2020 February; 14(2): pages 36-41 DOI: 10.22587/ajbas.2020.14.2.6
Original paper AENSI Publications
Australian Journal of Basic and Applied Sciences ISSN: 1991-8178, EISSN: 2309-8414
Journal home page: www.ajbasweb.com
A Generalization of Weierstrass Inequality with Some Parameters
Nizar Kh. Al-Oushousha, Rateb Al-btoushb, Moa’ath Oqielatc
a Department of Mathematics, Faculty of Science, Al-Balqa’ Applied University, Jordan b Department of Mathematics, Faculty of Science, Mutah University, Jordan
c Department of Mathematics, Faculty of Science, Al-Balqa’ Applied University, Jordan
Correspondence Author: Nizar Kh. Al-Oushoush, Department of Mathematics, Faculty of Science, Al-Balqa’ Applied University, Jordan E-mail:- [email protected]
Received date: 15 December 2019, Accepted date: 24 February 2020, Online date: 3 March 2020
Copyright: © 2020 Nizar Kh. Al-Oushoush et al., This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Keywords: Weierstrass Inequality, Bernoulli’s inequality, Mean Value Theorem (MVT) INTRODUCTION
Suppose
x
1
andn
. Then
(
1
x
)
n
1
nx
(1)(1) Is called Bernoulli’s inequality, which is essential in the analysis [4].
Suppose
0
x
i
1
,
i
1
,
2
,
3
,
,
n
, wheren
2
. The following inequalities
n
i
n
i i
i
x
x
1 1
1
)
1
(
(2)
n
i
n
i i
i
x
x
1 1
1
)
1
(
(3)
are well known as Weierstrass’s inequality [1] or Weierstrass’s Bernoulli’s inequality [3]. These inequalities are two of the most important inequalities in the supject of product polynomials. Many scientists studied this topic and a large number of documents
have been written on it [6],[9],[8]. For example, if
1
0
,
x
i
1
,
i
1
,
2
,
,
n
, and1
1
n
i i
; thenAbstract
n
i
n
i i i
i
x
x
i1 1
1
)
1
(
(4)
If
i
1
,
x
i
0
or
i
1
andx
i
0
,
for i1,2,,n then
n
i
n
i i i
i
x
x
i1 1
1
)
1
(
(5) Let
x
1,
x
2,
,
x
n be positive real numbers with1
1
n
i i
x
; H.Sh. Huang gave an inequality [2]
n
i
n i
i
n
n
x
x
1
)
1
(
)
1
(
(6)
More general, for
x
1,
x
2,
,
x
n be positive real numbers withx
k
k
n
n
i
i
1
,
1
where k,n; for
m
, ShanheWu and Huannan Shi [10] gave the following inequality
n
m m
m m n
k n k m n
i i n
i
n m m
m m m
i m
i
n
k
k
n
x
k
n
n
k
k
n
x
x
m m
m m
2 22
2 )
(
1 1
)
(
)
1
(
(7)
For more information on the weierstrass inequality, you can refer to [5,7] and the references therein .
In this paper, new generalizations of Weierstrass inequality are established by using principles of mathematical analysis as the principle of Mean Value Theorem.
