﻿ Proof of Newton's inequality and Maclaurin inequality Inequality Study 43rd file   Upload 2010-05-23
Proof of Newton's inequality and Maclaurin inequality
index 　 this 　 program 　 DocA 　 Push down diagram
XYGraph v2.3 - web page graph   ☜☞   donate   get code
The Cauchy-Schwarz Master Class   J. Michael Steele   ★★★★★
This file is personal home work. No one
proofread. Cannot promise correctness.
If you suspect any view point wrong,
please ask a math expert near by.
Freeman 2009-06-19-10-46

Please use MSIE browser to read this file.
Did not test other browser. This file is
written under MSIE 6.0

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

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

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

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

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

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

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

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

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

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

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

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

<a name="sideDraw">
P5(t), P4(t), P3(t), P2(t)

<a name="ch12b001">　Index begin　Index this file
2010-05-20-06-32 start
■　Define three corner E
Newton's inequality is
Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.2
for 0＜k＜n
Assume x=[a,b,c,d,e], n=5
Ek is averaged ek. 1≦k＜5
Ek(x)=ek(x)/bicof(5,k) ---eqn.BK001
ek is symmetric sum of five
numbers a,b,c,d,e in k-th power.
<a name="ch12b002">
If k=1, then
e1(x)=a+b+c+d+e ---eqn.BJ108
and
E1(x)=e1(x)/bicof(5,1) ---eqn.BJ123
Here
bicof(5,1)= 5!/[1!*(5-1)!]
bicof(5,1)= 5 ---eqn.BK002
Binomial coefficients is
defined at eqn.BJ128
<a name="ch12b003">
Ek2(x) lower right corner is k
Ek2(x) upper right corner is p=2
In the following discussion, we
will differentiate polynomial
several times and reduce its
order. Order of polynomial n
must be visible.
<a name="ch12b004">
Now use third corner, define
three corner E as following
nEkp(x) ---eqn.BK003
is averaged symmetric sum E of
n numbers x in k-th power seat,
whole term take p-th power.
Above is uppercase E. Same thing
apply to lowercase e.

<a name="ch12b005">　Index begin　Index this file
Example 1, if
x=[a,b,c,d,e]=[1,2,3,4,5] ---eqn.BK004
then n=5. For second seat k=2,
the square root value is
nEkp(x)=5E21/2(x) ---eqn.BK005
=[(a*b+a*c+b*c+a*d+b*d+c*d+a*e
+b*e+c*e+d*e)/bicof(5,2)]1/2
=[(1*2+1*3+2*3+1*4+2*4
+3*4+1*5+2*5+3*5+4*5)/10]1/2
=2.9154759474226503
2010-05-20-07-20 here
<a name="ch12b006">
Example 2, if
y=[f,g,h,i]=[6,7,8,9] ---eqn.BK006
then n=4. For second seat k=2,
the square value is
nEkp(y)=4E22(y) ---eqn.BK007
=[(+f*g+f*h+f*i+g*h+g*i+h*i)/bicof(4,2)]2
=[( 6*7+6*8+6*9+7*8+7*9+8*9)/6]2
=3117.361111111111

<a name="ch12b007">
y=[f,g,h,i] has n=4 information,
we need n value show up, please
see push down diagram. E's lower
left corner n number help us to
analysis. Following use eqn.BK003
expression for clear explanation.
2010-05-20-07-36 stop

<a name="ch12b008">　Index begin　Index this file
2010-05-20-10-34 start
■　Numerically test inequality

To prove Newton's inequality,
take two steps.
<a name="ch12b009">
First, relate higher polynomial
order nEk(x) to lower
polynomial order n-1Ek(x)
Second, prove any polynomial
high power end Newton's
inequality is true.
nEk(x) = n-1Ek(x) = n-2Ek(x)
please see Push down diagram.
Move down a column polynomial
reduce power n. zEk(x) unchange.
<a name="ch12b010">
Numerically test inequality help
us understand the problem.
If we have
x=[a,b,c,d,e]=[1,2,3,4,5] ---eqn.BK004
in hand then the polynomial
which generate symmetric sums
for this x is polynomial P5(t)
P5(t)=(t-a)*(t-b)*(t-c)
*(t-d)*(t-e) ---eqn.BK008

<a name="ch12b011">
In expanded form it is
P5(t)=  ---eqn.BK009
(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t
-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
<a name="ch12b012">
In nek(x) expression: (n=5)
P5(t)= //remind: 5e0(x)=1
=(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t*5e0(x)
-t*t*t*t*5e1(x)
+t*t*t*5e2(x)
-t*t*5e3(x)
+t*5e4(x)
-5e5(x) ---eqn.BK010

<a name="ch12b013">　Index begin　Index this file
In nEk(x) expression: (n=5)
P5(t)=  ---eqn.BK011
=(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t*5E0(x)*bicof(5,0)
-t*t*t*t*5E1(x)*bicof(5,1)
+t*t*t*5E2(x)*bicof(5,2)
-t*t*5E3(x)*bicof(5,3)
+t*5E4(x)*bicof(5,4)
-5E5(x)*bicof(5,5)

<a name="ch12b014">
For x=[a,b,c,d,e]=[1,2,3,4,5]
in partial numerical value A,
P5(t)=  ---eqn.BK012
=(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t*5E0(x)*1
-t*t*t*t*5E1(x)*5
+t*t*t*5E2(x)*10
-t*t*5E3(x)*10 //number is
+t*5E4(x)*5    //binomial
-5E5(x)*1   //coefficients

<a name="ch12b015">
In partial numerical value B,
P5(t)=   ---eqn.BK013
=(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t*1  //5E0(x)=1
-t*t*t*t*3*5  //5E1(x)=3
+t*t*t*8.5*10 //5E2(x)=8.5
-t*t*22.5*10  //5E3(x)=22.5
+t*54.8*5     //5E4(x)=54.8
-120          //5E5(x)=120
Red mark worth your attention.

