<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="ch11c001">　Index begin　Index this file
2010-05-07-21-26 start
■　Exercise 11.6 problem statement
　　textbook page 177
（Optimality?--It Depends on Context）

Many inequalities which cannot
be improved in general will
nevertheless permit improvements
under special circumstances.
<a name="ch11c002">
An elegant illustration of this
possibility was given in a 1991
American Mathematical Monthly 
problem posed by Walther Janous.
Reader were challenged to prove 
that for all 0＜x＜1 and all N≧1
one has the bound

<a name="ch11c004">
(a) Prove that a direct application
of Hardy's inequality provides a
similar bound where 4*log(2) is
replaced by 4. Since log2=0.693...
we then see this Janous's bound 
beats Hardy's in this particular
instance.
<a name="ch11c005">
(b) Prove Janous's inequality and 
show that one cannot replace
4*log(2) with a constant C＜4*log(2)
2010-05-07-21-46 stop





<a name="ch11c006">　Index begin　Index this file
2010-05-07-21-53 start
■　Exercise 11.6 hint
　　textbook page 272

The solution of this problem
is not easy, and here we follow
the one provided by Richberg 
(1993) that begins with the
observation that

<a name="ch11c012">
This bound and the fact that it
is sharp was already addressed
in Exercise 7.10, so the 
solution is complete.
2010-05-07-22-25 stop




2010-05-08-15-09 start
<a name="ch11c013">
■　Exercise 11.6 solution


First consider Hardy's inequality
eqn.11.9. Identify sequence as 
following: a1→1, a2→x, a3→x2
..., an→xn-1. Put the sequence
 1, x, x2, ..., xn-1 into eqn.11.9
get target inequality eqn.BJ001
with coefficient 4.

Above is Exercise 11.6 part (a)
<a name="ch11c014">
Below is part (b) Prove 
Janous's inequality eqn.BJ001

Observe the simple calculation
1+x+x2+x3+ ...+ xn-1
multiply by (1-x) get
1+x+x2+x3+ ...+ xn-1
 -x-x2-x3- ...- xn-1-xn
=========================
1  +0  +0  ...  +0 -xn ---eqn.BJ007
<a name="ch11c015">
We have 
  (1+x+x2+x3+ ...+ xn-1)(1-x)
 =1-xn ---eqn.BJ008
or
  (1+x+x2+x3+ ...+ xn-1)
 =(1-xn)/(1-x) ---eqn.BJ009
<a name="ch11c016">　Index begin　Index this file
If let x→x2, we have
  (1+x2+x4+ ...+ x2n-2)
 =(1-x2n)/(1-x2) ---eqn.BJ010

Put eqn.BJ009 to eqn.BJ001 less 
than side
Put eqn.BJ010 to eqn.BJ001 greater
than side, get modified target

<a name="ch11c018">
Exercise 11.6 hint observation
convert above red part to an
integration. Derive eqn.BJ002
as following.
Instead of
  (1+x+x2+x3+ ...+ xn-1)
 =(1-xn)/(1-x) ---eqn.BJ009
<a name="ch11c019">
we change x to st, get
  (1+(st)+(st)2+(st)3+ ...+ (st)N-1)
 =(1-(st)N)/(1-(st)) ---eqn.BJ012
's' and 't' are integration variable
and x＜s＜1, x＜t＜1. Write eqn.BJ012
in summation form as following
  ∑[j=1,N](st)j-1
 =(1-(st)N)/(1-(st)) ---eqn.BJ013

<a name="ch11c020">
Use eqn.BJ013, we change from
eqn.BJ002 left term to middle 
term (eqn.BJ002 line one).
Change also include switch from 
summation 1st, integration 2nd
to
integration 1st, summation 2nd.

<a name="ch11c021">　Index begin　Index this file
From eqn.BJ002 middle term to 
right term (second line) is
simple integration. This step
s and t can be separated. For
the double integration part 
only (one term of summation)
we have

<a name="ch11c023">
Each integration evaluation
as following
  ∫[s=x,1]d[sj-1+1]/(j-1+1)
 =[1j/(j)] - [xj/(j)]
 =[1-xj]/j ---eqn.BJ015
Two integration are identical.
<a name="ch11c024">
Double integration result is
[1-xj]2/j2. It is just one 
term of summation. Consider
total sum, we get eqn.BJ002
right end (second line)
This is eqn.BJ011 red term.
Put eqn.BJ002 back to eqn.BJ011
So our target inequality is 
equivalent to

<a name="ch11c026">　Index begin　Index this file
The blue (1-x)2 come from 
eqn.BJ011 black (1-x) and red 
power 2. In eqn.BJ016, x is 
integration lower bound, x is a
constant (s and t are variable)
x term can move in/out of 
integration freely. Because 
0＜x＜1, (1-x)2 and 1-x2N are 
both positive. Adjust x terms 
as following

<a name="ch11c028">
Because 0＜x＜1, we can cancel
red (1-x). The result is same 
as eqn.BJ003.
Without detail calculation, the
step at eqn.BJ003 puzzled for a 
while.
2010-05-08-17-17 here

<a name="ch11c029">
From eqn.BJ003 to eqn.BJ004,
dropped the term
  [1－(st)N]/[1－x2N] ---eqn.BJ018
Integration variables s and t
range from x≦s,t≦1, for N≧1
get xN≦sN
and xN≦tN, 
so  x2N≦sNtN
<a name="ch11c030">
then -x2N≧-sNtN
and 1-x2N≧1-sNtN
which give us
  1≧[1－(st)N]/[1－x2N] ---eqn.BJ019
eqn.BJ018 has variable s and t.
<a name="ch11c031">　Index begin　Index this file
but eqn.BJ018 is less than one.
Use 0.7 to represent eqn.BJ018.
eqn.BJ003 look like
  0.7*∫∫ds*dt/(1－st)
 ＜[4*log(2)](1-x)/(1+x) ---eqn.BJ020
Drop 0.7 from less than side,
increase less than side value,
risk reverse inequality. Example
0.7*5＜4 is true; 5＜4 is false.
<a name="ch11c032">
However, if drop 0.7 result
      ∫∫ds*dt/(1－st)
 ＜[4*log(2)](1-x)/(1+x) ---eqn.BJ021
is still true, then "has_0.7"
eqn.BJ020 is true for sure.
This logic give us
"This bound would follow from
 eqn.BJ004"

Next verify eqn.BJ005
2010-05-08-17-54 here

<a name="ch11c034">
2010-05-08-18-14 here
The term |[s=x,1]log(1－st) is
  log(1－1*t) - log(1－x*t) 
 =log(1－t) - log(1－x*t) 
 =log[(1－t)/(1－x*t)] ---eqn.BJ023

<a name="ch11c035">
eqn.BJ022 dt/(-t) has a negative 
sign. But eqn.BJ005 use dt/(+t)
Let eqn.BJ023 absorb a '-', then
The term -|[s=x,1]log(1－st) is
 =log[(1－x*t)/(1－t)] ---eqn.BJ024
Plug eqn.BJ024 into eqn.BJ022
and use dt/(+t), get

<a name="ch11c037">
eqn.BJ025 is different from hint
eqn.BJ005
2010-05-08-18-56 here

Exercise 11.6 is NOT DONE
Liu,Hsinhan still think how.
2010-05-08-18-58 stop

<a name="ch11c037a">
2010-05-17-22-00 in eqn.BJ011
eqn.BJ016 and eqn.BJ017 delete 'j'
from greater than side denominator,
In eqn.BJ011 copy quotient from less 
than side to greater than side,
LiuHH forget to delete j. 





<a name="ch11c038">　Index begin　Index this file
2010-05-08-21-06 start
■　Exercise 11.7 problem statement
　　textbook page 177
（Confirmation of the Obvious）

Show that if a1≧a2≧a3≧...
and if b1,b2,b3,... is any
rearrangement of the sequence
a1,a2,a3,..., then for each
N=1,2,... one has

<a name="ch11c040">
Thus, in the proof of Carleman's
inequality, one can assume
without lose of generality that
a1≧a2≧a3≧... since a
rearrangement does not change
the right side.
2010-05-08-21-22 stop





<a name="ch11c041">
2010-05-08-21-24 start
■　Exercise 11.7 hint
　　textbook page 273

This observation is painfully
obvious, but it seems necessary 
for completeness. The hypothesis 
give us the bounds b1≦a1, 
b2≦a2,...,bN≦aN; 
//LiuHH ●●change above silver to next
(b1b2)1/2≦(a1a2)1/2, ..., 
(∏[k=1,N]bk)1/N≦(∏[k=1,N]ak)1/N

<a name="ch11c042">
thus for all 1≦n≦N
we have //next line ---eqn.BJ026
  (b1b2...bn)1/n≦(a1a2...an)1/n
which is more than we need.
There are questions on infinite 
rearrangements which are subtle,
but this is not one of them.
2010-05-08-21-31 stop



<a name="ch11c043">　Index begin　Index this file
2010-05-08-22-01 start
■　Exercise 11.7 solution


This problem need just verbal 
explanation.
<a name="ch11c044">
Given a1≧a2≧a3≧...≧0
The condition '≧0' is important.
Otherwise if we have a sequence
  3＞2＞1＞-1＞-5
we have trouble at first time
meet negative, that is first
four elements geometric mean
  [3*2*1*(-1)]1/4
  =1.10668+i*1.10668 ---eqn.BJ027
we get complex number.

<a name="ch11c045">
Rearrangements means use same
set sequence but change to
different order. 
Assume we have a-sequence 
  5＞4＞3＞2＞1
if rearrange to b-sequence 
  2,4,3,5,1

<a name="ch11c046">　Index begin　Index this file
Gradually GM (geometric mean)
first comparison is
  a1=5 greater
  b1=2 smaller ---eqn.BJ028

Second comparison is // ---eqn.BJ029
  (a1*a2)1/2=(5*4)1/2 greater
  (b1*b2)1/2=(2*4)1/2 smaller
because 5＞2 (a1＞b1)

<a name="ch11c047">
Third comparison is // ---eqn.BJ030
  (a1*a2*a3)1/3=(5*4*3)1/3 greater
  (b1*b2*b3)1/3=(2*4*3)1/3 smaller
because 5＞2 (a1＞b1)

greater a1=5 show up at all
        downstream numbers
smaller b1=2 show up at all
        downstream numbers
We can even close our eye and
give a correct answer !!

