Journal of Inequalities in Pure and
Applied Mathematics
NEW INEQUALITIES BETWEEN ELEMENTARY
SYMMETRIC POLYNOMIALS
volume 4, issue 2, article 48,
2003.
TODOR P. MITEV
University of Rousse
Department of Mathematics
Rousse 7017, Bulgaria.
E-Mail: mitev@ami.ru.acad.bg
Received 7 November, 2002;
accepted 21 February, 2003.
Communicated by: C.P. Niculescu
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c 2000 Victoria University
ISSN (electronic): 1443-5756
116-02
Quit
Abstract
New families of sharp inequalities between elementary symmetric polynomials
are proven. We estimate σn−k above and below by the elementary symmetric
polynomials σn−k+1 , . . . , σn in the case, when x1 , . . . , xn are non-negative real
numbers with sum equal to one.
2000 Mathematics Subject Classification: 26D05.
Key words: Elementary symmetric polynomials.
Contents
1
2
3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
New Inequalities (Theorem 2.3 and Theorem 2.5) . . . . . . . . . 5
The Sharpness of the Inequalities (2.8) and (2.14) . . . . . . . . . 21
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
1.
Introduction
Let n ≥ 2 be an integer. As usual, we denote by σ 0 , σ 1 , . . . , σ n the elementary
symmetric polynomials of the variables x1 , . . . , xn .
In other words, σ 0 = σ 0 (x1 , . . . , xn ) = 1 and for 1 ≤ k ≤ n
X
xi1 . . . xik .
σ k = σ k (x1 , . . . , xn ) =
1≤i1 ≤···≤ik ≤n
The different σ 0 , σ 1 , . . . , σ n , are not comparable between them, but they are
connected by nonlinear inequalities.
To state them, it is more convenient to
consider their averages Ek = σ k nk , k = 0, 1, . . . , n.
There are three basic types of inequalities between the symmetric functions
with respect to the range of the variables x1 , . . . , xn .
For arbitrary real x1 , . . . , xn the following inequalities are known:
(1.1)
Ek2 ≥ Ek−1 Ek+1 ,
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
1 ≤ k ≤ n − 1,
(Newton-Maclaurin),
2
2
) ≥ (Ek+1 Ek+2 − Ek Ek+3 )2 ,
4(Ek+1 Ek+3 − Ek+2
)(Ek Ek+2 − Ek+1
k = 0, . . . , n − 3, (Rosset [4]),
JJ
J
II
I
Go Back
Close
as well as the inequalities of Niculescu [2]. A complete description about their
historical and contemporary stage of development can be found, for example,
in [1] and [2].
Suppose now that all xj , j = 1, . . . , n, are positive. Then the following
general result (see [1, Theorem 77, p. 64]) is known:
Quit
Page 3 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Theorem 1.1 ( Hardy, Littlewood, Pólya). For any positive x1 , . . . , xn and
positive α1 , . . . , αn , β 1 , . . . , β n the inequality
β
E1α1 · · · Enαn ≤ E1 1 · · · Enβ n
holds if and only if
αm + 2αm+1 + · · · + (n − m + 1)αn ≥ β m + 2β m+1 + · · · + (n − m + 1)β n
for each 1 ≤ m ≤ n.
For other results in this direction see [1].
The aim of this paper is to obtain new inequalities between σ 1 , . . . , σ n in the
case when x1 , . . . , xn are non-negative, (Theorem 2.3 and Theorem 2.5 below).
More precisely, we obtain the best possible estimates of σ k1 σ n−k from below and
0
above by linear functions of σ k−1
1 σ n−k+1 , . . . , σ 1 σ n . Since all these functions
are homogeneous with respect to (x1 , . . . , xn ) of the same order, we can set
σ 1 = x1 + · · · + xn = 1, then our inequalities give the best possible estimates
of σ n−k by linear functions of σ n−k+1 , . . . , σ n for k = 1, . . . , n − 1 in this
case (Theorem 3.1 and Theorem 3.2 below). Inequalities of this type for k =
n − 2 have been recently obtained by Sato [4], which can be obtained as a
consequence of Theorem 2.5 below.