<a name="ch12b016">
In complete numerical value,
P5(t)=  ---eqn.BK014
=(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t -t*t*t*t*15 +t*t*t*85
-t*t*225   +t*274  -120
2010-05-20-11-24 here

<a name="ch12b017">　Index begin　Index this file
Newton's inequality best view
is eqn.BK012. Newton said:
For sequence [1,2,3,4,5], its
symmetric sums has the following
relations //dropped n=5
Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.2
<a name="ch12b018">
Case n=5, complete set Newton's
for k=1:  E0(x)*E2(x)≦E12(x) ---eqn.BK015
for k=2:  E1(x)*E3(x)≦E22(x) ---eqn.BK016
for k=3:  E2(x)*E4(x)≦E32(x) ---eqn.BK017
for k=4:  E3(x)*E5(x)≦E42(x) ---eqn.BK018

<a name="ch12b019">
eqn.BK013 has numerical value
for x=[1,2,3,4,5], that is
E0(x)=1    ---eqn.BK019
E1(x)=3    ---eqn.BK020
E2(x)=8.5  ---eqn.BK021
E3(x)=22.5 ---eqn.BK022
E4(x)=54.8 ---eqn.BK023
E5(x)=120  ---eqn.BK024
<a name="ch12b020">
Check Newton's inequality as following
for k=1:    1*8.5 ≦   32
for k=2:    3*22.5≦ 8.52
for k=3:  8.5*54.8≦22.52
for k=4: 22.5*120 ≦54.82
<a name="ch12b021">
or
for k=1:    8.5 ≦   9
for k=2:   67.5 ≦  72.25
for k=3:  465.8 ≦ 506.25
for k=4: 2700.0 ≦3003.04
Newton's inequality is true for
above specific case.
2010-05-20-11-53 here

<a name="ch12b022">
Newton's inequality eqn.12.2 and
Maclaurin's inequality eqn.12.3
are close related.
It is easy to prove inequalities
at low power end (n=5, k=0,1) and
at high power end (n=5, k=4,5).
It is hard to prove inequalities
at middle power (n=5, k=2,3).
<a name="ch12b023">
Can we change hard to prove
middle power inequalities to
easy to prove high power end
inequality? In Newton's case
there is an easy way !!
<a name="ch12b024">　Index begin　Index this file
The key point is that
if we start from P5(t) and
generate P4(t), P3(t), P2(t),
P1(t) which are related to
P5(t) by differentiation,
<a name="ch12b025">
then we have //Prove
5Ek(x)=4Ek(x)=3Ek(x)
=2Ek(x)=1Ek(x)  ---eqn.BK025
(uEv end at u=v; eqn.BK025 whole
line enter front stage if k=1)
this is an unexpected and
surprised relation! It is true!
<a name="ch12b026">
With this relation in hand,
5E1(x)*5E3(x)5E22(x) ---eqn.BK026
we prove
3E1(x)*3E3(x)3E22(x) ---eqn.BK027
Lower right 1,3,2 is E's power.
Lower left 5,3=polynomial order
In Push down diagram eqn.BK026
is high order middle terms.
<a name="ch12b027">
eqn.BK027 is low  order end
terms. [3E3(x) is end]
eqn.BK025 promise that eqn.BK026
and eqn.BK027 have same value.
We prove much easy low
order end term inequality
conclude that high order
middle term inequality
is true !!
<a name="ch12b028">
LiuHH got above understanding
on 2010-05-19 after long reading
and long thinking. Textbook say
so, but no hands-on calculation
it is hard to believe we have
such nice/lucky eqn.BK025.
2010-05-20-12-30 stop

<a name="ch12b029">　Index begin　Index this file
2010-05-20-14-25 start
The following illustrate eqn.BK025
from different angle. Start from
eqn.BK011.
Differentiate P5(t) get P4(t)
Differentiate P4(t) get P3(t) etc.
<"ch12b030"> Push down diagram   P5 push down to P3
Key point: symbol E (colored) is unchanged in each column.
Prove P5(t)'s E1*E3E2*E2 at P3(t) where it is easier end.
 why hi? watch! watch! watch! key   pt t0=end P5(t)= 5!5E0t5 0!(5-0)! － 5!5E1t4 1!(5-1)! + 5!5E2t3 2!(5-2)! － 5!5E3t2 3!(5-3)! + 5!5E4t1 4!(5-4)! － 5!5E5t0 5!(5-5)! P4(t)= 4!4E0t4 0!(4-0)! － 4!4E1t3 1!(4-1)! + 4!4E2t2 2!(4-2)! － 4!4E3t1 3!(4-3)! + 4!4E4t0 4!(4-4)! I want number table ! P3(t)= 3!3E0t3 0!(3-0)! － 3!3E1t2 1!(3-1)! + 3!3E2t1 2!(3-2)! － 3!3E3t0 3!(3-3)! easy prove here at P3 ch12b031 Red blue purple are at end P2(t)= 2!2E0t2 0!(2-0)! － 2!2E1t1 1!(2-1)! + 2!2E2t0 2!(2-2)! 2E3 do NOT exist hard prove above at P5 t0 is constant end at t0
---eqn.BK028 　Group 1 push down;　Group 2 not push down.
width of above equation
<a name="ch12b032"> 　Index begin　Index this file
All red E's in one column have same value.
(although they belong to different polynomial)
All blue E's in one column have same value.
All purple E's in one column have same value.
Other E's not color, still in same column have
same value. All E0 are 1 by definition.
E5=a*b*c*d*e is highest order term.
Right end is E's high order terms, we use them.
Left end is t's high order terms, not use them.
In Ei*tj, i+j = n = constant in a row (equation).
<a name="ch12b033">
2010-05-20-15-31 here
Above is symbolic equation, no
numerical value tell us column
value are the same.
Below is numerical equation,
let sequence be
[a,b,c,d,e]=[1,2,3,4,5] ---eqn.BK029
Please goto Program Test Newton,
Maclaurin inequalities.
<a name="ch12b034">
Change from
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
to
var a=1,b=2,c=3,d=4,e=5
var outType=2 //change 0 to 2
Then click
[Test box3 command, output to box4]
<a name="ch12b035">
get the following
[[
E1 //length compare with volume wrong
3
E2 //upper case E* removed n,k factor
8.5
E3 //upper case E* use average value
22.5
E4 //but still wrong, here is outType=2
54.8
E5 //select outType=0 for correct answer
120
]]
From above assumed value, we have
the following numerical equations.
<a name="ch12b036"> 　Index begin　Index this file
Key point: numerical E is unchanged in each column.
 P5(t)= t5 －5*3*t4 +10*8.5*t3 －10*22.5*t2 +5*54.8*t1 －1*120*t0 P4(t)= t4 －4*3*t3 +6*8.5*t2 －4*22.5*t1 +1*54.8*t0 P3(t)= t3 －3*3*t2 +3*8.5*t1 －1*22.5*t0 P2(t)= t2 －2*3*t1 +1*8.5*t0
---eqn.BK030   Draw P5(t) to P2(t)   symbol equation
width of above equation
<a name="ch12b037">
Now let us look at the red terms
from P5(t)= t5－5*3*t4+10*8.5*t3+... ---eqn.BK031
to  P4(t)= t4－4*3*t3+6*8.5*t2+... ---eqn.BK032
Study how they change?
Why keep 8.5 unchanged?
From P5(t) to P4(t) is a
differentiation process
P4(t) = d[P5(t)]/dt ---eqn.BK033
P4(t) = d[t5－5*3*t4+10*8.5*t3+...]/dt ---eqn.BK034
= 5*t4－4*5*3*t3+3*10*8.5*t2+...
<a name="ch12b038">
Normalize 5*t4 to t4
P4(t)/5 = t4－4*3*t3+3*2*8.5*t2+...
= t4－4*3*t3+  6*8.5*t2+...
(eqn.BK030 dropped '/5')
Red multiply by 3, divide by 5
Red multiply by k, divide by n
Multiply by k due to d[tk]/dt=k*tk-1
Divide by n due to normalize P4(t)

