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

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

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

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

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

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

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

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

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

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

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

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

<a name="ch02c001">　Index begin　Index this file
2009-10-06-16-07 start
■　Exercise 2.1 problem statement
　　textbook page 31
（More from Leap-forward Fall-back Induction）

  Cauchy's leap-forward, fall-back induction 
can be used to prove more than just the AM-GM
inequality, in particular, it can be used to
show that Cauchy's inequality for n=2 implies
the general result. For example, by Cauchy's 
inequality for n=2 applied twice, one has

<a name="ch02c002">
 a1b1+a2b2+a3b3+a4b4
=(a1b1+a2b2)+(a3b3+a4b4)
≦(a12+a22)1/2(b12+b22)1/2+(a32+a42)1/2(b32+b42)1/2
≦(a12+a22+a32+a42)1/2*(b12+b22+b32+b42)1/2

<a name="ch02c003">
which is Cauchy's inequality for n=4 . 
Extend this argument to obtain Cauchy's
inequality for all n=2k and consequently
for all n. This may be the method by 
which Cauchy discovered his famous
inequality, even though in his textbook
he choose to present a different proof.
2009-10-06-16-27 here

<a name="ch02c004">
■　Exercise 2.1 discussion
Exercise 2.1 is same as textbook chapter
two page 21. Please read tute0011.htm

The key difference is that
Chapter one talk Cauchy's inequality, 
need two sequences.
Chapter two talk AM-GM inequality, need 
just one sequence.
Chapter two has Cauchy's leap-forward, 
and fall-back induction for one sequence.

<a name="ch02c005">
Exercise 2.1 require apply Cauchy's leap-
forward, fall-back induction for TWO
sequences under Cauchy's inequality,
not under AM-GM inequality.

Padding used in chapter two, change to 
set aj=bj=0 for n＜j≦2k
and apply 2k version.

LiuHH skip Exercise 2.1
2009-10-06-16-41 stop



<a name="ch02c006">　Index begin　Index this file
2009-10-06-18-56 start
■　Exercise 2.2 problem statement
　　textbook page 31
（Bernoulli and the Exponential Bound）

  Polya's proof of the AM-GM inequality 
used the analytic bound
  1+x ≦ ex for all x∈Real ---eqn.2.23
which is closely related to an inequality
of Jacob Bernoulli (1654-1705).
  1+n*x ≦ (1+x)n for all x∈[-1,∞)
           and all n=1,2,... ---eqn.2.24
<a name="ch02c007">
Prove Bernoulli's inequality (2.24) by
induction and show how it may be used to
prove that 1+x ≦ ex for all x∈Real 
Finally, by calculus or other means, 
prove one of the more general version 
of Bernoulli's inequality suggested by
Figure 2.2; for example, prove that
  1+p*x ≦ (1+x)p for all x≧-1
           and all p≧1 ---eqn.2.25
{ LiuHH note: in eqn.2.24 n is integer
              in eqn.2.25 p is real )
2009-10-06-19-12 here

<a name="ch02c008">
2009-10-06-20-58 start
■　Exercise 2.2 hint
　　textbook page 232
To prove the bound (2.24) by induction, 
  1+n*x ≦ (1+x)n for all x∈[-1,∞)
           and all n=1,2,... ---eqn.2.24
first
note that the case n=1 is trivial. Next, take
the bound for general n and multiply it by 1+x 
to get
 1+(n+1)x+n*x*x ≦ (1+x)n+1 ---eqn.AG001
<a name="ch02c009">
This is stronger than the bound (2.24) in the
case n+1, so the bound (2.24) holds for all 
n=1,2,... by induction. To show 1+x ≦ ex one
replaces x by x/n in Bernoulli's inequality
and lets n goto infinity. 
(above equation 2.24, n is integer 1,2,3,...
 below equation 2.25, p≧1 is real number.)
<a name="ch02c010">
Finally, to prove
the relation (2.25), one sets
 f(x)=(1+x)p-(1+px) ---eqn.AG002
then notes that f(0)=0, f'(x)≧0 for x≧0 and
f'(x)≦0 for -1＜x≦0  ---eqn.AG003
so 
 minx∈[-1,∞)f(x)=f(0)=0. ---eqn.AG004
2009-10-06-21-13 stop

<a name="ch02c011">　Index begin　Index this file
2009-10-07-11-55 start 
■　Exercise 2.2 solution


Exercise ask to prove Bernoulli inequality
equation (2.24) by induction 
  1+n*x ≦ (1+x)n for all x∈[-1,∞)
           and all n=1,2,... ---eqn.2.24
<a name="ch02c012">
Induction start from n=1, then eqn.2.24 become
  1+1*x ≦ (1+x)1 
that is
  1+x ≦ 1+x   ---eqn.AG005
equality is true for n=1.
Next, let us see n=2, eqn.2.24 become
  1+2*x ?≦? (1+x)2    ---eqn.AG006
"?≦?" indicate uncertain relation, wait for
proving. We know eqn.AG005 is true, since
given x∈[-1,∞), then 1+x∈[-1+1,∞+1) that 
is 1+x∈[0,∞) , 1+x is non-negative. We
<a name="ch02c013">
multiply eqn.AG005 with 1+x, less than sign
not changed for non-negative 1+x, result is
  (1+x)*(1+x) = (1+x)2
expand left hand side, get
   1+2*x+x*x = (1+x)2   ---eqn.AG007
Compare eqn.AG006 and eqn.AG007, we can drop
a non-negative x*x from eqn.AG007, change
exact equality to next rough inequality
   1+2*x ≦ (1+x)2   ---eqn.AG008
eqn.AG008 is true, so for n=2 case eqn.AG006
is also true. If x=0, eqn.AG008 (eqn.AG006)
equality become true.

