Notes on Inequalities∗
Fall 2006
1
Inequalities for Finite Sums
Inequalities are at the heart of analysis, and the underpinnings of many of the important
inequalities used in analysis lie in a few fundamental inequalities concerning generalized
means of finite sets (a) = {aj |j = 1, . . . , n} of nonnegative real numbers. The most basic
notion of Mean of a finite set of numbers is the rth -mean given in the following
Definition 1 (rth -Mean) Given a finite set of nonnegative numbers (a) and a nonzero real
number r, the rth -mean, Mr (a), of (a) is defined by
1/r

1 X r
Mr (a) := 
a
n j j
.
(1)
Important special cases are provided by the Arithmetic Mean, the Harmonic Mean and
the Geometric Mean
A(a) := M1 (a)
H(a) := M− (a)
1/n

G(a) := 
(2)
(3)
Y
aj 
.
(4)
j
Though referred to as a special case of (1), the geometric mean is actually the limit of (1)
as r → 0 as will be shown below.
Corresponding to the basic means Mr (a), one also defines weighted means Mr (a, p) for
positive weights pj > 0.
Definition 2 (weighted rth -mean) Given nonnegative numbers (a), a nonzero number r
and positive weights (p), the weighted rth -mean is defined to be

1/r
1 X
pj arj 
Mr (a, p) := 
P j
where P =
P
j
(5)
pj .
∗
These notes are adapted from the classic book Inequalities by G. H. Hardy, J. E. Littlewood and
G. Pólya, Cambridge University Press, 1967.
1
qj ,
By homogeneity, the weighted means (5) are equivalent to weighted means with weights
j qj = 1, that is
P

1/r
X
Mr (a, p) = Mr (a, q) =  qj arj 
j
with
qj = pj /P.
In similar fashion, one defines

G(a, p) := 
Y
1/P
pj 
aj
=
Y
j
q
aj j .
j
Exercise 1 Show that
min(a) ≤ Mr (a) ≤ max(a)
with strict inequality unless all aj are equal.
Exercise 2 Show that
min(a) ≤ G(a) ≤ max(a)
(6)
with strict inequality unless all aj are equal.
The previously advertised relationship between the geometric and rth -means is made
explicit in
Theorem 1
lim Mr (a, q) = G(a, q).
(7)
r→0
Proof: Using the identity
arj = er log(aj )
and the asymptotic approximation
ex = 1 + x + O(x2 ) as x → 0
one concludes that
Mr (a, q)
=
=
−→






X
exp  qj log(aj )
Y
q
aj j
j
=

X
1
exp  log 1 + r
qj log(aj ) + O(r2 )
r
j
j
=

X
1
exp  log  qj arj 
r
j
G(a, q).
2
Exercise 3 Show that
lim Mr (a, q) = max(a)
r→∞
lim Mr (a, q) = min(a).
r→−∞
Another fundamental inequality in analysis is given by
Theorem 2 (Cauchy’s Theorem) Given two nonnegative sets of numbers (a) and (b),
one has that

2



X
X
X

aj bj  ≤  a2j   b2j 
j
j
(8)
j
with strict inequality unless the sets (a) and (b) are proportional, i.e. unless there exists a
nonnegative numbers x and y, not both zero, such that xaj + ybj = 0 for j = 1, . . . , n.
Proof: Consider the quadratic form
X
(xaj + ybj )2 = x2
j
X
a2j + 2xy
j
X
aj b j + y 2
X
j
b2j ≥ 0
j
with strict inequality unless xaj + ybj = 0 for all j = 1, . . . , n. Hence, the discriminant of
the quadratic form must be nonpositive, i.e.
2

X




X
X
aj bj  −  a2j   b2j  ≤ 0,
j
j
j
which proves the theorem.
Exercise 4 Show that if r > 0, then
Mr (a, q) < M2r (a, q).
It is now convenient to prove the celebrated inequality between the algebraic and geometric means.
Theorem 3 (Inequality of the Means) Let a be nonnegative numbers and q positive
weights summing to 1. Then
G(a, q) ≤ A(a, q)
(9)
with strict inequality unless all the aj are equal.
Proof: The proof proceeds in cases.
Case 1. Assume equal weights qj = 1/n for j = 1, . . . , n. Then (9) is equivalent to
n

1X 
a1 a2 · · · an ≤ 
aj
= An .
n j
3
(10)
Rearrange the numbers, if necessary, so that a1 = max(a) and a2 = min(a). Next replace a1
and a2 by A and a1 + a2 − A. Then A(a1 + a2 − A) − a1 a2 = (a1 − A)(A − a2 ) ≥ 0. Thus
G is increased, but clearly A is the same. Label the new a, a2 = (a21 , . . . , a2n ) with a21 = A,
a22 = (a1 + a2 − A) and a2j = aj for j = 3, . . . n. Now construct a32 , . . . , a3n as before, i.e.
a32 = A, a33 = (max(a2j ) + min(a2j ) − A) with a3j = a2j for j = 4, . . . , n. Again G is increased
but A is unchanged. Repeating the argument at most (n − 1)-times, a1 , . . . , an becomes
A, . . . , A with G increasing at each stage but A unchanged. Clearly G at the last stage is A,
hence G ≤ A with strict inequality unles all aj are equal.
Case 2. The case of unequal weights is broken into two additional cases: all rational weights
and some irrational weights. For all rational weights one makes use of homogeneity of the
geometric and algebraic means to reduce the problem to Case 1. For the case of some
irrational weights, see the book Inequalities by Hardy, Littlewood and Polya.
An important application of the preceding theorem is
Theorem 4 (Hölder’s Inequality) If λ1 , . . . , λl > 0 with 1 =
l-sets of positive numbers, then
n X