<a name="ch12b039">
8.5 (=5E3=4E3) is unchanged
Red "+  6*" is bicof(4,2), just
right. All other nEk apply the
same reason, then no change.
n≧k OK, n＜k is undefined.
when n＝k, this term end here.
2010-05-20-16-30 stop

<a name="ch12b040">
2010-05-20-18-50 start
Above are observation, mainly
numerical check. But numerical
equation is not general, can
not be considered as proof. We
must use n,k symbol equation
to prove, because we can plug
any value in to n,k, then, it
cover all cases.

<a name="ch12b041">　Index begin　Index this file
■　Prove Ek(x1,...,xn)=Ek(y1,...,yn-1)
The following prove first step
of total two step proof.
That is if we differentiate a
n-th order polynomial to one
order lower, two polynomial
k-th seat two Ek have same
value, // eqn.12.6
nEk(x)=n-1Ek(x) ---eqn.BK035
See symbolic equation eqn.BK028
or numerical equation eqn.BK030
Numerical observation why unchange ?
is here
<a name="ch12b042">
Symbolic proof why unchange is
lectured in textbook page 180
to 181. Consider polynomial
P(t)=(t-x1)(t-x2)...(t-xn)
<a name="ch12b043">
 P(t) = k=n ∑ k=0 (-1)k ( n k ) Ek(x1,x2,...,xn) tn-k
---eqn.12.5
Why write polynomial this way?   Please read start
from Chapter 12 to at least Uppercase Ek(x) defined.
width of above equation
For n=5, expanded equation see eqn.BK011
where bicof(5,2) = bicof(n,k) is binomial coefficient
<a name="ch12b044">
2010-05-20-20-23 here
Use Q(t) as normalized derivative
of P(t). 'normalize' mean that
Q(t) highest power coefficient
is one, not n.
<a name="ch12b045">
Q(t) = {d[P(t)]/dt}/n = // ---eqn.BK036
 = n-1 ∑ k=0 (-1)k ( n k ) n-k n Ek(x1,x2,...,xn) tn-k-1 = n-1 ∑ k=0 (-1)k ( n-1 k ) * Ek(x1,x2,...,xn) tn-k-1
width of above equation
<a name="ch12b046">　Index begin　Index this file
n-k come from
d[tn-k]/dt = n-k*tn-k-1 ---eqn.BK037
n come from normalize
Q(t) = {d[P(t)]/dt}/n  ---eqn.BK038

<a name="ch12b047">
eqn.BK036 line one change to
line two as following
 ( n k ) n-k n = n! k!(n-k)! n-k n = (n-1)! k!(n-k-1)! = ( n-1 k )
---eqn.BK039
width of above equation

Red cancel result red; Blue cancel result blue.
<a name="ch12b048">
Above is coefficient analysis.
Not involve the root of Q(t).
But Ek of Q(t) is defined on
roots of Q(t).
See Lowercase ek(x) defined
and Uppercase Ek(x) defined
We take second, root-based
route to find Ek of Q(t).
Problem given {x1,x2,...,xn}
as roots of P(t).
<a name="ch12b049">
After differentiate P(t),
get Q(t), Q(t) has a set of
roots {y1,y2,...,yn-1}
Which should be different from
{x1,x2,...,xn}, unless there are
repeated root in {x1,x2,...,xn}.
Repeated root not change analysis.
Build Q(t) from n-1Ek(y) get
<a name="ch12b050">　Index begin　Index this file
Q(t)=P'(t)/n= ---eqn.BK040
=(t-y1)(t-y2)...(t-yn-1)
 = n-1 ∑ k=0 (-1)k ( n-1 k ) * Ek(y1,y2,...,yn-1) tn-k-1
---eqn.BK041 ; Equation build method see
Why write polynomial this way? at eqn.12.5
width of above equation
<a name="ch12b051">
eqn.BK041 and eqn.BK036 both
express Q(t), they are the same
equation from two different path
Equate Ek from eqn.BK036
and Ek from eqn.BK041 get
Ek(x1,x2,...,xn)
=Ek(y1,y2,...,yn-1) ---eqn.12.6
If use tute0043.htm express
method, eqn.12.6 is next
nEk(x)=n-1Ek(y) ---eqn.BK042
eqn.12.6 is our main result.
2010-05-20-21-38 stop

<a name="ch12b052">　Index begin　Index this file
2010-05-20-23-19 start
■　Why is it so remarkable?

