ON SOME INEQUALITIES IN NORMED ALGEBRAS
SEVER S. DRAGOMIR
School of Computer Science and Mathematics
Victoria University
PO Box 14428, Melbourne City
8001, VIC, Australia
EMail: sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/dragomir
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
Title Page
Contents
Received:
24 October, 2007
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Accepted:
29 October, 2007
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Communicated by:
S. Saitoh
2000 AMS Sub. Class.:
Primary 46H05; Secondary 47A10.
Key words:
Normed algebras, Unital algebras, Generalised triangle inequality.
Abstract:
Some inequalities
P in normed algebras that provides lower and upper bounds for
the norm of n
a x are obtained. Applications for estimating the quantities
−1 j=1 j j
x x ± y −1 y and y −1 x ± x−1 y for invertible elements x, y
in unital normed algebras are also given.
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Contents
1
Introduction
3
2
Inequalities for n Pairs of Elements
6
3
Inequalities for Two Pairs of Elements
11
4
Applications for Two Invertible Elements
15
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
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1.
Introduction
In [1], in order to provide a generalisation of a norm inequality for n vectors in a
normed linear space obtained by Pečarić and Rajić in [2], the author obtained the
following result:
(
)
n
n
X
X
|αk | (1.1)
max
xj −
|αj − αk | kxj k
k∈{1,...,n}
j=1
j=1
(
)
n
n
n
X
X
X
≤
αj xj ≤ min
|αk | xj +
|αj − αk | kxj k ,
k∈{1,...,n}
j=1
j=1
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
j=1
Title Page
where xj , j ∈ {1, . . . , n} are vectors in the normed linear space (X, k·k) over K
while αj , j ∈ {1, . . . , n} are scalars in K (K = C, R) .
For αk = kx1k k , with xk 6= 0, k ∈ {1, . . . , n} the above inequality produces the
following result established by Pečarić and Rajić in [2]:
(
" n
#)
n
X X
1
(1.2)
max
xj −
|kxj k − kxk k|
k∈{1,...,n}
kxk k j=1 j=1
(
" n
#)
n
n
X
X
X
1
xj ≤
xj +
|kxj k − kxk k|
,
≤ min
kxj k k∈{1,...,n} kxk k j=1
j=1
j=1
which implies the following refinement and reverse of the generalised triangle in-
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equality due to M. Kato et al. [3]:
#
"
n
X
xj (1.3)
min {kxk k} n − k∈{1,...,n}
kxj k j=1
#
"
n
n
n
X
X
X
x
j ≤
kxj k − xj ≤ max {kxk k} n − .
k∈{1,...,n}
kxj k j=1
j=1
j=1
The other natural choice, αk = kxk k , k ∈ {1, . . . , n} in (1.1) produces the result
n
(
)
n
X X
(1.4)
max
kxk k xj −
|kxj k − kxk k| kxj k
k∈{1,...,n}
j=1
j=1
n
n
(
)
n
X
X X
≤
kxj k xj ≤ min
kxk k xj +
|kxj k − kxk k| kxj k ,
k∈{1,...,n}
j=1
j=1
j=1
which in its turn implies another refinement and reverse of the generalised triangle
inequality:
P
Pn
n
2
j=1 kxj k − j=1 kxj k xj (1.5)
(0 ≤)
max {kxk k}
k∈{1,...,n}
P
n
Pn
n
2
n
X kx
k
−
kx
k
x
X
j
j
j
j=1
j=1
≤
kxj k − xj ≤
,
min
{kx
k}
k
j=1
j=1
k∈{1,...,n}
provided xk 6= 0, k ∈ {1, . . . , n} .
Inequalities in Normed
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Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
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In [2], the authors have shown that the case n = 2 in (1.2) produces the MaligrandaMercer inequality:
kx − yk + |kxk − kyk|
kx − yk − |kxk − kyk| x
y
(1.6)
≤
−
,
kxk kyk ≤
min {kxk , kyk}
max {kxk , kyk}
for any x, y ∈ X\ {0} .
We notice that Maligranda proved the right inequality in [5] while Mercer proved
the left inequality in [4].
We have shown in [1] that the following dual result for two vectors is also valid:
(1.7)
kx − yk
|kxk − kyk|
(0 ≤)
−
min {kxk , kyk} max {kxk , kyk}
x
y kx − yk
|kxk − kyk|
≤
≤
−
+
,
kyk kxk
max {kxk , kyk} min {kxk , kyk}
for any x, y ∈ X\ {0} .
