y2z3+z2x3+z2y3
 ≦xy4+xz4+yx4+yz4+zx4+zy4 ---eqn.12.16
and take inspiration from your
discoveries to generalize this
result as widely as you can.
2010-05-25-18-10 stop

<a name="ch12c003">
2010-05-25-18-20 start
eqn.12.16 has even power. Assume
that x,y,z are all the same 
physics quantity, for example
all be length. Then x2y3 and 
xz4 etc. all be length fifth 
power. AM-GM inequality relate
inequality between like term.
GM xz4 ≦ AM xz4 OK
GM x2y3 ≦ AM xz4 NO
We need power manipulation.
<a name="ch12c004">
We know GM≦AM. eqn.12.16 has 
x2y3+...≦xz4+... We manipulate 
power, hope express power 2,3 
in terms of power 1,4, that is
hope express x2y3 etc. in terms 
of xz4 etc. Assume
  a2b3=(ab4)ζ*(a4b)η ---eqn.BL001
for some pure number ζ and η
<a name="ch12c005">　Index begin　Index this file
Expand eqn.BL001 right side
  a2b3=aζ+4η*b4ζ+η ---eqn.BL002
Equate 'a' power get
  2=ζ+4η ---eqn.BL003
Equate 'b' power get
  3=4ζ+η ---eqn.BL004
Two equation, two unknown ζ,η.
Perfect. Solve for ζ,η get
  ζ=2/3 ---eqn.BL005
  η=1/3 ---eqn.BL006
<a name="ch12c006">
Now eqn.BL001 become
  a2b3=(ab4)2/3*(a4b)1/3 ---eqn.BL007
ζ=2/3 and η=1/3 is the only 
value which balance eqn.BL007.
That is (ab4)2/3*(a4b)1/3 also 
has length to fifth power.
Please check !
<a name="ch12c007">
Equality and inequality and
addition and subtraction must
relate two quantity with same
physics dimension.
<a name="ch12c008">
Multiplication and division
relate two different physics 
quantity and create new 
physics quantity. For example
d[length function]/d[time] 
get velocity. 
Velocity is not length.
Velocity is not time.
Velocity is length/time.

<a name="ch12c009">　Index begin　Index this file
OK, eqn.BL007 is here, 
what is next?
We need eqn.2.9
AM-GM Inequality is general for any
 irrational weights (page 23, eqn.2.9)
a1p1a2p2...anpn ≦ p1a1+p2a2+...+pnan ---eqn.2.9
ALERT: p1+p2+...+pn=1 ---eqn.AE20
<a name="ch12c010">
AM-GM Inequality require p1+...+pn 
sum to one. We have ζ=2/3 and
η=1/3 sum to one. OK
AM-GM Inequality require
 0≦p1,...,pn≦1
We have ζ=2/3 and η=1/3 both 
are in [0,1]. OK

<a name="ch12c011">
AM-GM Inequality GM (less than)
side has p1,...,pn in power
position. Converted eqn.BL007
has ζ and η in power position.
eqn.BL007 right side, ζ and η 
are p1, p2 (for GM).

<a name="ch12c012">
AM-GM Inequality AM (greater 
than) side has p1,...,pn in 
coefficient position. After 
apply (not yet now) AM-GM,
eqn.BL007 must have ζ and η 
in greater than side AM sum's 
coefficient position. That is 
based on AM-GM Inequality, 
eqn.BL001 and eqn.BL007 should 
have the following inequality
<a name="ch12c013"> 　Index begin　Index this file
 
a2b3 = (ab4)2/3*(a4b)1/3 ≦
2
3
 ab4 +
1
3
 a4b
---page 185
---line 4
---eqn.12.17
Equality is from eqn.BL007. Inequality is from eqn.2.9.
Blue a2b3 do not have power 2/3, 1/3. Blue term can
not apply AM-GM. After eqn.BL007 conversion
Red (ab4)2/3*(a4b)1/3 can apply AM-GM Inequality.
width <a name="ch12c014">
Exchange a,b we get the following
 
a3b2 ≦
2
3
 a4b +
1
3
 ab4
---page 185
---line 4 aux1
---eqn.BL008
width <a name="ch12c015">
Add a2b3 with a3b2 get
 
a2b3 + a3b2 ≦
2+1
3
 a4b +
1+2
3
 ab4
---page 185
---line 4 aux2
---eqn.BL009
width of above equation 
<a name="ch12c016">　Index begin　Index this file
Which is
  a2b3 + a3b2 ≦ a4b + ab4 ---eqn.BL010
Change a to x, change b to y:
  x2y3 + x3y2 ≦ x4y + xy4 ---eqn.BL011
Change a to y, change b to z:
  y2z3 + y3z2 ≦ y4z + yz4 ---eqn.BL012
Change a to z, change b to x:
  z2x3 + z3x2 ≦ z4x + zx4 ---eqn.BL013
Sum eqn.BL011, eqn.BL012, eqn.BL013
to get final answer eqn.12.16
2010-05-25-19-27 stop

<a name="ch12c017">　Index begin　Index this file
2010-05-25-20-21 start
■　Explore ζ,η, error analysis.
Problem 12.2 give eqn.12.16 
indicate inequality direction. 
That is very kind. What if 
problem give
  x2y3+x2z3+y2x3+y2z3+z2x3+z2y3
＜?＝?＞xy4+xz4+yx4+yz4+zx4+zy4 ---eqn.BL014
and if choose wrong start point,
<a name="ch12c018">
that is, instead of
  a2b3=(ab4)ζ*(a4b)η ---eqn.BL001
if we start from
  ab4=(a2b3)ζ*(a3b2)η ---eqn.BL015
what do we get? Explore !!
<a name="ch12c019">
Following is error on purpose.
Expand eqn.BL015 right side, get
  ab4=a2ζ+3η*b2η+3ζ ---eqn.BL016
Equate 'a' power get
  1=2ζ+3η ---eqn.BL017
Equate 'b' power get
  4=3ζ+2η ---eqn.BL018
<a name="ch12c020">　Index begin　Index this file
Two equation, two unknown ζ,η.
Perfect. Solve for ζ,η get
  ζ=2  ---eqn.BL019
  η=-1 ---eqn.BL020
Now eqn.BL015 become
  ab4=(a2b3)2*(a3b2)-1 ---eqn.BL021
Can we Apply AM-GM Inequality to
eqn.BL021 ?
<a name="ch12c021">
ζ=2 and η=-1 sum to one is OK
BUT AM-GM also require 0≦ζ,η≦1 
which is violated in this explore
process. We can NOT apply AM-GM 
Inequality.
2010-05-25-20-42 here