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 4 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
2.
New Inequalities (Theorem 2.3 and Theorem 2.5)
For the sake of completeness we give a straightforward proof of the following
proposition, which is a consequence of Theorem 1.1, cited in the introduction.
Here we suppose that x1 , . . . , xn are non-negative.
Proposition 2.1. Let x1 , . . . , xn be non-negative real numbers, n ≥ 2. Then for
1 ≤ p ≤ n − 1 we have
σ1σp ≥
(2.1)
Proof. Denote σ l,n =
is equivalent to
P
n(p + 1)
σ p+1 .
n−p
1≤i1 <···<il ≤n
xi1 xi2 · · · xil , 1 ≤ l ≤ n. Note, that (2.1)
Todor P. Mitev
σ 1,n σ p,n ≥
(2.2)
n(p + 1)
σ p+1,n .
n−p
First we shall check (2.2) for p = 1 and for p = n − 1.
(i) For p = 1 the inequality (2.2) reads
!2
n
X
X
2n
xi
≥
xi xj ,
n
−
1
i=1
1≤i<j≤n
(n − 1)
n
X
!2
xi
i=1
P
1≤i<j≤n (xi
Title Page
Contents
JJ
J
II
I
Go Back
Close
which is equivalent to
hence to
New Inequalities Between
Elementary Symmetric
Polynomials
− xj )2 ≥ 0.
X
2n
≥
xi xj ,
n − 1 1≤i<j≤n
Quit
Page 5 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
(ii) For p = n − 1 (2.2) is equivalent to σ 1,n σ n−1,n ≥ n2 σ n,n . If σ n,n = 0,
then (2.2) is obvious.
Let σ n,n 6= 0, then (2.2) is equivalent to n2 ≤
Pn
Pn 1 ( i=1 xi )
i=1 xi , which follow from AM-GM inequality.
We are going to prove (2.2) by recurrence with respect to n ≥ 2.
(iii) We already proved that (2.2) is true for n = 2.
(iv) Let (2.2) be true for n ≥ 2 and for each p, 1 ≤ p ≤ n − 1. Fix p,
2 ≤ p ≤ n − 1. We will prove, that
σ 1,n+1 σ p,n+1 ≥
(2.3)
(n + 1)(p + 1)
σ p+1,n+1 .
n+1−p
Since (2.3) is homogeneous, excluding the case x1 = · · · = xn = xn+1 =
0, we may assume, that σ 1,n+1 = 1.
Let x1 ≤ x2 ≤ · · · ≤ xn+1 . The following cases are possible:
1) Let xn+1 = 1. Then x1 = · · · = xn = 0 and (2.3) becomes an equality.
2) Let xn+1 =
1
.
n+1
Then x1 = · · · = xn = xn+1 =
1
n+1
and we obtain
(n + 1)(p + 1)
σ p,n+1 −
σ p+1,n+1
n+1−p
p+1
1
1
n+1
n+1
−
= 0,
=
p
p
(n + 1)
n + 1 − p p + 1 (n + 1)p
hence (2.3) becomes again an equality.
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 6 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
1
3) Let xn+1 ∈ n+1
; 1 . Substitute x1 + · · · + xn = 1 − xn+1 = σ 1,n = s,
n
. Then σ p,n+1 = σ p,n + (1 − s)σ p−1,n and σ p+1,n+1 =
with s ∈ 0; n+1
σ p+1,n + (1 − s)σ p,n . Hence (2.3) is equivalent to
σ p,n + (1 − s)σ p−1,n ≥
(n + 1)(p + 1)
[σ p+1,n + (1 − s)σ p,n ] ,
n+1−p
which is equivalent to
(2.4)
n+1−p
(1 − s)(n + 1 − p)
− (1 − s) σ p,n +
σ p−1,n
(n + 1)(p + 1)
(n + 1)(p + 1)
≥ σ p+1,n .