Motivated by the above results, the
Pnaim of the present paper is to establish lower
and upper bounds for the norm of j=1 aj xj , where aj , xj , j ∈ {1, . . . , n} are elements in a normed algebra (A, k·k) over the real or complex number field K. In
the case where (A, k·k) is a unital algebra and x, y are invertible, lower and upper
bounds for the quantities
−1 x x ± y −1 y and y −1 x ± x−1 y are provided as well.
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
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2.
Inequalities for n Pairs of Elements
Let (A, k·k) be a normed algebra over the real or complex number field K.
Theorem 2.1. If (aj , xj ) ∈ A2 , j ∈ {1, . . . , n} , then
(
!
)
n
n
X
X
(2.1)
max
xj −
kaj − ak k kxj k
ak
k∈{1,...,n} j=1
j=1
!
)
(
n
n
X
X
xj −
k(aj − ak ) xj k
≤ max
ak
k∈{1,...,n} j=1
j=1
n
X
≤
aj x j j=1
(
!
)
n
n
X
X
≤ min
xj +
k(aj − ak ) xj k
ak
k∈{1,...,n} j=1
j=1
(
!
)
n
n
X
X
≤ min
xj +
kaj − ak k kxj k .
ak
k∈{1,...,n} j=1
j=1
Proof. Observe that for any k ∈ {1, . . . , n} we have
!
n
n
n
X
X
X
aj x j = ak
xj +
(aj − ak ) xj .
j=1
j=1
j=1
Taking the norm and utilising the triangle inequality and the normed algebra proper-
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
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ties, we have
n
X
aj xj ≤ ak
j=1
≤ ak
≤ ak
! n
X
xj + (aj − ak ) xj j=1
j=1
!
n
n
X
X
xj +
k(aj − ak ) xj k
j=1
j=1
!
n
n
X
X
xj +
kaj − ak k kxj k ,
n
X
j=1
j=1
j=1
and utilising the continuity of the norm, we have
n
!
n
n
X
X
X
aj xj ≥ ak
xj −
(ak − aj ) xj j=1
j=1
j=1
! n
n
X
X
≥ ak
xj − (ak − aj ) xj j=1
j=1
!
n
n
X
X
≥ ak
xj −
k(ak − aj ) xj k
j=1
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
j=1
for any k ∈ {1, . . . , n} , which implies the second part in (2.1).
Observing that
!
n
n
n
X
X
X
aj x j = ak
xj −
(ak − aj ) xj
j=1
Inequalities in Normed
Algebras
j=1
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≥ ak
n
X
j=1
!
n
X
xj −
kak − aj k kxj k
j=1
for any k ∈ {1, . . . , n} , which implies the first part in (2.1).
Remark 1. If there exists r > 0 so that kaj − ak k ≤ r kak k for any j, k ∈ {1, . . . , n} ,
then, by the second part of (2.1), we have
" n
#
n
n
X
X X
(2.2)
aj xj ≤ min {kak k} xj + r
kxj k .
k∈{1,...,n}
j=1
j=1
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
j=1
Corollary 2.2. If xj ∈ A, j ∈ {1, . . . , n} , then
(
!
)
n
n
X
X
(2.3)
max
xj −
kxj − xk k kxj k
xk
k∈{1,...,n} j=1
j=1
(
!
) n
n
n
X
X X
≤ max
xj −
k(xj − xk ) xj k ≤ x2j xk
k∈{1,...,n} j=1
j=1
j=1
(
!
)
n
n
X
X
≤ min
xj +
k(xj − xk ) xj k
xk
k∈{1,...,n} j=1
j=1
(
!
)
n
n
X
X
≤ min
xj +
kxj − xk k kxj k .
xk
k∈{1,...,n} j=1
Inequalities in Normed
Algebras
j=1
Corollary 2.3. Assume that A is a unital normed algebra. If xj ∈ A are invertible
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for any j ∈ {1, . . . , n} , then
#
" n
n
X
−1 X
min xk kxj k − xj k∈{1,...,n}
j=1
j=1
n
n
X
X
−1 −1 ≤
xj kxj k − xj xj j=1
j=1
#
" n
n
X
−1 X
≤ max xk kxj k − xj .
k∈{1,...,n}
(2.4)
j=1
Now, assume that
min
k∈{1,...,n}
j=1
−1 −1 x = x . Then
k0
k
n
n
X
X
−1 −1 −1 x x − x kxj k
xj +
j
k0 k0
j=1
j=1
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
j=1
· 1 in (2.1) we get
Proof. If 1 ∈ A is the unity, then on choosing ak = x−1
k
)
(
n
n
X
X
−1 x x−1 − x−1 kxj k
(2.5)
max
xj −
j
k
k
k∈{1,...,n}
j=1
j=1
n
X
−1 x xj ≤
j
j=1
(
)
n
n
X
X
−1 x x−1 − x−1 kxj k .