//<a name="ch12c022">
//Test a set data, let 
x=1, y=3, z=6, 
//rewrite eqn.BL014 as
aa=+x*x*y*y*y+x*x*z*z*z+y*y*z*z*z+y*y*x*x*x+z*z*x*x*x+z*z*y*y*y
bb=+x*y*y*y*y+x*z*z*z*z+y*z*z*z*z+y*x*x*x*x+z*x*x*x*x+z*y*y*y*y
aa //aa is x^2*y^3 eqn.12.16 less than side
bb //bb is x^1*y^4 eqn.12.16 greater than side
//calculator local how to use
//complex2.htm#calculator output
aa
3204
bb
5760
Change x,y,z value, always get
  aa≦bb

<a name="ch12c023">
only eqn.12.16 is valid, 
ζ=2 and η=-1 violate 0≦ζ,η≦1
eqn.BL015 give us wrong answer.
2010-05-25-21-07 stop

<a name="ch12c024">　Index begin　Index this file
2010-05-26-16-08 start
■　What is convex hull?

Let us start from simplest one
dimension analysis. Assume two
reference points on x-axis
a locate at x=1
b locate at x=2
A third moving point p
p locate at x=1.7
<a name="ch12c025">
How to express p in terms of 
a and b? Assume
  p=ζ*a+η*b ---eqn.BL022
Here ζ,η and p are variables.
This is one degree of freedom
problem, let us assign
  ζ+η=1 ---eqn.BL023
<a name="ch12c026">
Eliminate η, eqn.BL022 become
  p=ζ*a+(1-ζ)*b ---eqn.BL024
Now given
  a=1  ---eqn.BL025
  b=2  ---eqn.BL026
  p=1.7 ---eqn.BL027
<a name="ch12c027">
solve for ζ get
  1.7=ζ*1+(1-ζ)*2 ---eqn.BL028
  1.7=ζ*1+2-2*ζ
  ζ(2-1)=2-1.7
  ζ=0.3 ---eqn.BL029
  η=1-ζ=0.7 ---eqn.BL030
<a name="ch12c028">
The relation
  1＜1.7＜2 or a＜p＜b
is p interpolation between a
and b.
If ζ=0, eqn.BL024 say p=b
If ζ=1, eqn.BL024 say p=a
<a name="ch12c029">　Index begin　Index this file
Key point is
Interpolation give us
  0≦ζ,η≦1 ---eqn.BL031
and
  ζ+η=1 ---eqn.BL032
Above is interpolation.
<a name="ch12c030">
Below is extrapolation.
Now given
  a=1  ---eqn.BL025
  b=2  ---eqn.BL026
  p=2.7 ---eqn.BL033 change !
<a name="ch12c031">
Solve eqn.BL024 for ζ get
  2.7=ζ*1+(1-ζ)*2 ---eqn.BL034
  2.7=ζ*1+2-2*ζ
  ζ(2-1)=2-2.7
  ζ=-0.7 ---eqn.BL035
<a name="ch12c032">
We still require
  ζ+η=1 ---eqn.BL031
then
  η=1-ζ=1.7 ---eqn.BL036
<a name="ch12c033">
Extrapolation give us 
  ζ≦0 and 1≦η ---eqn.BL037
and
  ζ+η=1 ---eqn.BL032
eqn.BL037 indicate extrapolation,
many mathematics theory not work.

<a name="ch12c034">　Index begin　Index this file
If a,b and p are not on x-axis
they are on an arbitrary straight
line. Above interpolation result
eqn.BL031, eqn.BL032 and
extrapolation result eqn.BL037, 
eqn.BL032 are still true.
You can verify.

<a name="ch12c035">
Next still do interpolation and
extrapolation. Change from 1D
straight line to 2D triangle.
Assume triangle vertices A,B,C
have the following coordinates
Treat x-axis as real axis and
treat y-axis as imag axis. 
<a name="ch12c036">
Use complex number expression.
  A=a+bi ---eqn.BL038
  B=c+di ---eqn.BL039
  C=e+fi ---eqn.BL040
A,B,C are base points. Next
define point P
  P=m+ni ---eqn.BL041
How to express P coordinates
in terms of A,B,C coordinates?
2010-05-26-16-48 here

<a name="ch12c037">
Try matrix expression
 [1] [1  1  1] [ζ]
 [m]=[a  c  e]*[η] ---eqn.BL042
 [n] [b  d  f] [γ]
expand eqn.BL042 get
  1 = ζ  + η + γ  ---eqn.BL043
  m = a*ζ+c*η+e*γ ---eqn.BL044
  n = b*ζ+d*η+f*γ ---eqn.BL045
2010-05-26-16-58 here
<a name="ch12c038">
Solve for ζ, η, γ
a*(eqn.BL043) get
  a = a*ζ+a*η+a*γ ---eqn.BL046
eqn.BL046-eqn.BL044 
eliminate a*ζ get
 a-m=a*η+a*γ-c*η-e*γ ---eqn.BL047
or
 a-m=(a-c)*η+(a-e)*γ ---eqn.BL048

<a name="ch12c039">　Index begin　Index this file
b*(eqn.BL043) get
  b = b*ζ+b*η+b*γ ---eqn.BL049
eqn.BL049-eqn.BL045 
eliminate b*ζ get
 b-n=b*η+b*γ-d*η-f*γ ---eqn.BL050
or
 b-n=(b-d)*η+(b-f)*γ ---eqn.BL051

<a name="ch12c040">
Next find η,γ
(a-c)*(eqn.BL051) get
 (a-c)*(b-n)=(a-c)*(b-d)*η+(a-c)*(b-f)*γ ---eqn.BL052

(b-d)*(eqn.BL048) get
 (b-d)*(a-m)=(a-c)*(b-d)*η+(a-e)*(b-d)*γ ---eqn.BL053
<a name="ch12c041">
eqn.BL052-eqn.BL053 to cancel
(a-c)*(b-d)*η, get
  (a-c)*(b-n)-(b-d)*(a-m)
=[(a-c)*(b-f)-(a-e)*(b-d)]*γ ---eqn.BL054

<a name="ch12c042">
  γ=[(a-c)*(b-n)-(a-m)*(b-d)]
   /[(a-c)*(b-f)-(a-e)*(b-d)] ---eqn.BL055
2010-05-26-17-15 here
Cyclic variables, get
  η=[(e-a)*(f-n)-(e-m)*(f-b)]
   /[(e-a)*(f-d)-(e-c)*(f-b)] ---eqn.BL056

<a name="ch12c043">
  ζ=[(c-e)*(d-n)-(c-m)*(d-f)]
   /[(c-e)*(d-b)-(c-a)*(d-f)] ---eqn.BL057
2010-05-26-17-29 here

