A MULTINOMIAL EXTENSION OF AN
INEQUALITY OF HABER
HACÈNE BELBACHIR
USTHB/ Faculté de Mathématiques
BP 32, El Alia, 16111 Bab Ezzouar
Alger, Algeria.
EMail: hbelbachir@usthb.dz
Received:
10 July, 2008
Accepted:
17 September, 2008
Communicated by:
L. Tóth
2000 AMS Sub. Class.:
05A20, 05E05.
Key words:
Haber inequality, multinomial coefficient, symmetric functions.
Abstract:
In this paper, we establish the following: Let a1 , a2 , . . . , am be non negative real
numbers, then for all n ≥ 0, we have
a + a + · · · + a n
X
1
1
2
m
.
ai11 ai22 · · · aimm ≥
n+m−1
m
m−1
i +i +···+i =n
1
2
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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m
The case m = 2 gives the Haber inequality. We apply the result to find lower
bounds for the sum of reciprocals of multinomial coefficients and for symmetric
functions.
Acknowledgements:
Multinomial Inequality of Haber
Special thanks to A. Chabour, S. Y. Raffed, and R. Souam for useful discussions.
The proof given in the remark is due to A. Chabour.
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Contents
1
Introduction
3
2
Main Result
5
3
Applications
9
Multinomial Inequality of Haber
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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1.
Introduction
In 1978, S. Haber [3] proved the following inequality: Let a and b be non negative
real numbers, then for every n ≥ 0, we have
n
a+b
1
n
n−1
n−1
n
(1.1)
a + a b + · · · + ab
+b ≥
.
n+1
2
Multinomial Inequality of Haber
Another formulation of (1.1) is
f (x, y) ≥ f 12 , 12 for all non negative numbers x, y satisfying x + y = 1,
where
f (x, y) =
X
i+j=n
i j
xy
b
a
and y =
.
with x =
a+b
a+b
In 1983 [5], A. Mc.D. Mercer, using an analogous technique, gave an extension
of Haber’s inequality for convex sequences.
Let (uk )0≤k≤n be a convex sequence, then the following inequality holds
(1.2)
n
n X
1 X
n
uk ≥
uk .
n + 1 k=0
k
k=0
In 1994 [1], using also the same tools, H. Alzer and J. Pečarić obtained a more
general result than the relation (1.2).
In 2004 [6], A. Mc.D. Mercer extended the result using an equivalent inequality
of (1.1) as a polynomial in x = ab , and deduced relations satisfying (1.2), see [1].
P
Let P (x) = nk=0 ak xk satisfy P (x) = (x − 1)2 Q (x) , where the coefficients
of Q (x) are real and non negative. Then if (uk )0≤k≤n is a convex sequence, we have
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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(1.3)
n
X
ak uk ≥ 0.
k=0
Our proposal is to establish an extension of the relation (1.1) to n real numbers.
Multinomial Inequality of Haber
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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2.
Main Result
In this section, we give an extension of the inequality given by the relation (1.1) for
several variables.
Theorem 2.1 (Generalized Haber inequality). Let a1 , a2 , . . . , am be non negative
real numbers, then for all n ≥ 0, one has
n
X
1
a1 + a2 + · · · + am
i1 i2
im
(2.1)
a1 a2 · · · am ≥
.
n+m−1
m
m−1
i +i +···+i =n
1
2
m
For another formulation of (2.1), let us consider the following homogeneous
polynomial of degree n
X
fm (x1 , x2 , . . . , xm ) =
xi11 xi22 · · · ximm
i1 +i2 +···+im =n
where x1 , x2 , . . . , xm are non negative real numbers satisfying the constraint x1 +
ai
x2 + · · · + xm = 1. By setting for all i = 1, . . . , m; xi = a1 +a2 +···+a
, the inequality
m
given by (2.1) becomes
(2.2)
fm (x1 , x2 , . . . , xm ) ≥ fm m1 , m1 , . . . , m1
Multinomial Inequality of Haber
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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Proof. Let (y1 , y2 , . . . , ym ) be the values for which fm is minimal. It is well known
that the gradient of fm at (y1 , y2 , . . . , ym ) is parallel to that of the constraint which is
(1, 1, . . . , 1) , one then deduces
∂fm
∂f
m
(x1 , . . . , xm )
=
(x1 , . . . , xm )
,
∂xα
∂xβ
xα =yα
xβ =yβ
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for all α, β, 1 ≤ α 6= β ≤ m, which is equivalent to




m
m
X
X
 Y ij  iα iβ −1
 Y ij  iα −1 iβ
xj  yα yβ ,
iβ 
xj  yα yβ =
iα 
i1 +···+im =n
i1 +···+im =n
j=1
j6=α,β
i.e.

