<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="ch12d001">　Index begin　Index this file
2010-05-31-10-09 start
■　Exercise 12.1 problem statement
　　textbook page 188
（On Polynomials with Positive Roots）

<a name="ch12d002">
Show that if the real polynomial
  P(x)=xn+a1xn-1+...+an-1x+an ---eqn.BM001
  P(x)=xn－a1xn-1+a2xn-2 ---eqn.BM002
  +...+(-1)n-1an-1x+(-1)nan
has only positive roots, then
one has the bound
  n*an≦a1an-1 ---eqn.BM003
  n*n*an≦a1an-1 ---eqn.BM004
2010-05-31-10-18 stop 

<a name="ch12d003">
eqn.BM002 guaranteed 
all ak to be positive.
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_Errata.pdf

(-1)n-1 corrected by CSMC_Errata.pdf
n* corrected by ByronSchmuland.txt
2010-05-31-10-29 here





<a name="ch12d004">　Index begin　Index this file
2010-05-31-10-34 start
■　Exercise 12.1 hint
　　textbook page 273

If the roots of P(x) are
  x1,x2,...,xn
then
  an-1/an=  ---eqn.BM005
  (+1/x1+1/x2+...+1/xn)
<a name="ch12d005">
and
  a1=x1+x2+...+xn ---eqn.BM006
So we have 
  (an-1/an)-1≦a1/n ---eqn.BM007
by the HM-AM inequality (8.14).
<a name="ch12d006">
This exercise offers a basic 
reminder: facts for polynomial
coefficients and facts for
symmetric sums are almost in a
one-to-one correspondence.
2010-05-31-10-42 stop



<a name="ch12d007">
2010-05-31-11-03 start
■　Exercise 12.1 discussion

Exercise 12.1 hint already 
solved problem. Here fill
in some details.
Use n=5 as an example. Assume
five roots are [a,b,c,d,e]
<a name="ch12d008">
Please be alert ak and 'a' 
are totally different. ak is
k-th element of vector a.
'a' is first element of 
vector [a,b,c,d,e].
ak is a symmetric sum.
a is a simple number.
Assume independent variable
is t. Then polynomial
  P5(t)=(t-a)*(t-b)*(t-c)
       *(t-d)*(t-e) ---eqn.BK008
Expanded P5(t) is eqn.BK009

<a name="ch12d009">　Index begin　Index this file
  an-1/an=  ---eqn.BM005
  (+1/x1+1/x2+...+1/xn)
is next
  an-1/an=  ---eqn.BM008
  (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)
 =1/e + 1/d + 1/c + 1/b + 1/a
<a name="ch12d010">
eqn.BM008 is same as eqn.BM005
eqn.BM008 is Harmonic Mean of
[a,b,c,d,e]
n/eqn.BM008 is Harmonic Mean of
[a,b,c,d,e] with equal weight 1/n.

<a name="ch12d011">
Next see a1. a1 is coefficient
of "-t*t*t*t*(a+b+c+d+e)"
  a1=(a+b+c+d+e) ---eqn.BM009
Please see eqn.BK009
  a1/n=(a+b+c+d+e)/n ---eqn.BM010
is Arithmetic Mean of [a,b,c,d,e]
<a name="ch12d012">
HM-AM inequality give us the 
answer. That is //n=5
  1/[(1/e+1/d+1/c+1/b+1/a)/n]
  ≦(a+b+c+d+e)/n
or //n=5
  n*n/[(1/e+1/d+1/c+1/b+1/a)]
  ≦(a+b+c+d+e)
<a name="ch12d013">
or
  n*n/[an-1/an]
  ≦a1
or
  n*n*an≦a1an-1 ---eqn.BM004

<a name="ch12d014">　Index begin　Index this file
Above is n=5 special case. It 
is NOT a proof.
Exercise 12.1 hint prove apply 
to any n value.
2010-05-31-11-25 here

<a name="ch12d015">
Newton's inequality eqn.12.2 and
Maclaurin's inequality eqn.12.3
both work with polynomial problem.
Can we use Newton or Maclaurin
to solve Exercise 12.1?

<a name="ch12d016">
Newton and Maclaurin use average 
result to compare magnitude.
Exercise 12.1 has a1/n , this
is average same as binomial 
coefficient n=bicof(n,n-1).
Exercise 12.1 has an-1/an , 
is this an average?
This question let LiuHH pay
attention to ByronSchmuland
correction.
2010-05-31-11-35 here

<a name="ch12d017">
2010-05-31-12-28
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/ByronSchmuland.txt
[[
188    Exercise 12.1 The bound is
 not correct without absolute value
 signs. Or divide through by a_n.
 Could replace n by n^2. 
]]

<a name="ch12d018">
2010-05-31-15-20 start
Exercise 12.1 has two changes.
First. eqn.BM001 is textbook
original equation. 
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_Errata.pdf
Change to eqn.BM002. 
Red text is change part.

<a name="ch12d019">
Second. eqn.BM003 is textbook
original equation. 
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/ByronSchmuland.txt
Change to eqn.BM004. 
Red text 'n*' is change part.
2010-05-31-15-39 here

<a name="ch12d020">　Index begin　Index this file
an-1/an is not an average!
Because no pk in eqn.BM005

2010-05-31-16-18 stop

<a name="ch12d023">
2010-05-31-19-05 start
LiuHH tried to use Newton or
Maclaurin's inequality to prove
Exercise 12.1. But it is not
as simple as HM-AM inequality.
HM-AM one step link head and
tail x1 and xn
Newton or Maclaurin go from
head to tail slow.

To save time, done Exercise 12.1 
here.


<a name="ch12d024">
■　Exercise 12.1 solution


Exercise 12.1 hint already 
solved problem.
2010-05-31-19-08 stop



<a name="ch12d025">　Index begin　Index this file
2010-05-31-20-03 start
■　Exercise 12.2 problem statement
　　textbook page 189
（Three Muirhead Short Stories）

<a name="ch12d026">
(a) Show that for nonnegative
    a,b and c, one has
  8abc≦(a+b)(b+c)(c+a) ---eqn.12.20
(b) Show that for real aj
    1≦j≦n, one has
  2∑[1≦j＜k≦n]ajak ---eqn.12.21
  ≦(n-1)∑[1≦j≦n]aj*aj
<a name="ch12d027">
(c) Show that for nonnegative
    aj, 1≦j≦n, one has

2010-05-31-20-18 stop





<a name="ch12d028">　Index begin　Index this file
2010-05-31-20-20 start
■　Exercise 12.2 hint
　　textbook page 274

(a) By expansion and simplification,
    we see that we need to prove
<a name="ch12d029">
  6abc≦ ac2+ab2+ba2+bc2
      +ca2+cb2 define= R ---eqn.BM012
and after setting a=x1, b=x2, 
and c=x3 we also have
  ∑[σ∈S(3)]xσ(1)xσ(2)xσ(3)=6abc ---eqn.BM013
and
  ∑[σ∈S(3)]xσ(1)xσ(2)2=R ---eqn.BM014
//'R' in eqn.BM014 see eqn.BM012
<a name="ch12d030">
Since
  (1,1,1)=(2,1,0)/6 +(2,0,1)/6 +...
   +(0,1,2)/6 ---eqn.BM015
we have (1,1,1) is in H[(2,1,0)]
so we may apply Muirhead's
inequality.

<a name="ch12d031">
(b) We have
  (1,1,0,...,0)=(2,0,0,...,0)/2
               +(0,2,0,...,0)/2 ---eqn.BM016
so we have
  (1,1,0,...,0)∈H[(2,0,0,...,0)] ---eqn.BM017
and by Muirhead's inequality
it suffice to note that

2010-05-31-20-41 here
<a name="ch12d033">　Index begin　Index this file
(c)Since the average
  {(1/2, 1/2, 0, ..., 0)+... ---eqn.BM020
  +(0, ..., 0, 1/2, 1/2)}/bicof(n,2)
equals (1/n, 1/n, ... , 1/n),
it suffices by Muirhead's 
inequality to note

2010-05-31-20-56 stop



<a name="ch12d035">
2010-05-31-21-44 start
■　Exercise 12.2 solution


(a) Reader can do expansion and
    simplification to eqn.12.20
and reach eqn.BM012.
<a name="ch12d036">
Muirhead's inequality main
point is power manipulation.
The problem equation
  6abc≦ ac2+ab2+ba2+bc2
      +ca2+cb2 define= R ---eqn.BM012
use a sequence = [a,b,c] three
elements. eqn.BM012 less than
side is 6abc. It is a symmetric
sum. Please see eqn.BL118 
<a name="ch12d037">
In eqn.BL118, change x to a
change y to b, change z to c
and set 
  α1=α2=α3=1 ---eqn.BM023
then we get 'abc' add six (3!)
times, it is 6abc in eqn.BM012
Please see (x,y,z) for three 
elements have 3!=6 choices.

<a name="ch12d038">　Index begin　Index this file
eqn.BM012 right side use the
power set
  β1=2 ---eqn.BM024
  β2=1 ---eqn.BM025
  β3=0 ---eqn.BM026
Complete six (3!) symmetric sum
of eqn.BM012 greater than side 
  a2b1c0+a2b0c1+a1b2c0
 +a1b0c2+a0b1c2+a0b2c1 ---eqn.BM027
eqn.BM027 main point is to
show zero-th power elements. 02
<a name="ch12d039">
Now change power 0 to pure 
number one. eqn.BM027 become 
  a2b1+a2c1+a1b2
 +a1c2+b1c2+b2c1 ---eqn.BM028
which is eqn.BM012 greater
than side.

<a name="ch12d040">
Before apply Muirhead's 
inequality we must show that 
α sequence and β sequence have
"α in β hull" relation.
Without in-hull relation
Muirhead will not work.
(α in β_container completely
 α's tail can not reach out
 side of β_container ! ill)
<a name="ch12d041">
First
  ∑αk=1+1+1=3
  ∑βk=2+1+0=3
  ∑αk=∑βk=3 ---eqn.BM029
Second
  max_1(αk)=1≦max_1(βk)=2 ---eqn.BM030
<a name="ch12d042">
Third
  max_2(αk)=1+1≦max_2(βk)=2+1 ---eqn.BM031
Above three relations confirmed
 α in β hull relation.