<a name="ch02c014">
Next push n=2 to n=3. But here, we need use
induction method to avoid unlimited cases.
Now assume
   1+n*x ≦ (1+x)n  ---eqn.AG009
is true (n=1 and n=2 are true)
we need to see (1+x)n+1 case.
Multiply eqn.AG009 by non-negative (1+x) get
  (1+n*x)*(1+x) ≦ (1+x)n*(1+x) ---eqn.AG010
Expand eqn.AG010 left hand side as below
  1+n*x+x+n*x*x ≦ (1+x)n+1
or
  1+(n+1)*x+n*x*x ≦ (1+x)n+1 ---eqn.AG011
<a name="ch02c015">
because n*x*x is non-negative, we can drop
n*x*x from eqn.AG011, less than sign unchanged
and enlarge inequality for x not=0 case.
result is
  1+(n+1)*x ≦ (1+x)n+1 ---eqn.AG012
eqn.AG012 indicate if n th level is true
then n+1 th level is also true. This condition
extend to infinity (induction method). That is
eqn.AG012 is true in general. 
But n=integer here ! and n=3.45 is undefined.
2009-10-07-12-27 here

<a name="ch02c016">
Next see Bernoulli's inequality eqn.2.24 or
   1+n*x ≦ (1+x)n  ---eqn.AG009
imply 1+x ≦ ex  ---eqn.AG013
In eqn.AG009, x∈[-1,∞) is arbitrary and
n=1,2,3,... are positive. The number x/n
also x/n∈[-1,∞) Let us replace x with x/n
in eqn.AG009, get
   1+n*(x/n) ≦ (1+(x/n))n
that is
   1+x ≦ (1+x/n)n  ---eqn.AG014
<a name="ch02c017">
from calculus, let n approach to infinity
(1+x/n)n approach e^x from below. 
where e=2.718281828459045...  That is
   1+x ≦ (1+x/n)n ＜ e^x ---eqn.AG015
Please see tute0011.htm#ch02a056
Click "Draw 603" button, 
red curve is y(x)=e^x
red dash straight line is y(x)=1+x
Red curve is always above red dash line.
That is graph statement of eqn.AG015.
2009-10-07-12-53 stop

<a name="ch02c018">　Index begin　Index this file
2009-10-07-14-39 start
Above is Bernoulli's inequality 
  1+n*x ≦ (1+x)n for all x∈[-1,∞) ---eqn.2.24
where n=1,2,... integer, n=3.45 is excluded.

Below generalize n to p and p≧1
p = arbitrary real number which is eqn.2.25
  1+p*x ≦ (1+x)p for all x≧-1
           and all p≧1 ---eqn.2.25
eqn.2.25 is a p-continuous function, we can
define eqn.2.25 greater than side minus
less than side as following
 g(x,p)=(1+x)p-(1+px) ---eqn.AG016
<a name="ch02c019">
Before prove, eqn.AG016 can be ≦0 or ≧0
Since we build eqn.AG016 as 
greater than side minus less than side,
we expect to prove eqn.AG016 ≧ 0
Problem give definition range, that is
 g(x,p)=(1+x)p-(1+px) ---eqn.AG016
      for all x≧-1 and all p≧1 
Whether there is a point that g(x0,p0)=0?
for this simple eqn.AG016, we can see if
  p=1
then
 g(x,1)=(1+x)1-(1+1*x) ≡ 0 ---eqn.AG017
for any x value.
<a name="ch02c020">
Second, if
  x=0
then
 g(0,p)=(1+0)p-(1+p*0) ≡ 0 ---eqn.AG018
for any p value.
If p=1, g(x,1) coincide with x-axis, that
is a no-interest condition. We skip p=1 
and define
 f(x)=(1+x)p-(1+px) ---eqn.AG019
same as textbook hint suggested eqn.AG002
in eqn.AG019, we are interested in p＞1
and treat p as a constant.

<a name="ch02c021">
We like to show in the whole range
 x≧-1 and all p≧1
must have eqn.AG019 ≧ 0
To do this, we need curve slope. 
Differentiate eqn.AG019 get
 f'(x)=p*(1+x)p-1-p ---eqn.AG020
when x=0,
 f'(0)=p*(1+0)p-1-p=p-p=0 ---eqn.AG021
when x＞0 and p＞1
 f'(x)=p*(1+x)p-1-p＞0 ---eqn.AG022
when -1＜x＜0 and p＞1
 f'(x)=p*(1+x)p-1-p＜0 ---eqn.AG023
(see [draw 606] green dash curve, which
 is f'(x) for p=3. 2009-10-11-09-48 note)

<a name="ch02c022">
eqn.AG022 f'(x)＞0 imply that tangent slope 
is positive. Since f(x=0)=0 and f'(x=0)=0
Positive slope imply when x value increase,
curve gradual rise above y=f(x)=0. That is
 f(x)＞0 for x＞0

eqn.AG023 f'(x)＜0 imply that tangent slope 
is negative. Since f(x=0)=0 and f'(x=0)=0
negative slope imply curve gradual drop 
from above y=f(x)＞0 to y=f(x=0)=0 . That is
 f(x)＞0 for -1＜x＜0

<a name="ch02c023">
Better view curve figure next. Please click
"Draw 606" button. 
2009-10-07-15-33 here

<a name="ch02c024">
2009-10-07-15-36 here
Draw 606 figure has five curves. Blue 
curve violate p≧1 condition on purpose
See what violation look like. Other four 
curves all comply with p≧1 condition.
All curve above x-axis represent f(x)＞0
[green dash curve is f'(x) for p=3 case]

We start at eqn.2.25 and p is real number
and p≧1 (broader than n=1,2,3 ... integer)

eqn.AG018, eqn.AG021, eqn.AG022, eqn.AG023
plus draw 606 figure, concluded that 
eqn.2.25 is true.
2009-10-07-15-43 stop



<a name="ch02c025">　Index begin　Index this file
2009-10-07-17-11 start
■　Exercise 2.3 problem statement
　　textbook page 32
（Bound by Pure Powers）
Separate cross product term x*y
to two unlinked terms x*x + y*y
You need this property!9810081026

  In the day-to-day work of mathematical 
analysis, one often uses the AM-GM inequality
to bound a product of a sum of products by
a simpler sum of pure powers. Show that for 
positive x, y, α and β one has

<a name="ch02c027">
2009-10-07-17-25 here
and, for a typical corollary, show that
one also has the more timely bound
 x2004*y+x*y2004≦x2005+y2005 ---eqn.AG024
2009-10-07-17-30 here



<a name="ch02c028">
2009-10-07-17-35 here
■　Exercise 2.3 hint
　　textbook page 232
To prove the bound (2.26) one takes 
 p1=α/(α+β), p2=β/(α+β), a1=xα+β, a2=yα+β
and applies the AM-GM bound (2.9).
To get the timely bound we specialize
(2.26) twice, once with α=2004 and β=1
and once with α=1 and β=2004. We then
sum the result bounds.
2009-10-07-17-43 stop






<a name="ch02c029">　Index begin　Index this file
2009-10-07-18-21 start
■　Exercise 2.3 solution


Put two equation side by side as following.