≤ min
xj +
j
k
k
k∈{1,...,n}
j=1
Inequalities in Normed
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n
!
n
n
X X
−1 X
−1 x kxj k .
= − xk0 kxj k − xj +
j
j=1
j=1
j=1
Utilising the second inequality in (2.5), we deduce
!
n
n
n
n
X
X
X
−1 X
−1 −1 x x kxj k − x xj kxj k − xj ≤
j
j
k0
j=1
j=1
j=1
j=1
Inequalities in Normed
Algebras
Sever S. Dragomir
and the first inequality in (2.4) is proved.
The second part of (2.4) can be proved in a similar manner, however, the details
are omitted.
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Remark 2. An equivalent form of (2.4) is:
P Pn −1 n
−1 x kxj k − j=1 xj xj j
j=1
(2.6)
max x−1
k
Contents
k∈{1,...,n}
P Pn n −1 −1 n
n
x
kx
k
−
x
x
j=1 j
X
j
j
j
j=1
X ≤
kxj k − xj ≤
,
min x−1
k
j=1
j=1
vol. 9, iss. 1, art. 5, 2008
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k∈{1,...,n}
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which provides both a refinement and a reverse inequality for the generalised triangle
inequality.
Close
3.
Inequalities for Two Pairs of Elements
The following particular case of Theorem 2.1 is of interest for applications.
Lemma 3.1. If (a, b) , (x, y) ∈ A2 , then
(3.1) max {ka (x ± y)k − k(b − a) yk , kb (x ± y)k − k(b − a) xk}
≤ kax ± byk ≤ min {ka (x ± y)k + k(b − a) yk , kb (x ± y)k + k(b − a) xk}
or, equivalently,
Sever S. Dragomir
1
(3.2)
{ka (x ± y)k + kb (x ± y)k − [k(b − a) yk + k(b − a) xk]}
2
1
+ |ka (x ± y)k − kb (x ± y)k + k(b − a) yk − k(b − a) xk|
2
≤ kax ± byk
1
≤ {ka (x ± y)k + kb (x ± y)k + [k(b − a) yk + k(b − a) xk]}
2
1
− |ka (x ± y)k + kb (x ± y)k − k(b − a) yk − k(b − a) xk| .
2
Proof. The inequality (3.1) follows from Theorem 2.1 for n = 2, a1 = a, a2 = b,
x1 = x and x2 = ±y.
Utilising the properties of real numbers,
1
1
[α + β − |α − β|] , max {α, β} = [α + β + |α − β|] ;
2
2
the inequality (3.1) is clearly equivalent with (3.2).
min {α, β} =
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α, β ∈ R;
The following result contains some upper bounds for kax ± byk that are perhaps
more useful for applications.
Theorem 3.2. If (a, b) , (x, y) ∈ A2 , then
(3.3)
kax ± byk ≤ min {ka (x ± y)k , kb (x ± y)k} + kb − ak max {kxk , kyk}
≤ kx ± yk min {kak , kbk} + kb − ak max {kxk , kyk}
and
(3.4)
kax ± byk ≤ kx ± yk max {kak , kbk} + min {k(b − a) xk , k(b − a) yk}
≤ kx ± yk max {kak , kbk} + kb − ak min {kxk , kyk} .
Proof. Observe that k(b − a) xk ≤ kb − ak kxk and k(b − a) yk ≤ kb − ak kyk, and
then
(3.5)
k(b − a) xk , k(b − a) yk ≤ kb − ak max {kxk , kyk} ,
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
Title Page
which implies that
min {ka (x ± y)k + k(b − a) yk , kb (x ± y)k + k(b − a) xk}
≤ min {ka (x ± y)k , kb (x ± y)k} + kb − ak max {kxk , kyk}
≤ kx ± yk min {kak , kbk} + kb − ak max {kxk , kyk} .