<a name="ch12d043">　Index begin　Index this file
In fact, we see //more, min
  α1=α2=α3=1 ---eqn.BM023
all elements are equal. In this
case, need just check total sum
  ∑αk=∑βk ---eqn.BM032
are equal. That is enough.

<a name="ch12d044">
After get α in β hull relation. 
We apply Muirhead's inequality
to eqn.BM012 two sides and
confirm the inequality is true.
Exercise 12.2 (a) solved.
2010-05-31-22-47 stop

<a name="ch12d045">
2010-06-01-06-48 start
Problem (b) target equation
eqn.12.21 has hidden cancellation.
During solution process, we need
to recover the cancelled terms.
Our tool is Muirhead inequality.
<a name="ch12d046">
First find out how many elements 
in the data sequence "real aj"
Its index range 1≦j≦n tell us
there are n elements. Then
power α-sequence must have n
power β-sequence must have n
elements.
<a name="ch12d047">
α-sequence is less than side
power sequence.
β-sequence is greater than side
power sequence. // again

<a name="ch12d048">　Index begin　Index this file
eqn.12.21 less than side has
product of aj and ak
j≠k. aj has power one, ak has 
power one. We know α-sequence 
should be [0,...,1,...,1,...0] 
or [1,1,0,...,0] two elements
are one, other n-2 elements 
are zero.

<a name="ch12d049">
Next find β-sequence. eqn.12.21 
greater than side has product 
of aj and aj. They have like
index. Then We know β-sequence 
should be [0,...,2,...0] or
[2,0,0,...,0] one element is 
two, other n-1 elements are 
zero.

<a name="ch12d050">
Muirhead require less than side
power α-sequence be in hull of
greater than side power β-seq.
"be in hull of" mean that α-seq
be interpolate by β-seq.
NOT extrapolate by β-seq.
<a name="ch12d051">
Express α-seq. in terms of β-seq.
the balance coefficients must all
be in [0,1] for interpolation.
A well example, an ill example.

<a name="ch12d052">
It is easy to see that
α-sequence [1,1,0,...,0] can be
written as
 [2,0,0,...,0]/2 + [0,2,0,...,0]/2
Two coefficients are 1/2 and 1/2
both in [0,1] both sum to one.
Muirhead requirement satisfied.
We can apply Muirhead inequality.

<a name="ch12d053">　Index begin　Index this file
Besides "be in hull of" check,
Muirhead has another point worth
attention. That is Muirhead sum
n! terms for n elements problem. 
<a name="ch12d054">
For illustration purpose, assume 
n=4. (4!=24) Assume elements are
[a,b,c,d]. ∑[1≦j＜k≦n]ajak in 
eqn.12.21 is next

<a name="ch12d055">
Symmetric Sums n=4, 
write α=[1,1,0,0] as next
α=[1,one,0,zero] 
'1,one,0,zero' are four different
elements 
(otherwise how to identify 24?)
+a1bonec0dzero  +a1boned0czero  
+a1coneb0dzero  +a1coned0bzero  
+a1doneb0czero  +a1donec0bzero  
//<a name="ch12d056">
+b1aonec0dzero  +b1aoned0czero  
+c1aoneb0dzero  +c1aoned0bzero  
+d1aoneb0czero  +d1aonec0bzero  

+b1conea0dzero  +b1donea0czero  
+c1bonea0dzero  +c1donea0bzero  
+d1bonea0czero  +d1conea0bzero  
//<a name="ch12d057">
+b1coned0azero  +b1donec0azero  
+c1boned0azero  +c1doneb0azero  
+d1bonec0azero  +d1coneb0azero  
2010-06-01-07-42 here

<a name="ch12d058">　Index begin　Index this file
Above is complete 24 terms 
that Muirhead worked out.
Now replace all zero power and
'0' power to one (delete)
and drop power 'one'
and drop power '1', get

<a name="ch12d059">
+ab  +ab  
+ac  +ac  
+ad  +ad  

+ba  +ba  
+ca  +ca  
+da  +da  
//<a name="ch12d060">
+bc  +bd  
+cb  +cd  
+db  +dc  

+bc  +bd  
+cb  +cd  
+db  +dc  

<a name="ch12d061">
Above 24 terms is 
1*∑[1≦j≠k≦n]ajak get 24 terms
compare with eqn.12.21 
2*∑[1≦j＜k≦n]ajak get
six terms then double.
How to get 24 and 6 ?
n!=4!=4*3*2*1=24 (for n=4)

<a name="ch12d061dbg1">
∑[1≦j＜k≦n]ajak has how many terms?

'j＜k' and n=4 has following
terms
2010-06-04-19-06 count start
∑[1≦j＜k≦4]aj*ak
aj can be a1,a2,a3, can not be a4
ak can be a2,a3,a4, can not be a1
aj has 3 choices
<a name="ch12d061dbg2">
if aj be a1, ak can be a2,a3,a4
if aj be a2, ak can be    a3,a4
if aj be a3, ak can be       a4
total six choices. 
Here 3+2+1=3*(3+1)/2=6 for n=4
<a name="ch12d061dbg3">
For n=any, ∑[1≦j＜k≦n]ajak
aj has n-1 choices. 
ak has n-1 to one choices
Total sum is 
  (n-1)+(n-2)+(n-3)+...+1
 =(n-1)*((n-1)+1)/2 ---eqn.BM301
get (n-1)*((n-1)+1)/2 terms !!
 1+2+...+n sum to n*(n+1)/2
 1+...+n-1 sum to (n-1)*((n-1)+1)/2
<a name="ch12d061dbg4">
2* in 2*∑[1≦j＜k≦n]ajak take care
'b*a' in 'a*b+b*a' since 'j＜k'
not allow 'b*a', but 'b*a' is
actually exist.
2010-06-04-19-10 count stop
2010-06-04-19-25 comment stop

<a name="ch12d061dbg5">
2010-06-04-21-10 comment start
Above is a comment to answer
∑[1≦j＜k≦n]ajak has how many terms?
where both aj and ak do not have
n choices. They have only n-1 
choices, get (n-1)*((n-1)+1)/2 
terms.
<a name="ch12d061dbg6">
It is possible view from another
direction. Ignore j＜k for a
while, consider only j≠k. In 
this case aj has n choices, ak 
has n-1 choices. Total n*(n-1)
choices. 
<a name="ch12d061dbg7">
Now take j＜k into
consideration. j＜k and j＞k 
have equal chance. For j＜k 
only, the choices is half of
n*(n-1), that is n*(n-1)/2
same as (n-1)*((n-1)+1)/2 .
2010-06-04-21-10 comment stop

2010-06-01-08-19 first writing stop

<a name="ch12d062">
2010-06-01-11-19 start
Textbook use eqn.BM018
[2(n-2)!]*∑[1≦j＜k≦n]ajak 
that is 
[2(n-2)!]*(+ab+ac+ad+bc+bd+cd)
Black terms no repeat, all
repeat show up at [2(n-2)!]

<a name="ch12d063">　Index begin　Index this file
Repeat come from like powers.
There are two sources.
α-sequence: [0,...,1,...,1,...0] 
or [1,1,0,...,0] 
Like power '1' cause 2! repeat.
Like power '0' has (n-2)! repeat.
Total repeat is 2!*(n-2)!
eqn.BM018 has the factor 2!*(n-2)!
Total n! terms, see here.

<a name="ch12d064">
Similarly, 
β-sequence [0,...,2,...0] or
[2,0,0,...,0] has (n-1) zeros
(n-1) permutation is (n-1)!
eqn.BM019 has the factor (n-1)!

<a name="ch12d065">
Apply Muirhead's inequality
must include all n! terms.
non-repeat terms multiply by 
repeat factor equal n! terms.
more. Muirhead give us
  eqn.BM018 ≦ eqn.BM019 
that is

<a name="ch12d067">
In eqn.BM033 cancel (n-2)!
get eqn.12.21
Exercise 12.2 (b) is done.
2010-06-01-11-47 stop

<a name="ch12d067dbg1">
2010-06-04-20-09 debug start
eqn.BM033 both side should have n!
terms. This is Muirhead signature!
eqn.BM033 right side has (n-1)!*n
terms which is n!. 
This n come from n in '∑[j=1,n]'

<a name="ch12d067dbg2">
eqn.BM033 left side has 
2!*(n-2)!*[(n-1)*((n-1)+1)/2] terms
'(n-1)*((n-1)+1)/2' please see
eqn.BM301 and see ch12d061dbg6
Calculation for
2!*(n-2)!*[(n-1)*((n-1)+1)/2]
give us n! 
Both side have n! terms. OK.
2010-06-04-20-09 debug stop

<a name="ch12d068">　Index begin　Index this file
2010-06-01-11-57 start
Exercise 12.2 (c) is same as 
(a) or (b). Above did detail
analysis. Now for (c) do quick
run.
<a name="ch12d069">
eqn.12.22 less than side has 
α sequence=[1/n,1/n,...,1/n]
it is constant vector, 
therefore smallest object.
we can put α sequence to 
any sum_to_one hull.
<a name="ch12d070">
β sequence=[1/2,1/2,0,...,0]
β seq. sum to one. Then 
Muirhead is ready to work.
eqn.12.22 less than side has 
repeat n!.
eqn.12.22 greater than side 
has repeat 2!*(n-2)!
Before cancel repeat: eqn.BM022 
After  cancel repeat: eqn.12.22 
<a name="ch12d071">
Muirhead say:
  (n!)*(a1*a2*...*an)1/n
  ≦  ---eqn.BM034
  [2!*(n-2)!]*√(ajak)
<a name="ch12d072">
Cancel (n-2)!, move n*(n-1) from
less than side numerator to
greater than side denominator
get eqn.12.22. Done.
2010-06-01-12-08 stop



<a name="ch12d073">　Index begin　Index this file
2010-06-01-12-32 start
■　Exercise 12.3 problem statement
　　textbook page 189
（The Homogenization Trick）

Show that if the positive 
quantities x, y and z satisfy
<a name="ch12d074">
the relation
  xyz=1 ---eqn.BM035
then one has the inequality
  x2+y2+z2≦x3+y3+z3  ---eqn.12.23
<a name="ch12d075">
The salient feature of this 
bound is that the left side
is homogeneous of order 2,
but the right side is 
homogeneous of order 3.
Somehow the constraint xyz=1 
must make up for this
incompatibility.

