Inequalities
An inequality is defined as “a statement of relative size or order of two
objects” [1]. Inequalities arose from the question what occurs when the two
objects or equations are not equal to each other. People use inequalities to
measure or compare two objects. Inequalities are applied in mathematics when
it is difficult to constrain quantities that are difficult to express in a formula. In
inequalities, there exists usually no one solution but a set of solutions. There are
basically two domains of inequalities. The first domain of inequalities is
connected to the shape of a triangle while the second domain is associated with
mean. We will examine the first domain related to the triangle.
Probably discovered centuries before, the person attributed with proving
the triangle inequality is Euclid.
Euclid of Alexandria as he was known during his time so as not to be confused
with Euclid of Megara who was a famous philosopher was “the most prominent
mathematician of antiquity best known for his treatise on mathematics The
Elements” [2]. Because little is known about Euclid many assumptions are made
about his life and work. It is thought that he lived during the lifetime of the first
Ptolemy and lived in Alexandria, Egypt.
One such assumption made about
Euclid’s life is that he must have been a pupil of Plato’s Academy in Athens due
to his vast knowledge of Eudoxus’ and Theaetetus’ geometry. Although The
Elements is Euclid’s most notable work, other works of Euclid’s include Data, On
Divisions, Optics, and Phaenomena. The following works of Euclid’s have all
been vanished: Conics, Book of Fallacies, Surface Loci, Porisms, and Elements
of Music.
Euclid’s most famous work The Elements consists of thirteen books. The
importance of The Elements may be contributed to the fact it is the first book that
1
creates an axiomatic system. The logical reasoning of the mathematics
demonstrated in The Elements was the first of its kind and later was extremely
influential in other areas such as science. The Elements is regarded as one of
the most successful textbooks ever written with the number of editions in print
second only to the Bible. Euclid commenced the work with five postulates and
definitions. It is thought Euclid used earlier resources when writing The Elements
because numerous definitions are introduced which were never applied
previously such as oblong, rhomboid, and rhombus. The subject of books one
through six is plane geometry. Topics of plane geometry discussed in books one
and two are “basic properties of triangles, parallels, parallelograms, rectangles,
and squares” [2]. The triangle inequality is discussed and proved in book one.
The remaining subject matter of books three through thirteen include properties
of circles, number theory, the Euclidean algorithm for computing the greatest
common divisor of two numbers, irrational numbers, and three dimensional
geometry. One of the greatest attributes of The Elements is how understandable
Euclid states the theorems and their proofs.
The triangle inequality is discussed under Proposition 20 in Book 1 of
Euclid’s The Elements. The following is a recreation by David Joyce a Professor
of Mathematics as Clark University of the triangle inequality as it appears in
Euclid’s The Elements [3].
D
A
B
C
In any triangle the sum of any two sides is greater than the remaining one.
Let ABC be a triangle.
We say that in the triangle ABC the sum of any two sides is greater than the
remaining one, that is, the sum of BA and AC is greater than BC, the sum of AB
and BC is greater than AC, and the sum of BC and CA is greater than AB.
Choose a point D such that DA equals to CA and then draw BD. Join DC.
Since DA equal AC, therefore the angle ADC also equals the angle ACD.
Therefore the angle BCD is greater than the angle ADC.
2
Since DCB is a triangle having the angle BCD greater than the angle BDC, and
the side opposite the greater angle is greater, therefore DB is greater than BC.
But DA equals AC, therefore the sum of BA and AC is greater than BC.
Similarly we can prove that the sum of AB and BC is also greater than CA, and
the sum of BC and CA is greater than AB.
Therefore in any triangle the sum of any two sides is greater than the remaining
one.
There are several proofs of the triangle inequality which we will progress
through during this paper. The first one to be discussed is the proof of the
triangle inequality for real numbers.
Triangle Inequality for Real Numbers
For any a, b,
ab  a  b
We know ab  a b is true because of these four cases:
Since ab  ab  a b , then 2ab  2 a b
1. a  0, b  0. Then ab  ab  a b
Thus, a 2  2ab  b 2  a 2  2 a b  b 2
2. a  0, b  0. Then ab  -(ab)  a(-b)  a b
3. a  0, b  0. Then ab  -(ab)  (a)b  a b
4. a  0, b  0. Then ab  ab  (a)( b)  a b
2
Because a  a 2 , then a 2  2ab  b 2  a  2 a b  b
which can be written as (a  b) 2  ( a  b ) 2 .
2
We can say that a  a 2 is true because
a
2
 a 2 and if we take the positive square
root,
a
2
 a 2 ; however, a  0 so
a  a 2 is true.
Thus,
(a  b) 2  ( a  b ) 2 .
Again since a  a 2 , then a  b  a  b .
This proves the triangle inequality for real numbers.
3
From real numbers, the natural progression is to move into the proof of the
triangle inequality for complex numbers. Below is the proof:
Triangle Inequality for Complex Numbers
The triangle inequality states if z and w are complex numbers then z  w  z  w
2
z  w  ( z  w)( z  w)
Because z z  z
2
by the following:
Let z = a + bi, then
z  a  bi
Thus z z  (a  bi)( a  bi)  a 2  b 2
But z  a 2  b 2 so z
2
Thus z z  a 2  b 2  z
2
z  w  ( z  w)( z  w)
Because Re( z )  z for all z  C
2
z  w  z z  ww  z w  w z
2
2
2
2
2
2
z  w  z  w  z w  ( z w)
z  w  z  w  2 Re( z w)
 a2  b2
2
By substitution of ( z  w) for ( z  w) because of the
following:
Let z  a  bi and w  c  di
Then zout
 wthe
 (aabove
 bi)  (c  di)  (a  c)  i(b  d )
Foiling
So ( z  w)  (a  c)  i(b  d )
Combine the last two terms
Also z  a  bi and w  c  di and so
because z  z  2 Re z . This is true because of
zw
 (a  bi)  (c  di)  (a  c)  i(b  d )  ( z  w)
the
following:
Let z  a  bi and z  a  bi
z  z  a  bi  a  bi  2a  2 Re( z)
2
2
2
2
2
2
z  w  z  w  2 zw
zw  z  w 2z w
Because Re( z )  z for all
z C
z w can be rewritten as z w because of the following :
zw  z w
zw
zw
zw
zw
2
 zw( zw)
2
 zw z w
2
 z zww
2
 z
2
w
2
Hence taking the square root of each sides
zw  z w
4
2
2
2
zw  z  w 2z w
w can be replaced with w because of the following :
Let z  a  bi
z  a 2  b 2 and z  a  bi so
z  a 2  (b) 2  a 2  b 2  z
2
z  w  ( z  w )2
Since z  w  0 and z  w  0