<a name="ch02c030">
and two terms AM-GM inequality
a1p1a2p2 ≦ p1a1+p2a2 ---eqn.2.9 dim2
Compare above two equations right hand
side. If we define
 p1=α/(α+β) ---eqn.AG025
 p2=β/(α+β) ---eqn.AG026
 a1 = xα+β ---eqn.AG027
 a2 = yα+β ---eqn.AG028
both eqn.2.26 and eqn.2.9 right hand side
are consistent. Plug eqn.AG025 to AG028
into eqn.2.9 left hand side, 
<a name="ch02c031">
get
a1p1=x(α+β)*α/(α+β)=xα ---eqn.AG029
a2p2=y(α+β)*β/(α+β)=yβ ---eqn.AG030
eqn.AG029 and eqn.AG030 right hand side
just match eqn.2.26 left hand side. Then
change of variable eqn.AG025 to AG028 is
consistent for both eqn.2.26 and eqn.2.9.
Above change of variable is our solution
for first part of Exercise 2.3

<a name="ch02c032">
Second part of our problem is to prove
 x2004*y+x*y2004≦x2005+y2005 ---eqn.AG024

In eqn.2.26 let α=2004 and β=1 get

<a name="ch02c033"> Change α and β role.
In eqn.2.26 let α=1 and β=2004 get

<a name="ch02c034">
2009-10-07-19-02 here
Now add eqn.AG031 and eqn.AG032 get
 x2004*y+x*y2004≦x2005+y2005 ---eqn.AG024
Second part solved.
2009-10-07-19-04 stop

<a name="ch02c035">
2009-10-07-19-09 start
Exercise 2.3 eqn.2.26 key point is
separate cross product term xα*yβ
to two unlinked terms xα+β + yα+β
(coefficients ignored)
Special case when α=β=1 eqn.2.26 become
humble bound
a*a+b*b ≧ 2*a*b  ---eqn.AD18
a*b are unlocked to a*a + b*b
2009-10-07-19-20 stop



<a name="ch02c036">　Index begin　Index this file
2009-10-07-20-39 start
■　Exercise 2.4 problem statement
　　textbook page 32
（A Canadian Challenge）

Participants in the 2002 Canadian Math 
Olympiad were asked to prove the bound

and to determine when equality can hold.
Can you meet the challenge?
2009-10-07-20-50 here

<a name="ch02c038">
2009-10-07-21-08
when you solve Exercise 2.4, there is one
condition added to the problem statement
 a, b, c are all positive real numbers.
otherwise, reader can not move denominator
from greater than side to less than side.
2009-10-07-21-11



<a name="ch02c039">
2009-10-07-20-52 here
■　Exercise 2.4 hint
　　textbook page 232

The target inequality is equivalent to
  a2bc+ab2c+abc2＜a4+b4+c4 ---eqn.AG034
A pure power bound. By the AM-GM inequality,
we have 
  a2bc=(a4)1/2(b4)1/4(c4)1/4
  a2bc≦(a4)/2+(b4)/4+(c4)/4 ---eqn.AG035
and analogous bound hold for ab2c and abc2.
The sum of these bounds yields the target
inequality.
2009-10-07-21-05 here

<a name="ch02c040">
2009-10-08-10-55 start (make up typing)
  Equality holds in the target inequality
if and only equality holds for both of our
applications of AM-GM bound. Thus, equality
holds in the target bound if and only if
a=b=c. Incidentally, three other solutions
of this problem are available on web site 
of the Canadian Mathematical Association.
2009-10-08-10-59 stop (make up typing)





<a name="ch02c041">　Index begin　Index this file
2009-10-07-21-17 here
■　Exercise 2.4 solution


Exercise 2.4 Hint already solved this 
problem. Please pay attention to the
following
 a2bc = a*a*b*c
 =(a4)1/4(a4)1/4(b4)1/4(c4)1/4
 ≦(a4)/4+(a4)/4+(b4)/4+(c4)/4 ---eqn.AG036
inequality in eqn.AG036 is four terms
AM-GM inequality.

<a name="ch02c042">
Do not be confused by fourth power, do next
change of variable and hide fourth power
define k = first  a4 , l = second a4
define m = b4, n = c4. eqn.AG036 become
 (k)1/4(l)1/4(m)1/4(n)1/4
 ≦(k)/4+(l)/4+(m)/4+(n)/4 ---eqn.AG036 aux1
or
 (k*l*m*n)1/4 ≦(k+l+m+n)/4 ---eqn.AG036 aux2
 Left hand side is geometric  mean and 
right hand side is arithmetic mean. 
Cristal clear.

<a name="ch02c043">
Similarly, other two inequalities are
 ab2c
 ≦(a4)/4+(b4)/4+(b4)/4+(c4)/4 ---eqn.AG037
 abc2
 ≦(a4)/4+(b4)/4+(c4)/4+(c4)/4 ---eqn.AG038
Add eqn.AG036, eqn.AG037, eqn.AG038 get
eqn.AG034, which is same as eqn.AG033
2009-10-07-21-28 here

<a name="ch02c044">
Next see when equality can hold.
Please see Equality GM=AM is easy
When sequence elements are all equal,
get GM=AM
eqn.AG036 is GM ≦ AM
sequence elements are a4, a4, b4, c4
When they are equal   a4 =a4 =b4 =c4
implies a=b=c.
2009-10-07-21-35 stop.