Utilising the second inequality in (3.1), we deduce (3.3).
Also, since ka (x ± y)k ≤ kak kx ± yk and kb (x ± y)k ≤ kbk kx ± yk , hence
ka (x ± y)k , kb (x ± y)k ≤ kx ± yk max {kak , kbk} ,
which implies that
min {ka (x ± y)k + k(b − a) yk , kb (x ± y)k + k(b − a) xk}
≤ kx ± yk max {kak , kbk} + min {k(b − a) xk , k(b − a) yk}
≤ kx ± yk max {kak , kbk} + kb − ak min {kxk , kyk} ,
and the inequality (3.4) is also proved.
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The following corollary may be more useful for applications.
Corollary 3.3. If (a, b) , (x, y) ∈ A2 , then
(3.6)
kax ± byk ≤ kx ± yk ·
kak + kbk
kxk + kyk
+ kb − ak ·
.
2
2
Proof. Follows from Theorem 3.2 by adding the last inequality in (3.3) to the last
inequality (3.4) and utilising the property that min {α, β} + max {α, β} = α + β,
α, β ∈ R.
The following lower bounds for kax ± byk can be stated as well:
Theorem 3.4. For any (a, b) and (x, y) ∈ A2 , we have:
(3.7)
max {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak max {kxk , kyk}
≤ max {ka (x ± y)k , kb (x ± y)k} − kb − ak max {kxk , kyk}
≤ kax ± byk
and
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Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
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(3.8) min {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak min {kxk , kyk}
≤ min {|kaxk − kayk| , |kbxk − kbyk|} − min {k(b − a) xk , k(b − a) yk}
≤ kax ± byk .
Proof. Observe that, by (3.5) we have that
max {ka (x ± y)k − k(b − a) yk , kb (x ± y)k − k(b − a) xk}
≥ max {kax ± ayk , kbx ± byk} − kb − ak max {kxk , kyk}
≥ max {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak max {kxk , kyk}
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and on utilising the first inequality in (3.1), the inequality (3.7) is proved.
Observe also that, since
ka (x ± y)k , kb (x ± y)k ≥ min {|kaxk − kayk| , |kbxk − kbyk|} ,
then
max {ka (x ± y)k − k(b − a) yk , kb (x ± y)k − k(b − a) xk}
≥ min {|kaxk − kayk| , |kbxk − kbyk|} − min {k(b − a) xk , k(b − a) yk}
≥ min {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak min {kxk , kyk} .
Inequalities in Normed
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Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
Then, by the first inequality in (3.1), we deduce (3.8).
Corollary 3.5. For any (a, b) , (x, y) ∈ A2 , we have
1
kxk + kyk
(3.9)
· [|kaxk − kayk| + |kbxk − kbyk|] − kb − ak ·
≤ kax ± byk .
2
2
The proof follows from Theorem 3.4 by adding (3.7) to (3.8). The details are
omitted.
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4.
Applications for Two Invertible Elements
In this section we assume that A is a unital algebra with the unity 1. The following
results provide some upper bounds for the quantity kkx−1 k x ± ky −1 k yk , where x
and y are invertible in A.
Proposition 4.1. If (x, y) ∈ A2 are invertible, then
(4.1) x−1 x ± y −1 y ≤ kx ± yk min x−1 , y −1 + x−1 − y −1 max {kxk , kyk}
Inequalities in Normed
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Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
and
(4.2)
−1 x x ± y −1 y ≤ kx ± yk max x−1 , y −1 + x−1 − y −1 min {kxk , kyk} .
Proof. Follows by Theorem 3.2 on choosing a = kx−1 k · 1 and b = ky −1 k · 1.
Corollary 4.2. With the above assumption for x and y, we have
−1 x x ± y −1 y (4.3)
kxk + kyk
kx−1 k + ky −1 k ≤ kx ± yk ·
+ x−1 − y −1 ·
.
2
2
Lower bounds for kkx−1 k x ± ky −1 k yk are provided below:
Proposition 4.3. If (x, y) ∈ A2 are invertible, then
(4.4) kx ± yk max x−1 , y −1 − x−1 − y −1 max {kxk , kyk}
≤ x−1 x ± y −1 y Title Page
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and
(4.5)
kx ± yk min x−1 , y −1 − x−1 − y −1 min {kxk , kyk}
≤ x−1 x ± y −1 y .