Take the square root of both sides yields z  w  z  w
(All of the above information came from Heriott Watt University Scholar Math
website [4])
Both of these proofs are very similar.
Each proof deals with
understanding that the absolute value of a number whether real or complex is
greater than or equal to that same number. However, the complex number proof
is a little more involved because one must also know and understand the
properties of complex numbers and their conjugates.
So far we have discussed with the triangle inequality only in a one
dimensional space. Now we will look at the triangle inequality in R n space. In
R n space, the triangle inequality is known as the Cauchy-Schwarz Inequality
in R n named after two famous mathematicians Augustin Louis Baron de Cauchy
and Herman Amandus Schwarz.
Augustin Cauchy (1789 - 1857) was born in Paris, France during the
French Revolution.
5
At an early age, Lagrange took an interest in young Cauchy’s mathematical
education and urged Cauchy’s father to enroll him in Ecole Centrale du Pantheon
to study languages. After Ecole Centrale du Pantheon, Cauchy entered Ecole
Polytechnique and then the engineering school Ecole des Ponts et Chaussees.
After graduating engineering school, Cauchy began work on the Ourcq Canal
project. For Cauchy’s first job in 1810, he worked “on port facilities for
Napoleon’s English invasion fleet” in Cherbourg [5]. In 1811, Cauchy proved
“that the angles of a convex polyhedron are determined by its faces” [5]. He left
his Cherbourg in 1813 for health reasons and returned to Paris where he applied
for several positions none of which he was offered until 1815 when he was
offered the assistant professorship of analysis at his alma mater Ecole
Polytechnique. During Cauchy’s period of unemployment, he devoted himself to
his mathematical research. From Ecole Polytechnique, Cauchy took a position at
the College de France in 1817. There he “lectured on methods of integration
which he had discovered, but not published, earlier” [5]. As well as defining an
integral, “Cauchy was the first to make a rigorous study of the conditions of
convergence of infinite series” [5]. Next, the deposed king Charles X of France
offered Cauchy the position of tutor to his grandson, the duke of Bordeaux. This
position allowed Cauchy to travel extensively. In exchange for his services, the
Charles X made Cauchy a baron. In 1838 upon returning to Paris, Cauchy
reclaimed his position at the Academy, but he was not allowed to teach there
because he refused to a take an oath of allegiance. After losing the position of
mathematics chair at the College de France, Cauchy’s mathematical output
included “work on differential equations and applications to mathematical
physics” [5]. Cauchy also wrote a four volume book entitled Exercises d’analyse
et de physique mathematique which discussed mathematical astronomy. Once
Louis Phillippe was overthrown, Cauchy returned to his position at Ecole
Polytechnique. An argument involving “a priority claim regarding a result on
6
inelastic shocks” with Duhamel overshadowed Cauchy’s remaining years [5]. On
May 23, 1857, Cauchy died rather calmly with is family by his side. Cauchy
wrote 789 mathematical papers and is associated with many items in
mathematics including the famous inequality we are studying. Some of these
items are the theory of series, calculus with the aid of limits and continuity, and
the “Cauchy-Kovalevskaya existence theorem that offers a solution of partial
differential equations” [5].
The second mathematician in our inequality is Hermann Schwarz (1843 1921).
He studied at Gewerbeinstitut, which is later to be known as the Technical
University of Berlin, with the goal of obtaining a chemistry degree. Not long after
coming to Gewerbeinstitut, Schwarz was convinced to change his program of
study to mathematics. He continued to study at Gewerbeinstitut until he attained
his doctorate. Schwarz researched minimal service areas which is a typical
problem of the calculus of variations. In 1865, he revealed a “minimal surface
has a boundary consisting of four edges of a regular tetrahedron,” which is
known today as Schwarz minimal surface [6]. Upon obtaining his certificate to
teach, Schwarz acquired positions at the University of Halle, Eidgenossische
Technische Hochschule, then at Gottingen University, and finally at the
University of Berlin in 1892. While at these various institutions, Schwarz
researched “the subjects of function theory, differential geometry, and the
calculus of variations” [7]. Another area Schwarz researched was conformal
mappings. His achievement in this area was connected with the Riemann
mapping theorem. Schwarz’s process incorporated the Dirichlet problem which
mapped “polygonal regions to a circle” [6]. By performing this method, Schwarz
was then able to provide a thorough proof of the Riemann mapping theorem.
Schwarz’ most significant work was answering the “question of whether a given
7
minimal surface really yields a minimal area” [6]. Another great achievement of
Schwarz’ was the Schwarz function. In this work, Schwarz characterized “a
conformal mapping of a triangle with arcs of circles as sides onto the unit disc”
[6]. The Schwarz function was one of the first examples of an automorphic
function which was a helped Klein and Poincare to originate the theory of
automorphic functions. According to O’Connor and Robertson, it is not surprising
that Schwarz discovered a “special case of the general result now known as the
Cauchy-Schwarz inequality.” A great deal of Schwarz’ work is characterized by
analyzing specialized and narrow questions and applying much more general
methods for solving them.
A third mathematician’s can be added to the Cauchy-Schwarz inequality.
His name is Viktor Bunyakovsky born in the Ukraine in 1804 (1804 – 1889).
Bunyakovsky was a pupil of Cauchy’s in Paris, where he remained until 1826.
Upon returning to St. Petersburg, Bunyakovsky became a central figure in
bringing mathematics to Russia and the Russian Empire. Bunyakovsky brought
to Russia the Cauchy’s theories as well as probabilistic theories of the French.
Bunyakovsky taught mathematics and mechanics at several institutions in St.
Petersburg including the First Cadet Corps, the Communications Academy, and
the Naval Academy. From 1846 to 1880, Bunyakovsky held a position at the
University of St. Petersburg. Bunyakovsky did most of his research at the St.
Petersburg Academy of Sciences. He held the position of vice-president of the