<a name="ch02c045">　Index begin　Index this file
2009-10-08-11-23 start
■　Exercise 2.5 problem statement
　　textbook page 32
（A Bound Between Differences）
  Show that for nonnegative x and y and
integer n one has
 n*[(x*y)(n-1)/2]≦(xn-yn)/(x-y) ---eqn.2.27

2009-10-08-11-27 here



<a name="ch02c046">
2009-10-08-11-30 start
■　Exercise 2.5 hint
　　textbook page 233
For all j and k, the AM-GM inequality 
gives us
  (xj+k*yj+k)1/2≦(xjyk+xkyj)/2 ---eqn.AG039
Setting  k=n-1-j  ---eqn.AG040
and summing over 0≦j＜n yields the bound
 n*[(x*y)(n-1)/2]≦xn-1+xn-2y+...+xyn-2+yn-1
that is
 n*[(x*y)(n-1)/2]≦(xn-yn)/(x-y) ---eqn.AG041
2009-10-08-11-47 stop







<a name="ch02c047">
2009-10-08-12-07 start
■　Exercise 2.5 solution

re-write
  (xj+k*yj+k)1/2≦(xjyk+xkyj)/2 ---eqn.AG039
as
  (xjxk*yjyk)1/2 ＝ (xjyk)1/2*(xkyj)1/2
consider (xjyk) as first term
consider (xkyj) as second term
Apply AM-GM inequality to two terms get
 (xjyk)1/2*(xkyj)1/2≦(xjyk+xkyj)/2
re-write left hand side get
  (xj+k*yj+k)1/2≦(xjyk+xkyj)/2 ---eqn.AG042
<a name="ch02c048">
here both j and k are variables, but
j,k are dependent. set
  k=n-1-j  ---eqn.AG040
eqn.AG042 become one variable j inequality
  (xj+n-1-j*yj+n-1-j)1/2≦(xjyn-1-j+xn-1-jyj)/2
or
  (xn-1*yn-1)1/2≦(xjyn-1-j+xn-1-jyj)/2 ---eqn.AG043
j=0:  (xn-1*yn-1)1/2≦(x0yn-1-0+xn-1-0y0)/2 ---eqn.AG044
j=1:  (xn-1*yn-1)1/2≦(x1yn-1-1+xn-1-1y1)/2 ---eqn.AG045
j=2:  (xn-1*yn-1)1/2≦(x2yn-1-2+xn-1-2y2)/2 ---eqn.AG046
...
j=k:  (xn-1*yn-1)1/2≦(xkyn-1-k+xn-1-kyk)/2 ---eqn.AG047
...
j=n-1:(xn-1*yn-1)1/2≦(xn-1yn-1-(n-1)+xn-1-(n-1)yn-1)/2 ---eqn.AG048
<a name="ch02c049">
Sum from eqn.AG044 to eqn.AG048 
j start from j=0, end at j=n-1, 
total n terms, NOT n-1 terms !!
get "n*", NOT "(n-1)*" 
 n*(xn-1*yn-1)1/2≦
   (yn-1+xn-1)/2
  +(x1yn-2+xn-2y1)/2 ---eqn.AG049 blue
  +(x2yn-3+xn-3y2)/2 ---eqn.AG049 red
  +...
  +(xkyn-1-k+xn-1-kyk)/2
  +...
  +(xn-3y2+x2yn-3)/2 ---eqn.AG049 red
  +(xn-2y1+x1yn-2)/2 ---eqn.AG049 blue
  +(xn-1+yn-1)/2 ---eqn.AG049
<a name="ch02c050">
Blue add blue, red add red, 
remove all '1/2' factor, simplify eqn.AG049
get
 n*[(x*y)(n-1)/2]≦xn-1+xn-2y+...+xyn-2+yn-1
that is
 n*[(x*y)(n-1)/2]≦(xn-yn)/(x-y) ---eqn.AG041
Problem solved.

<a name="ch02c051">
If n=odd=7 for example, when k=3
the middle term 
  +(xkyn-1-k+xn-1-kyk)/2
has no another term to add. But middle term
it self cancel out '1/2', since
  +(xkyn-1-k+xn-1-kyk)/2
= +(x3y7-1-3+x7-1-3y3)/2
= +x3y3

2009-10-08-12-50 stop



<a name="ch02c052">　Index begin　Index this file
2009-10-08-13-15 start
■　Exercise 2.6 problem statement
　　textbook page 33
（Geometry of the Geometric Mean）

  There is indeed some geometry behind the 
definition of the geometric mean. The key
relation were known to Euclid, although
there is no evidence that Euclid specifically
considered any inequalities. By appealing
to the geometry of Figure 2.3 prove that
h=√(ab) and thereby automatically deduce
that √(ab)≦(a+b)/2
2009-10-08-13-20 here



<a name="ch02c053">
■　Exercise 2.6 hint
　　textbook page 233
 Since α+β=π/2 we have γ=α and δ=β. so the
triangle ∆(ABD) and ∆(DBC) are similar. By
proportionality of the corresponding sides 
we have h:a=b:h and we find h2=ab.
just as required.
2009-10-08-13-26 stop

2009-10-08-18-27
<a name="ch02c054">
■　Exercise 2.6 solution


Please click above "Plot1" button.
Angle ∟ADC, ∟ABD, ∟CBD are all right angle.
Right triangle ⊿ABD and ⊿CBD are similar.
The ratio AB:BD and BD:BC are the same.
That is AB/BD = BD/BC 
which give BD*BD=AB*BC
and conclude h=√(ab) as required.
2009-10-08-18-33 stop



<a name="ch02c055">　Index begin　Index this file
2009-10-09-10-28 start
■　Exercise 2.7 problem statement
　　textbook page 33
（One Bounded Product Implies Another）

Show that for nonnegative x, y, and z one
has the implication
  1≦xyz  ==>  8≦(1+x)(1+y)(1+z) ---eqn.2.28
Can you also propose a generalization?