<a name="ch11c048">
If we have a-sequence 
  5＞4＞3＞2＞1
if rearrange to b-sequence 
  5,4,2,3,1
First two comparison are equal.
Start from third, 
gradually GMed a-sequence beat
gradually GMed b-sequence.

Summation of each gradually GMed
elements will not change answer.

Exercise 11.7 is done
2010-05-08-22-30 stop



<a name="ch11c049">　Index begin　Index this file
2010-05-09-10-26 start
■　Exercise 11.8 problem statement
　　textbook page 177
（Kronecker's Lemma）

<a name="ch11c050">
Prove that for any sequence 
a1,a2,...,an of real or 
complex numbers one has the 
inference
  ∑[n=1,∞]an/n converge
  implies //above & below ---eqn.11.25
  lim[n→∞](a1+a2+...+an)/n=0
Like Hardy's inequality, this 
result tells us how to convert 
one type of information about 
averages to another type of 
information. 
<a name="ch11c051">
This implication 
is particularly useful in 
probability theory where it 
is used to draw a connection 
between the convergence of 
certain random sums and the
famous law of large numbers.
2010-05-09-10-39 stop





<a name="ch11c052">　Index begin　Index this file
2010-05-09-12-04 start
■　Exercise 11.8 hint
　　textbook page 273

From the convergence of the sum,
we know that the sequence of
reminders
  rn=an+1/(n+1) +an+2/(n+2)
    +an+3/(n+3) +... ---eqn.BJ031
must converge to zero as n→∞.
<a name="ch11c053">
When we write these terms in
longhand  //below is ---eqn.BJ032
r0=a1 +a2/2 +a3/3 +... +an/n +rn
r1=    a2/2 +a3/3 +... +an/n +rn
r2=          a3/3 +... +an/n +rn
....    
rn-2=        an-1/(n-1) +an/n +rn
rn-1=                    an/n +rn
<a name="ch11c054">
We see they may be summed to 
yield the nice identity
  (a1+a2+...+an)/n= ---eqn.14.61
  -rn+(r0+r1+...+rn-1)/n
Which make the limit (11.25)
routine.
2010-05-09-12-21 stop



<a name="ch11c055">　Index begin　Index this file
2010-05-09-12-37 start
■　Exercise 11.8 solution


First see
[[
<a name="ch11c056">
From the convergence of the sum,
we know that the sequence of
reminders
  rn=an+1/(n+1) +an+2/(n+2)
    +an+3/(n+3) +... ---eqn.BJ031
must converge to zero as n→∞.
]]
<a name="ch11c057">
View from the reverse condition.
That is if
reminders
  rn=an+1/(n+1) +an+2/(n+2)
    +an+3/(n+3) +... ---eqn.BJ031
NOT converge to zero as n→∞.
can we have 
the convergence of the sum
?
<a name="ch11c058">
If reminder NOT converge to zero 
as n→∞. Then there exist a tiny
number δ such that all reminder
are greater than δ. In the sum
  ∑[n=1,∞]an/n 
each an＞δ. 
<a name="ch11c059">
This sum multiply
δ n times and divide by n, it is 
still a number greater than δ.
When n approach to infinity, sum
will diverge. For converge sum,
it is necessary reminders goto
zero.

<a name="ch11c060">　Index begin　Index this file
Next is sum of eqn.BJ032. It is 
a simple observation that the 
sum of all left side become
(r0+r1+...+rn-1)
The sum of all right end become
n*rn
The ak term sum column by column 
get (a1+a2+...+an)

<a name="ch11c061">
Put together and divide by n get
eqn.14.61.
Since remainder approach to zero
when n approach to infinity. The
equality of eqn.14.61 tell us 
that (a1+a2+...+an)/n
approach to zero too.
Exercise 11.8 is done
2010-05-09-12-57 stop

2010-05-09-15-57 done proofread
2010-05-09-16-09 done spelling check



========= Chapter eleven end here =========




<a name="ch12a001">　Index begin　Index this file
■■Chapter 12: Symmetric Sums

<a name="ch12note01">
2010-05-17-18-32 start
To begin a chapter, Liu,Hsinhan
always goto online search for
related paper, read more material
help understand new topic.
Chapter 12 spend more time online
spend more time to read, spend 
more time to think. Because find 
more question than other chapters.
<a name="ch12note02">
Why define
  Ek(x)=ek(x)/bicof(n,k) ---eqn.BJ301
?
Why take unequal power root
  [En(x)]1/n ≦ [En-1(x)]1/(n-1) ≦ ... 
  ≦ [E2(x)]1/2 ≦ E1(x) ---eqn.12.3
?
<a name="ch12note03">　Index begin　Index this file
All paper say: "Do this way",
but no one paper say why.
Liu,Hsinhan figure out reason
that is
Physics dimension consistency
and one to one
Please read, if you suspect any
view point wrong, please ask a
math/physics expert near by.

<a name="ch12note04">
If  f(x)=x*x ---eqn.BJ302
then f(x) is convex, but
log(f(x)) is log-concave!

What is the difference between
log-concavity and concavity?
There is log-concavity, whether
there is log-convexity? 

<a name="ch12note05">
exp(x) compete with log(x) even
log(exp(x)) is linear. then
log(exp(exp(x))) is 'log-convex'
If there is any importance about
'log-convex'
Study more, solve more problem,
but create more question too.

<a name="ch12note06">
Advanced paper authors can not
skip elementary work. Skyscraper
must build on solid foundation.
To understand Inequality of 
Newton and Maclaurin, LiuHH
start from elementary cases
roll up sleeve, and calculate
the detail for low power end 
and high power end cases.
<a name="ch12note07">　Index begin　Index this file
Because I can not find them
online and I want to know
what is going on. The work 
is here below.
Please read and verify, these
work may contain error, since
no one proofread for me.
2010-05-17-19-13 stop

<a name="ch12a002">
2010-05-13-12-06 start
If we have n numbers x1,x2,...,xn,
we can do symmetric summation.
Let ek(x1,x2,...,xn) be
a power k summation symbol.
x1,x2,...,xn each is power one.
for example observed length data.
Two of them multiply together is 
power two, example x1*x2+x3*x4+...
<a name="ch12a003">
Highest power we can reach with
n number is power n: x1*x2*...*xn
A number can not multiply twice 
either once or none. Five 
numbers can not have power six.
To simplify work, we write 
ek(x1,x2,...,xn) as ek(x) 
here x is an array of numbers
x1,x2,...,xn. x contain the
information: we have n elements.
<a name="ch12a004">
Symmetric summation as following
Power zero:
  e0(x) define= 1 ---eqn.BJ101
Power one:
  e1(x)=x1+x2+...+xn ---eqn.BJ102
<a name="ch12a005">　Index begin　Index this file
Power two longhand:
  e2(x)= ---eqn.BJ103
  x1*x2 + x1*x3 + ... + x1*xn
+ x2*x3 + x2*x4 + ... + x2*xn
+ ...
+ xn-1*xn
<a name="ch12a006">
Power two short symbol:
  e2(x)=∑[1≦j＜k≦n]xj*xk ---eqn.BJ104
Alert: the requirement j＜k in 
∑[1≦j＜k≦n] allow x1*x2 but 
not allow x2*x1. Power n:
  en(x)=x1*x2*...*xn ---eqn.BJ105
To gain solid feeling, assume
  n=5 ---eqn.BJ106
and five numbers a,b,c,d,e.
  x={a,b,c,d,e} ---eqn.BJ107

<a name="ch12a007">
The following will use them as
an example. For n=5
Power one symmetric sum is
  e1(x)=a+b+c+d+e ---eqn.BJ108
Chapter 12 topic is "Symmetric 
Sums" the non-symmetric sum
 a+b+c+d (missing 'e')
is NOT considered here.

<a name="ch12a008">　Index begin　Index this file
■　Polynomial has Symmetric Sums
Symmetric Sums appears naturally
in polynomial. Let us use P(t)
for polynomial, variable is t. 
Let x1,x2,...,xn be polynomial's
roots. We have
  P(t)=(t-x1)*(t-x2)*...*(t-xn) ---eqn.BJ109
Where x1,x2,...,xn are constants. 
<a name="ch12a009">
For easy work,
assume n=5 and a,b,c,d,e five 
roots. then
  P(t)=(t-a)*(t-b)*(t-c)
      *(t-d)*(t-e) ---eqn.BJ110

<a name="ch12a010">
(t-a)*(t-b) expansion is
  (t-a)*(t-b)=t*(t-b)-a*(t-b)
 =t*t-t*b-a*t+a*b
 =t*t-t*(a+b)+a*b ---eqn.BJ111

<a name="ch12a011">
(t-a)*(t-b)*(t-c) expansion is
  (t-a)*(t-b)*(t-c)
 =(t*t-t*a-t*b+a*b)*(t-c)
 =(t*t*t-t*t*a-t*t*b+t*a*b)
 -(c*t*t-c*t*a-c*t*b+c*a*b)
 =t*t*t-t*t*a-t*t*b-t*t*c
 +t*a*b+t*a*c+t*b*c-a*b*c ---eqn.BJ112

<a name="ch12a012">
(t-a)*(t-b)*(t-c)*(t-d) expansion is
  (t-a)*(t-b)*(t-c)*(t-d)
 =(t*t*t-t*t*a-t*t*b-t*t*c
  +t*a*b+t*a*c+t*b*c-a*b*c)*(t-d)

 =(t*t*t*t-t*t*t*a-t*t*t*b-t*t*t*c
  +t*t*a*b+t*t*a*c+t*t*b*c-t*a*b*c)
 -(d*t*t*t-d*t*t*a-d*t*t*b-d*t*t*c
  +d*t*a*b+d*t*a*c+d*t*b*c-d*a*b*c)

<a name="ch12a013">　Index begin　Index this file
(t-a)*(t-b)*(t-c)*(t-d)
 = t*t*t*t
  -t*t*t*a-t*t*t*b-t*t*t*c-t*t*t*d
  +t*t*a*b+t*t*a*c+t*t*b*c+t*t*a*d+t*t*d*b+t*t*d*c
  -t*a*b*c-t*a*b*d-t*a*c*d-t*b*c*d
  +a*b*c*d ---eqn.BJ113