Academy of Science until his death. Although not always credited, Bunyakovsky
is “best known for his discovery of the Cauchy-Schwarz inequality” some twenty
five years prior to Schwarz’ own discovery [8]. Other work of Bunyakovsky
includes “new proof of Gauss’ law of quadratic reciprocity” which is an area of
number theory, applied mathematics, and geometry [8]. Bunyakovsky’s work
Foundations of the mathematical theory of probability is credited with providing
8
the foundation of Russian probabilistic vocabulary. Over his life, Bunyakovsky
was published 150 times for his work in mathematics and mechanics.
Before tackling the Cauchy-Schwarz inequality, we need to understand
some of the vocabulary involved. The dimension in which the Cauchy-Schwarz
inequality is used is R n space also known as Euclidean n-space. In R n space
there are more than three dimensions. According to Anton and Rorres’
Elementary Linear Algebra [9], the dot product in R 2 or R 3 is referred to the
Euclidean inner product in R n which is defined for two vectors ( u  u1 , u 2 ,...., u n )
and v  v1 , v2 ,...., vn as u  v  u1v1  u 2 v2  ....  u n vn .
Another term is the
Euclidean norm or Euclidean length of a vector u = ( u1 , u 2 ,...., u n ) in R n
is u 
1
(u  u ) 2
 (u12 , u 2 2 ,..., u n 2 ) . There are some properties of length in R n that
we should also know. According to Elementary Linear Algebra [9], these
properties hold true if u and v are vectors in R n and k is any scalar, then:
(a) u  0
(b) u  0 if and only if u = 0
(c) ku  k u
(d) u  v  u  v .
The last property should look familiar because it is the triangle inequality. A
visualization of the triangle inequality with respect to vector length is:
v
u+v
u
9
Triangle Inequality for Euclidean n space
A proof of the triangle inequality for Euclidean n space is shown below and it
works in generality.
uv
2
 (u  v)  (u  v)  (u  u )  2(u  v)  (v  v)
uv
2
 u
2
 2(u  v)  v
uv
2
 u
2
 2u v  v
uv
2
 u
2
2u v  v
uv
2
 ( u  v )2
2
2
2
uv  u  v
Once we have found the norm or lengths of a certain object or vector, we
can now define a distance between two objects or vectors as follows. The
Euclidean distance between two points, u and v, in R n is defined
as d (u, v)  u  v  (u1  v1 ) 2 , (u 2  v 2 ) 2 ,..., (u n  v n ) 2 . The properties of distance
in R n are similar to properties of length. These following properties are true if
u,v, and w are vectors in R n and k is any scalar:
(a) d(u,v)  0
(b) d(u,v) = if and only if u = v
(c) d(u,v) = d(v,u) and
(d) d(u,v)  d(u,w) + d(w,v).
Once again, the last property should look familiar because it is the triangle
inequality only this time it is applied to distance. The proof works in complete
generality whenever we have a norm with the properties a – d.
We can then define the distance and the triangle inequality will follow:
d (u , v)  u  v  (u  w)  ( w  v)
d (u , v)  u  w  w  v
d (u , v)  d (u , w)  d ( w, v)
A visualization of this is also similar to the one for length with the exception that
the sides of the triangle are segments and not rays.
10
u+v
v
u
Both u  v  u  v
and d(u,v)  d(u,w) + d(w,v) generalize the results of
triangle inequality in Euclidean geometry. For length, u  v  u  v simply may
be restated that the sum of two sides of the triangle are greater than or equal to
the length of the third side. As for distance, d (u,v)  d(u,w) + d(w,v) may be
summed up by stating that the shortest distance between two points is a straight
line.
Now we are ready for the Cauchy-Schwarz Inequality in R n which states
that if u = ( u1 , u 2 ,...., u n ) and v= ( v1 , v2 ,...., vn ) are vectors in R n , then u  v  u v .
In other words, the absolute value of the inner product of u and v is less than or
equal to norm or length of u multiplied by the norm or length of v which is
symbolized
by u1v1  u 2 v 2  .....  u n v n  (u1  u 2  ....  u n
2
2
1
2 2
)
 (v1  v 2  ....  v n
2
2
1
2 2
)
.
The
proof of the Cauchy-Schwarz Inequality is shown below.
Cauchy-Schwarz Inequality
If u  0, then u  u  u  v  0 so the two sides are equal to each other.
Assume u  0 . Let a  u  u , b  2u  vand c  v  v and let t be any real number.
By the positivity axiom, the Euclidean inner product of any vector with itself is
always positive.
0  [(tu  v)  (tu  v)]  (u  u)t 2  2(u  v)t  (v  v) .
Using substitution, we can say that (u  u)t 2  2(u  v)t  (v  v)  at 2  bt  c .
The inequality implies that the quadratic polynomial at 2  bt  c has either no real
roots or a double root.
Thus, the discriminant must satisfy the condition of b 2  4ac  0 .
Expressing the coefficients a, b, and c in terms of vectors u and v gives
[2(u  v)]2  4(u  u)(v  v)  0 which may be simplified to 4(u  v) 2  4(u  u)(v  v)  0 .
This
statement
may
also
be
rewritten
2
2
as 4(u  v)  4(u  u)(v  v) which may be simplified to (u  v)  (u  u)(v  v) .
11
Taking
1
(u  u ) 2
the
square
root
of
both
sides
and
using
the
fact
that
 u yields u  v  u v .
From the Cauchy-Schwarz inequality which deals with vectors of a finite
number of components, we can now move into a discussion of a special type of
vector space called sequence space. A sequence space is a vector with an
infinite number of components which is represented by v = (v1 , v2 ,.....) that has no
ending. There are two inequalities which extend the Cauchy-Schwarz inequality
into sequence space.
The first of these inequalities is the Hölder inequality named after the
German born mathematician, Otto Hölder (1859 – 1937).
Hölder’s education includes studying at the polytechnic in Stuttgart and later at
the University of Berlin. While at the University of Berlin, Hölder became
interested in algebra because of the influence of Kronecker. In 1882, Hölder
gave his dissertation on “analytic functions and summation procedures by
arithmetic means” at the University of Tubingen [10]. In 1884, Hölder took the
position of lecturer at Gottingen. At Gottingen, Hölder discovered the inequality
which is now named for him, began his “work on the convergence of the Fourier
series,” and developed a fascination for the area of group theory” [10]. Hölder’s
lifetime work includes the Jordan-Hölder theorem which deals with the
“uniqueness of the factors groups in a composition series,” contributions to group
theory, the creation of inner and outer automorphisms.
12
A proof of Hölder’s inequality is given below as reconstructed by a
Professor Gabriel Nagy at Kansas State University who uses the method
induction [11].
Hölder’s Inequality states:
Let a1 , a2 ,.... , an , b1 , b2 ,. . ., bn be nonnegative numbers.
Let p, q > 1 be real
1
 n
p
1 1
 a p
a
b