<a name="ch02c056">
2009-10-09-10-37 here
■　Exercise 2.7 hint
　　textbook page 233
The product (1+x)(1+y)(1+z) expands as
(1+x)(1+y)(1+z)=1+x+y+z+xy+yz+zx+xyz ---eqn.AG050
and the AM-GM bound gives us
(x+y+z)/3≧(xyz)1/3≧1 ---eqn.AG051
that is (x+y+z) ≧ 3  ---eqn.AG052
and
(xy+yz+zx)/3≧{(xy){yz}{zx}}1/3=(xyz)2/3≧1 ---eqn.AG053
that is (xy+yz+zx) ≧ 3 ---eqn.AG054

<a name="ch02c057">
then eqn.AG050 become
(1+x)(1+y)(1+z)=1+(x+y+z)+[xy+yz+zx]+{xyz} 
  ≧ 1+(3)+[3]+{1} = 8
where
(3) use eqn.AG052
[3] use eqn.AG054
{1} use eqn.2.28 left side given condition
Add everything together get eqn.2.28 right
hand side. and problem solved.
2009-10-09-10-54 here

<a name="ch02c058">
With persistence, the same idea can be
used to show that for all nonnegative
ak, 1≦k≦n, one has the inference
  1≦∏[k=1,n]{ak} ＝＝＞ 
  2n≦∏[k=1,n]{1+ak} ---eqn.14.47
2009-10-09-11-01 here


<a name="ch02c059">
2009-10-09-11-25 start
■　Exercise 2.7 solution


LiuHH add detail in Exercise 2.7 hint.
About the generalization, observe
for n=5 case, expand
(1+a)(1+b)(1+c)(1+d)(1+e)
get coefficients
 +bino(5,0)
 +bino(5,1)+bino(5,2)
 +bino(5,3)+bino(5,4)
 +bino(5,5)
=1+5+10+10+5+1
=32=2^5
where bino(5,2)=5!/[2!*(5-2)!]

This is an observation, not a proof.
2009-10-09-11-27 stop



<a name="ch02c060">　Index begin　Index this file
2009-10-09-12-54 start
■　Exercise 2.8 problem statement
　　textbook page 33
（Optimality Principles for Products and Sums）

Given positive {ak: 1≦k≦n} and
positive c and d, we consider 
the maximization problem P1
  max{x1x2...xn: a1x1+a2x2+...+anxn=c} ---eqn.AG055
and the minimization problem P2
  min{a1x1+a2x2+...+anxn: x1x2...xn=d} ---eqn.AG056
<a name="ch02c061">
Show that for both of these problems the 
condition for optimality is given by the
relation
  a1x1=a2x2=...=anxn ---eqn.2.29
These optimization principles are extremely
productive, and they can provide useful
guidance even when they do not exactly
apply.





<a name="ch02c062">
2009-10-09-13-09 here
■　Exercise 2.8 hint
　　textbook page 233

The AM-GM inequality tells us
  {a1x1*a2x2*...*anxn}1/n
≦{a1x1+a2x2+...+anxn}/n ---eqn.AG057
and this yields a relation between the 
critical quantities of P1 and P2

<a name="ch02c064">
2009-10-09-13-25 here
We have equality here if and only if 
  a1x1=a2x2=...=anxn ---eqn.2.29
and nothing more is needed to confirm 
the stated optimality criterion.
2009-10-09-13-28 here




<a name="ch02c065">　Index begin　Index this file
2009-10-09-14-12 start
■　Exercise 2.8 solution


AM-GM inequality is for non-negative sequence
b1,b2,...,bn
Arithmetic mean 
AM = (b1+b2+ ... +bn)/n ---eqn.AE13
Geometric mean
GM = (b1*b2* ... *bn)1/n  ---eqn.AE14
have the following relation
AM ≧ GM

<a name="ch02c066">
Require all of its sequence elements be 
non-negative, otherwise Geometric mean
take 1/n th root of a negative number,
generate imaginary number, which we do
not want to involve in AM-GM inequality.

Now 
replace b1 by a1x1
replace b2 by a2x2
etc. we get eqn.AG059(AG057)

<a name="ch02c068">
2009-10-09-14-35 here
eqn.AG059 take n th power, and move 
a1*a2*...*a3 from less than side to
greater than side denominator place,
the result is eqn.AG058.

the maximization problem P1
  max{x1x2...xn: a1x1+a2x2+...+anxn=c} ---eqn.AG055
given constraint is
  a1x1+a2x2+...+anxn=c ---eqn.AG060
where c is a constant. Require maximize 
  OBJ1(x) = x1x2...xn ---eqn.AG061
<a name="ch02c069">
[[
Exercise 2.8 use two sequences
 a1,a2,...,an and x1,x2,...,xn
LiuHH make additional assumption that
sequence a1,a2,...,an is constant 
coefficients, and
sequence x1,x2,...,xn is variable 
sequence. If both sequences are variables, 
LiuHH do not know how to solve.
2009-10-09-14-49 here
]]
<a name="ch02c070">
Problem P1 given eqn.AG060. Go to
eqn.AG059, right hand side of eqn.AG059 is
constant (eqn.AG060) divided by constant n.
Left hand side of eqn.AG059 varies (xi vary)
eqn.AG059 guarantee that the vary value not
exceed the right hand side constant.
Problem P1 maximum possible value is when
GM=AM. Read from eqn.AG059, the equality
occurs when 
  a1x1=a2x2=...=anxn ---eqn.2.29
become true.
<a name="ch02c071">
Above is problem P1
Below is problem P2
The minimization problem P2
  min{a1x1+a2x2+...+anxn: x1x2...xn=d} ---eqn.AG056
problem P2 given constraint is
  x1x2...xn=d ---eqn.AG062
here 'd' is a constant.
problem P2 require minimize objective
function
<a name="ch02c072">
  OBJ2(x) = a1x1+a2x2+...+anxn ---eqn.AG063
Still go to eqn.AG059. This time eqn.AG059
left hand side 
  (a1x1*a2x2*...*anxn)1/n
is constant. Constraint by eqn.AG062.
eqn.AG059 says for problem P2, eqn.AG059
right hand side varies. But its value must
be greater than or equal to left hand side
constant value.  
<a name="ch02c073">
Minimum value is achieved when AM=GM, 
that is when
  a1x1=a2x2=...=anxn ---eqn.2.29
become true.

Here both problem P1 and problem P2
are solved.
2009-10-09-15-14 stop

<a name="20091011a01">
■　objective function and constraint

2009-10-11-15-35 start
When reader see next two problems
[[
the maximization problem P1
  max{x1x2...xn: a1x1+a2x2+...+anxn=c} ---eqn.AG055
and the minimization problem P2
  min{a1x1+a2x2+...+anxn: x1x2...xn=d} ---eqn.AG056
]]

<a name="20091011a02">
Put two equation side by side
  C=a1x1+a2x2+...+anxn ---eqn.AG201
  D=x1x2...xn ---eqn.AG202
P1 say maximize D, C is constraint.
P2 say minimize C, D is constraint.
<a name="20091011a03">
Do you have question?
Why two problems are so similar?
How do I know which one is constraint?
How do I know which one is objective
    function?
(an objective function is to be
 maximized or to be minimized.)
<a name="20091011a04">
The key difference is that 
constraint must equal to a constant
objective function must NOT equal to 
a constant.

