Irish Math. Soc. Bulletin 57 (2006), 69–79
69
Some Mean Inequalities
FINBARR HOLLAND
Dedicated to Trevor West on the occasion of his retirement.
Abstract. Let P denote the collection of positive sequences
defined on the set of natural numbers N. It is proved that if
x ∈ P, and s < 0, then
j
n
k
n
1 X 1 X 1/s s 1 X 1 X 1/s s
xj
≤
xk
,
n k=1 k j=1
n j=1 j k=1
n∈N
with equality if and only if x is a constant sequence. This
is a sharp refinement of an inequality discovered by Knopp
in 1928.
1. Introduction
When I received the invitation to participate in the Westfest, I was
in the throes of writing up a solution to the following problem, due to
Joel Zinn, which was posed in the American Mathematical Monthly,
and I offered to speak on this topic at the meeting in TCD to mark
Trevor’s retirement. I’m grateful to the organising committee of the
Westfest for giving me the opportunity to do so.
Problem 1 (Number 11145). Find the least c such that if n ≥ 1,
a1 , . . . , an > 0, then
n
X
k=1
k
Pk
1
j=1 aj
≤c
n
X
ak .
k=1
I propose to describe a method to handle a family of similar problems of which this, and classical ones due to Carleman, and Knopp,
are special cases.
70
Finbarr Holland
2. Background
We denote by P the collection of positive sequences x : N → (0, ∞).
Clearly, P is a convex set. It is closed under the usual pointwise operations of addition and multiplication, and ordered by the relation:
x ≤ y ⇐⇒ xn ≤ yn , ∀n ∈ N.
In particular, P is a commutative group under multiplication, with
the sequence vector e of ones acting as the identity. We’ll write 1/x
for the multiplicative inverse of x ∈ P:
(1/x)n =
1
, ∀n ∈ N.
xn
We recall a number of familiar functions that take P into itself:
n
1X
A : P → P; A(x)n =
xk , n = 1, 2, . . . ;
n
k=1
v
u n
uY
n
G : P → P; G(x)n = t
xk , n = 1, 2, . . . ;
k=1
H : P → P; H(x)n = Pn
n
1
k=1 xk
, n = 1, 2, . . . ;
min : P → P; min(x)n = min{xk : k = 1, 2, . . . , n}.
These are, respectively, the arithmetic, geometric, harmonic and
minimum means. (Weighted versions of these exist, but I’ll not have
any need to refer to them.)
It is a well-known fact [5] that
v
u n
n
uY
n
1X
n
t
≤
x
≤
xk , n = 1, 2, . . . .
min(x)n ≤ Pn
k
1
n
k=1 xk
k=1
k=1
Moreover, the inequalities are strict unless x is a constant sequence.
Equivalently,
min ≤ H ≤ G ≤ A.
It’s clear from the definitions that A, G, H and min are “homogeneous” in the sense that, if f ∈ {A, G, H, min}, then
f (λx) = λf (x),
∀x ∈ P, λ > 0.
Some Mean Inequalities
71
It’s perhaps less obvious, but nonetheless true, that they are superadditive: if f ∈ {A, G, H, min}, then
f (x) + f (y) ≤ f (x + y), ∀x, y ∈ P.
Hence they are also concave on P.
We also introduce a one-parameter family of functions {Mt : t > 0}
that leave P invariant. If x ∈ P, we define Mt (x) by
!t
n
Mt (x)n = Pn
, n = 1, 2, . . . ,
1
j=1 x1/t
j
so that Mt (x) = (H(x1/t ))t ,
min(x) ≤ Mt (x) ≤ G(x) ≤ A(x), ∀x ∈ P, ∀t > 0,
and
lim Mt (x) = min(x), lim Mt (x) = G(x), ∀x ∈ P.
t→∞
t→0+
3. An Inequality Between Compositions of Means
I’m interested in compositional relationships between these various
functions. I’ll describe the following result.
Theorem 1. Let t > 0. Then A ◦ Mt ≤ Mt ◦ A. Moreover, A ◦
Mt (x) = Mt ◦ A(x) if and only if x = λe for some λ > 0.
For instance, when t = 1, the claim is that A ◦ H ≤ H ◦ A.
Equivalently,
n
1X
k
Pk
n
j=1
k=1
1
xj
≤ Pn
k=1
n
Pk k
, n = 1, 2, . . .
j=1 xj
Even for small values of n this is already fairly challenging, as the
reader may discover for him or her self by considering the special
case n = 3.