What benefit we get from eqn.12.6?
Textbook page 181 explain as
following.
===textbook copy start===
Why is it so remarkable?
The left-hand side of the identity
(12.6) is a function of the n
vector x=(x1,x2,...,xn) while the
right side is a function of the
n-1 vector y=(y1,y2,...,yn-1)
<a name="ch12b053">
Thus if we can prove a relation
such as
0≦F(E0(y),E1(y),...,En-1(y)) ---eqn.BK043
for all y∈[a,b]n-1.
Then it follows that we also
have the relation
0≦F(E0(x),E1(x),...,En-1(x)) ---eqn.BK044
for all x∈[a,b]n.
<a name="ch12b054">
That is, any inequality -- or
identity -- which provides a
relation between the n-1
quantities E0(y),E1(y),...,En-1(y)
and which is valid for all
values of y∈[a,b]n-1 extends
automatically to a corresponding
relation for the n-1 quantities
E0(x),E1(x),...,En-1(x) which is
valid for all x∈[a,b]n.

<a name="ch12b055">
This presents a rare but valuable
situation where to prove a
relation for function of just n-1
variables. This observation can
be used in an ad hoc way to produce
many special identities which
<a name="ch12b056">
otherwise would be completely
baffling, and it can also be used
systematically to provide seamless
induction proofs for results such
as Newton's inequalities.
===textbook copy stop===
2010-05-20-23-39 stop

<a name="ch12b057">　Index begin　Index this file
2010-05-21-10-50 start
We had numerical example
x=[a,b,c,d,e]=[1,2,3,4,5] ---eqn.BK004
P5(t)=  ---eqn.BK014
=(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
=t*t*t*t*t -t*t*t*t*15 +t*t*t*85
-t*t*225   +t*274  -120
<a name="ch12b058">
That is
P5(t)=t5－15*t4+85*t3－225*t2+274*t－120 ---eqn.BK045
P4(t)=t4－12*t3+51*t2－90*t+54.8 ---eqn.BK046
P4(t) and P5(t) are related by
P4(t)=d[P5(t)]/dt ---eqn.BK047

<a name="ch12b059">
From one view point that P4(t) is
created from P5(t), no need go a
redundant loop find P4(t) roots
y and re-assemble P4(t) from y.

<a name="ch12b060">
From second view point that
n-1Ek(y) should be generated
by P4(t) roots, just like nEk(x)
are generated by P5(t) roots x.
Textbook equate two E's
Ek(x1,x2,...,xn)
=Ek(y1,y2,...,yn-1) ---eqn.12.6
<a name="ch12b061">
or in tute0043.htm notation
nEk(x)=n-1Ek(y) ---eqn.BK042
where n-1Ek(y) is created by
'redundant' step, that is n-1Ek(y)
is created by P4(t) roots y.

<a name="ch12b062">　Index begin　Index this file
The following verify that
P4(t) roots y create n-1Ek(y)
Starting point is
P4(t)=t4－12*t3+51*t2－90*t+54.8 ---eqn.BK046
end point is still eqn.BK046.

<a name="ch12b063">
http://freeman2.com/polyroot.htm#begin0
Click [04] button for 4th order
polynomial. Then fill in the
following coefficients
54.8  －90  51  －12  1
where 54.8 goto box marked [C00]
and in [C04] fill 1.
<a name="ch12b064">
Click [fast get roots] button
Box 1, answer has
[[
Below is answer, complex polynomial root ([1,2] same as 1+2i)
1st root is [1.355567131841731,0]
2nd root is [2.456088843227607,0]
3rd root is [3.543910070325765,0]
4th root is [4.644433954604896,0]

Below is answer too, complex number only, no text.
1.355567131841731+0i
2.456088843227607+0i
3.543910070325765+0i
4.644433954604896+0i
]]

<a name="ch12b065">
P4(t) roots y has
1.355567131841731
2.456088843227607
3.543910070325765
4.644433954604896

<a name="ch12b066">　Index begin　Index this file
Next goto(local)
http://freeman2.com/complex2.htm#calculator
in [Box3, input JS command]
fill
[[
f=1.355567131841731
g=2.456088843227607
h=3.543910070325765
i=4.644433954604896
e1=f+g+h+i
E1=e1/bicof(4,1)
e2=f*g+f*h+f*i+g*h+g*i+h*i
E2=e2/bicof(4,2)
e3=f*g*h+f*g*i+f*h*i+g*h*i
E3=e3/bicof(4,3)
e4=f*g*h*i
E4=e4/bicof(4,4)
E1
E2
E3
E4
]]
<a name="ch12b067">
Next click
[Test box3 command, output to box4]
Box 4 show up next
[[
E1
3
E2
8.5
E3
22.500000654128463
E4
54.80000354686018
]]
<a name="ch12b068">
Above is E1,E2,E3,E4 from P4(t)
roots y //P4=4th order polynomial
Please goto P5(t) E's
compare E1,E2,E3,E4 [P4 no E5]
Result show that E1,E2,E3,E4
from P4 and from P5 are the
same !!
<a name="ch12b069">
They should be identical,
because we walked a loop.
Above numerical experiments
confirmed textbook eqn.12.6
2010-05-21-11-46 stop

<a name="ch12b070">　Index begin　Index this file
2010-05-21-15-20 start
■　Prove high power end Newton
Inequality

We have eqn.12.6 in hand, for
example, a data set of ten
numbers
x=[x1,x2,...,x10] ---eqn.BK048
and ask to prove
10E3(x)*10E5(x)≦10E42(x) ---eqn.BK049
is true.
<a name="ch12b071">
eqn.12.6 tell us that use x to
build a tenth order polynomial
p10(t) use x as its roots.
Differentiate p10(t) to fifth
order polynomial p5(t). Then
prove next inequality eqn.BK050
<a name="ch12b072">
5E3(x)* 5E5(x)≦ 5E42(x) ---eqn.BK050
If eqn.BK050 is true, then
eqn.BK049 is true by eqn.12.6
Key point is red 5
10E5 is 10th order polynomial
middle term, hard to prove.
5E5 is 5th order polynomial
end term, easy to prove.
<a name="ch12b073">
eqn.BK050 is p5(t)'s high power
end inequality. This point can
be seen from the term 5E5(x)
for uEv(x), we must have u≧v.
If u=v, that is the end, and
no more. Because vEv(x) multiply
by tzero, vEv(x) is a constant.
Next differentiation v-1Ev(x)
is gone.  2E3 do NOT exist.

