Journal of Inequalities in Pure and
Applied Mathematics
A NEW ARRANGEMENT INEQUALITY
MOHAMMAD JAVAHERI
University of Oregon
Department of Mathematics
Fenton Hall, Eugene, OR 97403.
volume 7, issue 5, article 162,
2006.
Received 29 April, 2006;
accepted 17 November, 2006.
Communicated by: G. Bennett
EMail: javaheri@uoregon.edu
URL: http://www.uoregon.edu/∼javaheri
Abstract
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Victoria University
ISSN (electronic): 1443-5756
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Abstract
In this paper, we discuss the validity of the inequality
n
n
n
X
X
X
(1+a+b)/2
xi xai xbi+1 ≤
xi
i=1
i=1
!2
,
i=1
where 1, a, b are the sides of a triangle and the indices are understood modulo
n. We show that, although this inequality does not hold in general, it is true
when n ≤ 4. For general n, we show that any given set of nonnegative real
numbers can be arranged as x1 , x2 , . . . , xn such that the inequality above is
valid.
A New Arrangement Inequality
Mohammad Javaheri
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2000 Mathematics Subject Classification: 26Dxx.
Key words: Inequality, Arrangement.
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I would like to thank Omar Hosseini for bringing to my attention the case a = b = 1
of the inequality (1.2). I would also like to thank Professor G. Bennett for reviewing
the article and making useful comments.
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1
Main Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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5
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1.
Main Statements
Let a, b, x1 , x2 , . . . , xn be nonnegative real numbers. If a + b = 1 then, by the
Rearrangement inequality [1], we have
n
X
(1.1)
i=1
xai xbi+1
≤
n
X
xi ,
i=1
where throughout this paper, the indices are understood to be modulo n. In an
attempt to generalize this inequality, we consider the following
!2
n
n
n
X
X
X
a b
c
xi
xi xi+1 ≤
xi ,
(1.2)
i=1
i=1
i=1
b ≤ a + 1,
a ≤ b + 1,
1 ≤ a + b,
then the inequality (1.2) is true in the case of n = 4 (cf. Prop. 2.1). Moreover,
under the same conditions on a, b as in (1.3), we show that one can always find
a permutation µ of {1, 2, . . . , n} such that (cf. Prop. 2.4)
!2
n
n
n
X
X
X
(1.4)
xi
xaµ(i) xbµ(i+1) ≤
xci .
i=1
i=1
Mohammad Javaheri
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where c = (a + b + 1)/2. It turns out that if a + b 6= 1 then the inequality (1.2)
is false for n large enough (cf. Prop. 2.2). However, we show that if
(1.3)
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i=1
The conditions in (1.3) cannot be compromised in the sense that if for all nonnegative x1 , x2 , . . . , xn there exists a permutation µ such that the conclusion
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(1.4) holds, then a, b must satisfy (1.3). To see this, let x1 = x > 0 be arbitrary
and xi = 1, i = 2, . . . , n. Then, for any permutation µ, the inequality (1.4)
reads the same as:
(1.5)
(x + n − 1)(xa + xb + n − 2) ≤ (xc + n − 1)2 .
If the above inequality is true for all x and n, by comparing the coefficients of n
on both sides of the inequality (1.5), we should have xa + xb + x − 3 ≤ 2xc − 2.
Since x > 0 is arbitrary, 1, a, b ≤ c and conditions (1.3) follow.
The case of a = b = 1 of (1.2) is particularly interesting:
A New Arrangement Inequality
Mohammad Javaheri
(1.6)
n
X
i=1
xi
n
X
i=1
xi xi+1 ≤
n
X
!2
3/2
xi
.
i=1
There is a counterexample to (1.6) when n = 9, e.g. take
(1.7)
x1 = x9 = 8.5, x2 = x8 = 9,
x4 = x6 = 11.5, x5 = 12,
x3 = x7 = 10,
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and subsequently the inequality (1.6) is false for all n ≥ 9 (cf. prop. (2.2)).
Proposition 2.1 shows that the inequality (1.6) is true for n ≤ 4, and there
seems to be a computer-based proof [2] for the cases n = 5, 6, 7 which, if true,
leaves us with the only remaining case n = 8.
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2.