<a name="ch12d076">
It may be unclear how to exploit
the constraint xyz=1, but one
trick which works remarkably
often is to use the side condition
to construct a homogeneous 
problem which generalizes the
problem at hand. One then solves
the homogeneous problem with the
help of Muirhead's inequality
or related tools.
2010-06-01-12-41 stop





<a name="ch12d077">　Index begin　Index this file
2010-06-01-12-42 start
■　Exercise 12.3 hint
　　textbook page 274

Multiply the left side of the
bound (12.23) by (xyz)1/3 and
<a name="ch12d078">
consider the candidate inequality
  +x7/3y1/3z1/3
  +x1/3y7/3z1/3
  +x1/3y1/3z7/3
  ≦  ---eqn.14.62
  +x3+y3+z3
<a name="ch12d079">
This generalizes our original
problem in the sense that if
we can prove that the bound
(14.62) holds for all nonnegative
x,y,z then the bound (12.23) must
hold when xyz=1. Fortunately, the
<a name="ch12d080">
new bound (14.62) is a corollary 
of Muirhead's inequality and the
relationship
  (7/3,1/3,1/3)=(7/9)*(3,0,0)
  +(1/9)*(0,3,0)+(1/9)*(0,0,3) ---eqn.BM036
<a name="ch12d081">
Kedlaya (1999) presents several
more sophisticated examples of
the homogeneous device.
2010-06-01-12-53 stop



<a name="ch12d082">　Index begin　Index this file
2010-06-01-12-55 start
■　Exercise 12.3 solution


If the positive quantities x, 
y and z are observed physics
quantity, length for example.
<a name="ch12d083">
Then eqn.12.23 would be 
considered as an error. 
Because we can not compare
area with volume.
Constraint eqn.BM035 balance 
the physics dimension.
<a name="ch12d084">
Because xyz=1, then (xyz)1/3 is
still one in value and (xyz)1/3 
is length in dimension.
Multiply (xyz)1/3 to less than
side of eqn.12.23 balance
physics dimension. After
multiplication, original
eqn.12.23 become eqn.14.62.

<a name="ch12d085">
Before apply Muirhead's 
inequality, need check the
in-hull relation.

<a name="ch12d086">
eqn.14.62 less than side
power α sequence is
[7/3, 1/3, 1/3]
eqn.14.62 greater than side
power β sequence is
[3, 0, 0]
Both α and β sequence have
same number of elements 3,
which satisfy the first
necessary condition.

<a name="ch12d087">　Index begin　Index this file
α sum is 7/3 + 1/3 + 1/3 = 3
β sum is 3 + 0 + 0 = 3
α sum = β sum OK, satisfy the
second necessary condition.

<a name="ch12d088">
Next check third condition:
whether [7/3, 1/3, 1/3] is 
in hull of [3, 0, 0]?
Simple rule: If power sequence
is [totalSum, 0, ... 0], this 
n-1 zero power sequence is
<a name="ch12d089">
maximum possible for totalSum
That is 
[7/3, 1/3, 1/3] in hull of
[3, 0, 0] is true.

<a name="ch12d090">
Two condition satisfied,
Muirhead is ready to work.
Muirhead take just one step
say that eqn.14.62 is true.
Next replace (xyz)1/3 by one, 
get solution eqn.12.23
Exercise 12.3 is done.

<a name="ch12d090dbg1">
replace (xyz)1/3 by one is same 
as set light speed C to one and 
drop C=1 from equation. Create
big confusion for entry level 
reader.

<a name="ch12d091">
Reader please pay attention
to the following difference.

<a name="ch12d092">　Index begin　Index this file
LiuHH use the fact that
[totalSum, 0, ... 0] is
maximum possible hull for 
any totalSum power sequence.
<a name="ch12d093">
[totalSum/n, totalSum/n, ...
 totalSum/n] is minimum 
possible object for any 
totalSum power sequence.
Please see Maximum hull, 
minimum hull　not in hull

<a name="ch12d094">
The difference is that
Exercise 12.3 hint use
eqn.BM036 which is real
calculation. 
<a name="ch12d095">
Coefficients in eqn.BM036
are [7/9, 1/9, 1/9]
they sum to one, OK
they are in [0,1], OK
eqn.BM036 also satisfy
Muirhead.
2010-06-01-13-31 stop



<a name="ch12d096">　Index begin　Index this file
2010-06-01-16-00 start
■　Exercise 12.4 problem statement
　　textbook page 189
（Power Sum Inequality）

Show that for positive numbers
xk, 1≦k≦n, 
<a name="ch12d097">
the power sums defined by 
  Sm(x)=x1m+x2m+...+xnm ---eqn.BM037
satisfy the bounds
  Sm2(x)≦Sm-1(x)Sm+1(x) ---eqn.12.24
for all m=1,2,...
<a name="ch12d098">
These may remind us of Newton's
inequalities, but they are more
elementary. They also tell us
that the sequence {log[Sm(x)]) 
is convex, while Newton's 
inequalities tell us that 
{log[Em(x)]) is concave.
2010-06-01-16-09 stop





<a name="ch12d099">　Index begin　Index this file
2010-06-01-16-14 start
■　Exercise 12.4 hint
　　textbook page 275

By expanding the bound (12.24) 
we see after simplification 
that it is equivalent to the
assertion that

<a name="ch12d101">
but
  ( m , m ,0,...,0)=  ---eqn.BM039
  (m+1,m-1,0,...,0)/2
 +(m-1,m+1,0,...,0)/2
so the bound (12.24) follows from
Muirhead's inequality.
2010-06-01-16-27 stop



<a name="ch12d102">
2010-06-01-16-28 start
■　Exercise 12.4 solution


<a name="ch12d103">
Assume n=5 for easier expansion.
  Sm(x)  ---eqn.BM040
 =x1m+x2m+x3m+x4m+x5m

<a name="ch12d104">　Index begin　Index this file
  Sm(x)*Sm(x)  ---eqn.BM041
 =x1m*x1m+x1m*x2m+x1m*x3m+x1m*x4m+x1m*x5m
 +x2m*x1m+x2m*x2m+x2m*x3m+x2m*x4m+x2m*x5m
 +x3m*x1m+x3m*x2m+x3m*x3m+x3m*x4m+x3m*x5m
 +x4m*x1m+x4m*x2m+x4m*x3m+x4m*x4m+x4m*x5m
 +x5m*x1m+x5m*x2m+x5m*x3m+x5m*x4m+x5m*x5m

<a name="ch12d105">
  Sm-1(x)Sm+1(x)  ---eqn.BM042
 =x1m-1x1m+1+x1m-1x2m+1+x1m-1x3m+1+x1m-1x4m+1+x1m-1x5m+1
 +x2m-1x1m+1+x2m-1x2m+1+x2m-1x3m+1+x2m-1x4m+1+x2m-1x5m+1
 +x3m-1x1m+1+x3m-1x2m+1+x3m-1x3m+1+x3m-1x4m+1+x3m-1x5m+1
 +x4m-1x1m+1+x4m-1x2m+1+x4m-1x3m+1+x4m-1x4m+1+x4m-1x5m+1
 +x5m-1x1m+1+x5m-1x2m+1+x5m-1x3m+1+x5m-1x4m+1+x5m-1x5m+1

<a name="ch12d106">
Consider eqn.BM041?≦?eqn.BM042 
Red terms in eqn.BM041 and 
eqn.BM042 cancel out. 
j=k case not exist any more.
Remaining black term is
n=5 version of eqn.12.24
  Sm2(x)?≦?Sm-1(x)Sm+1(x) ---eqn.12.24

<a name="ch12d107">
Power of Sm(x)*Sm(x) is
(m,m,0,...,0)
Power of Sm-1(x)Sm+1(x) is
(m+1,m-1,0,...,0).
Power sequence (2m,0,0,...,0) 
do not exist. Because cancel 
out red terms.