<a name="ch12a014">
(t-a)*(t-b)*(t-c)*(t-d)*(t-e) expansion is
(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
 =(t*t*t*t
  -t*t*t*a-t*t*t*b-t*t*t*c-t*t*t*d
  +t*t*a*b+t*t*a*c+t*t*b*c+t*t*a*d+t*t*d*b+t*t*d*c
  -t*a*b*c-t*a*b*d-t*a*c*d-t*b*c*d
  +a*b*c*d)*(t-e)

 =t*(t*t*t*t
  -t*t*t*a-t*t*t*b-t*t*t*c-t*t*t*d
  +t*t*a*b+t*t*a*c+t*t*b*c+t*t*a*d+t*t*d*b+t*t*d*c
  -t*a*b*c-t*a*b*d-t*a*c*d-t*b*c*d
  +a*b*c*d)
 -e*(t*t*t*t
  -t*t*t*a-t*t*t*b-t*t*t*c-t*t*t*d
  +t*t*a*b+t*t*a*c+t*t*b*c+t*t*a*d+t*t*d*b+t*t*d*c
  -t*a*b*c-t*a*b*d-t*a*c*d-t*b*c*d
  +a*b*c*d)

<a name="ch12a015">
 =t*t*t*t*t
  -t*t*t*t*a-t*t*t*t*b-t*t*t*t*c-t*t*t*t*d
  +t*t*t*a*b+t*t*t*a*c+t*t*t*b*c+t*t*t*a*d+t*t*t*d*b+t*t*t*d*c
  -t*t*a*b*c-t*t*a*b*d-t*t*a*c*d-t*t*b*c*d
  +t*a*b*c*d
 -e*t*t*t*t
  +e*t*t*t*a+e*t*t*t*b+e*t*t*t*c+e*t*t*t*d
  -e*t*t*a*b-e*t*t*a*c-e*t*t*b*c-e*t*t*a*d-e*t*t*d*b-e*t*t*d*c
  +e*t*a*b*c+e*t*a*b*d+e*t*a*c*d+e*t*b*c*d
  -a*b*c*d*e

<a name="ch12a016">
(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
 =t*t*t*t*t  ---eqn.BJ114
  -t*t*t*t*a-t*t*t*t*b-t*t*t*t*c-t*t*t*t*d-t*t*t*t*e

  +t*t*t*a*b+t*t*t*a*c+t*t*t*b*c+t*t*t*a*d
  +t*t*t*d*b+t*t*t*d*c
  +t*t*t*a*e+t*t*t*b*e+t*t*t*c*e+t*t*t*d*e

  -t*t*a*b*c-t*t*a*b*d-t*t*a*c*d-t*t*b*c*d
  -t*t*a*b*e-t*t*a*c*e-t*t*b*c*e-t*t*a*d*e-t*t*d*b*e-t*t*d*c*e

  +t*a*b*c*d
  +t*a*b*c*e+t*a*b*d*e+t*a*c*d*e+t*b*c*d*e

  -a*b*c*d*e

<a name="ch12a017">
Re-write all five expansion as
(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
 =t*t*t*t*t  ---eqn.BJ115
 -t*t*t*t*(a+b+c+d+e)
 +t*t*t*(a*b+a*c+b*c+a*d+b*d
        +c*d+a*e+b*e+c*e+d*e)
 -t*t*(a*b*c+a*b*d+a*c*d+b*c*d+a*b*e
      +a*c*e+b*c*e+a*d*e+b*d*e+c*d*e)
 +t*(a*b*c*d+a*b*c*e+a*b*d*e
     +a*c*d*e+b*c*d*e)
 -a*b*c*d*e
2010-05-13-13-20 here

<a name="ch12a018">
Now use ek(x) symbol re-write
all expansion as
 P(t)= //remind: e0(x)=1
 =(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
 =t*t*t*t*t*e0(x)  ---eqn.BJ116
 -t*t*t*t*e1(x)
 +t*t*t*e2(x)
 -t*t*e3(x)
 +t*e4(x)
 -e5(x)

<a name="ch12a019">　Index begin　Index this file
■　Lowercase ek(x) defined
where //for n=5 special case
  e1(x)=(a+b+c+d+e) ---eqn.BJ117
  e2(x)=(a*b+a*c+b*c+a*d+b*d
        +c*d+a*e+b*e+c*e+d*e) ---eqn.BJ118
  e3(x)=(a*b*c+a*b*d+a*c*d+b*c*d+a*b*e
        +a*c*e+b*c*e+a*d*e+b*d*e+c*d*e) ---eqn.BJ119
  e4(x)=(a*b*c*d+a*b*c*e+a*b*d*e
        +a*c*d*e+b*c*d*e) ---eqn.BJ120
  e5(x)=a*b*c*d*e ---eqn.BJ121
Array x has a hidden condition: n=5
<a name="ch12a020">
Following we will compare the
magnitude among ek k=1,2,3,4,5.
Before do so, there are two 
important points to consider.
First: physics dimension 
       consistency
Second: one to one.

<a name="ch12a021">　Index begin　Index this file
■　Physics dimension consistency
First: physics dimension 
       consistency
In eqn.BJ116 we have
P(t)=t*t*t*t*t
 -t*t*t*t*e1(x)
 +t*t*t*e2(x)
 -t*t*e3(x)
 +t*e4(x)
 -e5(x)
<a name="ch12a022">
This equation tell us that
t*t*t*t*t and t*t*t*t*e1(x)
and t*t*t*e2(x) and t*t*e3(x)
and t*e4(x) and e5(x)
are of same physics dimension.
Otherwise we can not use addition
subtraction to relate unlike
physics quantities. 
<a name="ch12a023">
Do we ever 
see an equation like 
  time+area=distance-mass
NO !! NEVER !!
Because t*t*t*t*e1(x) and 
t*t*t*e2(x) are of same physics 
quanitiy, 
<a name="ch12a024">　Index begin　Index this file
we can compare t*t*t*t*e1(x) 
with t*t*t*e2(x). Both drop 
't*t*t', we can compare t*e1(x) 
with e2(x), BUT
we can NOT compare e1(x) with
e2(x). In general,
we can NOT compare ei(x) with
ej(x) for i≠j

<a name="ch12a025">
Our example P(t) start from
  P(t)=(t-a)*(t-b)*(t-c)
      *(t-d)*(t-e) ---eqn.BJ110
It tell us that a,b,c,d,e each
has same physics quantity as t.
(otherwise t and -a can not sit
 side by side)
Then all of t,a,b,c,d,e are of 
same physics quantity. 
<a name="ch12a026">
Assume they are distance. Then
  e1(x)=(a+b+c+d+e) ---eqn.BJ117
tell us that e1(x) is distance.
  e2(x)=(a*b+a*c+b*c+a*d+b*d
        +c*d+a*e+b*e+c*e+d*e) ---eqn.BJ118
tell us that e2(x) is area.
e3(x) is volume. etc.
<a name="ch12a027">
We can compare ei(x), ej(x) i≠j
only if AFTER we convert them 
to same physics quantity.
For example, length compare
with cubic root of a volume.
2010-05-13-13-55 stop

2010-05-13-14-59 start
Above is
First important point: 
physics dimension consistency

<a name="ch12a028">　Index begin　Index this file
■　one to one
Next is
Second important point: 
one to one.
Please see
  e1(x)=(a+b+c+d+e) ---eqn.BJ117
  e2(x)=(a*b+a*c+b*c+a*d+b*d
        +c*d+a*e+b*e+c*e+d*e) ---eqn.BJ118
.....
  e5(x)=a*b*c*d*e ---eqn.BJ121
where

<a name="ch12a030">
2010-05-13-15-24 here
e1(x) is length to first power
  and five terms.
e5(x) is length to fifth power
  and one term.
Let us change e5(x) to e5(x)1/5
Now e5(x)1/5 and e1(x) are both 
length to first power, can we
compare them?
Yes ! 
<a name="ch12a031">
length compare with length is OK.
But e5(x)1/5 has just one term
(x1*x2*x3*x4*x5)1/5
and e1(x) is sum of five terms
x1+x2+x3+x4+x5
The number of terms is a factor!
Why not remove this factor ?
<a name="ch12a032">　Index begin　Index this file
■　Uppercase Ek(x) defined
Define //for n=5 special case
  E1(x)=(x1+x2+x3+x4+x5)/bicof(5,1)
  E1(x)=e1(x)/bicof(5,1) ---eqn.BJ123
Similarly //Ek(x) power alert 1, 2
  E2(x)=e2(x)/bicof(5,2) ---eqn.BJ124
  E3(x)=e3(x)/bicof(5,3) ---eqn.BJ125
  E4(x)=e4(x)/bicof(5,4) ---eqn.BJ126
  E5(x)=e5(x)/bicof(5,5) ---eqn.BJ127
<a name="ch12a033">　Index begin　Index this file
bicof(n,k) is defined as

<a name="ch12a034">
ek(x) and Ek(x) are different.
Each ek(x) k=1,2,3,4,5 represent 
multiple terms, ei(x),ej(x) are 
not on equal foot. They have 
different number of terms.
(besides different physics
 dimension)
<a name="ch12a035">
Each Ek(x) k=1,2,3,4,5 represent 
ONE term, AVERAGED one term.
If we compare dimensionally 
consistent Ei(x)1/i,Ej(x)1/j i≠j
they are on equal foot: one term.
Factor of different number of 
terms is gone. Wonderful !
Alert: in Em(x)1/n both sub-note m
and sup-note 1/n represent physics 
dimension power. they cancel to m/n
<a name="ch12a036">
Above discussion is for special
case total number of data is n=5
For general case (n ＜＝＞ 5)
  ek(x)=∑[1≦i1＜i2＜...＜ik≦n]
         xi1*xi2*...*xik ---eqn.BJ129
Above equation is at textbook 
page 178 line 3. 
Alert i1＜i2＜...＜ik is correct
Alert i1≦i2≦...≦ik is typo.
Element xij appear just once.
<a name="ch12a037">
  P(t)=tn-e1(x)tn-1+e2(x)tn-2
  +...+(-1)k*ek(x)tn-k
  +...+(-1)n*en(x) ---eqn.12.1
Above equation is at textbook
page 178 line -6. 
<a name="ch12a038">　Index begin　Index this file
  Ek(x)=ek(x)/bicof(n,k) ---eqn.BJ130
Above equation is at textbook
page 179 line 3.
For the case n=5, eqn.BJ129
reduce to eqn.BJ117 to eqn.BJ121
where we can find solid feeling.
2010-05-13-16-42 stop



<a name="ch12a039">
2010-05-13-18-55 start
General equation eqn.12.1 and 
special case eqn.BJ116 both 
provide explanation. eqn.BJ116 
give us solid feeling. eqn.12.1 
cover general case (n ＜＝＞ 5)
We switch between the two which
ever is more convenient.
<a name="ch12a040">
Now let us see eqn.BJ116
 P(t)=(t-a)(t-b)(t-c)(t-d)(t-e)
 =t*t*t*t*t   -t*t*t*t*e1(x)
 +t*t*t*e2(x) -t*t*e3(x)
 +t*e4(x)     -e5(x)  ---eqn.BJ116B
Above is P(t) in ek(x) expression.
<a name="ch12a041">
Below is P(t) in Ek(x) expression.
Ek(x) is averaged ek(x) 
 P(t)=(t-a)(t-b)(t-c)(t-d)(t-e)
 =t*t*t*t*t   -5*t*t*t*t*E1(x)
 +10*t*t*t*E2(x) -10*t*t*E3(x)
 +5*t*E4(x)     -E5(x)  ---eqn.BJ131

<a name="ch12a042">
Let us see two end E's //define E
  E1(x)=(a+b+c+d+e)/5  ---eqn.BJ132
  E5(x)= a*b*c*d*e  ---eqn.BJ133
E1(x) and E5(x) are both
averaged value (for one term).