Proofs
Applying Jensen’s inequality [1, § 3.14] to the concave function log x gives
ur v s wt ≤ ru + sv + tw,
(2.1)
where u, v, w, r, s, t are nonnegative real numbers and r + s + t = 1. If, in
addition, we have r, s, t > 0 then the equality occurs iff u = v = w. However,
if t = 0 and r, s, w > 0 then the equality occurs iff u = v. We use this inequality
in the proof of the proposition below.
Proposition 2.1. Let a, b ≥ 0 such that a + 1 ≥ b, b + 1 ≥ a and a + b ≥ 1.
Then for all nonnegative real numbers x, y, z, t,
(2.2)
(x + y + z + t)(xa y b + y a z b + z a tb + ta xb ) ≤ (xc + y c + z c + tc )2 ,
where c = (a + b + 1)/2. The equality occurs if and only if {a, b} = {0, 1} or
x = y = z = t.
Proof. We apply the inequality (2.1) to
c
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c
c
u = (yz) , v = (xz) , w = (xy) ,
a
b
1
r =1− , s=1− , t=1− ,
c
c
c
(2.3)
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and obtain:
(2.4)
a
b
1
c
c
(yz) + 1 −
(xz) + 1 −
(xy)c .
x y z ≤ 1−
c
c
c
a b
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Notice that the assumptions on a, b in the lemma are made exactly so that r, s, t
are nonnegative. Similarly, by replacing z with t in (2.4), we have:
a
b
1
a b
c
c
(2.5)
x y t≤ 1−
(yt) + 1 −
(tx) + 1 −
(xy)c .
c
c
c
Next, apply (2.1) to
(2.6)
u = x2c ,
v = (xy)c ,
b
r =1− ,
c
w = 1,
b
s= ,
c
t = 0,
Mohammad Javaheri
and get
(2.7)
A New Arrangement Inequality
x
a+1 b
y ≤
b
1−
c
b
x2c + (xy)c .
c
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Similarly, by interchanging a and b, one has
a 2c a
(2.8)
xa y b+1 ≤ 1 −
x + (xy)c .
c
c
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Adding the inequalities (2.4), (2.5), (2.7) and (2.8) gives:
Close
1
3
(2.9) Sxa y b ≤ x2c + 4 −
c
c
(xy)c + 1 −
a
b
(yz)c + 1 −
(tx)c
c
c
a
b
c
+ 1−
(yt) + 1 −
(xz)c ,
c
c
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where S = x+y+z+t. There are three more inequalities of the form above that
are obtained by replacing the pair (x, y) by (y, z), (z, t) and (t, x). By adding
all four inequalities (or by taking the cyclic sum of (2.9)) we have
1 X 2c
2
2
(2.10) ST ≤
x + 4−
(xc + z c )(y c + tc ) + (xz)c + (yt)c ,
c
c
c
where ST stands for the left hand side of the inequality (2.2). The right hand
side of the above inequality is equal to
X 2 1
c
(2.11)
x
+
− 1 (xc + z c )2 + (y c + tc )2 − 2(xc + z c )(y c + tc ) ,
c
P
which is less than or equal to ( xc )2 , since c ≥ 1. This concludes the proof of
the inequality (2.2).
Next, suppose the equality occurs in (2.2)
(2.4) – (2.8)
P and
Pso the inequalities
P
are all equalities. If a = 0 then we have
x xb = ( xc )2 and so, by the
equality case of Cauchy-Schwarz, the two vectors (x, y, z, t) and (xb , y b , z b , tb )
have to be proportional. Then either b = c = 1 or x = y = z = t. Thus suppose
a, b 6= 0. Since c = a = b is impossible, without loss of generality suppose
that c 6= b. Since the inequality (2.7) must be an equality, x2c = xc y c (cf. the
discussion on the equality case of (2.1)). Similarly y 2c = y c z c , z 2c = z c tc and
t2c = tc xc . It is then not difficult to see that x = y = z = t.
Let N (a, b) denote the largest integer n for which the inequality (1.2) holds
for all nonnegative x1 , x2 , . . . , xn . By the above proposition, we have N (a, b) ≥
4.
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Proposition 2.2. Let a, b ≥ 0 such that a + b 6= 1. Then N (a, b) < ∞.
Moreover, if n ≤ N (a, b) then the inequality (1.2) is valid for all nonnegative
x1 , . . . , x n .