PRELIMINARIES AND LEMMAS
LEMMA1. Let
i
2
,
1
and0
x
i
1
fori1,2,,n. Then
n
i
n
i i i
i
i
x
x
1 1
(8)PROOF: If
x
i
0
, then (8) holds.If
0
x
i
1
and
i
2
then
x
i i
1
and since
1 then)
(
)
(
1 2 1 22 1 2
1
n n
n
n
x
x
x
x
x
x
thus we have
n
i
n
i i i
i i
x
x
1 1
.LEMMA2. Let
1
i
i
2
,
1
1
i
and
x
i
1
for i1,2,,n. Then
n
i
n
i
i i i
i
i
x
x
1 1
1
)
(
(9)i i
i i i
i
i
x
x
x
h
(
)
(
)
forx
i
1
Its clear that
h
(
x
i)
is continuous function forx
i
1
and differentiable forx
i
1
with
1 1 1
1 1
'
)
(
1
)
(
)
(
)
(
i i i
i i
i i
i i i
i i
i i i i
i i i
x
x
x
x
x
x
h
since
i1x
i
x
i
i
x
i1
then
1
1 1
i i
i
x
x
i
, it is mean that
1
)
(
11 1
1 1
i i i
i
i i
i i
i i
i
x
x
x
x
, we obtain that
0
)
(
'
x
h
. By using MVT,
c
i
(
1
,
x
)
such that1
)
1
(
)
(
)
(
'
i i i
x
h
x
h
c
h
, then
1
1
)
(
)
(
1 1
1 1
'
i
i i
i i i
i
i i
i i i
i i i
x
x
x
c
c
x
x
h
i i
i
i i
i
since
1
1 1
i i
i i
c
c
i
, thus
0
1
)
(
)
(
1 1
1 1
'
i i
i
i i
i i i
i i i
c
c
x
x
h
,its mean that
h
(
x
i)
is increasing function forx
i
1
, and also by MVT
0
1
)
(
'
i
i i
i i i
i i
x
x
x
x
h
i i
i
since
x
i
1
, then
i
i
i
i
i
i
0
i i
x
x
(10)since
1
i
i, then(
i
)
i
i
, from (10) we obtain
(
i
x
i)
i
i
x
ii (11)then
ni
i i n
i
i i
i
i
x
x
1 1
)
(
(12)
ni
i i n
i
i i
i
i
x
x
1 1
(13)Apply transitive property to (12) and (13) we obtain (9).
MAIN RESULTS
THEOREM 1. Let
1
i,
i
2
and0
x
i
1
, for i1,2,,n, then
ni
n
i i i
n
i
i i
i i
i
x
x
1 1
1
)
(
(14) and
ni
n
i i i
n
i
i i
i i
i
x
x
1 1
1
)
(
(15)PROOF: If
x
i
0
(14) holds.For
0
x
i
1
, define
ni i n
i i n
i
i i i
i i
i
x
x
x
f
1 1
1
)
(
)
(
(16)then
f
(
x
i)
is differentiable function with
n
i
i i j
j j i
i i
n
i
i i n
i
n
j
j j i
i i i
i
i j
j i
i
i j
j i
x
x
x
x
x
x
x
f
1
1 1
1
1
1
1 1
1 '
)
(
)
(
)
(
)
(
)
(
)
(
(17)
since
i
x
i
x
i and since
i
1
0
, then(
i
x
i)
i1
x
ii1, and since(
i
x
i)
i
1
then1
)
(
1
n
i
i i
i
x
, thus we obtain
(
)
(
)
10
1
1
i
i j
i i
i i n
j
i i i
i
i
x
x
x
(18)
(
)
(
)
0
1
1
1
1
n
i
i i n
j
j j i
i i
i
i j
j
i
x
x
x
(19)thus
f
'(
x
i)
0
for0
x
i
1
, thenf
(
x
i)
is increasing for0
x
i
1
, its mean thatf
(
x
i)
f
(
0
)
0
, thus
(
)
(
)
0
1 1
1
n
i i n
i i n
i
i i i
i i
i
x
x
x
f
for x[0,1)then
0
)
(
1 1
1
n
i i n
i i n
i
i i
i i
i
x
x
and because the inequality (14) holds when
x
i
0
, then
0
)
(
1 1
1
n
i i n
i i n
i
i i
i i
i
x
x
Which complete the proof of (14) of theorem 1.