<a name="ch12a043">　Index begin　Index this file
Assume t,a,b,c,d,e all be length
Let us take fifth root for E5(x)
E1(x) and [E5(x)]1/5 are both
averaged value plus both be 
length. They can be compared.

<a name="ch12a044">
E1(x) compare with [E5(x)]1/5 is
(a+b+c+d+e)/5 compare with 
(a*b*c*d*e)1/5 WHO ELSE !?
It is AM-GM inequality !!
(a+b+c+d+e)/5 is Arithmetic Mean 
(a*b*c*d*e)1/5 is Geometric Mean
<a name="ch12a045">
We do not do a thing, just write
down expression and apply AM-GM 
inequality get  GM≦AM  that is
  [E5(x)]1/5 ≦ E1(x) ---eqn.BJ134
See eqn.AE20 for general weight
or power mean style eqn.AZ028
Now the question is 
<a name="ch12a046">
What is the magnitude/order of
[E4(x)]1/4,[E3(x)]1/3,[E2(x)]1/2
They all be length to first power.
They all be averaged one term.
Alert: Ek is length to power k
      [Ek(x)]1/k is length power 1
      [Ek(x)]j is length power j*k
      Em(x)En(x) is length power m+n
<a name="ch12a047">
No question, no progress.
Good question, lead us to new
direction.
Thank Newton and thank Maclaurin
they solved problem for us.
2010-05-13-19-37 here

<a name="ch12a048">　Index begin　Index this file
■　Problem 12.1 (Inequality of
　　Newton and Maclaurin)
Show that for all x in Realn
one has Newton's inequality
  Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.2
and check that they imply 
<a name="ch12a049">
Maclaurin's inequality which 
asserts that
  [En(x)]1/n ≦ [En-1(x)]1/(n-1) ≦ ... 
  ≦ [E2(x)]1/2 ≦ E1(x) ---eqn.12.3
for all x=(x1,x2,...,xn)
Please see Ek(x) power alert 1, 2
<a name="ch12a050">
such that
  xk≧0 ---eqn.BJ135
for all 1≦k≦n
2010-05-13-19-54 stop

<a name="ch12a051">　Index begin　Index this file
2010-05-14-15-34 start
■　Concavity condition

Convex function is important 
in the theory of inequality.
Chapter six whole chapter 
discuss convexity. eqn.6.1
define convex function. 
<a name="ch12a052">
A convex function shape like a
bowl ╰╯. Curve center lower in
function value, ends higher.
(this word is not accurate)
A function can be whole domain
convex, like exp(x)

<a name="ch12a053">
Concave function is opposite 
to convex function. A concave 
function shape like a hat ╭╮
Curve center higher in function 
value, ends lower.
(this word is not accurate)
A function can be whole domain
concave, like log(x)

<a name="ch12a054">
Many function are partial convex
and partial concave. In this 
case domain definition is 
important. For example, sin(x)
is concave for x in [0,PI] and
convex for x in [PI,2PI].

<a name="ch12a055">
eqn.6.1 say: Select two points
x,y in a convex function domain
Draw a chord between two points
(x,f(x)), (y,f(y)) If chord is
always above function curve, 
then function in this domain
is defined to be convex.

<a name="ch12a056">　Index begin　Index this file
2010-05-14-16-06 here
Now let us goto concave case.
Please click Draw Prob1201
Red curve is main function
  f(x)=sin(x) ---eqn.BJ136
domain is x in [0,2.5]

<a name="ch12a057">
Follow eqn.6.1 definition for
convex function, now define
concave function.

<a name="ch12a058">
A function f:[a,b]→Real is said to be
concave function provided that for all
x,y∈[a,b] and all
  0≦p≦1 ---eqn.BJ137
one has //red '≧' is change part.
  f(px +(1-p)y)≧
  pf(x)+(1-p)f(y) ---eqn.BJ138
 xe=px +(1−p)y ∈ [a,b] ---eqn.BJ139

<a name="ch12a059">
Do we prove eqn.BJ138 ? 
NO ! we do not prove eqn.BJ138
Because eqn.BJ138 is our 
definition equation.

<a name="ch12a060">
Explain eqn.BJ138 as following.
In eqn.BJ138, [a,b] is a concave
domain. Function in [a,b] shape
like a hat ╭╮.
x,y∈[a,b] say: we are in [a,b]
For 0≦p≦1,
  xe=px +(1−p)y ---eqn.BJ140
is a point within [x,y], then
<a name="ch12a061">　Index begin　Index this file
xe is in concave domain [a,b]
f(px +(1-p)y) is a point on 
curve within concave domain 
pf(x)+(1-p)f(y) is a chord
from point [x,f(x)] for p=1
 to  point [y,f(y)] for p=0
<a name="ch12a062">
eqn.BJ138 simply say :
any points on curve f(xe)
is ≧ any points on chord.
If this condition satisfied
in a domain [a,b], we say
function is concave in [a,b]
2010-05-14-16-35 here

<a name="ch12a063">
In concave domain, a function
has the following properties.
(1) tangent f'(x) decrease 
    when x increase.
(2) Second derivative f''(x) is 
    negative in concave domain.
<a name="ch12a064">
(3) a tangent (straight line)
    is always above function
    curve.
(4) a chord (straight line) is
    always below function curve.
    This is definition condition.
2010-05-14-16-42 stop



<a name="ch12a065">　Index begin　Index this file
2010-05-14-20-23 start
Textbook use simplest example
problem. Let n=3 and set
  x=(x,y,z) ---eqn.BJ141
We have
  E1(x)=(x+y+z)/3 ---eqn.BJ142
  E2(x)=(xy+yz+xz)/3 ---eqn.BJ143
  E3(x)=(x*y*z) ---eqn.BJ144
<a name="ch12a066">
Assume x,y,z be length, then
E1 is length to power 1 length
E2 is length to power 2 area
E3 is length to power 3 volume
Convert them to length power 1
E1 no change 
  [E2(x)]1/2=[(xy+yz+xz)/3]1/2 ---eqn.BJ145
  [E3(x)]1/3=(x*y*z)1/3 ---eqn.BJ146
//Ek(x) power alert 1, 2
<a name="ch12a067">
Maclaurin's inequality say
  [E3(x)]1/3≦[E2(x)]1/2≦E1 ---eqn.BJ147
In expanded form, it is

<a name="ch12a068">
Less than end is Geometric Mean
Greater than end Arithmetic Mean
AM-GM inequality give us two end
relation. Maclaurin's inequality
eqn.12.3 tell us the middle term 
sit at right place. To prove 
Maclaurin's inequality start from
Newton's inequality eqn.12.2.
2010-05-14-20-56 here
<a name="ch12a069">　Index begin　Index this file
Assume data array x are all
positive. (negative1/2=complex)
(textbook use non-negative and
 exclude zero element)
In proof process, textbook use
log function, use log-concavity
property.

<a name="ch12a070">
Log function is monotone increase
function. If 0＜m＜n then
  log(m)＜log(n) ---eqn.BJ149
Log function change power to
coefficient, that is
  log(xp)=p*log(x) ---eqn.BJ150
<a name="ch12a071">
We use these two log properties
take log to Newton's inequality 
eqn.12.2 get the following

<a name="ch12a072">
When draw curve for Ek(x), use 
log(Ek(x)) instead. We see 
log(Ek(x)) is a concave function.
eqn.12.4 is a mathematical 
statement for concave function.
eqn.12.4 less than side is an
average function value on chord
<a name="ch12a073">
It average between k-1 and k+1.
eqn.12.4 greater than side is 
a point on curve. eqn.12.4 say:
point on curve value is greater
than point on chord value.
This is exactly concave function
specification.
2010-05-14-21-21 stop

<a name="ch12a074">　Index begin　Index this file
2010-05-16-11-07 start
The following is a short command
for total n=5 data.   local
Please goto the calculator page
http://freeman2.com/complex2.htm#calculator
Paste the code between '[[' and ']]'
to 'Box3, input JS command' 
<a name="ch12a075">
a,b,c,d,e value are the five elements
number sequence. You can change their
value. For output, you can select
from 0 to 8 for variable outType
In complex2.htm#calculator click 
[text box3 command, output to box4]
button. 
<a name="ch12a076">
outType=0 has answer for Maclaurin
          inequality
outType=1,2,3 has various error as
          comment stated.
outType=4 give polynomial coef.
outType=5 to 8 verify Newton's Ineq.