X
i1 +···+im =n


m
Y

Multinomial Inequality of Haber
Hacène Belbachir
i −1
i 
xjj  yαiα −1 yββ (iα yβ − iβ yα ) = 0,
vol. 9, iss. 4, art. 104, 2008
j=1
j6=α,β
which one can write as


n
X
X
i −1


yαiα −1 yββ (iα yβ − iβ yα ) 
r=0
j=1
j6=α,β
iα +iβ =r
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

X
i1 +···+im =n−r
ik 6=iα & ik 6=iβ
m
 Y ij 
xj  = 0.

j=1
j6=α,β
This last expression is a polynomial of several variables x1 , . . . , xj , . . . , xm
(j 6= α, β) which is null if all coefficients are zero. Then for yα = a and yβ = b, one
obtains for every r = 0, . . . , n
X
ai−1 bj−1 (ib − ja) = 0.
i+j=r
By developing the sum and gathering the terms of the same power, one obtains
r−1
X
i=0
(2i + 1 − r) ai br−1−i = 0.
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By gathering successively the extreme terms of the sum, we have
r+1
bX
2 c
(r − 2i − 1) ar−2i−1 − br−2i−1 a2i b2i = 0
i=0
which is equivalent to
(a − b)
r+1
bX
2 c r−2i−2
X
i=0
Multinomial Inequality of Haber
(r − 2i − 1) ak+2i br−k−2 = 0.
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
k=0
The double summation is positive, then one deduces that
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a = b ⇐⇒ yα = yβ .
The symmetric group Sm acts naturally by permutations over R [x1 , x2 , . . . , xm ]
and leaves invariant fm (x1 , x2 , . . . , xm ) and x1 + x2 + · · · + xm = 1. Finally, one
concludes that
1
y1 = y2 = · · · = ym = .
m
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Remark 1. We can prove the above inequality using:
1. induction over m exploiting Haber’s inequality and the well known relation
n
n − im
n
=
.
im
i1 , i2 , . . . , im
i1 , i2 , . . . , im−1
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2. the sectional method for the function
fm (x1 , x2 , . . . , xm ) =
X
xi11 xi22 · · · ximm
|i|=n
with the constraint x1 + x2 + · · · + xm = 1;
P
Let
a
,
a
,
.
.
.
,
a
and
b
,
b
,
.
.
.
,
b
be
real
numbers
such
that
1
2
m
1
2
m
i ai = 0 and
P
i bi = 1, bi > 0, and consider the curve
X
Φ (t) =
(a1 t + b1 )i1 (a2 t + b2 )i2 · · · (am t + bm )im .
Multinomial Inequality of Haber
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
|i|=n
We prove that b = (b1 , b2 , . . . , bm ) is a local minima for fm if and only if
b1 = b2 = · · · = bm = m1 .
Indeed, one has
i m am
i 1 a1 i 2 a2
+
+ ··· +
t + ···
b1
b2
bm
|i|=n
n+m−1
i
a
n+m−1
i
a
m
m
1
1
(
)
∼
+ ··· +
t + ··· .
(b1 · · · bm ) m−1
=
b1
bm
m−1
Φ (t) − Φ (0) ∼
=
X
bi11 bi22 · · · bimm
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then Φ (t) − Φ (0) ∼
= ct2 + · · · , c > 0, . . .
P
If not, we can choose a1 , a2 , . . . , am such that i abii 6= 0, . . .
N.B. The possible nullity of some bi ’s is not a problem.
If b1 = b2 = · · · = bm =
1
m
3. the Popoviciu’s Theorem given in [7].
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3.
Applications
In this section we apply the previous result to find lower bounds for the sum of
reciprocals of multinomial coefficient and for two symmetric functions.
1. Sum of reciprocals of multinomial coefficient.
Theorem 3.1. The following inequality holds
X
1
i1 +i2 +···+im =n
n
i1 ,...,im
Multinomial Inequality of Haber
Hacène Belbachir
n+m−1
≥
m−1
m! · mn
.
vol. 9, iss. 4, art. 104, 2008
Proof. It suffices to integrate each side of the inequality given by the relation
(2.2) :
fm (x1 , x2 , . . . , xm−1 , 1 − x1 − · · · − xm−1 ) ≥ fm