<a name="ch12c044">　Index begin　Index this file
  A=1.5+1.2i ---eqn.BL058
  B=0.8+4i   ---eqn.BL059
  C=3.2+5.2i ---eqn.BL060
A,B,C are base points. Next
define point P
  P=1.8+2.8i ---eqn.BL061

<a name="ch12c045">
a=1.5,b=1.2,c=0.8,d=4,e=3.2,f=5.2,m=1.8,n=2.8;
alpha=((c-e)*(d-n)-(c-m)*(d-f))/((c-e)*(d-b)-(c-a)*(d-f));
beta =((e-a)*(f-n)-(e-m)*(f-b))/((e-a)*(f-d)-(e-c)*(f-b));
gamma=((a-c)*(b-n)-(a-m)*(b-d))/((a-c)*(b-f)-(a-e)*(b-d));
alpha
beta 
gamma
2010-05-26-17-34 stop
ezGraph <a name="fig1202"> 2010-05-26-18-26
Output may contain error, Please verify first
Program environment is MSIE 6.0, please use MSIE
If you save this file tute0044.htm to your computer
and open local tute0044.htm, it can not draw figure.
You need also save http://freeman2.com/jsgraph2.js
to your computer stay in same folder as tute0044.htm
Graph area size, W: H:  
x min: , x max: ; y min: , y max: ;
x/y min/max : auto scale, user scale Click button
Box11
Box12

       



<a name="convex01"> 2009-12-04-20-30
Star that you can shape   　Index begin　Index this file
Number of star corners No chord
bottom to top ratio     



<a name="ch12c046">
2010-05-27-12-22 start
If given
  a=1  ---eqn.BL025
  b=2  ---eqn.BL026
  p=1.7 ---eqn.BL027
<a name="ch12c047">
Alert a＜p＜b, p is in [a,b].
we get interpolation result
  0≦ζ,η≦1 ---eqn.BL031
and
  ζ+η=1 ---eqn.BL032

<a name="ch12c048">
If given
  a=1  ---eqn.BL025
  b=2  ---eqn.BL026
  p=2.7 ---eqn.BL033
<a name="ch12c049">
Alert a＜b＜p. p is not in [a,b].
we get extrapolation result
  ζ=-0.7 ---eqn.BL035
  η=1-ζ=1.7 ---eqn.BL036
and
  ζ+η=1 ---eqn.BL032

<a name="ch12c050">　Index begin　Index this file
Major mathematical theory treat
interpolation, not extrapolation.
Interpolation relation
  a＜p＜b, p in [a,b]
is called 
"p is in convex hull of a,b"

<a name="ch12c051">
Above is interpolation in one 
dimension, need ζ,η two coef.
Interpolation in two dimension
is illustrated in ezGraph button
"Draw ∆ABC&P"
Need ζ,η,γ three coefficients.

<a name="ch12c052">
Triangle ABC three vertices A,B,C
are reference points. Any point 
P x/y coordinates can be expressed 
in vertices A,B,C x/y coordinates 
  Px=ζ*Ax+η*Bx+γ*Cx ---eqn.BL062
and
  Py=ζ*Ay+η*By+γ*Cy ---eqn.BL063
<a name="ch12c053">
ζ+η+γ always sum to one
  ζ+η+γ=1 ---eqn.BL064
If point P is inside ∆ABC then 
   0≦ζ,η,γ≦1
If point P is outside ∆ABC, 
   not all 0≦ζ,η,γ≦1
<a name="ch12c054">
If 0≦ζ,η,γ≦1, we can apply AM-GM
   inequality.
If point P is outside ∆ABC we can 
   not apply (If apply AM-GM, all
   AM＜＝＞GM are possible no help)

<a name="ch12c055">　Index begin　Index this file
In many cases, the interpolation
equation
  p=ζ*a+(1-ζ)*b ---eqn.BL024
is a POWER equation. That is
eventually, p is a power of
another term. For example, in
one dimensional case, 
<a name="ch12c056">
after we find ζ,η values
  ζ=2/3 ---eqn.BL005
  η=1/3 ---eqn.BL006
ζ,η interpolate 2 in [1,4]
  2=ζ+4η ---eqn.BL003
ζ,η interpolate 3 in [4,1]
  3=4ζ+η ---eqn.BL004
<a name="ch12c057">
This '2' and '3' are power of
next equation.
  a2b3=(ab4)2/3*(a4b)1/3 ---eqn.BL007
Because both 0≦ζ,η≦1, we can
apply AM-GM inequality to
right side of eqn.BL007.
2010-05-27-13-01 stop



<a name="ch12c058">　Index begin　Index this file
2010-05-27-14-22 start
Figure 12.2 is a figure of one
dimensional interpolation.
Boundary points are (1,4) and 
(4,1). Middle point is (2,3)
points (1,4) and (4,1) form
a convex hull.
<a name="ch12c059">
How can one dimensional domain
not be a convex hull? Is there
any example? Please go to
Star that you can shape and 
click "Draw ConvexA" button.
<a name="ch12c060">
Bottom of this figure has two
one dimensional domain. 
Red domain is continuous.
Red domain is convex hull.
Blue domain is discontinuous.
Blue domain is not convex hull.
<a name="ch12c061">
This figure was first done in
tute0022.htm#convex01 Now here
tute0044.htm re-use this figure
add convex hull red polygon go
around blue star.

<a name="ch12c062">
LiuHH had few questions:
"Besides convex hull,
 what other hull are they?"
"Why ignore other hull, only
 emphasize convex hull?"

<a name="ch12c063">　Index begin　Index this file
LiuHH figure out that 
Besides convex hull, the other 
hull is bone hull. (NO ONE use
this term 'bone hull' Alert !)
Bone hull is the blue star.
Blue bone hull go from star 
crest to star valley. Repeat.
<a name="ch12c064">
Red convex hull go from star 
crest to crest. Never go to
star valley. Please go to
Star that you can shape and 
click "Draw ConvexA" button.

<a name="ch12c065">
"Why ignore other hull, only
 emphasize convex hull?"
Because any point outside of
bone hull and inside of convex 
hull can be represented by
interpolation of star vertices
instead of extrapolation.
<a name="ch12c066">
Interpolation allow us apply 
most inequality theorems.
Convex hull is border line,
bone hull is not.