Proof. The proof is divided into two parts. First we show that the inequality
(1.2) cannot be true for all n. Proof is by contradiction. If a = b = 0 then (1.2)
is false for n = 2 (e.g. take x1 = 1, x2 = 2). Thus, suppose a + b > 0 and that
the inequality (1.2) is true for all n. Let f be a non-constant positive continuous
function on the interval I = [0, 1] such that f (0) = f (1). Let
i−1
(2.12)
xi = f
, yi = (xai xbi+1 )1/(a+b) , i = 1, . . . , n.
n
Since yi is a number between xi and xi+1 (possibly equal to one of them), by the
Intermediate-value theorem [3, Th 3.3], there exists ti ∈ Ii such that f (ti ) = yi .
By the definition of integral we have:
Z
Z
f (x)dx
(2.13)
I
I
1
n→∞ n2
f a+b (x)dx = lim
= lim
n→∞
1
n2
1
n→∞ n2
≤ lim
n
X
i=1
n
X
xi
xi
i=1
n
X
i=1
n
X
i=1
n
X
yia+b
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xai xbi+1
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!2
xci
A New Arrangement Inequality
Z
=
f c (x)dx
2
,
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J. Ineq. Pure and Appl. Math. 7(5) Art. 162, 2006
where we have applied the inequality (1.2) to the xi ’s. On the other hand, by
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the Cauchy-Schwarz inequality for integrals, we have
Z
2
Z
Z
a+b
1
a+b
f (x)dx f (x)dx ≥
f 2 (x)f 2 (x)dx
(2.14)
I
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I
Z
2
c
f (x)dx ,
=
I
a+b
with equality iff f and f
are proportional. The statements (2.13) and (2.14)
imply that the equality indeed occurs. Since a + b 6= 1 and f is not a constant
function, the two functions f and f a+b cannot be proportional. This contradiction implies that (1.2) could not be true for all n i.e. N (a, b) < ∞.
Next, we show that (1.2) is valid for all n ≤ N . It is sufficient to show that if
the inequality (1.2) is true for all ordered sets of k +1 nonnegative real numbers,
then it is true for all ordered sets of k nonnegative real numbers.
Let y1 , . . . , yk be nonnegative real numbers and set
(2.15)
S=
k
X
i=1
yi ,
A=
k
X
b
yia yi+1
,
i=1
P =
k
X
yic .
i=1
Without loss of generality we can assume P = 1. For each 1 ≤ i ≤ k, define
an ordered set of k + 1 nonnegative real numbers by setting:

1≤j ≤i+1
 yj
xj =

yj−1 i + 2 ≤ j ≤ k + 1
Applying the inequality (1.2) to x1 , . . . , xk+1 gives
(2.16)
(S + yi )(A + yia+b ) ≤ (P + yic )2 = 1 + yi2c + 2yic .
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Adding these inequalities for i = 1, . . . , k, yields:
X
(2.17)
kSA + S
yia+b + AS ≤ k + 2.
i
On the other hand, by the Rearrangement inequality [1] we have
(2.18)
k
X
b
yia yi+1
≤
k
X
i=1
yia+b ,
i=1
and the lemma follows by putting together the inequalities (2.17) and (2.18).
A New Arrangement Inequality
Mohammad Javaheri
The inequality (1.1) translates to N (a, b) = ∞ when a + b = 1. We expect
that N (a, b) → ∞ as a + b → 1. The following proposition supports this
conjecture. We define


!2 n
n
n
X

X
X
(2.19) An (a, b) = sup
xi
xai xbi+1 −
xci max xi = 1 .


1≤i≤n
i=1
i=1
i=1
This number roughly measures the validity of the inequality (1.2). Also let
n
(2.20)
1X t
x.
σt =
n i=1 i
By the Hölder inequality [1], if α, β > 0 and α + β = 1 then for any s, t > 0
we have:
(2.21)
σsα σtβ
≥ σαs+βt .
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Proposition 2.3. N (u, u) is a non-increasing function of u ≥ 1/2. Moreover,
for all n and a, b ≥ 0
lim An (a, b) = 0.
(2.22)
a+b→1
Proof. Suppose u > v > 1/2. We show that N (u, u) ≤ N (v, v). Without loss
of generality we can assume:
1
(2.23)
u−v < .