//<a name="ch12a077">
//[[
//2010-05-15-23-18
//small program to verify Newton Ineq
//and verify Maclaurin inequality.
//This program is designed for five
//data a,b,c,d,e only. case n=5
//you can change a,b,c,d,e values
var a=1,b=2.3,c=3.1,d=4.6,e=6.8
var outType=0 //you can select 0,1 to 8
//<a name="ch12a078">
//outType=0 verify Maclaurin inequality
//outType=1 to 3 are error analysis.
//outType=1 one to many, n,k factor involved
//outType=2 length compare with volume wrong
//outType=3 both one to many, length to volume
//outType=4 give polynomial coefficients
//the roots of this polynomial are a,b,c,d,e
//outType=5 E0*E2≦E1*E1 Newton Ineq n=5,k=1
//outType=8 E3*E5≦E4*E4 Newton Ineq n=5,k=4
//<a name="ch12a079">
//'e' is 5th data. 'e' is different from e_?
//e_0 to e_5 are Symmetric Sums.
e_0=1 // by definition
e_1=a +b +c +d +e
e_2=a*b+a*c+a*d+a*e+b*c+b*d+b*e+c*d+c*e+d*e
e_3=a*b*c+a*b*d+a*b*e+a*c*d+a*c*e+a*d*e+b*c*d+b*c*e+b*d*e+c*d*e
e_4=a*b*c*d+a*b*c*e+a*b*d*e+a*c*d*e+b*c*d*e
e_5=a*b*c*d*e
//e_1+e_2+e_3+e_4+e_5
//<a name="ch12a080">
E0=1 // by definition
E1=e_1/bicof(5,1)
E2=e_2/bicof(5,2)
E3=e_3/bicof(5,3)
E4=e_4/bicof(5,4)
E5=e_5/bicof(5,5)
//<a name="ch12a081">
E0p1=1
E1p1=E1
E2p1=pow(E2,1/2)
E3p1=pow(E3,1/3)
E4p1=pow(E4,1/4)
E5p1=pow(E5,1/5)
//<a name="ch12a082">
if(outType==0)
{
E1p1 //this is arithmetic mean, greatest
E2p1 //'p1' in 'E2p1' means power one
E3p1 //length compare with length is OK
E4p1 //averaged compare with averaged OK
E5p1 //this is geometric mean, smallest
}
//<a name="ch12a083">
else if(outType==1)
{
e_1p1=e_1
e_2p1=pow(e_2,1/2)
e_3p1=pow(e_3,1/3)
e_4p1=pow(e_4,1/4)
e_5p1=pow(e_5,1/5)
e_1p1 //one to many, n,k factor involved
e_2p1 //'p1' in 'e_2p1' means power one
e_3p1 //although length compare with length
e_4p1 //but n,k factor involved. trouble
e_5p1 //lower case e_* has n,k factor
}
//<a name="ch12a084">
else if(outType==2)
{
E1 //length compare with volume wrong
E2 //upper case E* removed n,k factor
E3 //upper case E* use average value
E4 //but still wrong, here is outType=2
E5 //select outType=0 for correct answer
}
//<a name="ch12a085">
else if(outType==3)
{
e_1 //error1: one to many, n,k involved
e_2 //error2: length compare with volume
e_3 //double error. here is outType=3
e_4 //select outType=0 for correct answer
e_5
} //2010-05-16-10-41
//<a name="ch12a086">
else if(outType==4)
{
+e_0 //polynomial x^5 coef.
-e_1 //polynomial x^4 coef.
+e_2 //polynomial x*x*x coef.
-e_3 //polynomial x*x coef.
+e_4 //polynomial x coefficient
-e_5 //polynomial constant
} //2010-05-16-11-24 
//Above outType=0 for Maclaurin inequality
//outType=1,2,3 for error analysis
//outType=4 for polynomial coefficient
//<a name="ch12a087">
//Below outType=5,6,7,8 for Newton inequality
else if(outType==5) //2010-05-16-12-18
{
E0*E2 //E0*E2≦E1*E1
E1*E1 //Newton inequality n=5,k=1
}
else if(outType==6)
{
E1*E3 //E1*E3≦E2*E2
E2*E2 //Newton inequality n=5,k=2
}
//<a name="ch12a088">
else if(outType==7)
{
E2*E4 //E2*E4≦E3*E3
E3*E3 //Newton inequality n=5,k=3
}
else if(outType==8)
{
E3*E5 //E3*E5≦E4*E4
E4*E4 //Newton inequality n=5,k=4
} //2010-05-16-12-23
//]]

2010-05-16-12-30 stop

<a name="ch12a089">　Index begin　Index this file
2010-05-16-13-30 start
■　Prove low power end Maclaurin
Before prove
Newton's inequality eqn.12.2 and
Maclaurin's inequality eqn.12.3
verify special case first. To
gain some understanding.
For the special case n=5
Maclaurin's inequality 
  E51/5≦E41/4≦E31/3≦E21/2≦E1 ---eqn.12.3
//Ek(x) power alert 1, 2
<a name="ch12a090">
low power end E21/2?≦?E1 is
  e_1=a +b +c +d +e ---eqn.BJ151
  e_2=a*b+a*c+a*d+a*e+b*c
     +b*d+b*e+c*d+c*e+d*e ---eqn.BJ152
  E1=(a +b +c +d +e)/5 ---eqn.BJ153
  E2=(a*b+a*c+a*d+a*e+b*c+b*d
     +b*e+c*d+c*e+d*e)/10 ---eqn.BJ154
<a name="ch12a091">
E21/2?≦?E1 is same as E12-E2?≧?0
and is same as 
  (a+b+c+d+e)*(a+b+c+d+e)/5/5
 -(a*b+a*c+a*d+a*e+b*c+b*d
  +b*e+c*d+c*e+d*e)/10
 ?≧?0 ---eqn.BJ155
<a name="ch12a092">　Index begin　Index this file
Now verify eqn.BJ155
ASSUME 
  0≦a≦b≦c≦d≦e ---eqn.BJ156

 50*eqn.BJ155=
  2*(a+b+c+d+e)*(a+b+c+d+e)
 -5*(a*b+a*c+a*d+a*e+b*c+b*d
    +b*e+c*d+c*e+d*e) ---eqn.BJ157
<a name="ch12a093">
 =2*(a*a+a*b+a*c+a*d+a*e
    +b*a+b*b+b*c+b*d+b*e
    +c*a+c*b+c*c+c*d+c*e
    +d*a+d*b+d*c+d*d+d*e
    +e*a+e*b+e*c+e*d+e*e)
 -5*(a*b+a*c+a*d+a*e+b*c+b*d
    +b*e+c*d+c*e+d*e) ---eqn.BJ158
<a name="ch12a094">
 =   2*a*a+2*a*b+2*a*c+2*a*d+2*a*e
    +2*b*a+2*b*b+2*b*c+2*b*d+2*b*e
    +2*c*a+2*c*b+2*c*c+2*c*d+2*c*e
    +2*d*a+2*d*b+2*d*c+2*d*d+2*d*e
    +2*e*a+2*e*b+2*e*c+2*e*d+2*e*e
    -5*a*b-5*a*c-5*a*d-5*a*e-5*b*c
    -5*b*d-5*b*e-5*c*d-5*c*e-5*d*e
     ---eqn.BJ159
<a name="ch12a095">
 =   2*a*a+
 2*a*b-5*a*b
+2*a*c-5*a*c
+2*a*d-5*a*d
+2*a*e-5*a*e
+2*b*a
+2*b*b
+2*b*c-5*b*c
+2*b*d-5*b*d
+2*b*e-5*b*e
+2*c*a
+2*c*b
+2*c*c
+2*c*d-5*c*d
+2*c*e-5*c*e
+2*d*a
+2*d*b
+2*d*c
+2*d*d
+2*d*e-5*d*e
+2*e*a
+2*e*b
+2*e*c
+2*e*d
+2*e*e ---eqn.BJ160

<a name="ch12a096">　Index begin　Index this file
 =  
 2*a*a
-3*a*b
-3*a*c
-3*a*d
-3*a*e
+2*b*a
+2*b*b
-3*b*c
-3*b*d
-3*b*e
+2*c*a
+2*c*b
+2*c*c
-3*c*d
-3*c*e
+2*d*a
+2*d*b
+2*d*c
+2*d*d
-3*d*e
+2*e*a
+2*e*b
+2*e*c
+2*e*d
+2*e*e ---eqn.BJ161

<a name="ch12a097">
 =  
 2*a*a
+2*b*b
+2*c*c
+2*d*d
+2*e*e

-3*a*b
-3*a*c
-3*a*d
-3*a*e
-3*b*c
-3*b*d
-3*b*e
-3*c*d
-3*c*e
-3*d*e

+2*a*b
+2*a*c
+2*b*c
+2*a*d
+2*b*d
+2*c*d
+2*a*e
+2*b*e
+2*c*e
+2*d*e ---eqn.BJ162
<a name="ch12a098">
 =  
+a*a
+b*b
+c*c
+d*d
+e*e

+a*a
+b*b
+c*c
+d*d
+e*e
<a name="ch12a099">
-a*b
-b*c
-c*d
-d*e
-e*a

-a*c
-b*d
-c*e
-d*a
-e*b ---eqn.BJ163

<a name="ch12a100">　Index begin　Index this file
eqn.BJ163 red terms are dot
product of array1 and array3
array1=[a,b,c,d,e]
array3=[a,b,c,d,e]

Blue term is next two array 
dot product
array1=[a,b,c,d,e]
array2=[b,c,d,e,a]

<a name="ch12a101">
Purple term is next two array 
dot product
array3=[a,b,c,d,e]
array4=[c,d,e,a,b]

<a name="ch12a102">
Since assume that
  0≦a≦b≦c≦d≦e ---eqn.BJ156
from rearrangement inequality
we conclude eqn.BJ163 ≧0
(first  red greater than blue,
 second red greater than purple)
Conclude for special case n=5 
Maclaurin's inequality low power 
end inequality eqn.BJ155 is true.
2010-05-16-14-36 stop

<a name="ch12a103">　Index begin　Index this file
2010-05-16-17-03 start
■　Middle power Maclaurin choked
Next verify special case n=5
Maclaurin's inequality 
  E51/5≦E41/4≦E31/3≦E21/2≦E1 ---eqn.12.3
E31/3?≦?E21/2 part is next

<a name="ch12a104">
  e_2=a*b+a*c+a*d+a*e+b*c
     +b*d+b*e+c*d+c*e+d*e ---eqn.BJ152
  e_3=a*b*c+a*b*d+a*b*e+a*c*d
     +a*c*e+a*d*e+b*c*d+b*c*e
     +b*d*e+c*d*e  ---eqn.BJ164

<a name="ch12a105">
  E2=(a*b+a*c+a*d+a*e+b*c+b*d
     +b*e+c*d+c*e+d*e)/10 ---eqn.BJ154
  E3=(a*b*c+a*b*d+a*b*e+a*c*d
     +a*c*e+a*d*e+b*c*d+b*c*e
     +b*d*e+c*d*e)/10  ---eqn.BJ165

<a name="ch12a106">
E31/3?≦?E21/2 is same as ---eqn.BJ166
E23-E32?≧?0
and is same as 
   [(a*b+a*c+a*d+a*e+b*c+b*d
    +b*e+c*d+c*e+d*e)/10]3
  -[(a*b*c+a*b*d+a*b*e+a*c*d
     +a*c*e+a*d*e+b*c*d+b*c*e
     +b*d*e+c*d*e)/10]2?≧?0 ---eqn.BJ167
<a name="ch12a107">
[a*b+a*c+a*d+a*e+b*c+b*d+b*e
+c*d+c*e+d*e]3 has 1000 terms
[a*b*c+a*b*d+a*b*e+a*c*d+a*c*e
+a*d*e+b*c*d+b*c*e+b*d*e+c*d*e]2
has 100 terms (then multiply 10).
Now verify eqn.BJ167
Still ASSUME 
  0≦a≦b≦c≦d≦e ---eqn.BJ156
..... work record omitted .....
.....  too  many  terms   .....