<a name="ch12d108">
Two power sequences have next
relation.
  ( m , m ,0,...,0)=  ---eqn.BM039
  (m+1,m-1,0,...,0)/2
 +(m-1,m+1,0,...,0)/2
the coefficient [1/2, 1/2]
sum to one and 0≦1/2, 1/2≦1

<a name="ch12d109">　Index begin　Index this file
Two power sequences satisfy
Muirhead's inequality requirement.
Muirhead one step conclude that
  Sm2(x)≦Sm-1(x)Sm+1(x) ---eqn.12.24
is true. 
Exercise 12.4 solved.
2010-06-01-16-59 stop

<a name="ch12d110">
But what is the cause and
consequences of log concavity
and log convexity?
Wait for future reading.
2010-06-01-17-03 stop

<a name="ch12d110dbg1">
2010-06-05-19-59 start
Brain exercise problem for 
reader.
Why use
  ( m , m ,0,...,0)=  ---eqn.BM039
  (m+1,m-1,0,...,0)/2
 +(m-1,m+1,0,...,0)/2
? If drop ",0,...,0" and use
  ( m , m )=  ---eqn.BM039 simpler
  (m+1,m-1)/2
 +(m-1,m+1)/2
would it be much easier for us?
Can reader find out the answer
for this question?
The answer show up in this page
many times!
2010-06-05-20-02 stop



<a name="ch12d111">　Index begin　Index this file
2010-06-01-18-09 start
■　Exercise 12.5 problem statement
　　textbook page 189
（Symmetric Problems & Symmetric
 Solutions）

Consider a real symmetric 
polynomial p(x,y) such that
p(x,y)→∞ as |x|→∞ and |y|→∞
<a name="ch12d112">
It is reasonable to suspect
that p attains its minimum
at  "symmetric point". That
is one might conjecture that
there is a t in Real such 
that 
  p(t,t)=min[x,y]p(x,y) ---eqn.BM043
<a name="ch12d113">
This conjecture was proved by
polynomial of degree three or
less by Victor Yacovlevich 
Bunyakovsky in 1854, some five 
years before the publication
of his famous Memoire on
integral inequalities.
<a name="ch12d114">
Bunyakovsky also provided a
counter example which shows 
that the conjecture is false
for a polynomial with degree 
four. Can you find such an
example?
2010-06-01-18-20 stop





<a name="ch12d115">　Index begin　Index this file
2010-06-01-18-22 start
■　Exercise 12.5 hint
　　textbook page 275

With surprising frequency, 
solver of this exercise find 
the same example discovered
<a name="ch12d116">
by Bunyakovsky (1854)
  p(x,y)=  ---eqn.BM044
  {x2+(1-y)2}{y2+(1-x)2}
Here one has p(1,0)=0 and 
p(0,1)=0 but otherwise p(x,y)
is strictly positive. Thus, 
<a name="ch12d117">
despite the symmetry of p,
the minimum of p is not on
the diagonal 
  D={(x,y):x=y} ---eqn.BM045
Incidentally, the problem
reminds us that whenever 
<a name="ch12d118">
we are in pursuit of some 
conjecture, it is important
to allocate time to the 
search for counter examples. 
One often discovers quite
quickly that the conjecture
must be refined -- or even
rejected.
2010-06-01-18-31 stop



<a name="ch12d119">
2010-06-01-18-32 start
■　Exercise 12.5 solution


This is a reading problem.
Done reading.
2010-06-01-18-33 stop



<a name="ch12d120">　Index begin　Index this file
2010-06-01-18-42 start
■　Exercise 12.6 problem statement
　　textbook page 190
（Symmetry -- Destroyed by Design）

Participants in the 1999 Canadian
Olympiad were asked to show that
<a name="ch12d121">
if x,y and z are nonnegative real
numbers for which
  x+y+z=1 ---eqn.BM046
then one has the bound
  f(x,y,z)=x2y+y2z+z2x≦4/27 ---eqn.BM047
<a name="ch12d122">
As a hint, first check by calculus
that f(x,y,z) is maximized on the 
set x+y=1 by taking
  x=2/3  ---eqn.BM048
and
  y=1/3  ---eqn.BM049
<a name="ch12d123">
so the crucial step is to show
that without loss of generality
one can assume that
  z=0  ---eqn.BM050
2010-06-01-18-49 stop





<a name="ch12d124">　Index begin　Index this file
2010-06-01-21-13 start
■　Exercise 12.6 hint
　　textbook page 275

First, by (cyclic) symmetry, we
can assume that x≧y and x≧z.
<a name="ch12d125">
This makes x "special" so it is
then natural to consider the
symmetry properties of y and z.
If we consider the difference
  f(x,y,z)-f(x,z,y)
  =(y-z)(x-y)(x-z) ---eqn.BM051
we see it is negative when y is
less than z, so we can assume
<a name="ch12d126">
without loss of generality that
y≧z. Finally, assuming x≧y≧z
we note
  f(x+z,y,0)-f(x,y,z) ---eqn.BM052
  =z2y+yz(x-y)+xy(y-z)≧0 //textbook line
  =z2y+yz(x-y)+xz(y-z)≧0 //LiuHH ●●change 201006021505
so we may also assume without loss
of generality that z=0. We can now
<a name="ch12d127">
finish with calculus as suggested
by the hint or, alternatively, we
can use the AM-GM inequality check
that for x+y=1 we have

<a name="ch12d129">　Index begin　Index this file
One lesson to take away from
this exercise is that it is
often possible to make step-
by-step progress by considering
how a function changes when
subjected to simple transfor-
mations such as the interchange 
of two variables.
2010-06-01-21-43 stop


<a name="ch12d130">
2010-06-02-15-33 start
■　Exercise 12.6 solution


LiuHH spend more time on 
Exercise 12.6.
<a name="ch12d131">
Problem given
  f(x,y,z) ---eqn.BM054
  =x*x*y+y*y*z+z*z*x
  x+y+z=1 ---eqn.BM055
then problem hint
[[
<a name="ch12d132">
As a hint, first check by calculus
that f(x,y,z) is maximized on the 
set x+y=1 by taking
  x=2/3  ---eqn.BM048
and
  y=1/3  ---eqn.BM049
]]
<a name="ch12d133">
This is a difficult step. Without
good reason, LiuHH will NOT set
z=0 at first step.

<a name="ch12d134">　Index begin　Index this file
Exercise 12.6 hint suggest
[[
by (cyclic) symmetry, we
can assume that x≧y and x≧z.
]]
<a name="ch12d135">
This is reasonable assumption.
"By (cyclic) symmetry",
We can assume one of (x,y,z)
be the maximum value element.
Here use x as maximum.

<a name="ch12d136">
For magnitude of y and z, 
textbook use f(x,y,z)-f(x,z,y)
as probe.

<a name="ch12d137">
  f(x,y,z)-f(x,z,y)
=xxy+yyz+zzx
-xxz-zzy-yyx
=(y-z)(x-y)(x-z)
because in
<a name="ch12d138">
R define= +xxy+yyz+zzx-xxz-zzy-yyx
let x=y, get R=0, R has (x-y) factor
let y=z, get R=0, R has (y-z) factor
let z=x, get R=0, R has (z-x) factor
then 
R1=(y-z)(x-y)(x-z)
or
R2=-(y-z)(x-y)(x-z)

<a name="ch12d139">　Index begin　Index this file
Set
x=0.1; y=0.3; z=1-x-y;
R0=x*x*y+y*y*z+z*z*x-x*x*z-z*z*y-y*y*x
R1=(y-z)*(x-y)*(x-z)
R2=-(y-z)*(x-y)*(x-z)
<a name="ch12d140">
Use one set data get
R0=R1=-R2
then R0=R1 is
  f(x,y,z)-f(x,z,y)
  =(y-z)(x-y)(x-z) ---eqn.BM051
<a name="ch12d141">
We assumed that x≧y and x≧z,
so (x-y)(x-z)≧0
Sign of f(x,y,z)-f(x,z,y) is
same as sign of (y-z).
We want f(x,y,z) be maximum,
we want f(x,y,z)-f(x,z,y)≧0
then we want (y-z)≧0
Up to here, determined the 
magnitude of x,y,z as x≧y≧z

<a name="ch12d142">
Use   f(x,y,z)-f(x,z,y)
is for detect
symmetry properties of y and z

<a name="ch12d143">
For what reason
[[
we note
  f(x+z,y,0)-f(x,y,z) ---eqn.BM052
]]
LiuHH is still not sure.

<a name="ch12d144">　Index begin　Index this file
Above
"We want f(x,y,z) be maximum"
do not apply to eqn.BM052 !!
Because here
  f(x+z,y,0)-f(x,y,z) ---eqn.BM052
  =z2y+yz(x-y)+xz(y-z)≧0 //LiuHH ●●change 201006021505
<a name="ch12d145">
If f(x,y,z) were maximum, then
eqn.BM052 is ≦0, not ≧0.
Red z was y in textbook page275
line -9. 
LiuHH change xy(y-z) to xz(y-z)
Reader can expand eqn.BM052 and
check which one is correct.
LiuHH MAY BE MISTAKE.

<a name="ch12d146">
After eqn.BM052, textbook say
[[
so we may also assume without loss
of generality that z=0.
]]
LiuHH do not understand why.

<a name="ch12d147">
Next come to eqn.BM053.
Why factor '1/2' is there?
eqn.BM053. use sequence [x,x,y]
GM side is (x*x*y)1/3
AM side is (x+x+2*y)/3
<a name="ch12d148">
GM power sum to one
   1/3+1/3+1/3=1 OK
AM coefficients sum to 4/3
  1/3+1/3+2/3 = 4/3 ≠ 1 WHY
there must be good reason.
But LihHH do not know what is
the reason at writing time.

Exercise 12.6 is NOT DONE
2010-06-02-16-15 stop



<a name="ch12d149">　Index begin　Index this file
2010-06-02-17-05 start
■　Exercise 12.7 problem statement
　　textbook page 190
（Creative Bunching）

A problem in the popular text 
Probability by Jim Pitman
requires one to show in 
essence that 
<a name="ch12d150">
if x,y and z are nonnegative 
real numbers for which 
  x+y+z=1 ---eqn.BM056
then
  1/4≦x3+y3+z3+6xyz ---eqn.BM057
Can you check the bound? Can
you check it in more than one 
way?
2010-06-02-17-09 stop





<a name="ch12d151">
2010-06-02-17-59 start
■　Exercise 12.7 hint
　　textbook page 276

Pitman solves his Problem 3.1.24
<a name="ch12d152">
by first expanding 1=(x+y+z)3
and then noting that it suffice
to show
  Q=x2y+x2z+y2x ---eqn.BM058
   +y2z+z2x+z2y≦1/4
when
  x+y+z=1 ---eqn.BM056
<a name="ch12d153">
If we write
  Q=x{x(y+z)} ---eqn.BM059
   +y{y(x+z)}+z{z(x+y)}
then it now suffice to notice 
that each of the three braced
expressions is bounded below
by 1/4 by the AM-GM inequality.
<a name="ch12d154">
Other solutions can be based
on the homogenization trick of
Exercise 12.3 or Schur's 
inequality (page 83), or the 
reduction device of Exercise 
12.6.
2010-06-02-18-17 stop