a1j
λ1
· · · alj
λl
≤
j=1
n
X
λ1

a1j  · · · 
j=1
n
X
P
i
λi and a1 , . . . , al are
λl
alj 
(11)
j=1
with strict inequality unless all sets of numbers ai are pairwise proportional.
Proof: The proof reduces to the previous theorem through the string
P j
P
(
j
a1j
λ1
· · · alj
a1j )λ1 · · · (
P
j
λl
alj )λl
=
X
j
≤
X
j
a1j
P 1
i ai
!λ1
al
··· P j l
i ai
!λl
al
a1
λ 1 P j 1 + · · · + λl P j l
i ai
i ai
!
= λ 1 + · · · + λl = 1
with strict inequality unless
a1j
alj
P 1 = ··· = P l
i ai
i ai
for all j = 1, . . . , n, which proves the theorem.
As a corollary of the preceding theorem one has
Theorem 5 (Hölder’s Theorem) If λ > 1, and a and b are sets of nonnegative numbers,
then



 0
1/λ
X
X
aj bj ≤  aλj 
j
1/λ
X 0

bλ 
j
j
j
with λ0 defined through the equation
1
1
+ 0 = 1.
λ λ
4
(12)
Proof: Exercise.
When 0 < λ < 1 one has
Theorem 6 If 0 < λ < 1, and a and b are sets of nonnegative numbers, then
X

1/λ 
1/λ0
X
X 0
aj bj ≥  aλj   bλj 
j
j
(13)
j
where λ0 is defined through
λ0 =
λ
1
=
< 0.
1 − 1/λ
λ−1
Proof: Apply the preceding theorem with l = 1/λ > 1, l0 = l/(l − 1) = 1/(1 − λ) > 1 and
uj = (aj bj )λ , vj = b−λ
j .
For complex numbers, one can easily prove
Theorem 7 If a and b are sets of n complex numbers, k > 1 and k 0 is such that
1
1
+ 0 =1
k k
then

1/k 
1/k0
X
X
X
0
aj bj ≤  |aj |k   |bj |k 
.
j
j
j
(14)
The next theorem proves the monotonicity of the weighted means.
Theorem 8 If r < s, and q are positive weights summing to one, then
Mr (a, q) ≤ Ms (a, q)
with strict inequality unless all aj are equal.
Proof: Assume first that 0 < r < s and let r = αs for 0 < α < 1. Define uj and vj by
u = qj asj and vj = qj . Then
s α 1−α
qj arj = qj asα
= uαj vj1−α
j = (qj aj ) qj
from which it follows that
X
uαj vj1−α

α 
1−α
X
X
≤  uj   vj 
j
j
j
or equivalently
X

r/s 
1−r/s
X
X
qj arj ≤  qj asj   qj 
j
j
j
which yields
X

r/s
X
qj arj ≤  qj asj 
j
j
as was required to be shown.
5
Exercise 5 Prove the previous theorem for the remaining cases r = 0, s > 0 and r < s ≤ 0.
A related inequality is given in the following
Theorem 9 (Jensen’s Inequality) If a are positive numbers and 0 < r < s, then

1/r

1/s
X
X

arj  ≥  asj 
j
j
with equality if and only if all aj but one are zero.
Proof: By homogeneity, we may assume j arj = 1. (Why?) Then each aj ≤ 1 from which
P
P
it follows that asj ≤ arj and j asj ≤ j arj = 1. One then concludes that
P

1/s

1/r
X
X

asj  ≤ 1 =  arj 
j
j
as required.
A simple argument leads to
Theorem 10 If a are nonnegative numbers then

1/r
X
lim  arj 
r→∞

1/r
X
lim  ar 
r→0
= max(aj )
j
j
j
= ∞.
j
Proof: Exercise.
As a corollary to (9), one can prove
Theorem 11 If α, β > 0 satisfy α + β > 1, then
X
aαj bβj
j

α 
β
X
X
≤  aj   b j 
j
j
with strict inequality unless all but perhaps one number from each set are zero.
Proof: Exercise.
Along another direction there is
Theorem 12 Suppose a, b, . . . , l are sets of n nonnegative numbers and r 6= 1. Then
Mr (a, q) + · · · + Mr (l, q) ≥ Mr (a + · · · + l, q) if r > 1
Mr (a, q) + · · · + Mr (l, q) ≤ Mr (a + · · · + l, q) if r < 1
with strict inequality unless all sets are proportional.
6
Proof: Define s = a + · · · + l and S = Mr (a, q). Then
Sr =
X
qj srj =
X
=
X
j
qj aj sr−1
+ ··· +
j
X
j
qj lj sr−1
j
j
1/r
1/r
(qj aj )(qj sj )r−1
+ ··· +
j
X
1/r
1/r
(qj lj )(qj sj )r−1 .
j
If r > 1, then defining r0 = r/(r − 1), one has

Sr ≤ 
X
1/r 
1/r0

1/r 
1/r0
X
X
X
qj arj   qj srj 
+ · · · +  qj ljr   qj srj 
j
=
j
j
j

1/r

1/r 
X
X


S r−1  qj arj  + · · · +  qj ljr  
j
j
which proves the theorem. The cases 0 < r < 1 and r < 0 follow as in previous inequalities.
Finally, as a corollary there is the famous
Theorem 13 (Minkowski’s Inequality)

1/r
X
 (aj + · · · + lj )r 
j

1/r
X
 (aj + · · · + lj )r 
j

1/r

1/r
X
X
≤  arj  + · · · +  ljr  if r > 1
j
j


1/r
1/r
X
X
≥  arj  + · · · +  ljr  if r < 1.
j
j
7