From (iv) we obtain σ p+1,n ≤
next inequality (if true):
n−p
sσ p,n .
n(p+1)
Then (2.4) follows from the
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
(2.5)
n−p
sσ p,n
n(p + 1)
n+1−p
(1 − s)(n + 1 − p)
≤
− (1 − s) σ p,n +
σ p−1,n ,
(n + 1)(p + 1)
(n + 1)(p + 1)
σ p−1,n
II
I
Go Back
Close
which is equivalent to
(2.6)
JJ
J
p[n(n + 2) − (n + 1)2 s]
≥
σ p,n .
n(n + 1 − p)(1 − s)
It follows from (iv) that σ p−1,n ≥
np
σ .
(n+1−p)s p,n
Quit
Page 7 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Hence (2.6), and consequently (2.5) and (2.4), follow from
np
p[n(n + 2) − (n + 1)2 s]
σ p,n −
σ p,n
(n + 1 − p)s
n(n + 1 − p)(1 − s)
p[(n + 1)s − n]2
=
σ p,n ≥ 0.
ns(n + 1 − p)(1 − s)
Since (2.2) is true for p = 1 and p = n according to (i) and (ii), then (2.2) is
fulfilled for n, n ≥ 2. Hence the proposition is proved.
Remark 2.1. It follows from the proof, that equality is achieved in the following
two cases:
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
1) x1 = x2 = · · · = xn = a ≥ 0.
2) n − p + 1 of x1 , . . . , xn are equal to 0 and the rest of them are arbitrary
non-negative real numbers.
Remark 2.2. (2.1) can be proven using Lemma 2.2 below, but in this way it will
be difficult to see when (2.1) turns into an equality.
From now on n will be a fixed positive integer. It will be assumed that at
least one of the non-negative numbers x1 , . . . , xn differs from zero.
Lemma 2.2. Let us assume that x1 , . . . , xn are non-negative real numbers (n ≥
2) and x1 + · · · + xn = σ 1 = 1. Then the function f (x1 , . . . , xn ) = a1 + a2 σ 2 +
· · · + an σ n (a1 , . . . , an are real numbers), achieves its maximum and minimum
at least in some of the points Pk,n k1 , . . . , k1 , 0, . . . , 0 , 1 ≤ k ≤ n (the first k
coordinates of Pk,n are equal to k1 , and the rest of them are equal to zero).
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 8 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Proof. The set An = {(x1 , . . . , xn )/xi ≥ 0, x1 + · · · + xn = 1} is compact and
f is continuous in it, hence f achieves its minimum and maximum values. We
rewrite f as follows:
f (x1 , . . . , xn ) = x1 x2 g(x3 , . . . xn ) + x1 h1 (x3 , . . . , xn )
+ x2 h2 (x3 , . . . , xn ) + t(x3 , . . . , xn ) + a1 .
As f is symmetric, then h1 ≡ h2 and therefore:
(2.7) f (x1 , . . . , xn )
= x1 x2 g(x3 , . . . xn ) + (x1 + x2 )h1 (x3 , . . . , xn ) + t(x3 , . . . , xn ) + a1 .
Let P (x01 , . . . , x0n ) be a point in which f achieves its minimum value. We consider the function F (x) = f (x, s − x, x03 , . . . , x0n ), s = x01 + x02 , for x ∈ [0; s]
(we assume, that s > 0). Obviously the minimum values of F and f are
equal and F achieves its minimum value for x = x01 . From (2.7) we obtain
that F (x) = αx(s − x) + sβ + γ = αx(s − x) + δ, where α, δ depend on
x01 , x02 , x03 , . . . , x0n , a1 , . . . , an .
The following three cases are possible:
(i) α = 0. Then F
(x) = const and we may assume that min F = F (0) or
s
min F = F 2 .