<a name="ch12a108">　Index begin　Index this file
During work need do multiply
cancel/collect etc
start from (for example)
[[
+a*b*a*b
+a*b*a*c
+a*b*a*d
+a*b*a*e
+a*b*b*c
+a*b*b*d
.....
]]
<a name="ch12a109">
middle step is
[[
+a*a*b*b
+a*a*b*c
+a*a*b*c
+a*a*b*d
+a*a*b*d
+a*a*b*e
.....
]]
<a name="ch12a110">
the following code
sort in each line, change 
from  abab  (+a*b*a*b)
 to   aabb  (+a*a*b*b)
to merge/cancel like term
need sort in-line.

<a name="ch12a111">　Index begin　Index this file
Following is working code, 
user define a2e
program output to // local
[[
aabb
aabc
aabd
aabe
abbc
..... etc. then sort in column. LiuHH 
..... use DOS prompt sort<i.txt>o.txt
]]

//<a name="ch12a112">
//2010-05-16-17-51
//[[ program sort in-line, 
//not sort in column. User define a2e
//'a2e' must string in one line. long long line OK
//required by complex2.htm, not required by javascript.
var a2e=['abab','abac','abad','abae','abbc','abbd','abbe','abcd','abce','abde','acab','acac','acad','acae','acbc','acbd','acbe','accd','acce','acde','adab','adac','adad','adae','adbc','adbd','adbe','adcd','adce','adde','aeab','aeac','aead','aeae','aebc','aebd','aebe','aecd','aece','aede','bcab','bcac','bcad','bcae','bcbc','bcbd','bcbe','bccd','bcce','bcde','bdab','bdac','bdad','bdae','bdbc','bdbd','bdbe','bdcd','bdce','bdde','beab','beac','bead','beae','bebc','bebd','bebe','becd','bece','bede','cdab','cdac','cdad','cdae','cdbc','cdbd','cdbe','cdcd','cdce','cdde','ceab','ceac','cead','ceae','cebc','cebd','cebe','cecd','cece','cede','deab','deac','dead','deae','debc','debd','debe','decd','dece','dede'];
//<a name="ch12a113">
//n=5 case, variable a,b,c,d,e
//"ae" is your rule, you can 
//change "ae" value to your need
var ae='abcde'; 
//do not change below 
var i,j,k;
var ouStr='';
for(i=0;i<a2e.length;i++)
{
for(k=0;k<ae.length;k++)
  {
for(j=0;j<a2e[0].length;j++)
//<a name="ch12a114">
if(a2e[i].charAt(j)==ae.charAt(k))
    {
ouStr+=a2e[i].charAt(j);
dummy=0; continue; //MUST use d=0 here
    } //if delete 'dummy=0;', program
  }   //print value of 'continue', error.
ouStr+='\n';
}//2010-05-16-18-01
ouStr //in-line sorted as 'ae' specified.
//]]


<a name="ch12a115">　Index begin　Index this file
2010-05-16-22-38
Spend a lot of time, but the
case of n=5, k=2 is hard to 
reach. For
    a*a*a*b*b*b 
.....
- 4*c*c*d*d*e*e
several hundred terms,
<a name="ch12a116">
not know how to use
rearrangement inequality
to show that
E31/3?≦?E21/2 
is true. After raise power to 
six power (like a*a*a*b*b*b)
and after cancel/collect like
term, 
<a name="ch12a117">
the last step is next
[[
+  a*a*a*b*b*b
+3*a*a*a*b*b*c
+3*a*a*a*b*b*d
+3*a*a*a*b*b*e
+3*a*a*a*b*c*c
+6*a*a*a*b*c*d
+6*a*a*a*b*c*e
+3*a*a*a*b*d*d
+6*a*a*a*b*d*e
+3*a*a*a*b*e*e
+  a*a*a*c*c*c
+3*a*a*a*c*c*d
+3*a*a*a*c*c*e
+3*a*a*a*c*d*d
+6*a*a*a*c*d*e
+3*a*a*a*c*e*e
+  a*a*a*d*d*d
+3*a*a*a*d*d*e
+3*a*a*a*d*e*e
+  a*a*a*e*e*e
+3*a*a*b*b*b*c
+3*a*a*b*b*b*d
+3*a*a*b*b*b*e
+3*a*a*b*c*c*c
+3*a*a*b*d*d*d
+3*a*a*b*e*e*e
+3*a*a*c*c*c*d
+3*a*a*c*c*c*e
+3*a*a*c*d*d*d
+3*a*a*c*e*e*e
+3*a*a*d*d*d*e
+3*a*a*d*e*e*e
+3*a*b*b*b*c*c
+6*a*b*b*b*c*d
+6*a*b*b*b*c*e
+3*a*b*b*b*d*d
+6*a*b*b*b*d*e
+3*a*b*b*b*e*e
+3*a*b*b*c*c*c
+3*a*b*b*d*d*d
+3*a*b*b*e*e*e
+6*a*b*c*c*c*d
+6*a*b*c*c*c*e
+6*a*b*c*d*d*d
+6*a*b*c*e*e*e
+6*a*b*d*d*d*e
+6*a*b*d*e*e*e
+3*a*c*c*c*d*d
+6*a*c*c*c*d*e
+3*a*c*c*c*e*e
+3*a*c*c*d*d*d
+3*a*c*c*e*e*e
+6*a*c*d*d*d*e
+6*a*c*d*e*e*e
+3*a*d*d*d*e*e
+3*a*d*d*e*e*e
+  b*b*b*c*c*c
+3*b*b*b*c*c*d
+3*b*b*b*c*c*e
+3*b*b*b*c*d*d
+6*b*b*b*c*d*e
+3*b*b*b*c*e*e
+  b*b*b*d*d*d
+3*b*b*b*d*d*e
+3*b*b*b*d*e*e
+  b*b*b*e*e*e
+3*b*b*c*c*c*d
+3*b*b*c*c*c*e
+3*b*b*c*d*d*d
+3*b*b*c*e*e*e
+3*b*b*d*d*d*e
+3*b*b*d*e*e*e
+3*b*c*c*c*d*d
+6*b*c*c*c*d*e
+3*b*c*c*c*e*e
+3*b*c*c*d*d*d
+3*b*c*c*e*e*e
+6*b*c*d*d*d*e
+6*b*c*d*e*e*e
+3*b*d*d*d*e*e
+3*b*d*d*e*e*e
+  c*c*c*d*d*d
+3*c*c*c*d*d*e
+3*c*c*c*d*e*e
+  c*c*c*e*e*e
+3*c*c*d*d*d*e
+3*c*c*d*e*e*e
+3*c*d*d*d*e*e
+3*c*d*d*e*e*e
+  d*d*d*e*e*e

<a name="ch12a118">
- 4*a*a*b*b*c*c
- 5*a*a*b*b*c*d
- 5*a*a*b*b*c*e
- 4*a*a*b*b*d*d
- 5*a*a*b*b*d*e
- 4*a*a*b*b*e*e
- 5*a*a*b*c*c*d
- 5*a*a*b*c*c*e
- 5*a*a*b*c*d*d
-24*a*a*b*c*d*e
- 5*a*a*b*c*e*e
- 5*a*a*b*d*d*e
- 5*a*a*b*d*e*e
- 4*a*a*c*c*d*d
- 5*a*a*c*c*d*e
- 4*a*a*c*c*e*e
- 5*a*a*c*d*d*e
- 5*a*a*c*d*e*e
- 4*a*a*d*d*e*e
- 5*a*b*b*c*c*d
- 5*a*b*b*c*c*e
- 5*a*b*b*c*d*d
-24*a*b*b*c*d*e
- 5*a*b*b*c*e*e
- 5*a*b*b*d*d*e
- 5*a*b*b*d*e*e
- 5*a*b*c*c*d*d
-24*a*b*c*c*d*e
- 5*a*b*c*c*e*e
-24*a*b*c*d*d*e
-24*a*b*c*d*e*e
- 5*a*b*d*d*e*e
- 5*a*c*c*d*d*e
- 5*a*c*c*d*e*e
- 5*a*c*d*d*e*e
- 4*b*b*c*c*d*d
- 5*b*b*c*c*d*e
- 4*b*b*c*c*e*e
- 5*b*b*c*d*d*e
- 5*b*b*c*d*e*e
- 4*b*b*d*d*e*e
- 5*b*c*c*d*d*e
- 5*b*c*c*d*e*e
- 5*b*c*d*d*e*e
- 4*c*c*d*d*e*e
]]
<a name="ch12a119">
(Cancellation may not help!)
Up to here, learned that 
direct approach to Maclaurin's 
inequality for middle terms
that is very hard to nearly
impossible.
Need indirect method.
<a name="ch12a120">
However, two end inequality
  E51/5≦E41/4 ---eqn.BJ168
and
  E21/2≦E1 ---eqn.BJ169
are possible to prove directly.
2010-05-16-22-56 stop