<a name="ch12b074">
For above reason, we only need
to prove high power end Newton
Inequality.
Newton's Inequality general form
is next
Ej-1(x1,x2,...,xn)*Ej+1(x1,x2,...,xn)≦
Ej2(x1,x2,...,xn) ---eqn.12.7
2010-05-21-15-52 here

<a name="ch12b075">
If we have one number in hand,
that is
x=[x1] ---eqn.BK051
There is nothing to do.

<a name="ch12b076">　Index begin　Index this file
■　Prove Newton Inequality n=2
Textbook page 182

If we have two data in hand,
that is
x=[x1,x2] ---eqn.BK052
Use x1,x2 as two roots, build
a polynomial P2(t).
P2(t)=(t-x1)*(t-x2) ---eqn.BK053
=t*t-t*(x1+x2)+x1*x2
=t*t-2*t*[(x1+x2)/2]+x1*x2
<a name="ch12b077">
On the other hand
P2(t)=t2*2E0(x1,x2)
-2*t1*2E1(x1,x2)
+1*t0*2E2(x1,x2) ---eqn.BK054
<a name="ch12b078">
For the creation of eqn.BK054
Chapter 12: Symmetric Sums
to at least
Uppercase Ek(x) defined.
eqn.BK053 and eqn.BK054 are
both expression of P2(t).
<a name="ch12b079">
We have
2E0(x1,x2)=1 ---eqn.BK055
2E1(x1,x2)=[(x1+x2)/2] ---eqn.BK056
2E2(x1,x2)=x1*x2 ---eqn.BK057
<a name="ch12b080">
For n=2 simplest case,
Newton's inequality is
E0(x1,x2)* E2(x1,x2)
≦ E12(x1,x2) ---eqn.12.8
Above is textbook equation.
Below is tute0043.htm equation.
2E0(x)* 2E2(x)≦ 2E12(x) ---eqn.BK058
Put eqn.BK055, eqn.BK056 and
eqn.BK057 into eqn.BK058, get
1*[x1*x2]?≦?[(x1+x2)/2]2 ---eqn.BK059
<a name="ch12b081">
We assume that x1＞0 and x2＞0
so that x1*x2＞0. Now
take square root of eqn.BK059
[x1*x2]1/2?≦?(x1+x2)/2 ---eqn.BK060
eqn.BK060 is certainly true.
<a name="ch12b082">
[x1*x2]1/2 is Geometric Mean
and
(x1+x2)/2 is Arithmetic Mean
GM≦AM is a matter of course.
(one year ago, LiuHH is inequality
stranger. Now LiuHH is sophomore)
2010-05-21-16-31 here

Above is n=2 case.
<a name="ch12b083">　Index begin　Index this file
■　Prove Newton Inequality n=3
Textbook page 182

Below is n=3 case.
We have three data in hand.
x=[x1,x2,x3] ---eqn.BK061
We use x build third degree
polynomial P3(t). To save work
express P3(t) with 3Ek(x)
directly
<a name="ch12b084">
P3(t)=1*t3*3E0(x1,x2,x3)
-3*t2*3E1(x1,x2,x3)
+3*t1*3E2(x1,x2,x3)
-1*t0*3E3(x1,x2,x3) ---eqn.BK062
<a name="ch12b085">
eqn.BK062 coefficient are
[1,-3,+3,-1]. Neglect +/-
[1,+3,+3,+1] is
bicof(3,0)=1
bicof(3,1)=3
bicof(3,2)=3
bicof(3,3)=1
<a name="ch12b086">
Why eqn.BK062 involve
binomial coefficients?
Lowercase ek(x) defined.
to
Uppercase Ek(x) defined.

<a name="ch12b087">
P3(t) eqn.BK062 contain two
Newton's inequalities. They are
E0(x1,x2,x3)* E2(x1,x2,x3)≦ E12(x1,x2,x3) ---eqn.12.9
E1(x1,x2,x3)* E3(x1,x2,x3)≦ E22(x1,x2,x3) ---eqn.12.10
Above is textbook equation.
Below is tute0043.htm equation.
3E0(x)* 3E2(x)≦ 3E12(x) ---eqn.BK063
3E1(x)* 3E3(x)≦ 3E22(x) ---eqn.BK064
<a name="ch12b088">　Index begin　Index this file
for n=3 case, eqn.BK063 do not
have 3E3(x). We can push eqn.BK063
to lower order polynomial. Compare
eqn.BK063 with eqn.BK058 below
3E0(x)* 3E2(x)≦ 3E12(x) ---eqn.BK063
2E0(x)* 2E2(x)≦ 2E12(x) ---eqn.BK058
<a name="ch12b089">
Lower right corner 0,2,1 are
the same in both equation.
Lower left  corner 3,2 tell us
that job of P3(t) hand over to
P2(t). Base on eqn.12.6 and
already proved n=2 case
eqn.BK058, we conclude that
eqn.BK063 is true.

<a name="ch12b090">
eqn.BK064 is different, because
it has the term 3E3(x). We can
not push eqn.BK064 to lower
order polynomial. Because
v-1Ev(x) is undefined.
v-1Ev(x) is non-exist.
<a name="ch12b091">
Lower left corner sub-note 'v-1'
means polynomial order is v-1.
Lower right corner sub-note 'v'
means Symmetric Sum has order v.
We have v-1 data in hand, we can
not create v-th multiplication.
Same argument in tute0042.htm
Five numbers can not have power
six.