(ii) α > 0. Then min F = F (0).
(iii) α < 0. Then min F = F 2s .
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 9 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Hence, as x01 and x02 were arbitrarily chosen then, for ∀i 6= j we may assume
that x0i = x0j or, at least one of them is equal to zero.
Let us choose a point P (x01 , . . . , x0n ), for which the number of coordinates p
which equal to zero is the highest possible and x01 ≥ x02 ≥ · · · ≥ x0n . If p = n −
1, then Lemma 2.2 is proven. Let 0 ≤ p ≤ n − 2, i.e. P (x01 , ..x0n−p , 0, . . . , 0),
x01 · · · x0n−p 6= 0. Then for the pairs (x0i , x0j ), 1 ≤ i < j ≤ n − p only case
(iii) is valid, from which Lemma 2.2 follows. Lemma 2.2 is true also for the
maximum value of f , since max f = min(−f ).
Remark 2.3. A result similar to Lemma 2.2 is proved by Sato in [4].
Theorem 2.3. Let n, k be integer numbers, 1 ≤ k ≤ n − 1. Then for arbitrary
non-negative x1 , . . . , xn , the following inequality is true:
(2.8) σ k1 σ n−k
≥
k
X
i=1
Todor P. Mitev
Title Page
(−1)i+1
n−k−1+i
i
(n − k + i)2 (n − k)i−2 σ k−i
1 σ n−k+i .
Proof. Since (2.8) is homogenous we may assume that x1 + · · · + xn = σ 1 = 1.
Then, according to Lemma 2.2 it suffices to prove, that f (Pm,n ) ≥ 0 for 1 ≤
m ≤ n, where
k X
n−k−1+i
f (x1 , . . . , xn ) = σ n−k +
(n − k + i)2 (k − n)i−2 σ n−k+i .
i
i=1
1
m
At the Pm,n point we have σ n−k+i = n−k+i
, hence
mn−k+i
(2.9)
New Inequalities Between
Elementary Symmetric
Polynomials
σ n−k+i 6= 0 if and only if
i ≤ m − n + k.
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 10 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
We consider the following three possible cases for m:
(i) m ≤ n − k − 1, k ≤ n − 2. Then obviously σ n−k = σ n−k+1 = · · · =
σ n = 0, hence f (Pm,n ) = 0.
(ii) m = n − k, k ≤ n − 1. From (2.9) we obtain σ n−k =
1
σ n−k+1 = · · · = σ n = 0, hence f (Pm,n ) = (n−k)
n−k > 0.
1
(n−k)n−k
and
(iii) m = n − k + p, 1 ≤ p ≤ k, k ≤ n − 1. From (2.9) and m = n − k + p we
obtain
k X
1
n−k+p
n−k−1+i
f (Pm,n ) =
+
n−k
i
(n − k + p)n−k
i=1
1
n−k+p
× (n − k + i)2 (k − n)i−2
n − k + i (n − k + p)n−k+i
p X
1
m
m−p−1+i
=
+
p
i
mm−p i=1
1
m
2
i−2
× (m − p + i) (p − m)
.
m−p+i
m−p+i m
Now from equality
m−p−1+i
m
m−1
p
(m − p + i) =
m
i
m−p+i
p
i
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 11 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
we obtain
1
m
f (Pm,n ) =
p
mm−p
p X m − 1 p 1
+
(m − p + i)2 (p − m)i−2 m−p−1+i .
p
i
m
i=1
This implies
m−p+1
m
f (Pm,n )
m−1
p
i−2
p X
m
p−m
m2
p
=
+ p(m − p + 1)
+
(m − p + i)
i
m−p
p − m i=2
m
p i−2
X p
p−m
= m(1 − p) +
(m − p)
i
m
i=2
i−2
p X
p−m
p
+
i
i
m
i=2
p
m2
p−m
p(p − m)
= m(1 − p) +
1+
−
−1
m−p
m
m
i−1
p mp X p − 1
p−m
+
.