Above are LiuHH guessed reason,
it could be wrong.
2010-05-27-14-50 here



<a name="ch12c067">　Index begin　Index this file
2010-05-27-14-55
■　Problem 12.3
　　(Muirhead's inequality)
Given that α in H(β)
(α in Hull of β)
where //well ill example. how
  α=(α1,α2,...,αn) ---eqn.BL065
and
  β=(β1,β2,...,βn) ---eqn.BL066
Show that for all positive
x1,x2,...,xn one has the bound
<a name="ch12c068">
　
∑
σ∈Sn
 xσ(1)α1 xσ(2)α2 ... xσ(n)αn ---eqn.12.18
≦
　
∑
σ∈Sn
 xσ(1)β1 xσ(2)β2 ... xσ(n)βn
width of above equation 
<a name="ch12c069">
Metaphor help to remember:
All α1,α2,...,αn stay in hull 
formed by β1,β2,...,βn
βi's are container
αj's are small things stored
in container.
<a name="ch12c070">
Container is greater than small 
things.
x's βi power (container power)
is greater than or equal to
x's αj power (small things power)

<a name="ch12c071">
Key point to watch:
Muirhead's inequality use three
sequences
α1,α2,...,αn
β1,β2,...,βn
and
x1,x2,...,xn
(Most other inequalities use
 two sequences.)
<a name="ch12c072">
Both αj and βi MUST BE pure 
number power index. 
xk is physics quantity. 
xk can be length or mass etc.
xk must be non-negative.
α, β can be fraction. Negative 
raise to fraction power get 
complex, which we want to avoid.
<a name="ch12c073">　Index begin　Index this file
Summation of αj MUST equal to
summation of βi.
  α1+α2+...+αn=β1+β2+...+βn ---eqn.BL067
Otherwise eqn.12.18 is something
like length compare with area.
This comparison is NOT accepted.
<a name="ch12c074">
If xk is length, if αj sum to one
then eqn.12.18 α power side is 
length.
If xk is length, if βi sum to two
then eqn.12.18 β power side is 
area.
Length can not compare with area,
We demand that SUM αj=SUM βi
2010-05-27-15-48 stop

<a name="ch12c075">　Index begin　Index this file
2010-05-27-17-58 start
■　A Quick Orientation
To explain Muirhead's inequality
textbook use Problem 12.2 as an
example. Three sequences are
  (α1,α2,α3)=(2,3,0) ---eqn.BL068
  (β1,β2,β3)=(1,4,0) ---eqn.BL069
  (x1,x2,x3)=(x,y,z) ---eqn.BL070
<a name="ch12c076">
First check necessary condition
SUM αj=SUM βi eqn.BL067. 
From eqn.BL068 SUM αj=2+3+0=5
From eqn.BL069 SUM βi=1+4+0=5
SUM αj=SUM βi is true.

<a name="ch12c077">
We need also check that one (αj)
is contained by the other (βi).
Domain αj=[0,3] //0=min(2,3,0)
Domain βi=[0,4] //4=max(1,4,0)
Domain βi is hull, domain αj is
small things in container (hull).

<a name="ch12c078">
Hull relation check is necessary.
For example if give ill problem
domain αj=[0,3]
domain βi=[2,4]
then αj elements 0,1 are
extrapolated by βi=[2,4], 
not interpolated.

<a name="ch12c079">
eqn.BL068 and eqn.BL069 passed
all checks. Now substitute
  (α1,α2,α3)=(2,3,0) ---eqn.BL068
  (β1,β2,β3)=(1,4,0) ---eqn.BL069
into Muirhead's Inequality
eqn.12.18
<a name="ch12c080">　Index begin　Index this file
First, do
  (α1,α2,α3)=(2,3,0) ---eqn.BL068
for
  (x1,x2,x3)=(x,y,z) ---eqn.BL070
Power α1=2 has three choices
either x or y or z.
<a name="ch12c081">
After one of (x,y,z) stay with α1
the second α2=3 has two choices 
out of three (x,y,z)
The last one α3=0 has one choice
Total possible permutation is
3*2*1 = 3! = 6 terms. 
<a name="ch12c082">
Power α1=2 first possibility
  x1α1x2α2x3α3=x2y3z0 ---eqn.BL071
 and //x1 take α1 as power
  x1α1x3α2x2α3=x2z3y0 ---eqn.BL072
Power α1=2 second possibility
  x2α1x3α2x1α3=y2z3x0 ---eqn.BL073
 and //x2 take α1 as power
  x2α1x1α2x3α3=y2x3z0 ---eqn.BL074
<a name="ch12c083">
Power α1=2 third possibility
  x3α1x1α2x2α3=z2x3y0 ---eqn.BL075
 and //x3 take α1 as power
  x3α1x2α2x1α3=z2y3x0 ---eqn.BL076
Above six equations permute 
a whole cycle for three 
variables problem. There
are total n!=3!=6 terms sum 
together. 
<a name="ch12c084">
Any term zero-th power is 
pure number one. 
(time/time=pure number 1)
Sum above six terms is in
textbook page 186, α sum
  ∑[σ∈S3]xσ(1)α1xσ(2)α2xσ(3)α3
 =x2y3+x2z3+y2z3+y2x3+z2x3+z2y3
 = ---eqn.BL077
<a name="ch12c085">　Index begin　Index this file
While the β sum is given by
  ∑[σ∈S3]xσ(1)β1xσ(2)β2xσ(3)β3
 =xy4+xz4+yz4+yx4+zx4+zy4
 = ---eqn.BL078
We apply Muirhead's inequality
and get answer that is (power)
α sum≦β sum eqn.12.16

<a name="ch12c086">
Constant requirement to α,β
is that summation of αj 
MUST equal to summation of βi.
Please see eqn.BL067
If any α or β are negative 
value, Muirhead's inequality 
still apply. Two attentions:
(1) α or β can be negative, 
    but pk in ∑α=∑βkpk must
    have 0≦p1,...,pn≦1
(2) ∑α must= ∑β NOT = 1 OK
    BUT ∑pk must = 1
<a name="ch12c087">
For example if
  (α1,α2,α3)=(1/2,1/2,0) ---eqn.BL079
  (β1,β2,β3)=(-1,2,0) ---eqn.BL080
β domain [-1,2] contain 
α domain [0,1/2], then for 
positive x,y,z Muirhead say:
  2[√(xy)+√(yz)+√(zx)]≦
  x*x/y + x*x/z + y*y/z + y*y/x
 +z*z/x + z*z/y  ---eqn.12.19
'√(mn)' come from α power 1/2
'm*m' (x*x) come from β power 2
'1/n' (1/y) come from β power -1
<a name="ch12c088">
eqn.12.19 less than side has 
a factor 2. Greater than side
no such factor. That is because
in eqn.BL079 α1=α2=1/2
This equality let us have 2!
(not just 2) repeats.
<a name="ch12c089">
IF eqn.BL079 is (1/3,1/3,1/3)
then we have six identical
terms (xyz)1/3, in this case 
instead of write six times
+(xyz)1/3, write (3!)*(xyz)1/3
2010-05-27-20-00 stop