<a name="ch12d155">　Index begin　Index this file
2010-06-02-19-24 start
■　Exercise 12.7 solution


Follow Exercise 12.7 hint,
first expanding 
  (x+y+z)3 ---eqn.BM060
  =x*x*x+y*y*y+z*z*z+6*x*y*z
  +3*x*x*y+3*y*y*z+3*x*z*z 
  +3*x*x*z+3*x*y*y+3*y*z*z
<a name="ch12d156">
Since  x+y+z=1 ---eqn.BM046
So eqn.BM060=13=1
Let eqn.BM060 multiply with 
eqn.BM057 less than side
create a all_power_3 equation
eqn.BM057 become
  13*1/4  ---eqn.BM061
  ?≦? x3+y3+z3+6xyz 
<a name="ch12d157">
or
  (x*x*x+y*y*y+z*z*z+6*x*y*z
  +3*x*x*y+3*y*y*z+3*x*z*z 
  +3*x*x*z+3*x*y*y+3*y*z*z)/4
  ?≦? x3+y3+z3+6xyz  ---eqn.BM062
<a name="ch12d158">
or
     x*x*x+  y*y*y+  z*z*z+6*x*y*z
  +3*x*x*y+3*y*y*z+3*x*z*z  ---eqn.BM063
  +3*x*x*z+3*x*y*y+3*y*z*z
  ?≦? 4*x3+4*y3+4*z3+24*xyz 
<a name="ch12d159">
or
  +3*x*x*y+3*y*y*z+3*x*z*z ---eqn.BM064 
  +3*x*x*z+3*x*y*y+3*y*z*z
  ?≦? 3*x3+3*y3+3*z3+18*xyz 
<a name="ch12d160">　Index begin　Index this file
or
  +x*x*y+y*y*z+x*z*z  ---eqn.BM065
  +x*x*z+x*y*y+y*z*z
  ?≦? x3+y3+z3+6xyz 
compare eqn.BM065 with eqn.BM057
  1/4≦x3+y3+z3+6xyz ---eqn.BM057
eqn.BM057 right side terms 
re-appear at eqn.BM065 right 
side.
<a name="ch12d161">
Exercise 12.7 hint suggest
the auxiliary problem is
Q=+x*x*y+y*y*z+x*z*z
  +x*x*z+x*y*y+y*z*z
  ?≦? 1/4  ---eqn.BM066
<a name="ch12d162">
or
Q=+x*x*y+x*x*z
  +y*y*z+x*y*y
  +x*z*z+y*z*z
  ?≦? 1/4  ---eqn.BM067
<a name="ch12d163">
or
Q=+x*{x*(y+z)}
  +y*{y*(z+x)}
  +z*(z*(x+y)}
  ?≦? 1/4  ---eqn.BM068
<a name="ch12d164">
Consider the red term x*(y+z)
For a two elements array 
[x, (y+z)], its AM-GM 
inequality is GM≦AM or
  √[x*(y+z)]≦[x+(y+z)]/2  ---eqn.BM069
<a name="ch12d165">　Index begin　Index this file
The given condition
  x+y+z=1 ---eqn.BM056
change eqn.BM069 to eqn.BM070
  √[x*(y+z)]≦[1]/2  ---eqn.BM070
Square eqn.BM070 get
   [x*(y+z)]≦1/4  ---eqn.BM071
<a name="ch12d166">
Similarly
   [y*(z+x)]≦1/4  ---eqn.BM072
   [z*(x+y)]≦1/4  ---eqn.BM073
Refer to eqn.BM059, we have
Q=x*[bound by 1/4] ---eqn.BM074
 +y*[bound by 1/4]
 +z*[bound by 1/4]
Q=(x+y+z)*[bound by 1/4] ---eqn.BM075
Q=1*[bound by 1/4] ---eqn.BM076

<a name="ch12d167">
auxiliary problem proved to be 
true, but 
Inequality in eqn.BM065 start
from eqn.BM061, this inequality
is wait_to_prove inequality.
It is not given condition.

<a name="ch12d168">
Auxiliary problem can not imply
main problem eqn.BM061 or
eqn.BM057
2010-06-02-20-35 stop



<a name="ch12d169">
2010-06-02-21-53 start
Use Schur's inequality get
success result.
Continue at
  +x*x*y+y*y*z+x*z*z  ---eqn.BM065
  +x*x*z+x*y*y+y*z*z
  ?≦? x3+y3+z3+6xyz 
<a name="ch12d170">　Index begin　Index this file
Move all terms to greater 
than side and write +6xyz
as +3xyz +3xyz 
find
   x*x*x+y*y*y+z*z*z
  +3*x*y*z
  -x*x*y-y*y*z-x*z*z
  -x*x*z-x*y*y-y*z*z
  +3*x*y*z ?≧?0 ---eqn.BM077
<a name="ch12d171">
or
  +x*x*x+x*y*z-x*x*y-x*x*z
  +y*y*y+x*y*z-y*y*z-y*y*x
  +z*z*z+x*y*z-z*z*x-z*z*y
  +3*x*y*z ?≧?0 ---eqn.BM078
<a name="ch12d172">
or
  +x*(x*x+y*z-x*y-x*z)
  +y*(y*y+x*z-y*z-y*x)
  +z*(z*z+x*y-z*x-z*y)
  +3*x*y*z ?≧?0 ---eqn.BM079
<a name="ch12d173">
or
  +x1*(x-y)*(x-z)
  +y1*(y-x)*(y-z)
  +z1*(z-x)*(z-y)
  +3*x*y*z ?≧?0 ---eqn.BM080
<a name="ch12d174">
eqn.BM080 is Schur's inequality
eqn.5.19 with α=1 plus 3*x*y*z
Schur's inequality is non negative.
3*x*y*z is also non negative.
eqn.BM080 ≧0 is true. Then the
starting point eqn.BM061 is true.
Exercise 12.7 is done.
2010-06-02-22-10 stop

<a name="ch12d175">　Index begin　Index this file
2010-06-02-22-42 start
Exercise 12.7 main problem is
eqn.BM057 Mark as A ?≧? B
where B is 1/4.

Exercise 12.7 auxiliary problem 
is eqn.BM058 Mark as B ?≧? C
where B is 1/4.

<a name="ch12d176">
String together is
 A ?≧? B ?≧? C ---eqn.BM081
In this case, prove B ≧ C 
do not help  A ?≧? B 

<a name="ch12d177">
If the relation is
 A ?≧? C ?≧? B ---eqn.BM082
After prove A ≧ C 
 and  prove C ≧ B 
then we can conclude A ≧ B 
is true.

<a name="ch12d178">
eqn.BM058 is in such no help
position. Is this right?
2010-06-02-22-51 stop





<a name="ch12d179">　Index begin　Index this file
2010-06-03-10-44 start
■　Exercise 12.8 problem statement
　　textbook page 190
（Weierstrass's Polynomial
 Product Inequality）

<a name="ch12d180">
Show that if the complex numbers
a1,a2,...,an and b1,b2,...,bn
satisfy 
  |aj|≦1 ---eqn.BM083
and
  |bj|≦1 ---eqn.BM084
<a name="ch12d181">
for all 1≦j≦n, then
  |a1a2...an-b1b2...bn|
 ≦∑[j=1,n]|aj-bj| ---eqn.12.25
2010-06-03-10-51 stop





<a name="ch12d182">
2010-06-03-10-56 start
■　Exercise 12.8 hint
　　textbook page 276

This elementary (but very useful!)
inequality serves as a reminder
that symmetry is often the key to
successful telescoping. 
<a name="ch12d183">
Here the telescoping identity
  a1a2...an-b1b2...bn ---eqn.BM085
 =∑[j=1,n]a1...aj-1(aj-bj)bj+1...bn
makes the Weierstrass Inequality
immediate. Naturally, generaliza-
tions of this identity lead one 
to more elaborate versions of
Weierstrass Inequality.
2010-06-03-11-04 stop


2010-06-03-11-13 start
<a name="ch12d184">　Index begin　Index this file
■　Exercise 12.8 solution


Exercise 12.8 hint suggest 
telescoping equation eqn.BM085
<a name="ch12d185">
Now set n=4 and see the
structure of eqn.BM085.
  a1a2a3a4-b1b2b3b4 ---eqn.BM086
 =a1a2a3a4+0+0+0-b1b2b3b4
<a name="ch12d186">
 =  //next line ---eqn.BM087
  a1a2a3a4
 -b1a2a3a4+b1a2a3a4 //telescoping is
 -b1b2a3a4+b1b2a3a4 //cancel middle
 -b1b2b3a4+b1b2b3a4 //keep begin/end
 -b1b2b3b4
<a name="ch12d187">
 =  //next line ---eqn.BM088
  a1a2a3a4-b1a2a3a4
 +b1a2a3a4-b1b2a3a4
 +b1b2a3a4-b1b2b3a4
 +b1b2b3a4-b1b2b3b4
<a name="ch12d188">
 =  //next line ---eqn.BM089
 +(a1-b1)a2a3a4 // j=1
 +b1(a2-b2)a3a4 // j=2
 +b1b2(a3-b3)a4 // j=3
 +b1b2b3(a4-b4) // j=4
<a name="ch12d189">　Index begin　Index this file
That is
  a1a2a3a4-b1b2b3b4 ---eqn.BM090
 =∑[j=1,4]a1...aj-1(aj-bj)bj+1...b4

Problem given
  |aj|≦1 ---eqn.BM083
and
  |bj|≦1 ---eqn.BM084

<a name="ch12d190">
Take absolute value for eqn.BM090
find
 |a1a2a3a4-b1b2b3b4| ---eqn.BM091
=|∑[j=1,4]a1...aj-1(aj-bj)bj+1...b4|

<a name="ch12d191">
Because each |aj|≦1 and |bj|≦1
Drop aj and bj from eqn.BM091
cause next inequality
 |a1a2a3a4-b1b2b3b4| ---eqn.BM092
≦∑[j=1,4]|(aj-bj)|
<a name="ch12d192">
It is like
  3.5=0.7*5 ---eqn.BM093
drop 0.7 get
  3.5≦5 ---eqn.BM094
Above is for n=4 special case.
It apply to n=any general case.
Exercise 12.8 is done.
2010-06-03-11-50 stop



<a name="ch12d193">　Index begin　Index this file
2010-06-03-15-55 start
■　Maximum hull, minimum hull
Textbook Chapter 13, page 191
defined majorization, and give
an example
      (1,1,1,1) (＜) (2,1,1,0)
 (＜) (3,1,0,0) (＜) (4,0,0,0) ---eqn.13.1
When we exam eqn.13.1, 

<a name="ch12d194">
First check that all number of 
elements are same. In eqn.13.1 
they are all four elements. OK.