numbers with the property   1 . Then
 j j  j 
p q
j 1

 j 1
Moreover,
one
has
equality
only
when
the
p
p
p
p
(a1 ,...., an ) and (b1 ,...., bn ) are proportional.
n
1
q

 bjq  .

 j 1


sequences
n
The proof of Hölder’s Inequality is as follows.
The case n = 1 is trivial.
Case n = 2
Assume (b1 , b2 )  (0,0) . Otherwise everything is trivial. Define the number
b1
r
.
1
(b1q  b2 q ) q
Notice that r  0,1
b1
The proof of why r 
1
q q
 1 follows as:
(b1q  b2 )
We were given that b1 and b2 are nonnegative numbers. Let us assume that the
second positive number b2
equals 0.
Thus the denominator
1
q
1
q q
b1
 1 . However, b2 must be a positive
b1
b1
 1.
number making 1 the greatest value r can be. Therefore, r 
1
becomes (b1  0)  (b1 )  b1 . So r 
q
(b1q  b2 q ) q
We also have
1
q q
b2
1
q q
 (1  r ) .
(b1q  b2 )
Notice also that, upon dividing
by (b1q
1
q q
 b2 ) , the desired inequality
13
1
p p
1
q q
1
q q
1
p p
a1b1  a2b2  (a1  a2 ) (b1  b2 ) reads a1r  a2 (1  r )  (a1  a2 ) .
It is obvious that this is an equality when a1  a2  0 . Assume (a1 , a2 )  (0,0) , and
p
q
p
1
q q
set up the function f (t )  a1t  a2 (1  t ) , t  0,1,
We now apply the Lemma of the following:
1 1
Let p, q >1 be such that   1 , and let u and v be two nonnegative numbers,
p q
at least one being non-zero. Then the function f: [0,1]  Real numbers defined
1
1
q q
 up q
by f (t )  ut  v(1  t ) , t  0,1 has a unique maximum point at s   p
 .
p
 u  v 
1
p p
The maximum value of f is max f (t )  (u p  v ) .
t0,1
The proof of this Lemma is as follows:
If v  0, then f (t )  tu, t  [0,1] ( with u  0), and in this case the Lemma is trivial.
1
q q
Likewise, if u  0, then f (t )  v(1  t ) , t  [0,1] (with v  0), and using the inequality
1
q q
(1  t )  1, t  (0,1]
We immediately get
f (t )  f (0), t  (0,1], and the Lemma again follows.
For the remainder of the proof we are going to assume that u , v  0 . We
concentrate on the first assertion. Obviously f is differentiable on (0, 1), so the
“candidates” for the maximum points are 0, 1, and the solutions of the equation
f ' (t )  0 .
1
 up q
Let s be defined as in s   p
 , so under the assumption that u , v  0 , we
p
 u  v 
clearly have 0<s<1. We are going to prove first that s is the unique solution in
(0, 1) of the equation f ' (t )  0 . We have
1
1
q q
1
f ' (t )  u  v  (1  t )
q
1
 q  t q 1
so the equation f ' (t )  0 reads
 t p
 , t  (0,1)
 u  v
1 t q 