<a name="ch12b092">
We can not push eqn.BK064 to lower
order polynomial. We must prove it
right here n=3. This part is done
in textbook page 183

<a name="ch12b093">　Index begin　Index this file
We need to find the definitions
3E1(x)=(x1+x2+x3)/3 ---eqn.BK065
3E2(x)=(x1*x2+x1*x3+x2*x3)/3 ---eqn.BK066
3E3(x)=x1*x2*x3 ---eqn.BK067
Above three equations are
Symmetric Sum of x1,x2,x3 three
data set.
<a name="ch12b094">
eqn.BK065 and eqn.BK066 divide
by three are taking average.
Please see one to one
Put eqn.BK065 to eqn.BK067 into
eqn.BK064. The equation to prove
is next
<a name="ch12b095">
 { x1+x2+x3 3 } {x1*x2*x3} ≦ { x1*x2+x1*x3+x2*x3 3 } 2
---page 182 ---line 18 ---eqn.12.11
width of above equation
<a name="ch12b096">
To prove high power end Newton
Inequality, it is always easier
to prove if use the reciprocal
of x. In this case we divide
whole equation by the n-th
multiplied term
nEn(x)=(x1*x2*...*xn) ---eqn.BK301
Inserted equation, use number 300+
<a name="ch12b097">
For this reason (divide by x),
we require that the observed data
x1,x2,...,xn all be positive.
Zero is not allowed.
<a name="ch12b098">　Index begin　Index this file
(x1*x2*x3)2 is expressed as 3E32(x)
Divide eqn.12.11 by (x1*x2*x3)2
get
 1 3 { 1 x1x2 + 1 x1x3 + 1 x2x3 } ≦ 1 9 { 1 x1 + 1 x2 + 1 x3 } 2
---page 182 ---line 21 ---eqn.BK068
width <a name="ch12b099">
Expand eqn.BK068 get
 9 3 { 1 x1x2 + 1 x1x3 + 1 x2x3 } ≦ 1 x1x1 + 1 x2x2 + 1 x3x3 + 2 x1x2 + 2 x1x3 + 2 x2x3
---page 182 ---mid step ---eqn.BK069
width <a name="ch12b100">
Cancel like term get
 1 x1x2 + 1 x1x3 + 1 x2x3 ≦ 1 x1x1 + 1 x2x2 + 1 x3x3
---page 182 ---line 23 ---eqn.BK070
width of above equation
<a name="ch12b101">
Whether eqn.BK070 is true?
From rearrangement inequality
view point, eqn.BK070 is true.
Because we can assume the order
0＜x1≦x2≦x3 ---eqn.BK071
then reciprocal give us
1/x1≧1/x2≧1/x3 ---eqn.BK072
<a name="ch12b102">
Consider
sequence_A=[1/x1, 1/x2, 1/x3]
sequence_B=[1/x2, 1/x3, 1/x1]
Rearrangement inequality say
sequence_A dot product sequence_A
is greater than or equal to
sequence_A dot product sequence_B

<a name="ch12b103">　Index begin　Index this file
Whether eqn.BK070 is true?
From Cauchy's inequality
view point, eqn.BK070 is true.
Because we can assume the
sequences
sequence_C=[1/x1, 1/x3, 1/x2]
sequence_D=[1/x2, 1/x1, 1/x3]
<a name="ch12b104">
sequence_C dot product sequence_D
get eqn.BK070 less than side
sequence_C dot product sequence_C
get square root of eqn.BK070
greater than side.
sequence_D dot product sequence_D
get square root of eqn.BK070
greater than side.
Net result is eqn.BK070.

<a name="ch12b105">
Whether eqn.BK070 is true?
From AM-GM inequality view point,
eqn.BK070 is true.
Because we can do three times
AM-GM inequality, then sum
them to eqn.BK070.
<a name="ch12b106">
That is
1/√(x1x1x2x2)≦[1/(x1x1)+1/(x2x2)]/2 ---eqn.BK073
1/√(x1x1x3x3)≦[1/(x1x1)+1/(x3x3)]/2 ---eqn.BK074
1/√(x2x2x3x3)≦[1/(x2x2)+1/(x3x3)]/2 ---eqn.BK075
Sum eqn.BK073 to eqn.BK075 get
eqn.BK070.

<a name="ch12b107">
Conclude:
For n=3 case, Newton Inequality
eqn.12.10
or eqn.12.11
or eqn.BK064
or eqn.BK070
is true.
2010-05-21-18-59 stop

<a name="ch12b108">　Index begin　Index this file
2010-05-21-22-30 start
■　Prove Newton Inequality n=any
Textbook page 183

Newton's inequality in general
case,
Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.2
for 0＜k＜n
<a name="ch12b109">
x contain hidden condition
that is x has n elements,
polynomial has degree=n.
Equation contain Ek+1(x)
which can not exceed En(x)
k+1≦n, then k＜n .
Equation contain Ek-1(x)
which can not be smaller than
E0(x), 0≦k-1 then 0＜k .

<a name="ch12b110">
If k+1＜n, we have
Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.12
for 1≦k＜n-1
eqn.12.12 condition 1≦k＜n-1
eqn.12.2  condition 0＜k＜n
The inequality in the group
k+1＜n can be push down to
lower order polynomial's
high end inequality. Done
proof in low order polynomial.

<a name="ch12b111">
If k+1=n, Ek+1(x)=En(x), we have
En-2(x)*En(x)≦En-12(x) ---eqn.12.13
this is a boundary inequality.
We can not push En(x) down to
lower order polynomial. Because
there is no more same E downward.
Must prove eqn.12.13 right here.