A more general weighted inequality of this kind was first postulated by Nanjundiah [13] in 1952, but he offered no proof, and indeed
his conjecture is not true generally. A special case of it was conjectured by myself [6] in 1992, and Kedlaya [8] supplied a proof of this
in 1994, namely that A ◦ G ≤ G ◦ A. In 1996, Mond and Pečarić [12]
proved an analogue of the inequality A ◦ H ≤ H ◦ A, the case t = 1,
for Hermitian matrices.
To establish the theorem, we begin by proving a lemma.
72
Finbarr Holland
Lemma 1. Let t > 0 and let p = t + 1. Let x ∈ P. Then, for each
n ≥ 1,
n
n
o
nX
X
p
t
n
ak = 1 .
Mt (x)n = n inf
xk ak : 0 < a ∈ R ,
k=1
k=1
n
Proof. Suppose 0 < a ∈ R , and
p/t. Then, by Hölder’s inequality,
1
=
n
X
Pn
k=1
ak = 1. Let q = p/(p−1) =
−1/p
(apj xj )1/p xj
j=1
≤
n
X
n
1/p X
apj xj
j=1
1
n
j=1
1/q
,
j=1
Hence
P
−q/p
xj
−q/p
xj
p/q ≤
n
X
apj xj .
j=1
Equality holds here if
aj =
It follows that
1
Pn
1
1
q/p
xj
P
n
j=1
t = P
n
= inf
, j = 1, 2, . . . , n.
1
−q/p
j=1
j=1 x1/t
j
−q/p
xj
xj
n
nX
p/q
xk apk : 0 < a ∈ Rn ,
k=1
n
X
o
ak = 1 .
k=1
The stated result follows.
An equivalent formulation is that, with p = t + 1,
n
n
nX
o
X
Mt (x)n = np−1 inf
xk apk : 0 < a ∈ Rn ,
ak = 1 .
k=1
(1)
k=1
Thus
Mt (x)n ≤ np−1
n
X
k=1
for all probability vectors a ∈ Rn .
xk apk
(2)
Some Mean Inequalities
73
Remark. Already this result reveals that Mt is super-additive and
hence concave.
The result we want to prove is the following: if x ∈ P,
!t
n
n
1X
k
1X
=
Mt (x)k ≤ Mt (A(x))n
Pk
1
n
n
j=1 1/t
k=1
k=1
xj
!t
n
=
, n = 1, 2, . . . ,
Pn
1
k=1 A(x)1/t
k
with equality if and only if x is a constant sequence.
Our idea is this: with n fixed, suppose a is a probability vector in
Rn . Then, by the previous lemma, with p = t + 1,
Mt (A(x))n
≤ np−1
n
X
apj A(x)j
j=1
= np−1
= np−1
j
n
X
apj X
j=1
n
X
j
xk
k=1
xk
k=1
n
X
apj
j=k
j
,
after interchanging the order of summation, and there is equality
here for a suitable a. But, also, if ui ∈ Ri is a probability vector,
Mt (x)i ≤ ip−1
i
X
upij xj , i = 1, 2, . . . , n,
j=1
whence
n
X
Mt (x)i
≤
i=1
n
X
ip−1
i=1
=
n
X
j=1
i
X
upij xj
j=1
xj
n
X
ip−1 upij .
i=j
So, we can accomplish our objective if, given a probability vector
a ∈ Rn , we can construct similar vectors ui ∈ Ri so that
n
X
i=j
ip−1 upij ≤ np
n
X
ap
k
k=j
k
, j = 1, 2, . . . , n.
(3)
74
Finbarr Holland
To reach our goal, and to show that these inequalities can be
solved, we construct a certain lower triangular row-stochastic matrix
from a probability vector. To this end, we use the following result
due to Kedlaya [8]:
Lemma 2. The rational numbers
i−1 n−i
αk (i, j) =
j−k
k−1
n−1
j−1
, 1 ≤ i, j, k ≤ n,
satisfy the following conditions
(1) αk (i, j) ≥ 0, for all i, j, k;
(2) αk (i, j) = 0 for all k > min (i, j);
(3) αk (i, j) = αk (j, i) for all i, j, k;
Pn
(4)
(i, j) = 1 for all i, j;
k=1 αkn
n
X
(5)
j , for 1 ≤ k ≤ j,
αk (i, j) =
0,
fork > j.