q
14
1
 t p
 0
u  v
1 t q 


q
1
u  t q

v  1  t q
p



p
 tq
u
   
q
v
1 t
tq 
u
 
v




p

up
p
up vp
u
1   
v
which gives t = s.
We have now shown that the “candidates” for the maximum point are 0, 1, and s.
Let us show that s is the only maximum point. For this purpose, we go back to
1
1
q q
1
p
 t
 , t  (0,1) and we observe that f ' is
 q  t q 1  u  v
1 t q 


also continuous on (0,1). Since
lim f ' (t )  u  0 and lim f ' (t )   ,
q
1
f ' (t )  u  v  (1  t )
q
t 0
t 1
and the equation f ' (t )  0 has exactly one solution in (0,1), namely s, this forces
f ' (t )  0, t  (0, s) and f ' (t )  0, t  ( s,1) . This means that f is increasing on [0,s]
and decreasing on [s,1], and we are proven that the maximum value of f is then
given by
max f (t )  f (s)  (u
t0,1
p
1
p p
v ) .
1
q q
1
p p
The Lemma immediately gives us a1r  a2 (1  r )  (a1  a2 ) .
Let us
examine when equality holds. If a1  a2  0 , the equality obviously holds, and in
this case (a1 , a2 ) is clearly proportional to (b1 , b2 ) . Assume (a1 , a2 )  (0,0) . Again
by
the
above
lemma,
we
know
that
equality
holds
1
q q
p
1
p p
in a1r  a2 (1  r )  (a1  a2 ) , exactly when
p
15
1
q
1

a
r p 1
 ,
p
a

a
2 
 1

p
equivalently
b1q
b1q  b2 q

that
b1
is
(b1  b2 )
q
a1 p
a1 p  a 2 p
q
 a1 p
 p

p
 a1  a 2 
q
1
q
b2 q
. Obviously this forces
b1q  b2 q

or
a2 p
a1 p  a 2 p
,
so indeed (a1 p  a2 p ) and (b1q  b2 q ) are proportional.
Having proven the case n = 2, we now must show with the proof of:
The implication: Case n=k  Case n = k+1
Start with two sequences (a1 , a2 ,....., ak , ak 1 ) and (b1 , b2 ,....., bk , bk 1 ) .
Define the numbers
1
p
1
q
 k
 k
a    a j p  and b    b j q  .



 j 1


 j 1

Using the assumption that the case n = k holds, we have
1
p
1
q
 k
 k
 a p    b q   a b  ab  a b .
a
b

 j j   j    j  k 1 k 1
k 1 k 1
j 1

 j 1
 j 1

Using the case n = 2 we also have
k 1
1
p p
ab  a k 1bk 1  (a p  a k 1 )  (b q  bk 1
combining with
1
1
p
k 1

j 1
 j 1
k
1
 k 1
p
)  a j p 



 j 1
1
q q
q
 k 1
  b jq  ,

 j 1


so
1
q
 k
   b j q   a k 1bk 1  ab  a k 1bk 1 we see

 j 1



 a j b j    a j p 

1
p
1
q

   b j q  holds for n = k+1.

 j 1
j 1


 j 1

Assume now we have equality. Then we must have equality in both
n
that the desired inequality
k 1

k

j 1
1
p
in ab  a k 1bk 1  (a p  a k 1
one
 a j b j    a j p 
n
1
q
 k
   b j q   a k 1bk 1  ab  a k 1bk 1

 j 1



 a j b j    a j p 
j 1
n
hand,
and
1
1
 k 1
 p  k 1 q  q
p
q

)  (b  bk 1 )   a j
  b j  .



