Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 162, 2006
A NEW ARRANGEMENT INEQUALITY
MOHAMMAD JAVAHERI
U NIVERSITY OF O REGON
D EPARTMENT OF M ATHEMATICS
F ENTON H ALL , E UGENE , OR 97403
javaheri@uoregon.edu
URL: http://www.uoregon.edu/~ javaheri
Received 29 April, 2006; accepted 17 November, 2006
Communicated by G. Bennett
A BSTRACT. In this paper, we discuss the validity of the inequality
!2
n
n
n
X
X
X
(1+a+b)/2
a b
xi
xi xi+1 ≤
xi
,
i=1
i=1
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.
Key words and phrases: Inequality, Arrangement.
2000 Mathematics Subject Classification. 26Dxx.
1. M AIN S TATEMENTS
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
(1.2)
xi
xi xi+1 ≤
xi ,
i=1
i=1
i=1
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
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.
126-06
2
M OHAMMAD JAVAHERI
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)
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
xci .
xaµ(i) xbµ(i+1) ≤
i=1
i=1
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 (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:
!2
n
n
n
X
X
X
3/2
(1.6)
xi
xi xi+1 ≤
.
xi
i=1
i=1
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,
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.
2. P ROOFS
Applying Jensen’s inequality [1, § 3.14] to the concave function log x gives
(2.1)
ur v s wt ≤ ru + sv + tw,
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
(2.3)
u = (yz)c , v = (xz)c , w = (xy)c ,
b
1
a
r =1− , s=1− , t=1− ,
c
c
c
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
http://jipam.vu.edu.au/
A N EW A RRANGEMENT I NEQUALITY
3
and obtain:
(2.4)
b
a
1
c
c
x y z ≤ 1−
(yz) + 1 −
(xz) + 1 −
(xy)c .
c
c
c
a b
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 ,
w = 1,
b
r =1− ,
c
b
s= ,
c
t = 0,
and get
(2.7)
x
a+1 b
y ≤
b
1−
c
b
x2c + (xy)c .
c
Similarly, by interchanging a and b, one has
a 2c a
x + (xy)c .
(2.8)
xa y b+1 ≤ 1 −
c
c
Adding the inequalities (2.4), (2.5), (2.7) and (2.8) gives:
3
a
b
1 2c
c
c
a b
(xy) + 1 −
(yz) + 1 −
(tx)c
(2.9)
Sx y ≤ x + 4 −
c
c
c
c
a
b
c
(yt) + 1 −
(xz)c ,
+ 1−
c
c
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
and
P Pin (2.2)
P so the inequalities (2.4) – (2.8) are all equalities. If a = 0 then we have
x xb = ( xc )2 and so, by the equality case of CauchySchwarz, 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.
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
http://jipam.vu.edu.au/
4
M OHAMMAD JAVAHERI
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 , . . . , xn .
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 Intermediatevalue theorem [3, Th 3.3], there exists ti ∈ Ii such that f (ti ) = yi . By the definition of integral
we have:
Z
Z
n
n
1 X X a+b
a+b
(2.13)
f (x)dx f (x)dx = lim 2
xi
yi
n→∞ n
I
I
i=1
i=1
n
n
1 X X a b
x
xi xi+1
i
n→∞ n2
i=1
i=1
!2 Z
2
n
1 X c
c
xi
=
f (x)dx ,
≤ lim 2
n→∞ n
I
i=1
= lim
where we have applied the inequality (1.2) to the xi ’s. On the other hand, by the CauchySchwarz inequality for integrals, we have
Z
2 Z
2
Z
Z
a+b
1
a+b
c
(2.14)
f (x)dx f (x)dx ≥
f 2 (x)f 2 (x)dx =
f (x)dx ,
I
I
I
I
with equality iff f and f a+b 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
yi ,
i=1
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 .
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
http://jipam.vu.edu.au/
A N EW A RRANGEMENT I NEQUALITY
5
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
k
X
(2.18)
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).
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
1X t
σt =
x.
n i=1 i
(2.20)
By the Hölder inequality [1], if α, β > 0 and α + β = 1 then for any s, t > 0 we have:
σsα σtβ ≥ σαs+βt .
(2.21)
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
u/v
We show that the nonnegative numbers yi = xi , 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
X
X
u+1/2v
u/v
u+1/2
(2.25)
xi
xi ≥
xi
xi .
i=1
i=1
i=1
i=1
To prove this, first let
(2.26)
u + 1/(2v) − u/v
,
u + 1/(2v) − 1
1
s = 1, t = u + .
2v
α=
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
β=
u/v − 1
,
u + 1/(2v) − 1
http://jipam.vu.edu.au/
6
M OHAMMAD JAVAHERI
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
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):
(1−α)/2 1−β/2
σ1+1/2v
σ1
(2.29)
≤ σu+1/2 .
Now we square the above inequality and multiply it with (2.28) to obtain:
2
2
≤ σu/v σu+1/2
,
σ1 σ1+1/2v
(2.30)
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 .
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.
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
i=1
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
xi ,
T =
n X
X
xai xbj .
i=1 j6=i
http://jipam.vu.edu.au/
A N EW A RRANGEMENT I NEQUALITY
7
Then ST = nσ1 (n2 σa σb − nσa+b ) = n3 σ1 σa σb − n2 σ1 σa+b . Now by the Cauchy-Schwarz
1/t
inequality [4], σc2 ≤ σ1 σa+b . On the other hand by the monotonicity of σt , we have σ1 ≤
1/c
a/c
b/c
σc , σa ≤ σc , σb ≤ σc , and so σ1 σa σb ≤ σc2 . It follows from these inequalities that
ST ≤ n2 (n − 1)σc2 .
(2.34)
Now for a permutation µ of 1, 2, . . . , n, let:
n
X
(2.35)
Aµ =
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 ) ∈
R let
n
µ · (x1 , . . . , xn ) = (xµ(1) , . . . , xµ(n) ).
(2.36)
n
Let R be a region in R that is invariant under the action of permutations (i.e. µ · R ⊆ R for all
µ). define:

!2 
X
n
n
n


X
X
xi
xai xbi+1 ≤
(2.37)
λ(R) = (x1 , . . . , xn ) ∈ R xci
.


i=1 i=1
i=1
By Proposition 2.4:
R⊆
(2.38)
[
µ · λ(R).
µ∈Sn
In particular, by taking the Lebesgue measure of the sides of the inclusion above, we get
vol R
.
(2.39)
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
vol R
(2.40)
vol λ(R) ≥
.
n−1
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
Aµm =
m=1
n−1 X
n
X
xami xbmi+m
m=1 i=1
=
n
X
j=1
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
xaj
n−1
X
m=1
=
n−1 X
n
X
xaj xbj+m
m=1 j=1
xbj+m =
n
X
j=1
xaj
X
xbi .
i6=j
http://jipam.vu.edu.au/
8
M OHAMMAD JAVAHERI
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).
R EFERENCES
[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 AND C.B. Jr MORREY, A First Course in Real Analysis, Springer (1997).
[4] M.J. STEELE, The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical
Inequalities, Cambridge University Press (2004).
J. Inequal. Pure and Appl. Math., 7(5) Art. 162, 2006
http://jipam.vu.edu.au/