4
By the definition of N = N (v, v), there must exist N + 1 nonnegative integers
x1 , . . . , xN +1 such that the inequality (1.2) is false and so
!2
N
+1
N
+1
N
+1
X
X
X
v+1/2
(2.24)
xi
xvi xvi+1 >
xi
.
i=1
i=1
i=1
We show that the nonnegative numbers yi =
i = 1, . . . , N + 1 give a
counterexample to (1.2) when a = b = u. In light of (2.24), one just needs to
show
!2 , N +1
!2 , N +1
N
+1
N
+1
X
X u/v
X
X
u+1/2v
u+1/2
(2.25)
xi
xi ≥
xi
xi .
i=1
i=1
u + 1/(2v) − u/v
,
u + 1/(2v) − 1
1
s = 1, t = u + .
2v
α=
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To prove this, first let
(2.26)
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u/v
xi ,
i=1
A New Arrangement Inequality
β=
u/v − 1
,
u + 1/(2v) − 1
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The numbers above are simply chosen such that α + β = 1 and αs + βt = u/v.
We briefly check that α, β > 0. The denominator of fractions above is positive,
since u + 1/(2v) ≥ (v + 1/v)/2 ≥ 1. This implies β > 0. Now the positivity
of α > 0 is equivalent to u(1 − v) < 1/2. If v ≥ 1 then u(1 − v) ≤ 0 < 1/2.
So suppose v ≤ 1. By using (2.23), we have:
1
3
1
1
(2.27)
u(1 − v) ≤ v +
(1 − v) = −v 2 + v + < ,
4
4
4
2
for all v ≥ 0. Now we can safely plug α, β, s, t in (2.21) and get
A New Arrangement Inequality
Mohammad Javaheri
β
σ1α σu+1/2v
≥ σu/v .
(2.28)
Next, let α0 = (1 − α)/2 and β 0 = 1 − β/2. Since α0 + β 0 = 1 and α0 , β 0 > 0,
we can use Hölder’s inequality (2.21) with α0 , β 0 instead of α and β (and the
same s, t as before) and get (this time α0 s + β 0 t = u + 1/2):
(2.29)
(1−α)/2 1−β/2
σ1+1/2v
σ1
≤ σu+1/2 .
Now we square the above inequality and multiply it with (2.28) to obtain:
(2.30)
2
2
σ1 σ1+1/2v
≤ σu/v σu+1/2
,
which is equivalent to the inequality (2.25). So far we have shown the existence
of a counterexample to (1.2) for a = b = u when n = N + 1. Then Prop. 2.2
gives N (u, u) ≤ N = N (v, v) and this concludes the proof of the monotonicity
of N .
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It remains to prove that An (a, b) converges to 0 as a+b → 1. To the contrary,
assume there exists > 0 and a sequence (aj , bj ) such that An (aj , bj ) > and
aj + bj → 1. Then by definition, for each j, there exists an n-tuple Xj =
(x1j , . . . , xnj ) such that max xij = 1 and
!2
n
n
n
X
X
X
cj
aj bj
≥ ,
xij
(2.31)
xij
xij xi+1j −
2
i=1
i=1
i=1
where cj = (aj + bj + 1)/2. Since Xj is a bounded sequence, it follows that,
along a subsequence jk , the Xjk ’s converge to some X = (x1 , . . . , xn ). On
the other hand, along a subsequence of jk (denoted again by jk ), ajk → a and
bjk → b for some a, b ≥ 0. Since aj + bj → 1, we have a + b = 1. By taking
the limits of the inequality (2.31) along this subsequence, we should have
!2
n
n
n
X
X
X
(2.32)
xi
xai xbi+1 −
xi
≥ > 0,
2
i=1
i=1
i=1
which contradicts the inequality (1.1). This contradiction establishes the equation (2.22).
The next proposition shows that the inequality (1.2) holds if one mixes up
the order of the xi ’s. The proof is simple and makes use of the monotonicity of
(σt )1/t where σt is defined by the equation (2.20). It is well-known that (σt )1/t
is a non-decreasing function of t [1, Th. 16].
Proposition 2.4. Let a, b, c be as in Proposition 2.1. Then for any given set of
n nonnegative real numbers there exists an arrangement of them as x1 , . . . , xn
such that the inequality (1.2) holds.