1 1
, , . . . , m1
m m
over the simplex
(
D=
xi , i = 1, . . . , m − 1 : xi ≥ 0,
m−1
X
)
xi ≤ 1 .
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i=1
The left hand side gives under the sum the Dirichlet function (or the generalized
beta function) and is equal to the reciprocal of a multinomial coefficient. For
the right hand side we are led to compute the volume of the simplex D which
1
is equal to m!
.
2. An identity due to Sylvester in the 19th century, see [2, Thm 5], states that
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Theorem 3.2. Let x1 , x2 , . . . , xm be independent variables. Then, one has in
R[x1 , x2 , . . . , xm ]
X
xk11 xk22
k1 +···+km =n
. . . xkmm
=
m
X
Q
i=1
xn+m−1
i
.
j6=i (xi − xj )
As corollary of this theorem and Theorem 2.1, one obtains the following lower
bound.
Corollary 3.3. Using the hypothesis of the above theorem, one has
n m
X
n+m−1
xn+m−1
x1 + x2 + · · · + xm
i
Q
≥
.
m
m−1
j6=i (xi − xj )
i=1
3. The third application is about the symmetric polynomials. We need the following result:
Theorem 3.4 ([2, Cor. 5] and [4, Th. 1]). Let x1 , x2 , . . . , xm be elements of
unitary commutative ring A with
X
Sk =
xi1 xi2 · · · xik ,
for 1 ≤ k ≤ m.
Multinomial Inequality of Haber
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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1≤i1 <i2 <···<ik ≤m
Then, for each positive integer n, one has
X
X k 1 + · · · + k m k1
km
x1 . . . xm =
(−1)n−k1 −···−km Sk11 . . . Skmm ,
k1 , . . . , k m
where the summations are taken over all m-tuples (k1 , k2 , . . . , km ) of integers
kj ≥ 0 satisfying the relations k1 + k2 + · · · + km = n for the left hand side
and k1 + 2k2 + · · · + mkm = n for the right hand side.
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This theorem and Theorem 2.1, give:
Corollary 3.5. Using the hypothesis of the last theorem, one has
S n
X k 1 + · · · + k m 1
1
n−k1 −···−km k1
km
(−1)
S1 . . . Sm ≥
.
n+m−1
k1 , . . . , k m
m
m−1
where the summation is being taken over all m-tuples (k1 , k2 , . . . , km ) of integers kj ≥ 0 satisfying the relation k1 + 2k2 + · · · + mkm = n.
Multinomial Inequality of Haber
Hacène Belbachir
vol. 9, iss. 4, art. 104, 2008
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References
[1] H. ALZER AND J. PEČARIĆ, On an inequality of A.M. Mercer, Rad Hrvatske
Akad. Znan. Umj. Mat. [467], 11 (1994), 27–30.
[2] H. BELBACHIR AND F. BENCHERIF, Linear recurrent sequences and powers
of a square matrix, Electronic Journal of Combinatorial Number Theory, A12,
6 (2006).
Multinomial Inequality of Haber
Hacène Belbachir
[3] H. HABER, An elementary inequality, Internat. J. Math. and Math. Sci., 2(3)
(1979), 531–535.
vol. 9, iss. 4, art. 104, 2008
[4] J. MCLAUGHLIN AND B. SURY, Powers of matrix and combinatorial identities, Electronic Journal of Combinatorial Number Theory, A13(5) (2005).
Title Page
[5] A. McD. MERCER, A note on a paper by S. Haber, Internat. J. Math. and Math.
Sci., 6(3) (1983), 609–611.
[6] A.McD. MERCER, Polynomials and convex sequence inequalities, J. Ineq. Pure
Appl. Math., 6(1) (2005), Art. 8. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=477].
[7] T. POPOVICIU, Notes sur les functions convexes d’ordre superieure (V), Bull.
de la sci. Acad. Roumaine, 22 (1940), 351–356.
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