 j 1


 j 1

the
equality
1
p p
1
q q
On the
in
16
1
 k
p
 a p
a
b

 j j  j 
j 1

 j 1
k 1
1
q
 k
   b j q   a k 1bk 1  ab  a k 1bk 1

 j 1


forces
(a1 p , a2 p ,...., ak p ) and (b1q , b2 q ,....., bk q ) to be proportional (since we assume the
case
n
=
k).
On
the
other
hand,
the
equality
in
1
p
1
q
 k 1
 k 1
ab  a k 1bk 1  (a p  a k 1 )  (b q  bk 1 )    a j p     b j q  forces



 j 1


 j 1

1
p p
(a p , ak 1 p ) and (b q , bk 1q ) to
1
q q
be
proportional
(by
 k


 k
Since a p    a j p  and b q    b j q  ,



 j 1


 j 1

the
it
case
n
is
=
2).
clear
that (a1 p , a2 p ,...., ak p , ak 1 p ) and (b1q , b2 q ,....., bk q , bk 1q ) are proportional.
1
 n
p
Thus Hölder’s inequality of  a j b j    a j p 


j 1

 j 1
generality.
n
1
q

   b j q  has been proven in

 j 1


n
We use Hölder’s inequality to prove the second inequality that extends the
Cauchy-Schwarz inequality into sequence space.
This inequality is the
Minkowski inequality named after the mathematician, Hermann Minkowski (1864
– 1909). The Minkowski inequality is the triangle inequality represented in
Lp space.
17
Minkowski was born the second son of Lewin Minkowski and Rachel Taubmann.
When he was eight, Minkowski and his family moved from Russia back to
Konigsberg, Germany where his father’s business was. Upon returning to
Konigsberg, Minkowski began his studies at the Gymnasium. It was here that
Minkowski’s interest in mathematics was first discovered. Minkowski studied at
the University of Konigsberg and the University of Berlin. It was at the University
of Konigsberg, he became friends with Hilbert. Early in his university studies,
Minkowski became fascinated by quadratic forms. So when the Academy of
Sciences revealed in 1881 “that the Grand Prix for mathematical science to be
awarded in 1883 would be for a solution to the problem of the number of
representations of an integer as the sum of five squares,” Minkowski was
intrigued [12]. The problem stemmed from Eisenstein’s 1847 “formula for the
number of such representations;” however, he never gave a proof of the
conclusion [12]. When the problem was announced, the Academy of Sciences
was ignorant of the fact that a mathematician by the name of Henry Smith had
already solved it. At the age of eighteen, Minkowski recreated Eisenstein’s
quadratic forms theory and produced a solution to the problem, launching
Minkowski’s mathematical career. The Academy of Sciences decided to award
the prize to both mathematicians. In 1885, Minkowski received his doctorate
which was an extension of the Grand Prix problem. Minkowski held positions at
several institutions which include the University of Bonn, Konigsberg,
Eidgenossicsche Polytechnikum Zurich, and finally the University of Gottingen. It
was at the University of Gottingen that Minkowski was reunited with Hilbert who
had had the position there created especially for him. Minkowski remained at the
18
University of Gottingen until his death.
One of Minkowski’s greatest
achievements was his idea of space-time continuum which influenced Einstein’s
theory of relativity significantly. Until Minkowski, space and time were thought to
be independent of each other; however, Minkowski joined the two together in four
dimensional space-time continuum. Another triumph of Minkowski was his
‘geometry of numbers,’ which led to his work “on convex bodies and to questions
about packing problems, the ways in which figures of a given shape can be
placed within another given figure” [12]. Sadly, Minkowski died from a ruptured
appendix in 1909 at the age of 44.
Now we are ready to tackle the proof of the Minkowski inequality in its
most general form for R n space according to Planet Math [13].
For p = 1 the result follows immediately from the triangle inequality, so we may
assume p>1.
We have
a k  bk
p
 a k  bk a k  bk

p 1

 a k  bk a k  bk
p 1
by the triangle
inequality. Therefore we have
p
a k  bk
 a k a k  bk
p 1
 bk a k  bk
p 1
when we distribute
ak  bk
p 1
to
ak
and bk .
Set q 
p
1 1
. Then   1 , so by the Hölder inequality we have
p 1
p q
1
n
 ak
k 0
a k  bk
p 1
n

pp
   a k    a k  bk

 k 0
  k 0
n
1
n
 bk
k 0
a k  bk
p 1
n
pp
   a k  bk
 
  k 0
 n
   bk
 k 0
1
( p 1) q  q



1
( p 1) q  q



Adding these two inequalities, dividing by the factor common to the right sides of
both, and observing that (p-1)q = p, we have
1


  a k  bk p 


 k 0

n
1
q
  ak
n

k 0
Finally, observe that 1 

 bk a k  bk

  a k  bk

 k 0
n
1
p q



p 1
1

 n
pp
   a k     bk

 k 0
 k 0

n
1
pp



1 1
 , and the result follows as required.
q p
19

  a k  bk

 k 0
n
  ak
n
1

p p




k 0

 bk a k  bk
p 1
1
 n
   a k
 k 0
1
q

  a k  bk p 


 k 0

n
 n
pp
    bk


 k 0

1
pp



Thus, the Minkowski inequality is proven in its most general form in R n space of
1
 n
  a k  bk

 k 0
 n
pp
    ak


 k 0

1
 n
pp
    bk


 k 0

1
pp
 .