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Proof. Equivalently, we show that if x1 , x2 , . . . , xn are nonnegative then there
exists a permutation µ of the set {1, 2, . . . , n} such that the inequality (1.4)
holds. Let
(2.33)
S=
n
X
xi ,
T =
i=1
n X
X
xai xbj .
i=1 j6=i
Then ST = nσ1 (n2 σa σb −nσa+b ) = n3 σ1 σa σb −n2 σ1 σa+b . Now by the CauchySchwarz inequality [4], σc2 ≤ σ1 σa+b . On the other hand by the monotonicity of
1/t
1/c
a/c
b/c
σt , we have σ1 ≤ σc , σa ≤ σc , σb ≤ σc , and so σ1 σa σb ≤ σc2 . It follows
from these inequalities that
(2.34)
ST ≤ n2 (n − 1)σc2 .
Now for a permutation µ of 1, 2, . . . , n, let:
(2.35)
Aµ =
n
X
xaµ(i) xbµ(i+1) .
i=1
We would like to show that SAµ ≤ (nσc )2 for some permutation µ. It is
sufficient to show that the average of SAµ over all permutations µ is less than
or equal to (nσc )2 . To show this, observe that the average of SAµ is equal to
ST /(n − 1) and so the claim follows from the inequality (2.34).
The symmetric group Sn acts on Rn in the usual way, namely for µ ∈ Sn
and (x1 , . . . , xn ) ∈ Rn let
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(2.36)
µ · (x1 , . . . , xn ) = (xµ(1) , . . . , xµ(n) ).
Let R be a region in Rn that is invariant under the action of permutations (i.e.
µ · R ⊆ R for all µ). define:

!2 
n
n
n


X
X X a b
c
(2.37)
λ(R) = (x1 , . . . , xn ) ∈ R xi
xi xi+1 ≤
xi
.


i=1
i=1
i=1
By Proposition 2.4:
(2.38)
R⊆
[
µ · λ(R).
Mohammad Javaheri
µ∈Sn
In particular, by taking the Lebesgue measure of the sides of the inclusion
above, we get
(2.39)
vol R
vol λ(R) ≥
.
n!
We prove a better lower bound for vol λ(R) when n is a prime number (similar
but weaker results can be proved in general).
Proposition 2.5. Let a, b be as in Proposition 2.1 and n be a prime number.
Let R ⊆ Rn+ be a Lebesgue-measurable bounded set that is invariant under the
action of permutations. Let λ(R) denote the set of all (x1 , . . . , xn ) ∈ R for
which the inequality (1.2) holds. Then
(2.40)
vol λ(R) ≥
A New Arrangement Inequality
vol R
.
n−1
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Proof. For m ∈ {1, 2, . . . , n − 1} let µm ∈ Sn and denote the permutation
µm (i) = mi,
(2.41)
where all the numbers are understood to be modulo n (in particular µm (n) = n
for all m). Now recall the definition of Aµ from the equation (2.35) and observe
that:
(2.42)
n−1
X
Aµ m =
m=1
n−1 X
n
X
xami xbmi+m
m=1 i=1
=
n
X
j=1
xaj
=
n−1 X
n
X
xaj xbj+m
m=1 j=1
n−1
X
xbj+m =
m=1
n
X
j=1
xaj
X
Mohammad Javaheri
xbi .
i6=j
Then, the same argument in the proof of Prop. 2.4 implies that, for some m ∈
{1, . . . , n − 1}, we have Aµm ≤ (nσc )2 . We conclude that
(2.43)
R⊆
n−1
[
µm · λ(R),
m=1
which in turn implies the inequality (2.40).
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cambridge University Press, 2nd ed. (1988).
[2] O. HOSSEINI, Private Communication, Fall 2006.
[3] M.H. PROTTER
Springer (1997).
AND
C.B. Jr MORREY, A First Course in Real Analysis,
[4] M.J. STEELE, The Cauchy-Schwarz Master Class: An Introduction to the
Art of Mathematical Inequalities, Cambridge University Press (2004).
A New Arrangement Inequality
Mohammad Javaheri
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J. Ineq. Pure and Appl. Math. 7(5) Art. 162, 2006
http://jipam.vu.edu.au