If objective function equal to a constant.
then how can we find its maximum/minimum?
function value is fixed !!

<a name="20091011a05">
If constraint function not equal to a 
constant.
then how can we relate two variables?
two variables can change independently,
since function value is NOT fixed !!

Come back to
How do I know which one is constraint?
How do I know which one is objective

<a name="20091011a06">
constraint must equal to a constant
objective function must NOT equal to 
a constant.

In P1, D is free, C is constant.
In P2, C is free, D is constant.

This is an important basic understanding
for optimization class students.
2009-10-11-15-52 stop



<a name="ch02c074">　Index begin　Index this file
2009-10-09-16-51 start
■　Exercise 2.9 problem statement
　　textbook page 34
（An Isoperimetric Inequality for the 3-Cube）

Show that among all boxes with a given 
surface area, the cube has the largest
volume. Since a box with edge lengths
a,b and c has surface area
 A=2ab+2bc+2ca  ---eqn.AG064
and since a cube with surface A has edge
length (A/6)1/2. The analytical task is 
to show
  abc ≦ (A/6)3/2 ---eqn.AG065
and to confirm that equality holds if 
and only if a=b=c  ---eqn.AG066



<a name="ch02c075">
2009-10-09-16-57 here
■　Exercise 2.9 hint
　　textbook page 234
By the AM-GM inequality, one has

<a name="ch02c076">
2009-10-09-17-10 here
and this gives the bound (2.9). Finally,
equality holds here if and only if 
  ab=ac=bc
This is possible if and only if a=b=c, so
the box of maximum volume for a given 
surface area is indeed the cube.
2009-10-09-17-12 here




<a name="ch02c077">
2009-10-09-17-23
■　Exercise 2.9 solution


Consider three elements sequence
(2ab, 2ac, 2bc) Apply 
AM-GM inequality to (2ab, 2ac, 2bc)
get red part of eqn.AG067.
Box total area A is 2ab+2ac+2bc, this
explain eqn.AG067 right equality.
eqn.AG067 left equality is just 
re-arrange a,b,c terms.

<a name="ch02c078">
In eqn.AG067, we should read the black
text, that is left end term and right 
end term. Given condition is
"given surface area" A.
Ask to show "cube has the largest volume"

eqn.AG067 left end is volume abc to 2/3
power. Volume is variable.

<a name="ch02c079">
eqn.AG067 right end is surface area A
Area is constant.

Maximum volume is reached when AM=GM
condition is satisfied. Is this a cube?

Please see Equality GM=AM is easy.
Now read eqn.AG067 red text part. If
 2ab=2ac=2bc ---eqn.AG068
is true, then AM=GM is true. How can
we get a=b=c from ab=ac=bc ?
<a name="ch02c080">
Because ab=ac=bc, so
 ab*ac = bc*bc ---eqn.AG068
cancel one bc, get
 a*a = b*c ---eqn.AG069
similarly
 b*b = a*c ---eqn.AG070
 c*c = a*b ---eqn.AG071

<a name="ch02c081">
eqn.AG069 to eqn.AG071 right hand side 
are equal (see eqn.AG068), then their
left hand side are equal too,
 a*a = b*b = c*c
we get
 a = b = c
This is a cubic box which has maximum
volume with given surface area.
2009-10-09-17-46 here

<a name="ch02c082">
■　Compare volume with surface area?!

We know 
volume can not compare with area,
 area  can not compare with length,
since volume/area/length are different
physics quantities.
In Exercise 2.9, we compare
  volume^(2/3) th power with area A
volume is length third power length^3
volume^(2/3) is (length^3)^(2/3). 
(length^3)^(2/3)=(length)^(3*2/3). We
have length^2 (area!) for volume^(2/3)
Area volume^(2/3) can be compared with
total surface area A.
So Exercise 2.9 do not have dimension
trouble.
2009-10-09-17-52 stop



<a name="ch02c083">　Index begin　Index this file
2009-10-09-19-00 start
■　Exercise 2.10 problem statement
　　textbook page 34
（Åkerberg's Refinement）

Show that for any nonnegative real numbers
a1,a2,...,an and n≧2
one has the bound

<a name="ch02c085">
2009-10-09-19-12 here
In a way, this relation is a refinement of
the AM-GM inequality since the AM-GM ineq.
follows immediately by iteration of the
bound (2.30) To prove the recurrence (2.30)
one might first show that
  y(n-yn-1)=ny-yn≦n-1 ---eqn.AG072
for all y≧0. The key is then to make a 
wise choice of y.
2009-10-09-19-17 here





<a name="ch02c086">
2009-10-09-19-20 start
■　Exercise 2.10 hint
　　textbook page 234
If we set p=n and x=y-1 in Bernoulli's 
inequality eqn.2.24 we find that 
 y(n-yn-1)≦n-1 ---eqn.AG073
and equality holds only for y=1. If we 
<a name="ch02c086b">
now choose y such that
  yn-1=an/a_bar ---eqn.AG074
where a_bar=(a1+a2+...+an)/n  ---eqn.AG075
then we have
 n-yn-1=(a1+a2+...+an-1)/a_bar ---eqn.AG076

and easy algebra (was arithmetic) takes
one the rest of the way to the recursion
formula.