<a name="ch12a121">　Index begin　Index this file
2010-05-17-10-30 start
■　Prove high power end Maclaurin
Now try prove end inequality
  E51/5≦E41/4 ---eqn.BJ170
for the case n=5
  e_4=a*b*c*d+a*b*c*e+a*b*d*e+a*c*d*e+b*c*d*e ---eqn.BJ171
  e_5=a*b*c*d*e ---eqn.BJ172
<a name="ch12a122">
  E4=e_4/bicof(5,4)
  E5=e_5/bicof(5,5)
  E4=(a*b*c*d+a*b*c*e+a*b*d*e+a*c*d*e+b*c*d*e)/5 ---eqn.BJ173
  E5=a*b*c*d*e ---eqn.BJ174
<a name="ch12a123">
The target inequality is eqn.BJ170
that is 
  (a*b*c*d*e)1/5?≦? ---eqn.BJ175
 [(a*b*c*d+a*b*c*e+a*b*d*e
  +a*c*d*e+b*c*d*e)/5]1/4
<a name="ch12a124">
eqn.BJ175 take fourth power, let
greater than side be power one
  (a*b*c*d*e)4/5?≦? ---eqn.BJ176
  (a*b*c*d+a*b*c*e+a*b*d*e
   +a*c*d*e+b*c*d*e)/5
<a name="ch12a125">
Let eqn.BJ176 divide by a*b*c*d*e
get //here require abcde all＞0
  (a*b*c*d*e)4/5/(a*b*c*d*e)?≦? ---eqn.BJ177
  (a*b*c*d+a*b*c*e+a*b*d*e
   +a*c*d*e+b*c*d*e)/(a*b*c*d*e)/5
which simplify to
  1/[(a*b*c*d*e)1/5] ?≦? ---eqn.BJ178
  (1/e + 1/d +1/c +1/b +1/a)/5
<a name="ch12a126">
eqn.BJ178 is true or not true ?
TRUE !! Because if we take the
sequence_5
  [1/a, 1/b, 1/c, 1/d, 1/e]
eqn.BJ178 less than side is 
Geometric Mean of sequence_5
eqn.BJ178 greater than side 
is Arithmetic Mean of seq_5.

<a name="ch12a127">
For the special case n=5,
Maclaurin's inequality 
  E51/5≦E41/4≦E31/3≦E21/2≦E1 ---eqn.12.3
High power end E51/5≦E41/4 proved
to be true by AM-GM inequality.
Low power end E21/2≦E1 proved to
be true by rearrangement ineq.
Middle inequalities tried but
hard to draw conclusion.
2010-05-17-11-01 here

Above is Maclaurin's inequality.
<a name="ch12a128">　Index begin　Index this file
■　Low power end Newton=Maclaurin
Below is Newton's inequality
  Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.2
LiuHH still hope to gain some
understanding for special and 
simpler cases.
Now see n=5 and k=1 low power 
end Newton's inequality
//Ek(x) power alert 1, 2
<a name="ch12a129">
eqn.12.2 become
  E0(x)*E2(x)?≦?E12(x) ---eqn.BJ179
where
  E0(x)=1 ---eqn.BJ180
  E1(x)=(a +b +c +d +e)/5 ---eqn.BJ181
  E2(x)=(a*b+a*c+a*d+a*e+b*c+b*d
        +b*e+c*d+c*e+d*e)/10 ---eqn.BJ182
<a name="ch12a130">
Target eqn.BJ179 become
  1*(a*b+a*c+a*d+a*e+b*c+b*d
        +b*e+c*d+c*e+d*e)/10
  ?≦? [(a +b +c +d +e)/5]
      *[(a +b +c +d +e)/5]
Move everything to greater than
side, get 
<a name="ch12a131">
  target01 define=
  [(a +b +c +d +e)/5]2
 -(a*b+a*c+a*d+a*e+b*c+b*d
   +b*e+c*d+c*e+d*e)/10
  ?≧? 0 ---eqn.BJ183
eqn.BJ183 is same as eqn.BJ155
already proved to be true

<a name="ch12a132">　Index begin　Index this file
■　Prove high power end Newton
Now see n=5 and k=4 high power 
end Newton's inequality, 
eqn.12.2 become
  E3(x)*E5(x)?≦?E42(x) ---eqn.BJ184
(there is NO n=5 and k=5 case
 because k+1=5+1=6 is undefined
 for n=5. simply no n=6 term)
<a name="ch12a133">
E3(x)=(a*b*c+a*b*d+a*b*e
    +a*c*d+a*c*e+a*d*e+b*c*d
    +b*c*e+b*d*e+c*d*e)/10 ---eqn.BJ185
E4(x)=(a*b*c*d+a*b*c*e+a*b*d*e
    +a*c*d*e+b*c*d*e)/5 ---eqn.BJ186
E5(x)= a*b*c*d*e ---eqn.BJ187
<a name="ch12a134">
eqn.BJ184 become
  target02 define=
  [(a*b*c*d+a*b*c*e+a*b*d*e
    +a*c*d*e+b*c*d*e)/5]2
  -a*b*c*d*e*[(a*b*c+a*b*d+a*b*e
    +a*c*d+a*c*e+a*d*e+b*c*d
    +b*c*e+b*d*e+c*d*e)/10]
  ?≧? 0 ---eqn.BJ188
Next is expansion, cancel, collect
terms.
2010-05-17-11-33 here

<a name="ch12a135">
  target02 = ---eqn.BJ189
(
+a*b*c*d*a*b*c*d+a*a*b*c*d*b*c*e+a*b*c*d*a*b*d*e+a*b*c*d*a*c*d*e+a*b*c*d*b*c*d*e
+a*b*c*e*a*b*c*d+a*b*c*e*a*b*c*e+a*b*c*e*a*b*d*e+a*b*c*e*a*c*d*e+a*b*c*e*b*c*d*e
+a*b*d*e*a*b*c*d+a*b*d*e*a*b*c*e+a*b*d*e*a*b*d*e+a*b*d*e*a*c*d*e+a*b*d*e*b*c*d*e
+a*c*d*e*a*b*c*d+a*c*d*e*a*b*c*e+a*c*d*e*a*b*d*e+a*c*d*e*a*c*d*e+a*c*d*e*b*c*d*e
+b*c*d*e*a*b*c*d+b*c*d*e*a*b*c*e+b*c*d*e*a*b*d*e+b*c*d*e*a*c*d*e+b*c*d*e*b*c*d*e
)/25
+
(
-a*b*c*d*e*a*b*c-a*b*c*d*e*a*b*d-a*b*c*d*e*a*b*e-a*b*c*d*e*a*c*d-a*b*c*d*e*a*c*e
-a*b*c*d*e*a*d*e-a*b*c*d*e*b*c*d-a*b*c*d*e*b*c*e-a*b*c*d*e*b*d*e-a*b*c*d*e*c*d*e
)/10

<a name="ch12a136">
  50*target02 = ---eqn.BJ190
2*(
+a*b*c*d*a*b*c*d+a*a*b*c*d*b*c*e+a*b*c*d*a*b*d*e+a*b*c*d*a*c*d*e+a*b*c*d*b*c*d*e
+a*b*c*e*a*b*c*d+a*b*c*e*a*b*c*e+a*b*c*e*a*b*d*e+a*b*c*e*a*c*d*e+a*b*c*e*b*c*d*e
+a*b*d*e*a*b*c*d+a*b*d*e*a*b*c*e+a*b*d*e*a*b*d*e+a*b*d*e*a*c*d*e+a*b*d*e*b*c*d*e
+a*c*d*e*a*b*c*d+a*c*d*e*a*b*c*e+a*c*d*e*a*b*d*e+a*c*d*e*a*c*d*e+a*c*d*e*b*c*d*e
+b*c*d*e*a*b*c*d+b*c*d*e*a*b*c*e+b*c*d*e*a*b*d*e+b*c*d*e*a*c*d*e+b*c*d*e*b*c*d*e
)
+
5*(
-a*b*c*d*e*a*b*c-a*b*c*d*e*a*b*d-a*b*c*d*e*a*b*e-a*b*c*d*e*a*c*d-a*b*c*d*e*a*c*e
-a*b*c*d*e*a*d*e-a*b*c*d*e*b*c*d-a*b*c*d*e*b*c*e-a*b*c*d*e*b*d*e-a*b*c*d*e*c*d*e
)

<a name="ch12a137">　Index begin　Index this file
  50*target02 = ---eqn.BJ191
+2*a*a*b*b*c*c*d*d
+2*a*a*b*b*c*c*d*e
+2*a*a*b*b*c*c*d*e
+2*a*a*b*b*c*c*e*e
+2*a*a*b*b*c*d*d*e
+2*a*a*b*b*c*d*d*e
+2*a*a*b*b*c*d*e*e
+2*a*a*b*b*c*d*e*e
+2*a*a*b*b*d*d*e*e
+2*a*a*b*c*c*d*d*e
+2*a*a*b*c*c*d*d*e
+2*a*a*b*c*c*d*e*e
+2*a*a*b*c*c*d*e*e
+2*a*a*b*c*d*d*e*e
+2*a*a*b*c*d*d*e*e
+2*a*a*c*c*d*d*e*e
+2*a*b*b*c*c*d*d*e
+2*a*b*b*c*c*d*d*e
+2*a*b*b*c*c*d*e*e
+2*a*b*b*c*c*d*e*e
+2*a*b*b*c*d*d*e*e
+2*a*b*b*c*d*d*e*e
+2*a*b*c*c*d*d*e*e
+2*a*b*c*c*d*d*e*e
+2*b*b*c*c*d*d*e*e
-5*a*a*b*b*c*c*d*e
-5*a*a*b*b*c*d*d*e
-5*a*a*b*b*c*d*e*e
-5*a*a*b*c*c*d*d*e
-5*a*a*b*c*c*d*e*e
-5*a*a*b*c*d*d*e*e
-5*a*b*b*c*c*d*d*e
-5*a*b*b*c*c*d*e*e
-5*a*b*b*c*d*d*e*e
-5*a*b*c*c*d*d*e*e