<a name="ch12c090">　Index begin　Index this file
2010-05-27-21-09 start
■　Proof of Muirhead's inequality

Muirhead's Inequality is 
eqn.12.18. Left side is sum 
of product of xσ(i)αi permute
a whole cycle. We assumed 
that α power can be written
as interpolation of β power.
(α is in β convex hull) 
<a name="ch12c091">
We can write the following
equality (power of xσ(i))
  (α1,α2,...,αn) ---eqn.BL081
 =∑[τ∈Sn]pτ(βτ(1),βτ(2),...,βτ(n))
where //　α,β vs. pk
  pτ≧0 ---eqn.BL082
and 
  ∑[τ∈Sn]pτ=1 ---eqn.BL083
<a name="ch12c092">
Interpolation, extrapolation
both satisfy eqn.BL083.
Only interpolation satisfy 
eqn.BL082.
When solve problem, do not 
stop at ∑pτ=1, must make 
sure each individual 0≦pτ≦1
<a name="ch12c093">
eqn.BL081 is from assumption.
eqn.BL081 is power equation
for data array x1,x2,...,xn
xk to α power can not apply 
AM-GM inequality.
xk to β power can apply AM-GM 
inequality.
Because β power has eqn.BL082
and eqn.BL083, α power not.
<a name="ch12c094">　Index begin　Index this file
Let us put power equation
eqn.BL081 on top of data 
array. We find
xσ(1)α1xσ(2)α2...xσ(n)αn ---eqn.BL084
=∏[τ∈Sn]{xσ(1)βτ(1) * xσ(2)βτ(2) * ... * xσ(n)βτ(n)}pτ
Above '∏' for n=3 case is eqn.BL120 right side.
eqn.BL084 is one term of total n! terms. For n=3
detail calculation can be found at eqn.BL119
<a name="ch12c095"> 2010-05-28-11-20 add start
Why '∏'? Example of eqn.BL084 in two variables
is eqn.BL007. One term a2b3 = (a1b4)2/3*(a4b)1/3
Power αi is small thing 2,3; power βj is hull 1,4;
power pτ is power 2/3 and 1/3. For one a2b3
2/3 + 1/3 sum to 1 once. But xσ(k)αk permute
n times. Total 2/3 + 1/3 sum to 1 take place
n times. Above simple example n=2. Other one is
a3b2 = (a1b4)1/3*(a4b)2/3 Check power balance.
a: 3= 1*1/3 +4*2/3 ; b: 2= 4*1/3 +1*2/3
2010-05-28-11-26 add stop 
<a name="ch12c096">
eqn.BL084 is an equality. 
After permutation, there are 
total n! terms like eqn.BL084.
Now we apply AM-GM inequality 
to eqn.BL084 β-term side.
<a name="ch12c097">
Because eqn.BL082 and eqn.BL083
serve β-terms (not α-terms). 
AM-GM inequality is valid.
Reader need review eqn.AZ028 
2010-05-27-22-10 stop

<a name="ch12c098">
2010-05-28-11-36 start
Next step is sum eqn.BL084 
for all permutation, get 
textbook page 188 line two 
equation left side.
<a name="ch12c099"> 　Index begin　Index this file
∑[σ∈Sn]{xσ(1)α1*xσ(2)α2*...*xσ(n)αn} ---eqn.BL085
=∑[σ∈Sn]∏[τ∈Sn]{xσ(1)βτ(1) * xσ(2)βτ(2) * ... * xσ(n)βτ(n)}pτ
Blue pτ is GM power. Red pτ is AM coefficient.
Blue ∏ is GM product. Red ∑ is AM summation.
Blue = before AM-GM. Red ≦ after AM-GM.
Real expansion for one term and n=3 case is here
<a name="ch12c100">
Next step, apply AM-GM inequality to β-term side.
eqn.BL085 right hand side is GM (Geometric Mean)
After apply AM-GM inequality, GM power pτ change
to AM (Arithmetic Mean) coefficient pτ.
<a name="ch12c101">
Freshman level reader, please pay special attention to
the change from GM power pτ to AM coefficient pτ.
Please read AM-GM inequality chapter 2.
Next step is eqn.BL085 right side
GM power pτ to AM coefficient pτ and
GM product symbol ∏ to AM summation symbol ∑
<a name="ch12c102"> eqn.BL086 is important step.
This step apply AM-GM inequality and equality sign
in eqn.BL085 become inequality sign in eqn.BL086
∑[σ∈Sn]{xσ(1)α1*xσ(2)α2*...*xσ(n)αn} ---eqn.BL086
≦∑[σ∈Sn]∑[τ∈Sn]pτ{xσ(1)βτ(1) * xσ(2)βτ(2) * ... * xσ(n)βτ(n)}
Blue pτ is GM power. Red pτ is AM coefficient.
Blue ∏ is GM product. Red ∑ is AM summation.
Blue = before AM-GM. Red ≦ after AM-GM.
<a name="ch12c103"> 　Index begin　Index this file
AM coefficient pτ is not a function of σ. Move pτ
out of σ summation. get next equation
∑[σ∈Sn]{xσ(1)α1*xσ(2)α2*...*xσ(n)αn} ---eqn.BL087
≦∑[τ∈Sn]pτ∑[σ∈Sn]{xσ(1)βτ(1) * xσ(2)βτ(2) * ... * xσ(n)βτ(n)}
The term //real calculation for n=3 is at eqn.BL122
∑[τ∈Sn]pτ=1 ---eqn.BL088 ; see eqn.BL083
sum to one and not show up in equation.
<a name="ch12c104">
τ factor disappear. Change βτ(k) to βk get
　　∑[σ∈Sn]{xσ(1)α1 * xσ(2)α2 * ... * xσ(n)αn} ---eqn.BL089
≦∑[σ∈Sn]{xσ(1)β1 * xσ(2)β2 * ... * xσ(n)βn}
eqn.BL089 is Muirhead's inequality eqn.12.18.
Problem 12.3　Muirhead's inequality is solved.
2010-05-28-14-40 stop 
<a name="ch12c105">　Index begin　Index this file
2010-05-28-18-05 start
■　How do I know given two 
　　sequences α and β have 
　　"α in H(β)" relation?
　　Guide line
Problem 12.3　Muirhead's 
inequality first word is
"Given that α in H(β)"
"α in H(β)" is defined in 
Chapter 13. Here Chapter 12
discuss Muirhead's inequality,
need peek the future work.
<a name="ch12c106del">
2010-06-09-15-07 start
alert the following is wrong. 
Correct argument is
interpolation and majorization 
are NOT related. For example
α sequence=[5,3,3,2,2]
β sequence=[5,4,3,2,1]
α is in β hull, but the following
permuted matrix calculation give
answer out side of [0,1].
2010-06-09-15-11 stop