i−1
p − m i=2
m
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 12 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Substituting i = j + 1 we obtain:
mm−p+1
f (Pm,n )
m−1
p
m2 p p p(m − p)
+
= m(1 − p) +
−1
m−p m
m
j
p−1 mp X p − 1
p−m
+
j
p − m j=1
m
m 2 p p
m2
= m(1 − p) +
+ mp −
m−p m
m−p
"
#
p−1
p−m
mp
+
1+
−1
p−m
m
m 2 p p
m2
mp p p−1
mp
=m+
−
+
−
= 0.
m−p m
m−p p−m m
p−m
From (i) – (iii) it follows that Theorem 2.3 is true.
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Remark 2.4. Theorem 2.3 for k = 1 is equivalent to Proposition 2.1 in the case
when p = n − 1.
Remark 2.5. It is easy to verify, that (2.8) is equivalent to
i−1
k 1X k
k−n
k
(n − k + i)
E1 En−k ≥
E1k−i En−k+i .
i
n i=1
n
Close
Quit
Page 13 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
We define the sequence of real numbers {αm,l }, m ∈ N, l ∈ N as follows:
1
ll
=0
α1,l =
(2.10)
(2.11)
(2.12)
∀l ∈ N,
for
αm,l
for
m ≥ l ≥ 2 or m > 1, l = 1,
m
X
l
l
m
l
l = l α1,l−m +
lm−j α1+j,l−m+j
m
m−j
j=1
for 1 ≤ m ≤ l − 1.
More precisely, the numbers αm,l can be defined recurrently (excluding the
cases when: m > 1, l = 1 or m ≥ l ≥ 2) as follows:
1) We get α1,l for l ≥ 1 from (2.10).
Todor P. Mitev
Title Page
l
1
l = ll α1,l−1 + α2,l .
3) Then we determine α3,l for l ≥ 4 from 2l l2 = ll α1,l−2 + 1l lα2,l−1 +α3,l .
4) Then we determine α4,l for l ≥ 5 from 3l l3 = ll α1,l−3 + 2l l2 α2,l−2 +
l
lα3,l−1 + α4,l and so on.
1
2) Then we determine α2,l for l ≥ 3 from
New Inequalities Between
Elementary Symmetric
Polynomials
For example, the values of αm,l for m ≤ 5, l ≤ 6 are given in Table 1.
The sequence {αm,l } has interesting properties. For example one can prove,
that in the case when αm,l 6= 0: sgn αm,l = 1 for m even and sgn αm,l = −1 for
m odd, m ≥ 3.
We are going to prove the following property of the sequence {αm,l }:
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 14 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Proposition 2.4. For each integer number n, n ≥ 2 we have:
2
n+1
n
(2.13)
αn,n+1 = (−1)
.
2
Proof. We will prove (2.12) by induction.
2
(i) We show, that α2,3 = (−1)2 2+1
, (see Table 1).
2
(ii) Let (2.13) hold true for α2,3 , . . . , αn−1,n .
(iii) Using (2.12) for l = n + 1 and m = n − 1, (2.10) for l = 2 and (ii) we
obtain
n+1
(n + 1)n−1
2
2
n−2 j+2
(n + 1)n+1 X n + 1
j+1
=
+
(−1)
(n+1)n−1−j +αn,n+1 .
j+2
4
2
j=1
Substituting j = i − 1, this implies
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
αn,n+1 =
n+1
2
Close
(n + 1)n−1
n−1
(n + 1)n+1 1 X
−
−
4
4 i=2
Quit
n+1
i+1
(−1)i (i + 1)2 (n + 1)n−i .