i=1
Given a probability vector a ∈ R2 , construct the n × n matrix
A = [aij ] by
n
X
aij =
αj (i, k)ak , 1 ≤ i, j ≤ n.
k=1
Then each row of A is a probability vector, because
n
n X
n
X
X
aij =
αj (i, k)ak
j=1
=
=
j=1 k=1
n
n
X
X
ak
k=1
n
X
αj (i, k)
j=1
ak (by 4.)
k=1
=
1
for all i. Also, aij = 0 for all j > i. Thus A is a lower triangular
row-stochastic matrix.
But, for each pair of indices i, j, aij is a convex combination of
a1 , a2 , . . . , an , and so, if p ≥ 1,
n
X
apij ≤
αj (i, k)apk , i, j = 1, 2, . . . , n.
k=1
Some Mean Inequalities
75
Hence
n
X
ip−1 apij
=
i=j
n
X
ip−1 apij
i=1
≤ np−1
n X
n
X
αj (i, k)apk
i=1 k=1
= np−1
= n
n
X
apk
k=1
n
X
apk
p
k=j
k
n
X
αj (i, k)
i=1
(by 5.).
Looking back at (3) we now see that we can solve this by selecting
uij = aij , j = 1, 2, . . . , i.
We’re now ready to provide a proof of the theorem.
Fix x ∈ P, and a positive integer n. Let a be a probability vector
in Rn . Choose the corresponding lower triangular row-stochastic
matrix A = [aij ] as above. By Lemma 1, if 1 ≤ i ≤ n,
Mt (x)i ≤ ip−1
i
X
apij xj .
j=1
Hence
n
X
Mt (x)i
≤
i=1
n
X
ip−1
i
X
i=1
=
≤
n
X
j=1
n
X
j=1
xj
n
X
= np
ip−1 apij
i=j
xj np
j=1
= np
apij xj
n
X
k=1
n
X
k=1
n
X
ap
k
k=j
apk
k
k
1X
xj
k j=1
apk A(x)k .
76
Finbarr Holland
Whence
n−t−1
n
X
Mt (x)i
i=1
is a lower bound for the set
n
n
nX
o
X
apk A(x)k : 0 < a ∈ Rn ,
ak = 1 ,
k=1
k=1
−t
whose infimum is n Mt (A(x))n . Hence
n
1X
Mt (x)i ≤ Mt (A(x))n ,
n i=1
and we’re done, apart from dealing with the case of equality, which
is easily settled.
4. A Number of Corollaries
We deduce a number of special cases of Theorem 1.
Corollary 1.
A ◦ min ≤ min ◦A.
This is obtained by letting t → 0+ .
Corollary 2 (Kedlaya).
A ◦ G ≤ G ◦ A.
i.e., ∀x ∈ P,
n
1X
G(x)i ≤ G(A(x))n , ∀x ∈ P, ∀n ≥ 1;
n i=1
or, more explicitly,
v
n uY
X
u i
1
i
t
xj ≤
n i=1 j=1
v
u n
j
uY 1 X
n
t
xi .
j i=1
j=1
This is obtained by letting t → ∞. This implies Carleman’s
classical inequality [3, 5, 7]:
||G(x)||1 ≤ e||x||1 , ∀x ∈ P.
Corollary 3 (Mond & Pečarić).
A ◦ H ≤ H ◦ A.
Some Mean Inequalities
77
This is obtained by letting t = 1. It says that, ∀x ∈ P,
n
1X
k
n
≤ Pn
, n = 1, 2, . . .
Pk
1
Pk k
n
k=1
j=1
k=1
xj
j=1
xj
Since the sequence
k
X
xj , k = 1, 2, . . .
j=1
is strictly increasing we deduce that the right-hand side does not
exceed
Pn
Pn
n j=1 xj
2 j=1 xj
Pn
,
=
n+1
k=1 k
whence
n
n
n
X
X
k
2n X
≤
x
<
2
xj ,
Pk
j
1
n + 1 j=1
j=1
j=1
k=1
xj
which gives a solution to Zinn’s Monthly problem. Moreover, the
constant 2 cannot be replaced by a smaller number, as can be seen
by taking xj = 1/j, j = 1, 2, . . . , n. Thus, if x ∈ P ∩ `1 , so does
H(x), and ||H(x)||1 < 2||x||1 .