<a name="ch02c087">
As a sidebar, one should note that the
recursion also follows from the weighted
AM-GM inequality
  x1/nz(n-1)/n≦x/n + z*(n-1)/n ---eqn.AG077
by taking x=an ---eqn.AG078
 and 
 z=(a1+a2+...+an-1)/(n-1) ---eqn.AG079
2009-10-09-19-32 stop

<a name="ch02c087b">
2009-10-11-18-55 note
eqn.AG079 in textbook is y=....
LiuHH change to  z=....
because in eqn.AG074 y is defined
with different meaning.
2009-10-11-18-56 stop




<a name="ch02c088">　Index begin　Index this file
2009-10-09-20-50 start
■　Exercise 2.10 solution


Jacob Bernoulli inequality is
  1+n*x ≦ (1+x)n for all x∈[-1,∞)
           and all n=1,2,... ---eqn.2.24
Exercise 2.10 hint suggest use
  x=y-1  y∈[0,∞) ---eqn.AG080
Substitute eqn.AG080 into eqn.2.24 get
   1+n*(y-1) ≦ (1+y-1)n
or 1+n*y-n ≦ yn
or n*y-yn ≦ n-1
or y*(n-yn-1) ≦ n-1 ---eqn.AG072

<a name="ch02c089">
We are working with the following 
sequence a1,a2,...,an

Full sequence Arithmetic mean is
 a_bar=(a1+a2+...+an)/n  ---eqn.AG075
a_bar is AM_of_[n]_terms
after drop an, partial sequence
Arithmetic mean is
 z=(a1+a2+...+an-1)/(n-1) ---eqn.AG079

<a name="ch02c090">
Exercise 2.10 hint suggest choose y 
such that
  yn-1=an/a_bar ---eqn.AG074
eqn.AG072 use n-yn-1 , evaluate n-yn-1
below.
  n-yn-1=n-an/a_bar
  =n-an/[(a1+a2+...+an)/n]
  =n-n*an/(a1+a2+...+an)
  =n*{1-an/(a1+a2+...+an)}
  =n*{[a1+a2+...+an - an]/(a1+a2+...+an)}
  =n*{[a1+a2+...+an-1]/(a1+a2+...+an)}
  =[a1+a2+...+an-1]/[(a1+a2+...+an)/n]
  n-yn-1=[a1+a2+...+an-1]/a_bar ---eqn.AG076

<a name="ch02c091">
Put eqn.AG076 to eqn.AG072 get
  y*(n-yn-1) ≦ n-1 ---eqn.AG072
  y*[a1+a2+...+an-1]/a_bar ≦ n-1
  y*{[a1+a2+...+an-1]/[n-1]}/a_bar ≦ 1
that is
  y*{AM_of_[n-1]_terms}/{AM_of_[n]_terms} ≦ 1 ---eqn.AG081
<a name="ch02c092">
eqn.AG074 is yn-1, eqn.AG081 is y only.
now rise eqn.AG081 to (n-1) th power

<a name="ch02c093">
use eqn.AG074 for yn-1 get

see eqn.AG075, a_bar is AM_of_[n]_terms
eqn.AG083 become

eqn.AG084 and target eqn.2.30 are the
same. Problem solved.
2009-10-09-21-46 stop

<a name="ch02c094">
2009-10-09-21-48 
Next iteration is the following equation

2009-10-09-21-55 here
Continue this way, left hand side change
to a1*a2*...*an

<a name="ch02c095">
right hand side is the very beginning 
eqn.2.30 (not eqn.2.30 aux) it is
full sequence AM rise to n th power.
When iteration is done, whole equation 
take n th root, recover 
 GM=[a1*a2*...*an]1/n
and recover
 AM=[a1+a2+...+an]/n to first power.
Result is 
 GM ≦ AM
2009-10-09-21-59 stop



<a name="ch02c096">　Index begin　Index this file
2009-10-10-10-56 start
■　Exercise 2.11 problem statement
　　textbook page 34
（Superadditivity of the Geometric Mean）

Show that for nonnegative ak and bk
 1≦k≦n , one has

<a name="ch02c098">
2009-10-10-11-12 here
This inequality of H. Minkowski asserts 
that the geometric mean is a superadditive
function of its vector of arguments. Show
that this inequality follows from the AM-
GM inequality and determine the circum-
stances under which one can have equality.

<a name="ch02c099">
For a generic hint, consider the possibility
of dividing both sides by the quantity on
the right side. Surprisingly often one finds
that an inequality may become more evident
if it is placed in a "standard form" which
asserts that a given algebraic quantity is
bounded by one.
2009-10-10-11-21 here





<a name="ch02c100">
2009-10-10-11-34 start
■　Exercise 2.11 hint
　　textbook page 234

Following the hint, one finds from the
AM-GM inequality that

<a name="ch02c104">
2009-10-10-12-01 here
and the proof is complete. The division
device is decisive here, and as the
introduction to the exercise suggests,
this is not an isolated instance.
2009-10-10-12-03 here




<a name="ch02c105">
■　Exercise 2.11 solution


Prof. J. Michael Steele explained so 
clearly, what can I say? Done.
2009-10-10-12-20 stop



<a name="ch02c106">　Index begin　Index this file
2009-10-10-14-19 start
■　Exercise 2.12 problem statement
　　textbook page 35
（On Approximate Equality in the AM-GM Bound）

If the nonnegative real numbers a1,a2,...,
an are all approximately equal to a 
constant λ, then it is easy to check that 
the arithmetic mean A and the geometric 
mean G are approximately equal. There are
several ways to frame a converse to this 
observation, and this exercise considered
an elegant method first proposed by George
Polya.

<a name="ch02c107">
Show that if one has the inequality
  0 ＜ (A-G)/A ＝ ε ＜ 1 ---eqn.2.32
then one has the bound 
  ρ0≦ak/A≦ρ1 for all k=1,2,...,n ---eqn.2.33
where ρ0∈(0,1] and ρ1∈[1,∞) are the
two roots of the equation
  x/[ex-1]=(1-ε)n ---eqn.2.34
As figure 2.4 suggest, one key to this
result is the observation that the map
x→x/[ex-1] is monotone increasing
on [0,1] and monotone decreasing on
[1,∞)
2009-10-10-14-37 stop

<a name="ch02c108">
2009-10-10-17-00 start
■　Exercise 2.12 hint
　　textbook page 234

As figure 2.4 suggests, we have the bound
  f(x)=x/exp(x-1) ≦ 1 ---eqn.AG088
for all x ≧ 0. In fact, we used this 
bound long ago (page 24); it was the key
to Polya's proof of the AM-GM inequality.
If we now write
<a name="ch02c108b">
  ck=ak/A ---eqn.AG089
[[ 
2009-10-11-19-25 LiuHH note
eqn.AG089 is normalization. 
ak is original sequence elements.
'A' is original sequence Arithmetic Mean 
]]
then we have
<a name="ch02c109">
  c1+c2+...+cn=n ---eqn.AG090
and from this fact, we see that for each k
we have

<a name="ch02c111">
2009-10-10-17-31 here
Since  ε ＝ (A-G)/A ---eqn.AG092
and    ck=ak/A ---eqn.AG089
we have for all k=1,2,...,n that

<a name="ch02c112">
2009-10-10-17-45 here
Now the bounds
  ρ0≦ak/A≦ρ1 for all k=1,2,...,n ---eqn.2.33
are immediately from the definition of
ρ-, ρ+ together with the fact that f is
strictly increasing on [0,1) and strictly 
decreasing on (1,∞).