The proof of the Minkowski inequality given above can easily be proven in
the same manner for Lp space. In order to understand what Lp space is we need
familiarize ourselves with some vocabulary. Lp space can be defined as “spaces
of p-power integrable functions, and corresponding sequence spaces” [14].
1
According to Mathworld, the Lp norm of a function f is
f
Lp

 f

X
p
p
 on a


1
p

p
measure space X. Lp functions are function for which   f  converges. A


X

measure space is a measurable space possessing a nonnegative measure.
Examples of measure spaces include -dimensional Euclidean space with
Lebesgue measure and the unit interval with Lebesgue measure (i.e., probability)
as defined by Mathworld [15]. Norms if we recall are the lengths of vectors.
Thus, by utilizing counting measure, Lp norm is a generalization of vector pnorms. A counting measure is defined as “a measure u is called a counting
n if A has exactly n elements
measure on X if u ( A  B)  
” [16]. A p-norm is
 otherwise
“defined as x
 ( x1
p
p
1
p p
 ....  xn ) , p  1, x  R n ” and denoted by “  p ” [17].
The three most common p-norms are 1-norm, 2-norm, and  -norm.
norms can be defined as the following:
1 – norm: x  x1  x2  .....  xn
These
1
2 – norm:
x
 - norm:
x
2


2
x1  x2
2
 . .. . .  x n
2
 xT x
 max xi
1i 
20
We are already familiar with the 2 – norm because it is known as the Euclidean
vector norm which generates the “Euclidean distance between any two vectors”
[17]. In finite dimensional spaces, the three p-norms above as well as all pnorms are equivalent. Lp space is an example of a Banach space as long as p ≠
2 that has applications in finite element analysis. A special case of Lp space is
when p = 2 which represents a Hilbert space. Another unique case of Lp space
is l p space. In l p space, the measure space is discrete and “the measure used
in the integration in the definition is a counting measure.” The definition of l p (S)
is a “set of sequences x  xi , i in S, for which the quantity x
p

 xi
p
is
iS
finite” [14].
We now turn to the second domain of inequalities which focuses on
means. There are several types of means: geometric, arithmetic, and harmonic
just to name a few. Their applications can be seen in the areas of statistics,
genetics, the stock market, computer science, etc.
The first inequality we will discuss is the arithmetic mean versus the
geometric mean (AM-GM inequality). Before we jump into the inequality itself, let
us define arithmetic mean and geometric mean. The arithmetic mean is
identified as “sum of all the members of a set divided by the number of items in
the set” while the geometric mean can be described as “the product of all the
members of the set, raised to a power equal to the reciprocal of the number of
members” [18] and [19]. Mathematically both means may be represented by the
following:
x  x 2  ......  x n
The arithmetic mean which is denoted by a  1
, where
n
x1 , x2 ,......, xn are the members of a set.
The geometric mean which is denoted by
g  n x1  x2  .....  xn , where
x1 , x2 ,......, xn are the members of a set.
Now let us attempt the arithmetic mean versus the geometric mean
inequality. The person credited with originally proving this inequality can be
traced to Augustin Louis Cauchy who we have already discussed earlier in his
paper, Cours d'analyse.
The arithmetic mean versus the geometric mean inequality states for any positive
x  x 2  ..... .  x n
real numbers, x1 , x2 ,......, xn , a  1
 g  n x1  x2  .....  xn and if
n
x1  x2  ......  xn n
x1  x2  ......  xn then
 x1  x2  .....  xn .
n
21
The proof of the arithmetic mean vs. the geometric mean as reconstructed by
Korovkin in Inequalities [20] follows:
Using g  n x1  x2  .....  xn
If we divided both sides by g, then
x
x x x
g n x1  x2  .....  xn

 n 1  2  3 ,......, n
g
g
g g g
g
1 n
x
x1 x 2 x3

  ......  n
g g g
g
This can be rewritten as
x
x x x
1  1  2  3  ......  n
g g g
g
The above step provides that we prove “As the product of n positive numbers
equals 1, then their sum is not less than n.” This can be stated mathematically as
x
x1 x 2 x3


 ......  n  n
g
g
g
g
The proof of if the product of n positive numbers equals 1 their sum is not less
than n is as follows:
Suppose for n  k  2 that x1  x2  x3  ......  xk  k is true if x1 x2 ..... xk  1 .
We need to prove for n  k  1.
First, x1 x2 ..... xk xk 1  1
We have 2 cases:
1.) x1  x2  .....  xk  xk 1  1
If all the terms are equal to 1, then their sum equals k  1 .
x1  x2  .....  xk  xk 1  k  1
2.) Terms are not all equal. There will be terms less than 1 and greater than 1.
x1  1 and x k 1  1
( x1 xk 1 ) x2 x3 ........xk  1
If y1  x1 xk 1 , we may rewrite the above y1 x2 x3 ........xk  1
So by our assumption their sum is not less than k
y1 x2 x3 ........xk  k
22
But
x1  x 2  ..  x k  x k 1  k  x k 1  y1  x1
x1  x 2  ..  x k  x k 1  ( y1  x 2  ..  x k )  x k 1  x1  y1 and
k  x k 1  y1  x1  (k  1)  x k 1  y1  x1  1
So,
( y1  x 2  ..  x k )  x k 1  x1  y1  (k  1)  x k 1  y1  x1  1
Substitute x1 xk 1 in for y1
x1  x2  ...  xk 1  (k  1)  xk 1  x1 xk 1  x1  1
The right hand side of the inequality can be written as
k  1  ( xk 1  1)(1  x1 )
So the inequality now can be stated as
x1  x2  ...  xk 1  k  1  ( xk 1  1)(1  x1 )
Because we assumed that x1  1 and x k 1  1, then ( xk 1  1)(1  x1 )  0 .
So then (k  1)  ( xk 1  1)(1  x1 )  k  1
Thus, x1  x2  ....  xk  xk 1  k  1 .
Recall the last step we did in our AM-GM
x
x1 x 2 x3