<a name="ch12d195">
Second check that each sequence
sum to same total number.
(1,1,1,1) sum to 1+1+1+1=4 ---eqn.BM095
(2,1,1,0) sum to 2+1+1+0=4 ---eqn.BM096
(3,1,0,0) sum to 3+1+0+0=4 ---eqn.BM097
(4,0,0,0) sum to 4+0+0+0=4 ---eqn.BM098
They all sum to four. OK.

<a name="ch12d196">
Third check partial sum has 
uniform inequality. For this 
purpose, each sequence need
be arranged 
<a name="ch12d197">
from maximum to minimum order
(1,1,1,1)  ---eqn.BM099 (four lines)
(2,1,1,0) 
(3,1,0,0) 
(4,0,0,0) 
Above descending order, OK
<a name="ch12d198">　Index begin　Index this file
Next random order,  NO
(1,1,1,1) 
(1,2,1,0) 
(0,1,0,3)   ---eqn.BM100 (four lines)
(0,4,0,0) 

<a name="ch12d199">
Now refer to OK arrangement
sq1=(1,1,1,1) ---eqn.BM101
sq2=(2,1,1,0) ---eqn.BM102
sq3=(3,1,0,0) ---eqn.BM103
sq4=(4,0,0,0) ---eqn.BM104

<a name="ch12d200">
Partial sum1 is
sq1_Psum1=(1) ---eqn.BM105
sq2_Psum1=(2) ---eqn.BM106
sq3_Psum1=(3) ---eqn.BM107
sq4_Psum1=(4) ---eqn.BM108
We have 4＞3＞2＞1 ---eqn.BM109
that is
　 sq4_Psum1 ≧ sq3_Psum1 ---eqn.BM110
≧ sq2_Psum1 ≧ sq1_Psum1 OK

<a name="ch12d201">
Partial sum2 is
sq1_Psum2=(1+1) ---eqn.BM111
sq2_Psum2=(2+1) ---eqn.BM112
sq3_Psum2=(3+1) ---eqn.BM113
sq4_Psum2=(4+0) ---eqn.BM114
We have 4+0≧3+1＞2+1＞1+1 ---eqn.BM115
that is
　 sq4_Psum2 ≧ sq3_Psum2 ---eqn.BM116
≧ sq2_Psum2 ≧ sq1_Psum2 OK

<a name="ch12d202">
Partial sum3 is
sq1_Psum3=(1+1+1) ---eqn.BM117
sq2_Psum3=(2+1+1) ---eqn.BM118
sq3_Psum3=(3+1+0) ---eqn.BM119
sq4_Psum3=(4+0+0) ---eqn.BM120
We have
 4+0+0≧3+1+0≧2+1+1＞1+1+1 ---eqn.BM121
that is
　 sq4_Psum3 ≧ sq3_Psum3 ---eqn.BM122
≧ sq2_Psum3 ≧ sq1_Psum3 OK

<a name="ch12d203">　Index begin　Index this file
Partial sum4 is total sum for
n=4 case. Which must be equal.

<a name="ch12d204">
After check and pass all above
exam, we say the sequences
sq1=(1,1,1,1) ---eqn.BM101
sq2=(2,1,1,0) ---eqn.BM102
sq3=(3,1,0,0) ---eqn.BM103
sq4=(4,0,0,0) ---eqn.BM104
have majorization relation,
<a name="ch12d205">
write as
  sq4(≧)sq3(≧)sq2(≧)sq1 ---eqn.BM123
where '(≧)' is unicode '≽'
LiuHH computer MSIE 6.0 do 
not display '≽', use '(≧)' 
as replacement.

<a name="ch12d206">
From above example, we know that
the sequence
sqA=(totalSum, 0, 0, ... 0, 0)  ---eqn.BM124
is maximum hull. Because sqA from
start to end, use totalSum to 
compare, totalSum is maximum 
possible number.

<a name="ch12d207">
On the other hand, the sequence
sqB=(totalSum/n, totalSum/n, ...
     totalSum/n, totalSum/n)  ---eqn.BM125
is minimum hull. Because sqB from
start to end, use Arithmetic mean 
totalSum/n to compare, this is 
minimum possible sequence.
<a name="ch12d208">　Index begin　Index this file
From start to end, sqB use 
smallest sum to compare, OR
from start to end, sqB hide
maximum possible NOT to compare.
Above red totalSum/n,totalSum/n,
... is smallest sum to compare.
Above blue totalSum/n,totalSum/n
... is hide maximum possible 
NOT to compare.

Other sequence are in between.

<a name="ch12d209">
Re-visit the example sequences.
For totalSum=4, totalElement=4 
for positive integer sequence
(1,1,1,1) is smallest possible
(2,1,1,0) next to smallest
(3,1,0,0) next to greatest
(4,0,0,0) is greatest possible

<a name="ch12d210">
A long handle spoon can not
put in a short neck bottle.
A short neck bottle can not
put in a long handle spoon.
No one contain the other is
possible. For example //ill
  sqC=[3,3,0] ---eqn.BM126
  sqD=[4,1,1] ---eqn.BM127

<a name="ch12d211">
Before apply Muirhead inequality
it is important to check the
in_hull (in_bottle) condition.
2010-06-03-16-57 stop

<a name="ch12d212">　Index begin　Index this file
2010-06-03-18-50 start
■　Muirhead: Root Mean Square ≧ AM

The following is a reading of
[[
2010-05-23-22-37 LiuHH access
http://2000clicks.com/MathHelp/IneqMuirheadsInequality.aspx
]]
Example 4: RMS ≥ AM ≥ GM ≥ HM
MathHelp use three elements
non-negative real x,y,z
<a name="ch12d213">
LiuHH use five elements
non-negative real 
  r=[a,b,c,d,e] ---eqn.BM128
Reader please pay attention
to the difference.
<a name="ch12d214">
For non-negative real numbers
a,b,c,d,e, use Muirhead prove 
that the RMS Mean is greater than
 or equal to the Arithmetic Mean, 
<a name="ch12d215">
the Arithmetic Mean is greater
 than or equal to the Geometric
 Mean, 
the Geometric Mean is greater
 than or equal to the Harmonic
 Mean. i.e.,
<a name="ch12d216">
Root Mean Square ≧ AM
  sqrt((a2+b2+c2+d2+e2)/5)
 ≧ (a+b+c+d+e)/5   ---eqn.BM129
AM ≥ GM :
  (a+b+c+d+e)/5≧(abcde)1/5 ---eqn.BM130
GM ≥ HM :
  (abcde)1/5≧5/(1/a+1/b+1/c+1/d+1/e) ---eqn.BM131

<a name="ch12d217">　Index begin　Index this file
Now start prove RMS ≥ AM
Root Mean Square ≧ Arithmetic Mean 
Given non-negative real 
  r=[a,b,c,d,e] ---eqn.BM128
RMS = sqrt((a2+b2+c2+d2+e2)/5)  ---eqn.BM132
 AM = (a+b+c+d+e)/5 ---eqn.BM133
<a name="ch12d218">
We have five elements.
Muirhead's inequality work
for symmetric sum. That means 
that Muirhead will build 5!
(n! in general) terms. 5!=120
RMS has a2+b2 Addition separate
two terms. RMS build square for 
each element of r.
<a name="ch12d219">
This observation suggest use
β1 power sequence = [2,0,0,0,0]
β1 has five elements, because
given data r has five elements.
β1 use one '2' and four '0's, 
because RMS use '+b2+' Again, 
addition separate two terms.
<a name="ch12d220">
β1 seq. is power of [a,b,c,d,e]
Define β1 seq. as
  β1 = [2,0,0,0,0] ---eqn.BM134
which give us
  a2b0c0d0e0+a0b2c0d0e0
 +a0b0c2d0e0+a0b0c0d2e0
 +a0b0c0d0e2+... ---eqn.BM135
It simplify to
  RMS0=(n-1)!(a2+b2+c2+d2+e2) ---eqn.BM136
<a name="ch12d221"> 01
Please do not say eqn.BM135 is 
redundant step ! If we invite 
Muirhead help us, especially
in this detail analysis, we 
must follow Muirhead's guide 
line. eqn.BM135 let us see
Muirhead's structure clearly !

<a name="ch12d222">　Index begin　Index this file
eqn.BM135 has 5!=120 terms. 
DO NOT FORGET THIS POINT !
For better illustration, let 
us set
  w=x=y=z=0 ---eqn.BM137

<a name="ch12d223">
a2b0c0d0e0 in eqn.BM135 has 
next hidden play
 a2bwcxdyez ---eqn.BM138
+a2bxcydzew
+a2byczdwex
+a2bzcwdxey
+...
<a name="ch12d224">
Remember w=x=y=z=0, above terms
ALL SHOW UP AS a2.
For n=5 problem and four zero
β1 power sequence = [2,0,0,0,0] ---eqn.BM134
there are (n-1)!=(5-1)!=24 terms
of a2! That is right!
<a name="ch12d225">
That is Muirhead signature !
Usually we write as (4!)*a2
instead of write '+a2' 24 times
2010-06-03-19-50 here

Above is RMS.
<a name="ch12d226">
Below is AM
 AM = (a+b+c+d+e)/5 ---eqn.BM133
Observe eqn.BM133, we may want
to use α1_ill power sequence
  α1_ill = [1,0,0,0,0] ---eqn.BM139
If we use α1_ill seq. Muirhead 
is not happy. Because 
<a name="ch12d227">　Index begin　Index this file
Muirhead first rule is 
α sequence and β sequence and
data r have same number of 
elements. First point is OK.
Muirhead second rule is 
α sequence total sum and 
β sequence total sum are equal.
<a name="ch12d228">
Please pay attention to the
difference
α seq. is less than side power
β seq. is greater than side power.