Page 15 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Now from the equalities n+1
(i + 1) = ni (n + 1) and ni i = n−1
n
i+1
i−1
we obtain:
(n + 1)n−1
n+1
αn,n+1 =
(n + 1)n−1 −
2
4
n−1
n+1X n
−
(−1)i (i + 1)(n + 1)n−i
i
4 i=2
"
i #
n−1 X
2n
−1
(n + 1)n+1
n
−1−
(i + 1)
=
i
4
n+1
n+1
i=2
"
i
n−1 (n + 1)n+1 n − 1 X n
−1
=
−
i
4
n + 1 i=2
n+1
i #
n−1 X
−1
n−1
−n
.
i−1
n+1
i=2
Substituting i = k + 1 we obtain
n
−1
−1
(n + 1)n+1 n − 1
αn,n+1 =
− 1+
+1+n
4
n+1
n+1
n+1
n
k+1 #
n−2
X
−1
−1
n−1
−n
+
k
n+1
n+1
k=1
n
n
n
n
−1
(n + 1)n+1
−
+
=
4
n+1
n+1
n+1
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 16 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
"
n−1
n−1 #)
n
−1
−1
+
1+
−1−
n+1
n+1
n+1
n
n
n
n
−1
(n + 1)n+1
=
−
+
4
n+1
n+1
n+1
n−1
n−1 #
n
n
n
n
−1
+
−
−
n+1 n+1
n+1 n+1 n+1
2
(n + 1)n+1
1
n
n+1
n
n
(−1)
+
= (−1)
.
=
4
(n + 1)n (n + 1)n
2
From (i), (ii) and (iii) it follows that (2.13) is true for each n ≥ 2.
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Theorem 2.5. Let n and k be fixed integer numbers for which 1 ≤ k ≤ n − 2.
Then for arbitrary non-negative x1 , . . . , xn , the following inequality is fulfilled:
(2.14)
σ k1 σ n−k
≤
α1,n−k σ n1
+
k
X
α1+i,n−k+i σ k−i
1 σ n−k+i ,
i=1
where {αm,l } are defined from (2.10)-(2.12).
Contents
JJ
J
II
I
Go Back
Close
Quit
Proof. (2.14) is homogenous, therefore we may assume, that x1 + · · · + xn =
σ 1 = 1. Then according to Lemma 2.2 it is sufficient to prove, that
(2.15)
f (Pm,n ) ≥ 0,
for each m, 1 ≤ m ≤ n,
Page 17 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
where
f (x1 , . . . , xn ) = α1,n−k +
k
X
α1+i,n−k+i σ n−k+i − σ n−k .
i=1
Obviously at the point Pm,n we have σ q =
(2.16)
σ q 6= 0
m
q
if and only if
1
mq
for 1 ≤ q ≤ n, hence
q ≤ m.
We consider the following three possible cases for m:
(i) m ≤ n − k − 1. Then from (2.16) and (2.10) we obtain f (Pm,n ) =
1
α1,n−k = (n−k)
n−k > 0.
(ii) m = n − k. Then from (2.16) and (2.10) we obtain f (Pn−k,n ) = α1,n−k −
1
= 0.
(n−k)n−k
(iii) m = n − k + p, where 1 ≤ p ≤ k. From (2.16) it follows
f (Pm,n ) = α1,n−k +
k
X
α1+i,n−k+i
n−k+p
n−k+i
1
(n − k + p)n−k+i
i=1
1
n−k+p
−
n−k
(n − k + p)n−k
1
(n − k + p)n−k+p α1,n−k
=
(n − k + p)n−k+p
k X
n−k+p
+
(n − k + p)p−i α1+i,n−k+i
n−k+i
i=1
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 18 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
−
n−k+p
n−k
However, n−k+p
6= 0 for i ≤ p, and
n−k+i
to (2.10), and we get
p
(n − k + p)
1
(n−k+p)n−k+p
.
= α1,n−k+p according
(2.17) f (Pm,n ) = α1,n−k+p (n − k + p)n−k+p α1,n−k
p X
n−k+p
+
(n − k + p)p−i α1+i,n−k+i
p−i
i=1
n−k+p
p
−
(n − k + p) .
p
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Obviously α1,n−k = α1,(n−k+p)−p and α1+i,n−k+i = α1+i,(n−k+p)−p+i . Then
the right hand side of (2.17) is equal to zero according (2.12) for l = n − k + p
and m = p.