 ......  n  n
g
g
g
g
Multiply both sides by g which denotes the geometric mean
x
x
x 
x
g  1  2  3  ......  n   g  n
g
g
g 
g
inequality
was
x1  x 2  .....  x n  g  n
Then divide both sides by n
x1  x2  .....  xn g  n

n
n
The left hand side of the inequality equals the arithmetic mean.
Thus, a  g .
So without loss of generality the arithmetic mean is greater than the geometric
mean.
The AM-GM inequality can be represented geometrically as well using the
Pythagorean Theorem. For this version of the inequality, there are only two real
numbers a and b. Below is an illustration of a right triangle with the lengths of the
ab
ab
legs equaling
and x .The hypotenuse length equals
.
2
2
23
a-b
a+b
2
2
x
Using the Pythagorean Theorem, we know that the sum of the squares of the two
legs of a right triangle equals the square of the hypotenuse.
2
2
a b
ab
  x 2  
 .
Thus, 
 2 
 2 
a 2  2ab  b 2
a 2  2ab  b 2
 x2 
4
4
2
ab
ab
 and 

 2 
 2 
2
Foiling 
x2 
a 2  a 2  2ab  2ab  b 2  b 2
4
a 2  2ab  b 2
Subtract
from both sides of the equation
4
x2 
4ab
4
Combine like terms
x 2  ab
Simplify
x  ab
Take the square root of both sides
Because the hypotenuse is always the longest side of a right triangle, then
ab
 ab . The left side of the equation represents the arithmetic means while
2
the right hand side represents the geometric mean. Thus the arithmetic mean is
greater than the geometric mean.
If the arithmetic mean is always greater than the geometric mean, then
why would one use the geometric mean we ask? The geometric mean is slightly
higher than the arithmetic mean when one is working with percentages and rates
of change such as the yearly growth rate of a company’s profits over a certain
number of years. To know when it is appropriate to use the geometric mean, one
24
should ask the question, “if all the quantities had the same value, what would that
value have to be in order to achieve the same product" [19].
An extension to the AM-GM inequality is the arithmetic mean-geometric
mean-harmonic mean inequality. First we need to know that the harmonic mean
n
h

for the real numbers, x1 , x2 ,......, xn , is
1
1
1 . Archytas of

 ..... 
x1 x 2
xn
Tarentum, a Greek mathematician, is the person credited with coining the term
harmonic mean. Originally called the sub-contrary mean, Archytas renamed it
the harmonic mean because “the ratio proved to be useful for generating
harmonious frequencies on string instruments” [21]. A useful application of the
harmonic mean is “a measure of central tendency for data which consists of rates
or frequencies” [21]. The harmonic mean is always less than the arithmetic
mean and the geometric mean; thus the arithmetic-geometric-harmonic means
inequality exists.
The proof of the arithmetic-geometric-harmonic
reconstructed by Planet Math [22] is shown below.
means
inequality
as
The arithmetic-geometric-harmonic means inequality states for any positive real
numbers,
then
x1 , x2 ,......, xn ,
x  x 2  ......  x n
n
a 1
 g  n x1  x 2  .....  x n  h 
.
1
1
1
n

 ..... 
x1 x 2
xn
Using what we have already proven about the arithmetic and geometric means.
We will now prove the geometric mean is greater than the inequality.
Let t i be
1
.
xi
t1  t 2  ....  t n n
 t1t 2t3 ... t n . The left hand side denotes the arithmetic
n
mean and the right hand side represents the geometric mean.
From
Substitute
1
for t i .
xi
1
1
1
1


 .... 
x1 x2 x3
xn
1 1 1
1
 n    . .. .. 
n
x1 x2 x3
xn
25
When you multiply both sides of the equation by the reciprocal of the left hand
n
side,
you obtain
1
1
1
1


 .... 
x1 x2 x3
xn
1 n
1 1 1
1
n
   ..... 

1
1
1
x1 x2 x3
xn 1


 .. .. 
x1 x2 x3
xn
1 1 1
1
   .. . . . 
x1 x2 x3
xn
1
n
.

1
1
1
1
1 1 1
1


 . .. . 
   ..... 
x1 x2 x3
xn
x1 x2 x3
xn
Then divide both sides by
n
n
By
the
properties
of
exponents,
we
know
1 1 1
1

  ..... 
 ( x1  x 2  x3  ....  x n ) 1 . So the denominator of the left hand
x1 x 2 x3
xn

side can be rewritten as ( x1  x2  x3  ....  xn

1
1 n
)
. Again by the laws of exponents
1
the denominator of the left hand side can be modified to
1
( x1  x2  x3  .... xn )
which can be changed to ( x1  x2  x3  .....  xn
1
)n

n
 n x1  x2  x3  .....  xn .
n
. In other words the geometric mean
1
1
1
1


 . .. . 
x1 x2 x3
xn
is always greater than or equal to the harmonic mean.
Thus, n x1 x2 x3 .... xn 
Thus,
our
proof
is
complete
without
loss
of
x  x 2  ......  x n
n
a 1
 g  n x1  x 2  .....  x n  h 
.
1
1
1
n

 ..... 
x1 x 2
xn
generality,
26