Proof. The first inequality in (4.4) follows from the second inequality in (3.7) on
choosing a = kx−1 k · 1 and b = ky −1 k · 1.
We know from the proof of Theorem 3.4 that
Inequalities in Normed
Algebras
Sever S. Dragomir
(4.6)
max {ka (x ± y)k − k(b − a) yk , kb (x ± y)k − k(b − a) xk}
≤ kax ± byk .
If in this inequality we choose a = kx−1 k · 1 and b = ky −1 k · 1, then we get
−1 x x ± y −1 y n
≥ max x−1 kx ± yk − x−1 − y −1 kyk ,
o
−1 y kx ± yk − x−1 − y −1 kxk
≥ kx ± yk min x−1 , y −1 − x−1 − y −1 min {kxk , kyk}
and the inequality (4.5) is obtained.
Corollary 4.4. If (x, y) ∈ A2 are invertible, then
(4.7)
kx ± yk ·
kxk + kyk
kx−1 k + ky −1 k − x−1 − y −1 ·
2
2 −1 ≤ x x ± y −1 y .
vol. 9, iss. 1, art. 5, 2008
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Remark 3. We observe that the inequalities (4.3) and (4.7) are in fact equivalent with:
−1 −1 kx−1 k + ky −1 k x± y
y − kx ± yk ·
(4.8) x
2
kxk + kyk
≤x−1 − y −1 ·
.
2
Now we consider the dual expansion kky −1 k x ± kx−1 k yk, for which the following upper bounds can be stated.
Inequalities in Normed
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Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
Proposition 4.5. If (x, y) are invertible in A, then
(4.9) y −1 x ± x−1 y ≤ kx ± yk min x−1 , y −1 + x−1 − y −1 max {kxk , kyk}
and
(4.10)
−1 y x ± x−1 y ≤ kx ± yk max x−1 , y −1 + x−1 − y −1 min {kxk , kyk} .
In particular,
(4.11) y −1 x ± x−1 y kxk + kyk
kx−1 k + ky −1 k + x−1 − y −1 ·
.
≤ kx ± yk ·
2
2
The proof follows from Theorem 3.2 on choosing a = ky −1 k·1 and b = kx−1 k·1.
The lower bounds for the quantity kky −1 k x ± kx−1 k yk are incorporated in:
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Proposition 4.6. If (x, y) are invertible in A, then
(4.12) kx ± yk max x−1 , y −1 − x−1 − y −1 max {kxk , kyk}
≤ y −1 x ± x−1 y and
(4.13)
kx ± yk min x−1 , y −1 − x−1 − y −1 min {kxk , kyk}
≤ y −1 x ± x−1 y .
Inequalities in Normed
Algebras
Sever S. Dragomir
vol. 9, iss. 1, art. 5, 2008
In particular,
(4.14)
kxk + kyk
kx−1 k + ky −1 k kx ± yk ·
− x−1 − y −1 ·
2
2 ≤ y −1 x ± x−1 y .
Remark 4. We observe that the inequalities (4.11) and (4.14) are equivalent with
−1
−1 −1 −1 kx
k
+
ky
k
(4.15) y x ± x y − kx ± yk ·
2
kxk + kyk
≤ x−1 − y −1 ·
.
2
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References
[1] S.S. DRAGOMIR, A generalisation of the Pečarić-Rajić inequality in normed
linear spaces, Preprint. RGMIA Res. Rep. Coll., 10(3) (2007), Art. 3. [ONLINE:
http://rgmia.vu.edu.au/v10n3.html].
[2] J. PEČARIĆ AND R. RAJIĆ, The Dunkl-Williams inequality with n elements in
normed linear spaces, Math. Ineq. & Appl., 10(2) (2007), 461–470.
Inequalities in Normed
Algebras
Sever S. Dragomir
[3] M. KATO, K.-S. SAITO AND T. TAMURA, Sharp triangle inequality and its
reverses in Banach spaces, Math. Ineq. & Appl., 10(3) (2007).
vol. 9, iss. 1, art. 5, 2008
[4] P.R. MERCER, The Dunkl-Williams inequality in an inner product space, Math.
Ineq. & Appl., 10(2) (2007), 447–450.
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[5] L. MALIGRANDA, Simple norm inequalities, Amer. Math. Monthly, 113
(2006), 256–260.
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