<a name="ch12a138">
  50*target02 = ---eqn.BJ192
+2*a*a*b*b*c*c*d*d

+2*a*a*b*b*c*c*d*e
+2*a*a*b*b*c*c*d*e-5*a*a*b*b*c*c*d*e

+2*a*a*b*b*c*c*e*e

+2*a*a*b*b*c*d*d*e
+2*a*a*b*b*c*d*d*e-5*a*a*b*b*c*d*d*e

<a name="ch12a139">
+2*a*a*b*b*c*d*e*e
+2*a*a*b*b*c*d*e*e-5*a*a*b*b*c*d*e*e

+2*a*a*b*b*d*d*e*e

+2*a*a*b*c*c*d*d*e
+2*a*a*b*c*c*d*d*e-5*a*a*b*c*c*d*d*e

+2*a*a*b*c*c*d*e*e
+2*a*a*b*c*c*d*e*e-5*a*a*b*c*c*d*e*e

<a name="ch12a140">
+2*a*a*b*c*d*d*e*e
+2*a*a*b*c*d*d*e*e-5*a*a*b*c*d*d*e*e

+2*a*a*c*c*d*d*e*e

+2*a*b*b*c*c*d*d*e
+2*a*b*b*c*c*d*d*e-5*a*b*b*c*c*d*d*e

+2*a*b*b*c*c*d*e*e
+2*a*b*b*c*c*d*e*e-5*a*b*b*c*c*d*e*e

<a name="ch12a141">
+2*a*b*b*c*d*d*e*e
+2*a*b*b*c*d*d*e*e-5*a*b*b*c*d*d*e*e

+2*a*b*c*c*d*d*e*e
+2*a*b*c*c*d*d*e*e-5*a*b*c*c*d*d*e*e

+2*b*b*c*c*d*d*e*e

<a name="ch12a142">　Index begin　Index this file
  50*target02 = ---eqn.BJ193
+2*a*a*b*b*c*c*d*d
+2*a*a*b*b*c*c*e*e
+2*a*a*b*b*d*d*e*e
+2*a*a*c*c*d*d*e*e
+2*b*b*c*c*d*d*e*e
-a*a*b*b*c*c*d*e
-a*a*b*b*c*d*d*e
-a*a*b*b*c*d*e*e
-a*a*b*c*c*d*d*e
-a*a*b*c*c*d*e*e
-a*a*b*c*d*d*e*e
-a*b*b*c*c*d*d*e
-a*b*b*c*c*d*e*e
-a*b*b*c*d*d*e*e
-a*b*c*c*d*d*e*e
2010-05-17-12-36 here
<a name="ch12a143">
can not use AM-GM, 
because no GM here
rearrangement is possible

+a*b*c*d *a*b*c*d no e
+a*b*c*e *a*b*c*e no d
+a*b*d*e *a*b*d*e no c
+a*c*d*e *a*c*d*e no b
+b*c*d*e *b*c*d*e no a ---eqn.BJ194

<a name="ch12a144">
-a*b*c*e *a*b*c*d d/e = d and e are single
-a*b*c*d *a*b*d*e c/e
-a*b*d*e *a*b*c*e c/d
-a*b*c*d *a*c*d*e b/e
-a*b*c*e *a*c*d*e b/d
-a*b*d*e *a*c*d*e b/c
-a*b*c*d *b*c*d*e a/e
-a*b*c*e *b*c*d*e a/d ---eqn.BJ195
-a*b*d*e *b*c*d*e a/c
-a*c*d*e *b*c*d*e a/b correct 9905171319

Above/below b/c, a/d etc are
auxiliary notes, not part of
equation.

<a name="ch12a145">
  50*target02 = ---eqn.BJ196
2*(
+a*b*c*d *a*b*c*d e/e
+a*b*c*e *a*b*c*e d/d
+a*b*d*e *a*b*d*e c/c
+a*c*d*e *a*c*d*e b/b
+b*c*d*e *b*c*d*e a/a
  )
-a*b*c*e *a*b*c*d d/e
-a*b*c*d *a*b*d*e c/e
-a*b*d*e *a*b*c*e c/d
-a*b*c*d *a*c*d*e b/e
-a*b*c*e *a*c*d*e b/d
-a*b*d*e *a*c*d*e b/c
-a*b*c*d *b*c*d*e a/e
-a*b*c*e *b*c*d*e a/d
-a*b*d*e *b*c*d*e a/c
-a*c*d*e *b*c*d*e a/b

<a name="ch12a146">
2010-05-17-13-33
each group has
aa,bb,cc,dd,ee
Negative terms split to two
groups.

-a*b*d*e *a*b*c*e c/d ---eqn.BJ197
-a*b*c*d *a*c*d*e b/e
-a*b*d*e *a*c*d*e b/c
-a*b*c*d *b*c*d*e a/e
-a*b*c*e *b*c*d*e a/d

-a*b*c*e *a*c*d*e b/d ---eqn.BJ198
-a*b*c*e *a*b*c*d d/e
-a*b*c*d *a*b*d*e c/e
-a*b*d*e *b*c*d*e a/c
-a*c*d*e *b*c*d*e a/b

<a name="ch12a147">　Index begin　Index this file
Positive term and negative term
divide into two groups

[[ group A before reorder
+a*b*c*d *a*b*c*d e/e ---eqn.BJ199
+a*b*c*e *a*b*c*e d/d
+a*b*d*e *a*b*d*e c/c
+a*c*d*e *a*c*d*e b/b
+b*c*d*e *b*c*d*e a/a

-a*b*c*e *a*c*d*e d/b
-a*b*c*e *a*b*c*d d/e
-a*b*c*d *a*b*d*e e/c
-a*b*d*e *b*c*d*e c/a
-a*c*d*e *b*c*d*e b/a
]]

<a name="ch12a148">
[[ group B before reorder
+a*b*c*d *a*b*c*d e/e ---eqn.BJ200
+a*b*c*e *a*b*c*e d/d
+a*b*d*e *a*b*d*e c/c
+a*c*d*e *a*c*d*e b/b
+b*c*d*e *b*c*d*e a/a

-a*b*d*e *a*b*c*e c/d
-a*b*c*d *a*c*d*e e/b
-a*b*d*e *a*c*d*e c/b
-a*b*c*d *b*c*d*e e/a
-a*b*c*e *b*c*d*e d/a
]]

<a name="ch12a149">
2010-05-17-13-46 next is match group
[[ group A after reorder
+a*b*c*d *a*b*c*d e/e ---eqn.BJ201
+a*b*c*e *a*b*c*e d/d
+a*b*d*e *a*b*d*e c/c
+a*c*d*e *a*c*d*e b/b
+b*c*d*e *b*c*d*e a/a

-a*b*c*d *a*b*d*e e/c
-a*b*c*e *a*b*c*d d/e
-a*b*d*e *b*c*d*e c/a
-a*c*d*e *a*b*c*e b/d
-b*c*d*e *a*c*d*e a/b
]]

<a name="ch12a150">
2010-05-17-13-52 next is match group
[[ group B after reorder
+a*b*c*d *a*b*c*d e/e ---eqn.BJ202
+a*b*c*e *a*b*c*e d/d
+a*b*d*e *a*b*d*e c/c
+a*c*d*e *a*c*d*e b/b
+b*c*d*e *b*c*d*e a/a

-a*b*c*d *a*c*d*e e/b
-a*b*c*e *b*c*d*e d/a
-a*b*d*e *a*b*c*e c/d
-a*c*d*e *a*b*d*e b/c
-b*c*d*e *a*b*c*d a/e
]]

<a name="ch12a151">
2010-05-17-13-54 here
Let us see
[[ group A after reorder
left seq * right seq.

+a*b*c*d *a*b*c*d e/e ---eqn.BJ203
+a*b*c*e *a*b*c*e d/d
+a*b*d*e *a*b*d*e c/c
+a*c*d*e *a*c*d*e b/b
+b*c*d*e *b*c*d*e a/a

-a*b*c*d *a*b*d*e e/c
-a*b*c*e *a*b*c*d d/e
-a*b*d*e *b*c*d*e c/a
-a*c*d*e *a*b*c*e b/d
-b*c*d*e *a*c*d*e a/b
]]

<a name="ch12a152">　Index begin　Index this file
The line
-a*b*c*d *a*b*d*e e/c
say that 
in total five variables a,b,c,d,e
 left half 'a*b*c*d' missing e
right half 'a*b*d*e' missing c
Positive terms has
e/e, d/d, c/c, b/b, a/a
Sum total 2 a, 2 b, 2 c, 2 d, 2 e

<a name="ch12a153">
Negative terms has
e/c, d/e, c/a, b/d, a/b
Sum total 2 a, 2 b, 2 c, 2 d, 2 e

Since we assume
  0≦a≦b≦c≦d≦e ---eqn.BJ156

<a name="ch12a154">
then e/e is product of two
missing 'e' that is product of
(a*b*c*d)*(a*b*c*d)
it has smallest value
Similarly
a/a is product of two
missing 'a' that is product of
(b*c*d*e)*(b*c*d*e)
it has greatest value.

<a name="ch12a155">
Positive terms has
e/e, d/d, c/c, b/b, a/a
left sequence * right sequence
(  e,d,c,b,a  * e,d,c,b,a )
both are in increase order

Negative terms has
e/c, d/e, c/a, b/d, a/b
left sequence * right sequence
(  e,d,c,b,a  * c,e,a,d,b )
left sequence  is increase order
right sequence is non-increase order

<a name="ch12a156">
By rearrangement inequality
Positive terms is greater than
Negative terms.
then 
Positive terms + Negative terms
is greater than or equal to zero

<a name="ch12a157">　Index begin　Index this file
This reason apply to both 
group A and group B.

The special case n=5 and k=4 for
high power end Newton's inequality
is true.

<a name="ch12a158">
Laborious work, no advance theory,
everyone understandable. But 
there are infinite many special 
cases to care. and non-low-end, 
non-high-end middle inequalities 
are hard to analysis.
After taste the flavor, last 
solid proof must cover general
all possible cases.
2010-05-17-14-09 stop

<a name="ch12a159">
2010-05-18-05-45 start
At the step eqn.BJ193, if divide
whole equation by aabbccddee,
then get
  2/a/a + 1/b/b + 2/c/c + 2/d/d + 2/e/e
 -1/d/e - 1/c/e - 1/c/d - 1/b/e - 1/b/d
 -1/b/c - 1/a/e - 1/a/d - 1/a/c - 1/a/b
 ?≧?0 ---eqn.BJ204
The analysis is same as that from
eqn.BJ193 to eqn.BJ203
<a name="ch12a160">
Use rearrangement for two arrays
  [1/a, 1/b, 1/c, 1/d, 1/e]
  [1/b, 1/d, 1/a, 1/e, 1/c]
same for
  [1/a, 1/b, 1/c, 1/d, 1/e]
  [1/e, 1/c, 1/d, 1/a, 1/b]
get same conclusion: high power 
end Newton's inequality is true.
2010-05-18-05-56 stop

2010-05-18-11-15 done first  proofread
2010-05-18-16-42 done second proofread
2010-05-18-18-29 done spelling check



<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