<a name="ch12c106">
2010-06-09-15-05 change to silver
indicate wrong argument.
A simple example.
If α power is
  α1=[3,2,1] ---eqn.BL090
If β power is
  β1=[4,1,1] ---eqn.BL091
Let us find interpolation
coefficients p1, p2, p3
for α1 and β1.
<a name="ch12c107">
First let 3 (from [3,2,1])
express by β1=[4,1,1]
  3=p1*4+p2*1+p3*1 ---eqn.BL092
Second let 2 ([3,2,1]) be
  2=p1*1+p2*1+p3*4 ---eqn.BL093
[4,1,1] permute to [1,1,4]
Third let 1 ([3,2,1]) be
  1=p1*1+p2*4+p3*1 ---eqn.BL094
[4,1,1] permute to [1,4,1]
Above three equations, left
side change number, right
side change permutation.
<a name="ch12c108">
eqn.BL092 to eqn.BL094 write
in matrix form it is next
  [3] [4  1  1] [p1]
  [2]=[1  1  4]*[p2] ---eqn.BL095
  [1] [1  4  1] [p3]
Matrix row is permutation of
β1=[4,1,1]
<a name="ch12c109">
Solve for p1, p2, p3 get
  p1=2/3 ---eqn.BL096
  p2=0   ---eqn.BL097
  p3=1/3 ---eqn.BL098
Three p's sum to one. OK
Three p's all be in [0,1] OK.

<a name="ch12c110">　Index begin　Index this file
Next is similar/ill example.
If α power is
  α2=[3,3,0] ---eqn.BL099
If β power is
  β2=[4,1,1] ---eqn.BL100
Let us find interpolation
coefficients q1, q2, q3
for α2 and β2.
<a name="ch12c111">
In matrix form it is next
  [3] [4  1  1] [q1]
  [3]=[1  1  4]*[q2] ---eqn.BL101
  [0] [1  4  1] [q3]
Left column is vector of α2=[3,3,0]
<a name="ch12c112">
Solve for q1, q2, q3 get
  q1=+2/3 ---eqn.BL102
  q2=-1/3 ---eqn.BL103
  q3=+2/3 ---eqn.BL104
<a name="ch12c113">
Three q's sum to one. OK. But
q2=-1/3 is out of [0,1] NO !
q2=-1/3 is extrapolation,
q2=-1/3 is NOT interpolation.
2010-06-09-15-05 change to silver
indicate wrong argument.

<a name="ch12c113a">
2010-07-04-15-23 start
Example
  α1=[3,2,1] ---eqn.BL090
  β1=[4,1,1] ---eqn.BL091
correct answer is
2/3  1/3  0  
1/3  2/3  0  
 0    0   1  

<a name="ch12c113b">
Wrong answer is
  p1=2/3 ---eqn.BL096
  p2=0   ---eqn.BL097
  p3=1/3 ---eqn.BL098

<a name="ch12c113c">
Please goto (local)
http://freeman2.com/jsmajor2.htm
Click example button [13]
Click run button [Run Stoch]
Click "Main result at box 7"
goto box 7 to get correct answer.
More at jsmajor2.htm#docG000
and at tute0047.htm#ch13b001
2010-07-04-15-32 stop

<a name="ch12c114">
The power set
  α1=[3,2,1] ---eqn.BL090
  β1=[4,1,1] ---eqn.BL091
is qualified for Muirhead's 
inequality.

<a name="ch12c115">　Index begin　Index this file
BUT Muirhead's inequality do 
not apply to the power set
  α2=[3,3,0] ---eqn.BL099
  β2=[4,1,1] ---eqn.BL100

<a name="ch12c116">
Then how do I know 
"Given that α in H(β)" is
true or not?
What is the difference between
eqn.BL090 and eqn.BL099 ??

<a name="ch12c117">
Textbook Chapter 13, page 191
say the following.
2010-05-28-18-35 here
===textbook copy start===
<a name="ch12c118">
Two Bare-Bones Definitions
Given an n-tiple 
  γ=(γ1,γ2,...,γn) ---eqn.BL105
we let γ[j], 1≦j≦n, denote
the j-th largest of the n 
coordinates, so
  γ[1]=max{γj: 1≦j≦n} ---eqn.BL106
and in general one has
  γ[1]≧γ[2]≧...≧γ[n] ---eqn.BL107
<a name="ch12c119">
Now for any pair of nonnegative
real n-tiples
  α=(α1,α2,...,αn) ---eqn.BL108
and
  β=(β1,β2,...,βn) ---eqn.BL109
<a name="ch12c120">　Index begin　Index this file
we say that α is majorized by β 
and write α(＜)β provided that 
α and β satisfy the following
system of n-1 inequalities:
  α[1]≦β[1] ---eqn.BL110
  α[1]+α[2]≦β[1]+β[2] ---eqn.BL111
   ... ≦ ...
  α[1]+α[2]+...+α[n-1]
 ≦β[1]+β[2]+...+β[n-1] ---eqn.BL112
<a name="ch12c121">
Together with the final
equality
  α[1]+α[2]+...+α[n]
 =β[1]+β[2]+...+β[n] ---eqn.BL113
Thus, for example, we have 
the majorization
      (1,1,1,1) (＜) (2,1,1,0)
 (＜) (3,1,0,0) (＜) (4,0,0,0) ---eqn.13.1
and since the definition of
<a name="ch12c122">
the relation α(＜)β depends
only on the corresponding
ordered value, {α[j]} and {β[j]} 
we could as well write the
chain (13.1) as
      (1,1,1,1) (＜) (0,1,1,2)
 (＜) (1,3,0,0) (＜) (0,0,4,0) ---eqn.BL114
===textbook copy stop===
2010-05-28-18-57 here
//'(＜)'=unicode '≺', LiuHH
//computer no such font.
//2010-05-30-13-26

<a name="ch12c123">
Now let us see
What is the difference between
eqn.BL090 and eqn.BL099 ??

<a name="ch12c124">　Index begin　Index this file
The well behave sequences
  α1=[3,2,1] ---eqn.BL090
  β1=[4,1,1] ---eqn.BL091
are already in descending order
First: 4≧3 is true. next
   4+1≧3+2 is true, next total
 4+1+1=3+2+1 is true.