Now 
  β1     = [2,0,0,0,0] ---eqn.BM134
  α1_ill = [1,0,0,0,0] ---eqn.BM139
α1_ill  total sum is 1
β1 seq. total sum is 2
they are not equal.
<a name="ch12d229">
To satisfy Muirhead, we need use
  α1_OK = [1,0,1,0,0] ---eqn.BM140
Immediate question is that
power sequence [1,0,1,0,0] will
create a1b0c1d0e0, which is ac !
But 
<a name="ch12d230">
 AM = (a+b+c+d+e)/5 ---eqn.BM133
is power one equation, AM is not 
power two. (a*c is power two)
Good question. But Muirhead second 
rule is here, no choice. Go ahead,
see if there is an open path.

<a name="ch12d231">
Up to here determined use
  β1 = [2,0,0,0,0] ---eqn.BM134
  α1 = [1,1,0,0,0] ---eqn.BM141
Here [1,1,0,0,0] or [1,0,1,0,0]
or [0,0,1,0,1] they are the same.
Muirhead is going to permute n!
times for symmetric sum. Any
permutation show up once.

<a name="ch12d232">　Index begin　Index this file
Muirhead third rule is partial sum
inequality, 
Please see textbook at ch12c120
Please see notes at ch12d196

<a name="ch12d233">
Check
  β1 = [2,0,0,0,0] ---eqn.BM134
  α1 = [1,1,0,0,0] ---eqn.BM141
Partial sum1 is
β1_Psum1=(2) ---eqn.BM142
α1_Psum1=(1) ---eqn.BM143
2＞1 OK
<a name="ch12d234">
Partial sum2 is
β1_Psum2=(2+0) ---eqn.BM144
α1_Psum2=(1+1) ---eqn.BM145
2≧2 OK
The following partial sum add 
only zero. No contribution. 
Two OK conclude that
  α1 = [1,1,0,0,0]
is in hull of
  β1 = [2,0,0,0,0]
<a name="ch12d235">
Muirhead's pre-condition all
satisfied.
Muirhead say
  (4!)*(a2+b2+c2+d2+e2)
  ≧ ---eqn.BM146
  (2!)*(3!)*(ab+ac+ad
  +ae+bc+bd+be+cd+ce+de)
<a name="ch12d236">
Please count greater than 
side, it has n!=5!=120 terms.
Please count less than side, 
it has n!=5!=120 terms.

<a name="ch12d237">　Index begin　Index this file
Greater than side has 
(a2+b2+c2+d2+e2) five terms
multiply by (4!)=4*3*2*1=24
5*24 = 120 perfect !!

<a name="ch12d238">
Less than side has 
(ab+ac+ad+ae+bc+bd+be+cd+ce+de)
ten terms, multiply by (2!)*(3!)
(2!)*(3!)*10=120 perfect !!

<a name="ch12d239">
(4!) come from four identical
number '0' in β1 = [2,0,0,0,0]
Four identical number permute
get 4!=24 identical terms.

(2!)*(3!) come from [1,1,0,0,0]
with similar reason.

<a name="ch12d240">
Reader should pay attention to
non-zero permutation '1,1'.
They are a*b+b*a written as
2*a*b, where b*a is absorbed by
(2!). This is a simplification
step, let us write shorter
equation. What follows is that
<a name="ch12d241">
use of (2!) do not allow the
appearance of b*a. We write
only a*b. This consideration
show up at equation, for
example, eqn.12.22 eqn.BM018
eqn.BM033 etc. as j＜k in 
1≦j＜k≦n
2010-06-03-20-51 here

<a name="ch12d242">　Index begin　Index this file
OK! master Muirhead's work is 
done. Muirhead give us
  (4!)*(a2+b2+c2+d2+e2)
  ≧ ---eqn.BM146
  (2!)*(3!)*(ab+ac+ad
  +ae+bc+bd+be+cd+ce+de)
Clean up, and convert to 
RMS≧AM is our business.
We do not bother master.

<a name="ch12d243">
But, how ?
First easy job is cancel.
Write eqn.BM146 as eqn.BM147
next //(4!)/[(2!)*(3!)]=2
  2*(a2+b2+c2+d2+e2)
  ≧ ---eqn.BM147
  ab+ac+ad+ae+bc+bd+be+cd+ce+de
AM is 
 AM = (a+b+c+d+e)/5 ---eqn.BM133
<a name="ch12d244">
RMS eqn.BM132 use a2+b2 etc
We square AM too. AM square is
  (a+b+c+d+e)(a+b+c+d+e)/25 ---eqn.BM148
 =(aa+ab+ac+ad+ae
  +ba+bb+bc+bd+be
  +ca+cb+cc+cd+ce
  +da+db+dc+dd+de
  +ea+eb+ec+ed+ee)/25
<a name="ch12d245">
Write it in three groups
AM_sq ---eqn.BM149
 =(aa+bb+cc+dd+ee
+ab+ac+ad+ae+bc+bd+be+cd+ce+de
+ba+ca+cb+da+db+dc+ea+eb+ec+ed)/25

<a name="ch12d246">
First AM_sq group is 
  aa+bb+cc+dd+ee ---eqn.BM150
which match RMS eqn.BM132.

<a name="ch12d247">　Index begin　Index this file
Second AM_sq group is 
  +ab+ac+ad+ae+bc+bd+be+cd+ce+de ---eqn.BM151
which show up under 1≦j＜k≦n

Third AM_sq group is 
  +ba+ca+cb+da+db+dc+ea+eb+ec+ed ---eqn.BM152
which hide under 1≦j＜k≦n

<a name="ch12d248">
Group two and group three have
same value. 
a*b is in group two
b*a is in group three

<a name="ch12d249">
Start from eqn.BM147,
To create AM_square need hidden
group three show up.
eqn.BM147 multiply by two to 
match this requirement. We get
  2*2*(a2+b2+c2+d2+e2)
  ≧ ---eqn.BM153
  ab+ac+ad+ae+bc+bd+be+cd+ce+de
 +ba+ca+cb+da+db+dc+ea+eb+ec+ed

<a name="ch12d250">
Next add one array of
  aa+bb+cc+dd+ee ---eqn.BM150
to match AM_square group one
requirement. eqn.BM153 become
  2*2*(a2+b2+c2+d2+e2)
  +(a2+b2+c2+d2+e2)
  ≧ ---eqn.BM154
  ab+ac+ad+ae+bc+bd+be+cd+ce+de
 +ba+ca+cb+da+db+dc+ea+eb+ec+ed
 +(a2+b2+c2+d2+e2)

<a name="ch12d251">
But eqn.BM148 tell us that
AM_square need '/25'. Modify 
eqn.BM154 as next eqn.BM155
  5*(a2+b2+c2+d2+e2)
  /25
  ≧ ---eqn.BM155
 (ab+ac+ad+ae+bc+bd+be+cd+ce+de
 +ba+ca+cb+da+db+dc+ea+eb+ec+ed
 +a2+b2+c2+d2+e2)
 /25

<a name="ch12d252">　Index begin　Index this file
Now eqn.BM155 less than side
is AM_squared exactly.
Whole equation eqn.BM155 take
root square get
  √[(a2+b2+c2+d2+e2)/5]
  ≧ ---eqn.BM156
 (a+b+c+d+e)/5
<a name="ch12d253">
We get RMS≧AM for the case n=5
successfully. However, this is 
not general, it is for n=5
special case. 
Convert to a proof, change 5 to 
n and insert many dot dot dot.
2010-06-03-21-25 stop

<a name="ch12d254">
2010-06-03-22-31 start
Above show RMS≧AM from
Muirhead's inequality.

To show AM≧GM 
Arithmetic Mean ≧ Geometric Mean
from Muirhead's inequality.
MathHelp give just one line

This is easiest case. It is done 
at tute0044.htm#ch12c148

<a name="ch12d255">　Index begin　Index this file
■　Muirhead: GM≧HM
Next show that GM ≥ HM
Geometric Mean ≧ Harmonic Mean
Still use non-negative real 
  r=[a,b,c,d,e] ---eqn.BM128
GM is //eqn.AE14
  gm=(a*b*c*d*e)1/5 ---eqn.BM157
HM is //eqn.8.13
  hm=1/(0.2/a +0.2/b ---eqn.BM158
       +0.2/c +0.2/d +0.2/e)
Here 0.2=1/5, use equal weight.
<a name="ch12d256">
eqn.BM158 is not compatible
with Muirhead. Change form
as following
  hm=(abcde)/[(abcde)*(0.2/a +0.2/b
       +0.2/c +0.2/d +0.2/e)] ---eqn.BM159
or
  hm=(5abcde)/[bcde +acde
        +abde +abce +abcd] ---eqn.BM160
<a name="ch12d257">
GM ≥ HM is next
  (a*b*c*d*e)1/5 ---eqn.BM161
 ≧(5abcde)/[bcde+acde+abde+abce+abcd]
or
  (a*b*c*d*e)1/5*[bcde ---eqn.BM162
       +acde+abde+abce+abcd]
 ≧(5abcde)
<a name="ch12d258">
Now hope to prove GM ≧ HM
  a1/5*(b*c*d*e)6/5
 +b1/5*(a*c*d*e)6/5
 +c1/5*(b*a*d*e)6/5
 +d1/5*(b*c*a*e)6/5
 +e1/5*(b*c*d*a)6/5
 ≧(5abcde) ---eqn.BM163
<a name="ch12d259">
eqn.BM163 derived from GM ≧ HM
eqn.BM163 fit Muirhead's inequality
pattern better. 