<a name="ch12b112">　Index begin　Index this file
We write eqn.12.13 in longhand
and use x^j remind us that this
xj is a missing term.
Why missing? The only one not
missing any is nEn(x). All
elements multiply together. The
next one nEn-1(x) has one missing
in symmetric sum.
<a name="ch12b113">
Example, assume x=[a,b,c,d,e]
5E5(x)=a*b*c*d*e no one missing.
5E4(x)=b*c*d*e+a*c*d*e
+a*b*d*e+a*b*c*e
+a*b*c*d
In which
"b*c*d*e" missing an 'a'
"a*c*d*e" missing a  'b'
etc.
<a name="ch12b114">
If we write "b*c*d*e" as
"a^*b*c*d*e", a^=pure number 1
at the step whole equation
divide by "a*b*c*d*e" , the
missing one "x^j" show up at
denominator.
Use "a^" or use "x^j" help
us do bookkeeping right.
2010-05-21-23-22 stop

<a name="ch12b115">
2010-05-22-09-26 start
We begin prove eqn.12.13 which
is the high power end Newton's
inequality. eqn.12.13 use the
following symmetric sums
<a name="ch12b116">
En(x)=x1*x2*...*xn ---eqn.BK076
 En-1(x)= j=n ∑ j=1 x1x2...x^j...xn /( n n-1 ) ---eqn.BK077 En-2(x)= ∑ 1≦j＜k≦n x1...x^j...x^k...xn /( n n-2 ) ---eqn.BK078
Consider red terms as PURE NUMBER ONE.
'－1' in 'n－1' exclude one missing element "x^j".
'－2' in 'n－2' exclude two missing "x^j" and "x^k".

width <a name="ch12b117">
Binomial coefficient is defined here
Substitute En(x), En-1(x), En-2(x) into eqn.12.13 get
 2 n(n-1) { ∑ 1≦j＜k≦n x1...x^j...x^k...xn } x1*x2*...*xn ≦ { 1 n j=n ∑ j=1 x1x2...x^j...xn } 2
---eqn.12.14   Highest order Newton's inequality It is
highest order because x1*x2*...*xn no missing element
width of above equation
<a name="ch12b118">
2010-05-22-10-28 here
For n elements sequence, we build
n-th order polynomial. There are
n-1 Newton's inequality. They are
<a name="ch12b119">　Index begin　Index this file
nEn-2(x)*nEn(x)≦nEn-12(x) ---eqn.12.13
nEn-3(x)*nEn-1(x)≦nEn-22(x) ---eqn.BK079
nEn-4(x)*nEn-2(x)≦nEn-32(x) ---eqn.BK080
..... //Red are boundary terms
<a name="ch12b120">
nE2(x)* nE4(x)≦ nE32(x) ---eqn.BK081
nE1(x)* nE3(x)≦ nE22(x) ---eqn.BK082
nE0(x)* nE2(x)≦ nE12(x) ---eqn.BK083
<a name="ch12b121">
Boundary Newton's inequality
eqn.12.13 high power end and
eqn.BK083 low  power end are
easy to prove. Because there
are less terms involved.
bicof(10,10)=1, bicof(10,9)=10;
bicof(10,1)=10, bicof(10,0)=1.
<a name="ch12b122">
Middle Newton's inequality has
many terms and hard to sort out
a form for inequality argument.
Please see a hard work example
at here and get trouble at here.
bicof(10,5)=252, bicof(10,4)=210;
after square 252*252=63504 terms
<a name="ch12b123">　Index begin　Index this file
Now we prove boundary Newton's
inequality eqn.12.14, it is much
easier than the middle Newton's.
2010-05-22-11-09 here

<a name="ch12b124">
We assume all data x1,x2,...,xn
are positive numbers, no zero.
We divide eqn.12.14 by square
of x1*x2*...*xn convert high
power (n-th power x1*x2*...*xn)
to reciprocal of low power 1/x1
or 1/(x2*x3) etc. This division
let x^j change from invisible
one to visible 1/xj.
After division we get
<a name="ch12b125">
 2 n(n-1) { ∑ 1≦j＜k≦n x1...x^j...x^k...xn x1*x2*...*xn } ≦ { 1 n j=n ∑ j=1 x1x2...x^j...xn x1*x2*...*xn } 2
---eqn.BK084   x hat x^j stand out in this equation. See
x hat as ONE, next step x^j show up at denominator.

width <a name="ch12b126">
 2 n(n-1) ∑ 1≦j＜k≦n 1 xjxk ≦ { 1 n j=n ∑ j=1 1 xj } 2
---eqn.12.15   Was x hat x^j, drop hat here.
Reciprocal of highest order Newton's inequality

width of above equation
<a name="ch12b127">
2010-05-22-11-43 here
How to prove eqn.12.15?
Textbook page 183 explain as
following.
===textbook copy start===
<a name="ch12b128">　Index begin　Index this file
We could now stick with the
pattern that worked for H3,
but there is a more graceful
way to finish which is almost
staring us in the face.
<a name="ch12b129">
If we adopt the language of
symmetric function, the
target bound (12.15) may be
written more systematically as
<a name="ch12b130">
E0(1/x1,1/x2,...,1/xn)     //E0 defined to be 1
*E2(1/x1,1/x2,...,1/xn)     //E2=∑[1/(xixj)]/bicof(n,2)
≦E12(1/x1,1/x2,...,1/xn)  //E1=∑[1/xk]/bicof(n,1)
Above is ---eqn.BK085
<a name="ch12b131">
and one now sees that this
inequality is covered by the
first bound of the group
(12.12). Thus the proof of
Newton's inequality is
complete.
===textbook copy stop===
2010-05-22-11-54 here

<a name="ch12b132">　Index begin　Index this file
First time read here, LiuHH
was very puzzle.
Why?!
Newton's inequality is designed
for x. Now the data become
y=[1/x1,1/x2,...,1/xn] ---eqn.BK086
How can I use x equation to
prove y property?

<a name="ch12b133">
After more think, find out why.

Newton's inequality is designed
for x. That is very true.
x and y are different, that is
also true.
<a name="ch12b134">
We need to see what limitation
applied to data x ?
Review from the beginning,
Newton's inequality require
x has n elements, otherwise
it is another set Newton.
<a name="ch12b135">
Require x has all positive
elements, so that division
of x1*x2*...*xn is possible.
That is all.
<a name="ch12b136">
Then y satisfy
these two conditions too. So
x's Newton inequality apply
to y perfect. However,
<a name="ch12b137">
we must differentiate n-th
order polynomial Pn(t) all
the way down to P2(t).
['2' in E2(y) demand P2(t)]
This is not a problem, as
long as the proof is DONE.