<a name="ch02c113">
This solution was given by Gabor Szego in
1914 in response to a question posed by 
George Polya. It is among the earliest
of their many joint efforts; at the time
Szego was just 19.
2009-10-10-17-51 stop






<a name="ch02c114">　Index begin　Index this file
2009-10-10-19-40 start
■　Exercise 2.12 solution


the inequality
  0 ＜ (A-G)/A ＝ ε ＜ 1 ---eqn.2.32
is always true for positive elements 
sequence. If allow elements to be zero
then if sequence has one zero element
 GM=(0*other_elements)^(1/n) = 0
<a name="ch02c115">
in this case ε ＝ (A-G)/A ＝ (A-0)/A ＝ 1 
On the other hand, if sequence all 
elements are constant, then AM=GM
in this case ε ＝ (A-G)/A ＝ (0)/A ＝ 0
Exercise 2.12 is interested in the case
of elements all be nearly constant.
Certainly exclude zero elements.

<a name="ch02c116">
Exercise 2.12 given condition is
  0 ＜ (A-G)/A ＝ ε ＜ 1 ---eqn.2.32
ask to prove the consequence
the bound 
  ρ0≦ak/A≦ρ1 for all k=1,2,...,n ---eqn.2.33
where ρ0∈(0,1] and ρ1∈[1,∞)

<a name="ch02c117">
Exercise 2.12 hint suggest consider
  f(x)=x/exp(x-1) ≦ 1 ---eqn.AG088
eqn.AG088 is same as 
  x≦exp(x-1) ---eqn.2.10 page 24
eqn.AG088 is drawn at fig2.4.

<a name="ch02c118">
What is 'x' in eqn.AG088 ?
Compare eqn.AG091 right hand side with
eqn.AG088, so x is ck, defined to be
  ck=ak/A ---eqn.AG089
ak is one element in target sequence
a1,a2,...,ak,...,an
<a name="ch02c119">
'A' in eqn.AG089 is Arithmetic mean 
 A = (a1+a2+...+an)/n ---eqn.AG094
n is total number of elements. Then 
we know ck is one element in normalized 
sequence.
SUM[ck] = SUM[ak/A] = SUM[ak]/A
SUM[ck] = n*A/A = n ---eqn.AG095
eqn.AG095 and 
  c1+c2+...+cn=n ---eqn.AG090
is same thing. 

<a name="ch02c120">
AM for normalized sequence ck is
  (c1+c2+...+cn)/n
AM for normalized sequence is n/n = 1
2009-10-10-20-18 here

GM for normalized sequence ck is
eqn.AG091 . 

eqn.AG091 write normalized GM which
is bounded by normalized AM=1
  f(x)=x/exp(x-1) ≦ 1 ---eqn.AG088

<a name="ch02c121">　Index begin　Index this file
eqn.AG091 'α＝' move ck out of
normalized GM .

eqn.AG091 '≦' apply eqn.AG088 (same 
as eqn.2.10 ) to cj , there are total
n-1 application, exclude ck.

<a name="ch02c122">
eqn.AG091 'β＝' convert ∏{exp(cj - 1)}
to exp(1 - ck). Why it is so?
Assume n=5, assume k=3, then
  ck*∏{exp(cj - 1)}
=
  c3*exp(c1-1)*exp(c2-1)*exp(c4-1)*exp(c5-1)
=
  c3*exp(c1-1+c2-1+c4-1+c5-1)
<a name="ch02c123">
= // red  terms are adding zero +/- c3
  c3*exp(c1+c2+c3+c4+c5-c3-4-1+1)
= // blue terms are adding zero +1/-1
  c3*exp(5-5  -c3+1)
= // normalized ck sum to 5
  c3*exp(1-c3)
= // since k=3
  ck*exp(1-ck)
<a name="ch02c124">
that is
  ck*∏{exp(cj - 1)} = ck*exp(1-ck) ---eqn.AG096
Above calculation shows eqn.AG091 'β＝' 
is correct.

eqn.AG091 'γ＝' turn exp(1-ck) from
numerator to denominator exp(ck-1).

eqn.AG091 'δ＝' write eqn.AG091 right 
side in eqn.AG088 form.

<a name="ch02c125">
Up to here, eqn.AG091 is cleared.
eqn.AG091 use ck, ck is normalized
sequence, eqn.AG089 relate ck with
original sequence elements ak
From eqn.AG091 use eqn.AG089 recover ak
we get eqn.AG093 inequality part.

<a name="ch02c126">
eqn.AG093 equality part use eqn.AG092.
Since a1*a2*...*an = GM^n
eqn.AG093 middle part is GM^n/AM^n 
or (GM/AM)^n. But GM/AM is (1-ε) because
 ε ＝ (A-G)/A  ---eqn.AG092
 1-ε＝1-[(A-G)/A]＝(A-A+G)/A=G/A ---eqn.AG097

<a name="ch02c127">
Up to here cleared eqn.AG093 
eqn.AG093 is same as eqn.2.34
  x/[ex-1]=(1-ε)n ---eqn.2.34
Please go to fig2.4, click [Plot2]
  x/[ex-1] is blue curve.
  (1-ε)n is horizontal red line.

<a name="ch02c128">　Index begin　Index this file
eqn.AG093 say function value (blue curve)
is greater than or equal to (1-ε)n (red
line)
When ε approach to zero, (1-ε)n approach
to one. All ak/A squeeze to narrow space
and this is data nearly constant situation.

Exercise 2.12 spend longer time to digest.
2009-10-10-21-46 stop
2009-10-11-19-58 done proofread
2009-10-11-20-26 done spelling check

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

<a name="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