<a name="ch12c125">
The ill behave sequences
  α2=[3,3,0] ---eqn.BL099
  β2=[4,1,1] ---eqn.BL100
are already in descending order
First: 4≧3 is true. next
   4+1≧3+3 is FALSE, next total
 4+1+1=3+3+0 is true.
The FALSE 4+1≧3+3 give us
trouble.

<a name="ch12c126">
Above main point is how to
evaluate the word
"Given that α in H(β)"
We need Chapter 13 concept
majorization.
2010-05-28-19-06 stop

<a name="ch12c127">　Index begin　Index this file
2010-05-28-22-18 start
■　Compare Muirhead with Newton

Muirhead inequality and Newton
inequality both belong to 
Chapter 12: Symmetric Sums.
Both Muirhead and Newton use
symmetric sums. Besides this 
one similar point, other
properties are different.

<a name="ch12c128">
Compare 01.
Number of sequence.
Newton use one sequence as 
polynomial roots and create
a polynomial.
<a name="ch12c129">
Muirhead use three sequences.
One sequence is observed data
which is physics quantity.
Two sequences are power of the 
data sequence. Two power seq. 
must be pure numbers, and
must have equal sum.

<a name="ch12c130">
Compare 02.
Polynomial link.
Newton inequality born from
polynomial.
Muirhead inequality has no
thing to do with polynomial.

<a name="ch12c131">
Compare 03.
Power manipulation.
Newton inequality has no
power manipulation.
Muirhead inequality main
work is power manipulation.

<a name="ch12c132">　Index begin　Index this file
Compare 04.
Power reduction.
Newton inequality has power 
reduction. Ek start from k=n, 
it has length to n-th power.
End at E0 which is defined 
to be pure number 1. En,Ek,E0 
still add in one equation.
<a name="ch12c133">
tn-k make up the physics
dimension difference. That is
  Ent0+Ektn-k+...+E0tn
They still keep even power.
Muirhead inequality do not
have power reduction.
SUM αj=SUM βi=constant power

<a name="ch12c134">
Compare 05.
Excluded case.
Newton inequality has no
excluded case. Given n real
numbers, Newton always build 
polynomial with n real roots.
<a name="ch12c135">
Muirhead inequality has
excluded case. If given two 
sequences α and β they do not
have [α in H(β)] relation
then Muirhead inequality do
not apply.
2010-05-28-22-50 stop

<a name="ch12c136">
2010-05-29-06-28 start
Compare 06.
Factorial usage
Newton use binomial coefficients
bicof(n,k)
Muirhead use factorial n!
n=3 case example
2010-05-29-06-31 stop

<a name="ch12c137">　Index begin　Index this file
2010-05-29-15-20 start
■　Verify Muirhead's Ineq. for n=3

Muirhead inequality manipulate
power. Given that α in H(β)
  α=(α1,α2,...,αn) ---eqn.BL065
  β=(β1,β2,...,βn) ---eqn.BL066
For all positive x1,x2,...,xn 
we have eqn.12.18
<a name="ch12c138">
The following change n elements
to n=3 elements. Smaller n is 
easier for real calculation.
Also change from x sequence
to r sequence. Use x as r seq.
element. That is use
  α=(α1,α2,α3) ---eqn.BL115
  β=(β1,β2,β3) ---eqn.BL116
  r=(x,y,z) ---eqn.BL117

<a name="ch12c139">
Whole set Muirhead's Inequality
is next
+xα1*yα2*zα3 +xα1*yα3*zα2
+yα1*zα2*xα3 +yα1*zα3*xα2
+zα1*xα2*yα3 +zα1*xα3*yα2 ---eqn.BL118
//2010-05-29-16-53 follow eqn.BL085
<a name="ch12c140">
Above is whole set six terms, below is one term.
+xα1*yα2*zα3
= ---eqn.BL119 //because α in hull of β
+xp1*β1+p2*β2+p3*β3
*yp1*β2+p2*β3+p3*β1
*zp2*β1+p3*β2+p1*β3
<a name="ch12c141">
Next re-group like pk together.
+xα1*yα2*zα3 ---eqn.BL120 (eqn.BL085)
=[xβ1yβ2zβ3]p1*[xβ2yβ3zβ1]p2*[xβ3yβ1zβ2]p3
<a name="ch12c142"> 　Index begin　Index this file
Next step is the central step. Inequality is born here.
Apply AM-GM inequality to eqn.BL120 get
+xα1*yα2*zα3 ---eqn.BL121 (eqn.BL086)
≦p1*[xβ1yβ2zβ3] +p2*[xβ2yβ3zβ1] +p3*[xβ3yβ1zβ2]
<a name="ch12c143">
There is not enough pk, can not sum pk to one here.
After permutation, there are six equation like eqn.BL121.
2010-05-29-17-42 here
<a name="ch12c144">
eqn.BL118 has six terms. From eqn.BL119 to
eqn.BL121 is expansion of just first term in
eqn.BL118. If expand (can you expand one
term?) other five terms and collect term in
favor of sum pk to one. The result is
//LiuHH get eqn.BL122 from one calculated term
//and five permutation terms. Any error? Alert !
<a name="ch12c145">
eqn.BL118 sum of six terms ≦ ---eqn.BL122
+(p1+p2+p3)*[xβ1yβ2zβ3]
+(p1+p2+p3)*[xβ2yβ3zβ1]
+(p1+p2+p3)*[xβ3yβ1zβ2]
+(p1+p2+p3)*[xβ1zβ2yβ3]
+(p1+p2+p3)*[xβ2zβ3yβ1]
+(p1+p2+p3)*[xβ3zβ1yβ2]
<a name="ch12c146"> 　Index begin　Index this file
Apply the condition
(p1+p2+p3)=1 ---eqn.BL123
get the final Whole set Muirhead's Inequality for n=3
+xα1*yα2*zα3 +xα1*yα3*zα2 +yα1*zα2*xα3
+yα1*zα3*xα2 +zα1*xα2*yα3 +zα1*xα3*yα2
≦ ---eqn.BL124 <a name="ch12c147">
+xβ1*yβ2*zβ3 +xβ2*yβ3*zβ1 +xβ3*yβ1*zβ2
+xβ1*zβ2*yβ3 +xβ2*zβ3*yβ1 +xβ3*zβ1*yβ2
2010-05-29-18-08 stop 

<a name="ch12c148">　Index begin　Index this file
2010-05-29-21-33 start
■　From Muirhead to AM-GM
To prove Muirhead's inequality
need use AM-GM inequality. 
Can we do the reverse?
Can we prove AM-GM by using
Muirhead? Yes, it is possible.