To prove (15) : If
x
i
0
(15) holdsFor
0
x
i
1
, define
ni i n
i i n
i
i i i
i i
i
x
x
x
g
1 1
1
)
(
)
(
(20)Its clear that
g
(
x
i)
is continuous for0
x
i
1
and differentiable function with
n
i
n
j j i
i n
i
n
j
j j i
i i i
i j
j i
i j
j
i
x
x
x
x
x
g
1 1 1
1 1
1 '
)
(
)
(
)
(
(21)
n
i
n
j i i
i n
j
j j i
i i i
i j
j i
i j
j
i
x
x
x
x
x
g
1 1
1
1 1 '
)
(
)
(
)
(
(22)
since
(
i
x
i)
i1
x
ii1 and(
i
x
i)
i
x
ii, then
(
)
(
)
0
1 1
1
1
n
j j i
i n
j
j j i
i i
i j
j i
i j
j
i
x
x
x
x
(23)then
(
)
(
)
0
1 1
1
1
1
n
i
n
j j i
i n
j
j j i
i i
i j
j i
i j
j
i
x
x
x
x
then
g
'(
x
i)
0
for0
x
i
1
, theng
(
x
i)
is increasing on0
x
i
1
, its mean thatg
(
x
i)
g
(
0
)
0
, then we lead the proof of inequality (15).Remark 1:If we put
i
1
,
1 and
i
1
in (14), we obtain the famous Weierstrass’s inequality (2).Remark 2:If we put
i
1
,
1 and
i
in (14), we obtain the famous Weierstrass’s inequality (4).THEOREM 2: Let
1
i
i,
2 andx
i
1
, then
ni
n
i i i
n
i
i i
i i
i
x
x
1 1
1
)
(
(25)
Proof: Substitute
x
i
1
in the left hand side of (25) and letn i
n n
i i
I
(
)
(
)
1(
)
2(
)
3(
)
3 2
1 1
Since
1
i
i, for i1,2,,n, theni i n
n i
i i
i n
n n
i i
i n
n
C
C
C
I
3 2 1 3
2 1
3 3 2 2 1 1
1 2 2 1
2 1
.. 1 3
2 1
3 2
1
)
,
(
)
,
(
)
,
(
)
(
)
(
)
)(
)(
(
(26) Where
C
j,
j
1
,
2
,
,
2
n2are coefficients of
i and
, and since all ofC
j
1
for2
2
,
,
2
,
1
nj
, then
ni j
i j n
i i
i n
i i
i
C
I
1 2
2
1
1 .. 1
)
,
(
1
ni n
i i
i
1 1
1
Which complete the proof of theorem 2.
THEOREM 3. Let
1
i
, 1(
)
1i i i
i
x
x
i
,
i
x
i
2
andx
i
1
for i1,2,,n , then
ni
n
i i i
n
i
i i
i i
i
x
x
1 1
1
)
(
(27)Proof: Define
ni i n
i i n
i
i i i
i i
i
x
x
x
g
1 1
1
)
(
)
(
then
g
(
x
i)
is differentiable function with
n
i
n
j j i
i n
i
n
j
j j i
i i i
i j
j i
i j
j
i
x
x
x
x
x
g
1 1 1
1 1
1 '
)
(
)
(
)
n
i
n
j i i
i n
j
j j i
i i
i j
j i
i j
j
i
x
x
x
x
1 1
1
1 1
)
(
)
(
since
(
i
x
i)
i1
x
ii1 and using Lemma 2, we get0
)
(
)
(
1 1
1
1
n
j j i
i n
j
j j i
i i
i j
j i
i j
j
i
x
x
x
x
then
0
)
(
)
(
1 1
1
1
1
n
i
n
j j i
i n
j
j j i
i i
i j
j i
i j
j
i
x
x
x
x
then
g
'(
x
i)
0
, forx
i
1
, theng
(
x
i)
is incrasing forx
i
1
, thus
0
)
(
)
1
(
)
(
1 1 1
n
i
n
i i n
i i i
i i
g
x
g
for
x
i
(
1
,
)
, which leads to inequality(27).Acknowledgments
.
We would like to thank the editor Prof.Abdel Rahman Al-Tawaha and referees for their valuable comments
REFERENCES
D.S.Mitrinovic, P.M.Vasic, Analytic Inequalities,Springer-Verlag, New York, 1970, pp.210-211.
H.Sh. Huang, On inequality
n
i
n i
i
n
n
x
x
1
)
1
(
)
1
(
Shuxue Tongxun, 5(1987), pp.5--7. (Chinese).Shi, H.N., 2008. Generalizations of Bernoulli’s inequality with applications. J. Math. Inequal, 2(1), pp.101-107.
Ji Chang Kuang, Applied Inequalities (Chang yong bu deng shi) 4thed, Shandong Press of science and technology, Jinan, China,
2004.(Chinese).
Pečarić, J.E. and Klamkin, M.S., 1988. Extensions of the Weierstrass product inequalities III. SEA Bull. Math, 11, pp.123-126.. Luis F.Moreno, An Invitation to Real Analysis, The Mathematical Association of America, 17 May, 2015.
M. S. Klamkin, Extensions of the Weierstrass Product Inequalities II, Amer.Math.Monthly 82(1975)741-742.
Klamkin, M.S. and Newman, D.J., 1970. Extensions of the Weierstrass product inequalities. Mathematics Magazine, 43(3), pp.137-141.
Wu, S., 2005. Some results on extending and sharpening the Weierstrass product inequalities. Journal of mathematical analysis and applications, 308(2), pp.689-702.