Corollary 4. ∀t > 0 and ∀x ∈ P,
n
X
Mt (x)i ≤
i=1
n1+1/t
Pn
1/t
j=1 j
!t
n
X
xk , n = 1, 2, . . .
k=1
and the inequality is strict unless n = 1.
Proof.
nMt (A)n
= n
!t
1
p
Pn
j=1
−1/t
Aj
t

= np  Pn
1
j 1/t
1/t
j=1 (Pj
k=1 xk )
≤ n
p
n
X
xk
k=1
=
n1+1/t
Pn
1/t
j=1 j
!t
1
Pn
j=1
!t

j 1/t
n
X
k=1
xk .
78
Finbarr Holland
Since
1
n1+1/t
lim Pn
=1+ ,
1/t
n→∞
t
j=1 j
a simple consequence of the fact that, with s = 1/t,
Pn
n
s
1X j s
j=1 j
=
( )
ns+1
n j=1 n
is a Riemann sum for the integral
Z 1
xs dx =
0
1
,
1+s
Corollary 4 implies a result of Knopp [10] to the effect that
||Mt (x)||1 ≤ (1 + 1/t)t ||x||1 , ∀x ∈ P.
(4)
5. Companion Results When t < 0
The means Mt also make sense when t < 0. Similar methods to those
employed in the previous section lead to the following statement.
Theorem 2. If −1 < t < 0, then A ◦ Mt ≥ Mt ◦ A. Moreover,
A ◦ Mt (x) = Mt ◦ A(x) if and only if x = λe for some λ > 0.
Letting p = −1/t, we can recast this in terms of p: If p ≥ 1, then,
for all x ∈ P, and all n ≥ 1,
!p !1/p
!1/p
n
k
n
k
1X 1X p
1X 1X
xi
≤
x
.
(5)
n
k i=1
n
k i=1 i
k=1
k=1
There is equality only when x is a constant sequence. This is a substantial improvement of a very well-known result due to Hardy [4, 5],
which states that, if x ∈ `p , then A(x) ∈ `p , and
p
||A(x)||p ≤
||x||p .
p−1
Inequality (5) was found by Bennett [2], who pointed out that the
reversed inequality holds when 0 < p < 1. A stronger form of (5)
was established by B. Mond and J. E. Pečarić [11], and a weighted
version of their result was outlined by Kedlaya [9]. But results of
this kind were announced much earlier by Nanjundiah [13], though
he appears not to have published a proof.
Some Mean Inequalities
79
References
[1] G. Bennett, Factorizing the Classical Inequalities, Mem. Amer. Math. Soc.
120 (1996), no. 576, viii+130pp.
[2] G. Bennett, Summability matrices and random walk, Houston J. Math 28
(2002), no. 4, 865–898.
[3] T. Carleman, Sur les functions quasi-analytiques, Fifth Scand. Math. Congress (1923), 181–196.
[4] G. H. Hardy, Notes on some points of the integral calculus (LX), Messenger
of Math. 54 (1925), 150–156.
[5] G. H. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, Cambridge
University Press, 1934.
[6] F. Holland, On a mixed arithmetic-mean, geometric-mean inequality,
Math. Competitions 5 (1992), 60–64.
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history, proofs and some new generalizations, J. Ineq. Pure & Appl. Math.
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[8] K. Kedlaya, Proof of a mixed arithmetic-mean, geometric-mean inequality,
Amer. Math. Monthly 101 (1994), 355–357.
[9] K. Kedlaya, A weighted mixed-mean inequality, Amer. Math. Monthly 106
(1999), 355–358.
[10] K. Knopp, Über Reihen mit positiven Gliedern, J. London Math. Soc., 3
(1928), 205-211.
[11] B. Mond and J. E. Pečarić, A mixed means inequality, Austral.Math. Soc.
Gaz., 23(2) (1996), 67–70.
[12] B. Mond and J. E. Pečarić, A mixed arithmetic-mean-harmonic-mean
inequality, Lin. Algebra Appl. 237/238 (1996), 449–454.
[13] T. S. Nanjundiah, Sharpening of some classical inequalities, Math. Student
20 (1952), 24–25.
Finbarr Holland,
Mathematics Department,
University College,
Cork, Ireland
f.holland@ucc.ie
Received on 8 June 2006.