Therefore f (Pm,n ) = 0 in this case.
It follows from (i), (ii) and (iii) that (2.15) is true, and hence (2.14) is also
true.
Contents
JJ
J
II
I
Go Back
Remark 2.6. Theorem 2.5 is true as well for k = n − 1, since both sides of
(2.14) are equal in this case, which follows from (2.11).
Close
Remark 2.7. An analogue of Theorem 2.5 for k = 0 is the inequality between
the arithmetic and geometric means.
Page 19 of 25
Quit
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Corollary 2.6. Let An , Gn , Hn be the classical averages of the positive real
numbers x1 , . . . , xn (n ≥ 2). Then the following inequality is true:
"
n−1
n−1 #
nAn
1
1
1
n
+ n− 1+
≥
.
(2.18)
(n − 1)Gn
Gn
n−1
An
Hn
Proof. (2.18) follows from:
nGnn
, σ n = Gnn ,
Hn
1
nn
2
=
,
α
=
n
−
2,n
(n − 1)n−1
(n − 1)n−1
σ 1 = nAn , σ n−1 =
α1,n−1
and from Theorem 2.5 for k = 1.
1
σ 1n−2 σ 2 ≤ σ n1 +
4
i=1
Todor P. Mitev
Title Page
Corollary 2.7 (Explicit expression of Theorem 2.5 for k = n − 2). For each
integer number n (n ≥ 3) we have:
n−2
X
New Inequalities Between
Elementary Symmetric
Polynomials
(−1)i+1
i+2
2
2
σ n−2−i
σ 2+i .
1
Proof. It follows from Proposition 2.4 and from Theorem 2.5 for k = n − 2.
Remark 2.8. Corollary 2.7 is the principle result in [4].
Remark 2.9. Corollary 2.7 shows that Theorem 2.5 for k = n − 2 is equivalent
to Theorem 2.3 in the case when k = n − 1.
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 20 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
3.
The Sharpness of the Inequalities (2.8) and (2.14)
The following two theorems prove that the estimates in Theorem 2.3 and Theorem 2.5 are, in a certain sense, the best possible.
Theorem 3.1. Let n and k, 1 ≤ k ≤ n − 1 be integers. Let the real numbers
β 1 , . . . , β k have the property (3.1). We say that the real numbers β 1 , . . . , β k
have the property (3.1) if for any non-negative real numbers x1 , . . . , xn with a
sum equal to one the following inequality is fulfilled:
σ n−k ≥
(3.1)
k
X
β i σ n−k+i .
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
i=1
Then for arbitrary non-negative real numbers x1 , . . . , xn with sum equal to one
the following inequality is fulfilled:
Title Page
Contents
(3.2)
k
X
JJ
J
β i σ n−k+i
i=1
≤
k
X
i=1
i+1
(−1)
n−k−1+i
i
(n − k + i)2 (n − k)i−2 σ n−k+i
II
I
Go Back
Close
Proof. Set f1 = f1 (x1 , . . . , xn ) = σ n−k −
Pk
i=1
β i σ n−k+i and
k X
n−k+i
f2 = f2 (x1 , . . . , xn ) = σ n−k +
(n−k+i)2 (k−n)i−2 σ n−k+i .
i
i=1
Quit
Page 21 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Then (3.2) is equivalent to f1 − f2 ≥ 0. On the other hand, according to Lemma
2.2, it is sufficient to verify this inequality at the points Pm,n . We have at these
points:
(i) For 1 ≤ m ≤ n − k − 1, k ≤ n − 2 apparently f1 = f2 = 0, hence
f1 − f2 = 0.
(ii) For m = n − k, k ≤ n − 1 we obtain f1 = f2 =
0.