<a name="ch12c149">
We get solid feeling if use
numerical n example. Instead 
of eqn.BL065, eqn.BL065 and 
x1,x2,...,xn.
Now assume n=5 and define
  α=[α1,α2,α3,α4,α5] ---eqn.BL125
  β=[β1,β2,β3,β4,β5] ---eqn.BL126
  x=[a,b,c,d,e] ---eqn.BL127
<a name="ch12c150">
What is the value of α's ?
What is the value of β's ?
For given sequence [a,b,c,d,e]
Its AM, GM are
  am=(a+b+c+d+e)/5 ---eqn.BL128
  gm=(a*b*c*d*e)1/5 ---eqn.BL129
<a name="ch12c151">
Muirhead's Inequality is 
eqn.12.18
One sequence with five elements
Muirhead use 5!=120 terms. But
am has five terms a,b,c,d,e.
(four additions)
gm has one term (a*b*c*d*e)1/5
(no addition)
<a name="ch12c152">
They are far away from 120 terms!!
How to determine [α1,α2,α3,α4,α5]?
How to determine [β1,β2,β3,β4,β5]?
A term link by multiplication,
A term complete by addition.
<a name="ch12c153">　Index begin　Index this file
Muirhead one term is
  +aα1*bα2*cα3*dα4*eα5 ---eqn.BL130
We can re-write gm eqn.BL129
as
  +a1/5*b1/5*c1/5*d1/5*e1/5 ---eqn.BL131
Usually we use α power for less
than side. GM is less than side
in AM-GM inequality. Compare
eqn.BL130 with eqn.BL131, we 
find
  α=[α1,α2,α3,α4,α5] ---eqn.BL125
  α=[1/5,1/5,1/5,1/5,1/5] ---eqn.BL132
<a name="ch12c154">
Next how about am eqn.BL128
First term is am1=a/5 ---eqn.BL133
Muirhead one term for β is
  +aβ1*bβ2*cβ3*dβ4*eβ5 ---eqn.BL134
How to express a/5 in the 
form of eqn.BL134?
'/5' is division, not power.
Forget '/5' for a while. 
<a name="ch12c155">
Check power first. If equate
'a' (from 'a/5') power with
eqn.BL134, we must have
β1=1, and other β's=0
<a name="ch12c156">
For 'a' (from 'a/5') we need
  β=[β1,β2,β3,β4,β5] ---eqn.BL126
  β=[1,0,0,0,0] ---eqn.BL135
Similarly,
For 'b' (from 'b/5') we need
  β=[0,1,0,0,0] ---eqn.BL136
<a name="ch12c157">
For 'c' (from 'c/5') we need
  β=[0,0,1,0,0] ---eqn.BL137
For 'd' (from 'd/5') we need
  β=[0,0,0,1,0] ---eqn.BL138
For 'e' (from 'e/5') we need
  β=[0,0,0,0,1] ---eqn.BL139

<a name="ch12c158">　Index begin　Index this file
Where to find a '/5' for am?
Muirhead use permutation. We 
need apply permutation to 
eqn.BL128 and eqn.BL129

<a name="ch12c159">
GM use eqn.BL125 and eqn.BL132
for one GM term, we have
eqn.BL130. Put eqn.BL132 
α value to eqn.BL130 we get
eqn.BL129 exactly. Permute 
eqn.BL130, there are 5!=120
terms. They are identical !
<a name="ch12c160">
Because 
  α1=α2=α3=α4=α5=1/5 ---eqn.BL140
Instead of write eqn.BL129
120 times, we write
Muirhead's less than side as
 (5!)*(a*b*c*d*e)1/5 ---eqn.BL141
Here n!=5!=5*4*3*2*1=120

<a name="ch12c161">
Above is GM. Next see AM.
Muirhead's greater than side.
It is still 5!=120 terms.
But there are five different 
flavor. They are eqn.BL135
to eqn.BL139. 
<a name="ch12c162">　Index begin　Index this file
Put power on
top of data a,b,c,d,e, they
are (five flavors)
  a1*b0*c0*d0*e0 ---eqn.BL142
  a0*b1*c0*d0*e0 ---eqn.BL143
  a0*b0*c1*d0*e0 ---eqn.BL144
  a0*b0*c0*d1*e0 ---eqn.BL145
  a0*b0*c0*d0*e1 ---eqn.BL146
<a name="ch12c163">
eqn.BL142 has four zero power, 
it permuted (n-1)!=(5-1)!=4!=24
times. We do not write eqn.BL142
24 times, we write as
  (4!)*a ---eqn.BL147
Zero-th power are dropped here.
<a name="ch12c164">
Similarly,
  (4!)*b ---eqn.BL148
  (4!)*c ---eqn.BL149
  (4!)*d ---eqn.BL150
  (4!)*e ---eqn.BL151
Add eqn.BL147 to eqn.BL151 get
  (4!)*(a+b+c+d+e) ---eqn.BL152
eqn.BL152 is Muirhead's greater 
than side 120 terms !!

<a name="ch12c165">
Muirhead tell us that
 (5!)*(a*b*c*d*e)1/5
≦(4!)*(a+b+c+d+e) ---eqn.BL153
eqn.BL153 is complete equation
of Muirhead inequality for
AM-GM inequality.
<a name="ch12c166">
eqn.BL153 solved the puzzle.
Where '/5' come from?
'/5' come from (4!)/(5!)

Prove AM-GM by using Muirhead
is done.

<a name="ch12c167">　Index begin　Index this file
Please pay special attention
to that for 
  ∑[k]αk=1 ---eqn.BL154
  ∑[k]βk=1 ---eqn.BL155
case
GM α=[1/n,1/n,...,1/n] ---eqn.BL156
has most average distribution.
GM is majorized by any other 
power distribution.
(GM is smaller than any other) 

<a name="ch12c168">
AM β=[1,0,0,...0] ---eqn.BL157
has most un-even distribution.
AM majorize any other power 
distribution. (need eqn.BL154)
(AM is greater than any other) 

<a name="ch12c169">
Muirhead's Inequality has power
distribution in between AM and
GM, therefore we have
  GM≦Muirhead≦AM ---eqn.BL158

Do you agree?
2010-05-29-22-57 stop

2010-05-30-13-55 done  first proofread
2010-05-30-18-41 done second proofread
2010-05-30-19-13 done spelling check



<a name="docB001">
2010-06-13-23-08 start
"Update 2010-06-14" change
from [for any pair of real]
to [for any pair of nonnegative
    real]
Just find this change is listed 
in errata page.
2010-06-13-23-10 stop




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2009-06-19-10-48
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