Problem 12.1 is solved.
2010-05-22-12-13 stop

<a name="ch12b138">　Index begin　Index this file
2010-05-22-16-01 start
■　Prove Maclaurin Inequality

Maclaurin's inequality asserts
that
[En(x)]1/n ≦ [En-1(x)]1/(n-1) ≦ ...
≦ [E2(x)]1/2 ≦ E1(x) ---eqn.12.3
<a name="ch12b139">
Here we still use eqn.12.6
With the property
nEk(x)=n-1Ek(y) ---eqn.BK042
where 1≦k＜n
we only prove high power end
Maclaurin Inequality.
<a name="ch12b140">
Middle power Maclaurin Ineq.
go to lower order polynomial's
high power end, like Newton did.
Please see Push down diagram
Newton use 3 E, Maclaurin use 2.
uEv(x) u=v indicate reach end.
Diagram colored for Newton but
same reason apply to Maclaurin.
<a name="ch12b141">
Let us exam high power end
Maclaurin Inequality
[En(x)]1/n ?≦? [En-1(x)]1/(n-1) ---eqn.BK087
The requirement to x is that
x has n elements and all
of x elements are positive.
If we normalize x such that
Geometric Mean of x is one
x1*x2*...*xn=1 ---eqn.BK088
<a name="ch12b142">　Index begin　Index this file
Normalize x still satisfy
two requirements.
eqn.BK087 raise power to (n-1)
eqn.BK087 become
1 ?≦? En-1(x) ---eqn.BK089
Divide eqn.BK089 by
x1*x2*...*xn=1, we have
<a name="ch12b143">
 { x1*x2*...*xn x1*x2*...*xn } 1/n ?≦? 1 n j=n ∑ j=1 x1x2...x^j...xn x1*x2*...*xn
---eqn.BK090   x1*x2*...*xn=1
width <a name="ch12b144">
which reduce to
 { 1 x1*x2*...*xn } 1/n ?≦? 1 n [ 1 x1 + 1 x2 +...+ 1 xn ]
---eqn.BK091   x1*x2*...*xn=1
width of above equation
<a name="ch12b145">
eqn.BK091 is AM-GM inequality.
for the sequence
y=[1/x1,1/x2,...,1/xn] ---eqn.BK086
which is true.
Proof of Maclaurin Inequality
is done.
2010-05-22-16-44 stop

<a name="ch12b146">　Index begin　Index this file
2010-05-22-17-07 start
■　'high power end'? why it is t0?
The term 'high power end' is
Sometime it may be confuse.
Please see polynomial in terms
of E's. In one equation, each
Eu*tv must have u+v=constant.
Please see related topic.
<a name="ch12b147">
'high power end' for E, u=n is
' low power end' for t, v=0 .
' low power end' for E, u=0 is
'high power end' for t, v=n .
<a name="ch12b148">
'high power end' is reserved
for E, not for t. If you see
notes say 'high power end',
look for E, do not look at t!

<a name="ch12b149">
Polynomial  P6(t) differentiate
once become P5(t)
Eu*tv u+v=constant 6.
change to
Eu*tw u+w=constant 5.
tv reduce power by one.
Eu is constant (no t),
Eu is unchanged.
2010-05-22-17-25 stop

<a name="ch12b150">　Index begin　Index this file
2010-05-22-18-30 start
■　Equality in Newton and Maclaurin

Newton's inequality eqn.12.2.
Maclaurin's inequality eqn.12.3
equality condition is
x1=x2=...=xn ---eqn.BK092

<a name="ch12b151">
For n=3 case, Prove Newton's
Inequality, used three methods
From rearrangement inequality
From Cauchy's inequality
From AM-GM inequality.

<a name="ch12b152">
We explain equality condition
from three methods too.
The basic point is that we are
given one data set
x=[x1,x2,...,xn] ---eqn.BK093
we can not use xn+1.
<a name="ch12b153">
Use numerical example for easy
illustration. Assume given
seq_A1=[1,2,3,4,5] ---eqn.BK094

<a name="ch12b154">　Index begin　Index this file
Rearrangement inequality need
just one data set. Second seq.
is re-arranged from first.
Define
seq_B1=[3,2,4,1,5] ---eqn.BK095

<a name="ch12b155">
Rearrangement inequality get
seq_A1*seq_B1 ≦ seq_A1*seq_A1 ---eqn.BK096
1*3+2*2+3*4+4*1+5*5  //48≦
≦1*1+2*2+3*3+4*4+5*5  //55
How can we do re-arrangement
still get equality?
The only answer is that given
seq_A2=[2,2,2,2,2] ---eqn.BK097
All elements are the same.
Any seq_B2 is still [2,2,2,2,2]

<a name="ch12b156">
The condition
x1=x2=...=xn ---eqn.BK092
let us get equality in
Newton's inequality eqn.12.2.
and
Maclaurin's inequality eqn.12.3

<a name="ch12b157">
Next see Cauchy's inequality
Cauchy require two sequences.
If we have
seq_A1=[1,2,3,4,5] ---eqn.BK094
seq_C1=[2,4,6,8,10] ---eqn.BK098
Two sequence are proportional.
Cauchy will give us equality!
<a name="ch12b158">
But, we discuss Symmetric Sums
'6,8,10' are not given !
seq_C1 is forbidden. Then
create second sequence from
given sequence is same as
re-arrangement. Same result
eqn.BK092 let Cauchy give us
equality.

<a name="ch12b159">　Index begin　Index this file
Third, see AM-GM inequality.
We applied three times AM-GM
inequality. From eqn.BK073 to
eqn.BK075. Each one equality
condition is
xi=xj ---eqn.BK099
Link one by one, all xk are
equal. x1=x2,x2=x3,x3=x4 etc.
Again we get eqn.BK092 Please
see Equality GM=AM is easy
2010-05-22-19-12 stop

2010-05-23-18-20 done first  proofread
2010-05-23-21-23 done second proofread
2010-05-23-21-46 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

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