1
,
(n−k)n−k
hence f1 − f2 =
(iii) For 1 ≤ n − k < m ≤ n from the proof of Theorem 2.3 it follows, that
f2 = 0. As f1 ≥ 0 according to (3.1), hence f1 − f2 ≥ 0.
New Inequalities Between
Elementary Symmetric
Polynomials
From (i), (ii) and (iii) it follows that f1 − f2 ≥ 0 in each point Pm,n and we
complete the proof of the theorem.
Todor P. Mitev
Theorem 3.2. Let n and k be integers, 1 ≤ k ≤ n − 2. Let the real numbers
γ 1 , . . . , γ k+1 have the property (3.3). We say that the real numbers γ 1 , . . . , γ k+1
have the property (3.3) if for any non-negative real numbers x1 , . . . , xn with
sum equal to one, the following inequality is fulfilled:
Title Page
σ n−k ≤ γ 1 +
(3.3)
k
X
γ i+1 σ n−k+i .
Then for any non-negative real numbers x1 , . . . , xn with sum equal to one the
following inequality is fulfilled:
(3.4)
α1,n−k +
i=1
α1+i,n−k+i σ n−k+i ≤ γ 1 +
k
X
i=1
JJ
J
II
I
Go Back
i=1
k
X
Contents
Close
Quit
Page 22 of 25
γ 1+i σ n−k+i .
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Proof. Set
f1 = f1 (x1 , . . . , xn ) = γ 1 +
k
X
γ 1+i σ n−k+i − σ n−k
i=1
and
f2 = f2 (x1 , . . . , xn ) = α1,n−k +
k
X
α1+i,n−k+i σ n−k+i − σ n−k .
i=1
Then (3.4) is equivalent to f1 − f2 ≥ 0. We are going to check this inequality
at the points Pm,n . From (3.3) at Pn−k,n it follows, that
(3.5)
γ1 ≥
1
= α1,n−k .
(n − k)n−k
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
We consider the possible cases for m:
(i) 1 ≤ m ≤ n − k − 1. Then f1 − f2 = γ 1 − α1,n−k ≥ 0 at Pm,n according
to (3.5).
(ii) n − k ≤ m ≤ n. Then f1 ≥ 0 at Pm,n according to (3.3) and from the
proof of Theorem 2.5 it follows that f2 = 0, therefore f1 − f2 ≥ 0.
JJ
J
II
I
Go Back
Close
Quit
From (i) and (ii) we obtain, that f1 −f2 ≥ 0 in each point Pm,n (1 ≤ m ≤ n).
Applying Lemma 2.2 we complete the proof of Theorem 3.2.
Page 23 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
Table 1:
l
1
2
3
4
5
6
α1,l
1
1/4
1/27
1/256
1/3125
1/46656
α2,l
0
0
9/4
176/27
3275/256
65844/3125
α3,l
0
0
0
-4
-775/27
-6579/64
α4,l
0
0
0
0
25/4
316/3
α5,l
0
0
0
0
0
-9
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 24 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au
References
[1] G. HARDY, J.E. LITTLEWOOD AND G.PÓLYA,Inequalities, Cambridge
Mathematical Library 2nd ed., 1952.
[2] C.P. NICULESCU, A new look at Newton’s inequalities, J. Inequal. Pure
and Appl. Math., 1(2) (2000), Article 17. [ONLINE: http://jipam.
vu.edu.au/v1n2/014_99.html]
[3] S. ROSSET, Normalized symmetric functions, Newton inequalities and a
new set of stronger inequalities. Amer. Math. Soc., 96 (1989), 815–820.
[4] N. SATO, Symmetric polynomial inequalities, Crux Mathematicorum with
Mathematical Mayhem, 27 (2001), 529–533.
New Inequalities Between
Elementary Symmetric
Polynomials
Todor P. Mitev
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 25 of 25
J. Ineq. Pure and Appl. Math. 4(2) Art. 48, 2003
http://jipam.vu.edu.au