<a name="ch12d260">　Index begin　Index this file
Greater than side power form 
β sequence. eqn.BM163 greater 
than side power suggest β seq.
  β2=[6/5,6/5,6/5,6/5,1/5] ---eqn.BM164

<a name="ch12d261">
Less than side power form 
α sequence.
eqn.BM163 less than side power
suggest α sequence 
  α2=[1,1,1,1,1] ---eqn.BM165
<a name="ch12d262">
Please pay attention to next 
point.
When work with AM≧GM, GM has 
α sequence [1/n,1/n,...,1/n]
the smallest possible.
When work with GM≧HM, HM has 
α sequence [1,1,.....,1]
the smallest possible.
GM has β sequence eqn.BM164.

<a name="ch12d263">
  β2=[6/5,6/5,6/5,6/5,1/5] ---eqn.BM164
and
  α2=[1,1,1,1,1] ---eqn.BM165
they have same number of 
elements five.
they have same number of 
total sum five.
β2 partial sum is greater than
α2 partial sum . (α2 is minimum)
<a name="ch12d264">
Muirhead's inequality requirements
are satisfied.
Muirhead say 
β2 power symmetric sum is 
greater then
α2 power symmetric sum.

<a name="ch12d265">　Index begin　Index this file
That is
  (4!)*[a1/5*(b*c*d*e)6/5
       +b1/5*(a*c*d*e)6/5
       +c1/5*(b*a*d*e)6/5
       +d1/5*(b*c*a*e)6/5
       +e1/5*(b*c*d*a)6/5]
 ≧(5!)*(abcde) ---eqn.BM166
<a name="ch12d266">
Although Muirhead did not
give us eqn.BM163,
Muirhead give us eqn.BM166
instead. That is good enough.
<a name="ch12d267">
eqn.BM166 cancel (4!) get
eqn.BM163.
eqn.BM163 is derived from 
eqn.BM161 which is GM ≥ HM
then Muirhead proved GM ≥ HM.
Job is done.
2010-06-03-23-18 stop

<a name="ch12d268">　Index begin　Index this file
2010-06-04-12-05 start
■　SAKARDU - 07.10.2006 Problem.5

2010-05-15-11-44 LiuHH access next page
http://www.sms.edu.pk/journals/mathtrack/mtvol3/mt3_2.pdf
save as sms.edu.pk-mt3_2.pdf
The following is working record
only. No proof of anything.
Record has a small program help
to find answer.

<a name="ch12d269">
2010-06-04-11-02 read sms.edu.pk-mt3_2.pdf
[[
SAKARDU - 07.10.2006.

Problem.5. Prove that, for any
 positive integer n ≥ 5, the
 set Xn = {1, 2, . . . , n} 
can be partitioned into two
 subsets Sn and Pn, such that
 the sum of the elements of Sn
 be equal to the product of the
 elements of Pn.
]]

<a name="ch12d270">
2010-06-04-11-03
[1,2,3,4,5]      =[1*2*4]+[3+5]        //8
[1,2,3,4,5,6]    =[1*2*6]+[3+4+5]      //12
[1,2,3,4,5,6,7]  =[1*3*6]+[2+4+5+7]    //18
[1,2,3,4,5,6,7,8]=[1*3*8]+[2+4+5+6+7]  //24
2010-06-04-11-13

<a name="ch12d271">
2010-06-04-11-18
[[
Solution. The naive attempt of trying
 Pn = {p} (one element only) fails
miserably, as (1 + 2 + · · · + n) − p
 = n(n + 1)/2 − p > p for any 1 ≤ p ≤ n.
The (less) naive attempt of trying
 Pn = {p, q} (two elements only) 
<a name="ch12d272">
fails for more subtle reasons, as 
(1 + 2 + · · · + n) − p − q 
= n(n + 1)/2 − p − q = pq is
equivalent to 
n(n + 1)/2 + 1 = (p + 1)(q + 1),
 which does not always have a
solution in positive integers,
 with 1 ≤ p, q ≤ n.
<a name="ch12d273">　Index begin　Index this file
Trying to find ”by hand” solutions for
 small values of n yields P5 = {1, 2, 4},
P6 = {1, 2, 6}, P7 = {1, 3, 6},
 P8 = {1, 3, 8}. This encourages us 
to try Pn = {1, p, q}, which after
 similar to the above calculations
 leads to n(n + 1)/2 =
(p + 1)(q + 1). Now, for n even we
 can take Pn = {1, (n − 2)/2, n},
 while for
n odd we can take
 Pn = {1, (n − 1)/2, (n − 1)}.
]] //why solution this way?

<a name="ch12d274">
2010-06-04-11-29
[1,2,3,4,5]      =[1*2*4]+[3+5]  
[1,2,3,4,5,6]    =[1*2*6]+[3+4+5]
[1,2,3,4,5,6,7]  =[1*3*6]+[2+4+5+7]   
[1,2,3,4,5,6,7,8]=[1*3*8]+[2+4+5+6+7] 

<a name="ch12d275">
[1*2*4]+[3+5]       =8   bicof(5,2)=10 =n*(n-1)/2 //n=5
[1*2*6]+[3+4+5]     =12  bicof(6,2)=15 =n*(n-1)/2 //n=6
[1*3*6]+[2+4+5+7]   =18  bicof(7,2)=21 =n*(n-1)/2 //n=7
[1*3*8]+[2+4+5+6+7] =24  bicof(8,2)=28 =n*(n-1)/2 //n=8
2010-06-04-11-34

<a name="ch12d276">
2010-06-04-11-36 use n(n+1)/2 
[1*2*4]+[3+5]       =8   n*(n+1)/2 =15 //n=5
[1*2*6]+[3+4+5]     =12  n*(n+1)/2 =21 //n=6
[1*3*6]+[2+4+5+7]   =18  n*(n+1)/2 =28 //n=7
[1*3*8]+[2+4+5+6+7] =24  n*(n+1)/2 =36 //n=8
2010-06-04-11-38
bicof(n,2) is better than n*(n+1)/2 

<a name="ch12d277">　Index begin　Index this file
2010-06-04-11-42
1+2+3+4+5+6+7+8=36

n*(n+1)/2 -1-x-y = 1*x*y
if n=even,
 x=n/2 - 1
 y=n
if n=odd,
 x=(n-1)/2
 y=n-1

//<a name="ch12d278">
//2010-06-04-11-48
//[[  program start
//calculator local how to use
//SAKARDU - 07.10.2006 Problem.5
var n=12 //you can change n to any n≧5
//Please do not change code below
var x,y,sum0,prod;
var partialSum='';
var partialPro='1';
if(n%2){x=(n-1)/2; y=n-1;}
else   {x=n/2 - 1; y=n;}
sum0=n*(n+1)/2 -1-x-y 
prod=1*x*y
sum0 //partial sum
prod //partial product
//<a name="ch12d279">
for(var i=2;i<=n;i++){
if(i==x)partialPro+='*'+i;
else if(i==y)partialPro+='*'+i;
else partialSum+='+'+i;
}
partialSum
partialPro
eval(partialSum)
eval(partialPro)
//calculator local how to use
//]] program end
//2010-06-04-11-00

<a name="ch12d280">
for n=21
complex2.htm#calculator
output next
[[
sum0 //partial sum
200
prod //partial product
200
<a name="ch12d281">　Index begin　Index this file
partialSum
+2+3+4+5+6+7+8+9+11+12+13+14+15+16+17+18+19+21
partialPro
1*10*20
eval(partialSum)
200
eval(partialPro)
200
]]
2010-06-04-12-20 stop

2010-06-05-11-56 done first proofread
2010-06-05-21-11 done second proofread
2010-06-05-21-37 done spelling check

========= Chapter twelve end here =========

<a name="ch12d282">
"Update 2010-06-07" is next

2010-06-06-23-32 start
It is easy to explain 
SAKARDU - 07.10.2006 Problem.5
Problem statement is here 
Problem answer is here 

<a name="ch12d283">
The sum (1 + 2 + · · · + n) has
an equation expression as next
  ∑[i=1,n]i = n*(n+1)/2 ---eqn.BM167
from experiment find the following
relation
  n*(n+1)/2 -x-y-1 = x*y*1 ---eqn.BM168
<a name="ch12d284">
eqn.BM168 say:
from (1, 2, · · ·, n) remove
x,y,1 and the sum of the rest
is equal to the product of
x,y,1. Problem is to find x and
y two unknowns with one equation.
Re-write eqn.BM168 as
  n*(n+1)/2 = x*y +x +y +1 ---eqn.BM169
or
  n*(n+1)/2 = (x+1)*(y+1) ---eqn.BM170
<a name="ch12d285">
What x, y value will satisfy
eqn.BM170 ? Initially, may 
wonder for a while. But !
Look ! answer is right in
front of our eye !! Since
  n*(n+1)/2 = (x+1)*(y+1) 
<a name="ch12d286">
use '*' to cut equation
(this is second condition)
get
  n = (x+1) ---eqn.BM171
and
  (n+1)/2 = (y+1) ---eqn.BM172
eqn.BM171 and eqn.BM172 is 
good for n=odd. So that
(n+1)/2 is a whole number.

<a name="ch12d287">
Because we require integer
answer, if n=even, n/2 is 
a whole number. The other 
solution is
  n/2 = (x+1) ---eqn.BM173
and
  (n+1) = (y+1) ---eqn.BM174

<a name="ch12d288">
To find answer, 
solve for x from (x+1)
solve for y from (y+1)
if n=odd
  x = n-1 ---eqn.BM175
  y = (n-1)/2 ---eqn.BM176
if n=even
  x = n/2-1 ---eqn.BM177
  y = n     ---eqn.BM178
Problem is solved.
2010-06-06-23-50 stop




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

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

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