MORE ON REVERSE TRIANGLE INEQUALITY IN
INNER PRODUCT SPACES
A. H. ANSARI AND M. S. MOSLEHIAN
Received 8 February 2005 and in revised form 17 May 2005
Refining some results of Dragomir, several new reverses of the generalized triangle inequality in inner product spaces are given. Among several results, we establish some reverses for the Schwarz inequality. In particular, it is proved that if a is a unit vector in
a real or complex inner product space (H; ·, ·), r,s > 0, p ∈ (0,s], D = {x ∈ H, rx −
sa ≤ p}, x1 ,x2 ∈ D − {0}, and αr,s = min{(r 2 xk 2 − p2 + s2 )/2rsxk : 1 ≤ k ≤ 2}, then
(x1 x2 − Rex1 ,x2 )/(x1 + x2 )2 ≤ αr,s .
1. Introduction
It is interesting to know under which conditions the triangle inequality went the other
way in a normed space X; in other
words, we would
like to know if there is a positive
constant c with the property that c nk=1 xk ≤ nk=1 xk for any finite set x1 ,...,xn ∈ X.
Nakai and Tada [7] proved that the normed spaces with this property are precisely those
of finite dimensional.
The first authors investigating reverse of the triangle inequality in inner product spaces
were Diaz and Metcalf [2] by establishing the following result as an extension of an inequality given by Petrovich [8] for complex numbers.
Theorem 1.1 (Diaz-Metcalf theorem). Let a be a unit vector in an inner product space
(H; ·, ·). Suppose the vectors xk ∈ H, k ∈ {1,...,n} satisfy
Re xk ,a
0≤r ≤ x k ,
k ∈ {1,...,n}.
(1.1)
Then
n
n
,
xk ≤ x
r
k
k =1
k =1
Copyright © 2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:18 (2005) 2883–2893
DOI: 10.1155/IJMMS.2005.2883
(1.2)
2884
Reverse triangle inequality
where equality holds if and only if
n
xk = r
k =1
n
xk a.
(1.3)
k =1
Inequalities related to the triangle inequality are of special interest (cf. [6, Chapter
XVII]). They may be applied to get interesting inequalities in complex numbers or to
study vector-valued integral inequalities [4, 5].
Using several ideas and following the terminology of [4, 5], we modify or refine some
results of Dragomir and ours [1] and get some new reverses of triangle inequality. Among
several results, we show that if a is a unit vector in a real or complex inner√product
space (H; ·, ·), xk ∈ H − {0}, 1 ≤ k ≤ n, α = min{xk : 1 ≤ k ≤ n}, p ∈ (0, α2 + 1),
max{xk − a : 1 ≤ k ≤ n} ≤ p, and β = min{(xk 2 − p2 + 1)/2xk : 1 ≤ k ≤ n}, then
n
n
n
1−β
x k − ≤
xk xk ,a .
Re
β
k =1
k =1
k =1
(1.4)
We also examine some reverses for the celebrated Schwarz inequality. In particular, it is
proved that if a is a unit vector in a real or complex inner product space (H; ·, ·), r,s > 0,
p ∈ (0,s], D = {x ∈ H, rx − sa ≤ p}, x1 ,x2 ∈ D − {0}, and αr,s = min{(r 2 xk 2 − p2 +
s2 )/2rsxk : 1 ≤ k ≤ 2}, then
x1 x2 − Re x1 ,x2
≤ αr,s .
2
x1 + x2 (1.5)
Throughout the paper, (H; ·, ·) denotes a real or complex inner product space. We
use repeatedly the Cauchy-Schwarz inequality without mentioning it. The reader is referred to [3, 9] for the terminology on inner product spaces.
2. Reverse of triangle inequality
We start this section by pointing out the following theorem of [1] which is a modification
of [5, Theorem 3].
Theorem 2.1. Let a1 ,...,am be orthonormal vectors in the complex inner product space
(H; ·, ·). Suppose that for 1 ≤ t ≤ m, rt ,ρt ∈ R, and that the vectors xk ∈ H, k ∈ {1,...,n}
satisfy
0 ≤ rt2 xk ≤ Re xk ,rt at ,
0 ≤ ρt2 xk ≤ Im xk ,ρt at ,
1 ≤ t ≤ m.
(2.1)
Then
m
t =1
rt2 + ρt2
n
1/2 n
x k ≤ xk k =1
k =1
(2.2)
A. H. Ansari and M. S. Moslehian 2885
and the equality holds in (2.2) if and only if
n
xk =
k =1
n
m
xk rt + iρt at .
(2.3)
t =1
k =1
The following theorem is a strengthen of [5, Corollary 1] and a generalization of [1,
Theorem 2].
Theorem 2.2. Let a be a unit vector in the complex inner product space (H; ·, ·). Suppose
that the vectors xk ∈ H − {0}, k ∈ {1,...,n} satisfy
max rxk − sa : 1 ≤ k ≤ n ≤ p,
max r xk − is a : 1 ≤ k ≤ n ≤ q,
(2.4)
where r,r ,s,s > 0 and
p ≤ (rα)2 + s2
1/2
q ≤ (r α)2 + s2
,
1/2
α = min xk : 1 ≤ k ≤ n .
,
(2.5)
Let
2
2
r 2 xk − p2 + s2
αr,s = min
:1≤k≤n ,
2rsxk r 2 xk − q2 + s2
βr ,s = min
:1≤k≤n .
2r s xk Then
1/2
α2r,s + βr2 ,s
n
n
x k ≤ xk k =1
(2.6)
(2.7)
k =1
and the equality holds if and only if
n
xk = αr,s + iβr ,s
k =1
n
xk a.
(2.8)
k =1
Proof. From the first inequality above, we infer that
rxk − sa,rxk − sa ≤ p2 ,
2
r 2 xk + s2 − p2 ≤ 2Re rxk ,sa .
(2.9)
Then
2
r 2 xk + s2 − p2 xk ≤ Re xk ,a .
2rs xk
(2.10)
Similarly,
2
r 2 xk − q2 + s2 xk ≤ Im xk ,a ,
2r s xk
(2.11)
2886
Reverse triangle inequality
consequently,
αr,s xk ≤ Re xk ,a ,
(2.12)
βr ,s xk ≤ Im xk ,a .
Applying Theorem 2.1 for m = 1, r1 = αr,s , and ρ1 = βr ,s , we deduce the desired inequal
ity.
The next result is an extension of [1, Corollary 3].
Corollary 2.3. Let a be a unit vector in the complex inner product space (H; ·, ·). Suppose that xk ∈ H, k ∈ {1,...,n}, max{rxk − sa : 1 ≤ k ≤ n} ≤ r, max{rxk − isa : 1 ≤
k ≤ n} ≤ s, where r > 0, s > 0, and α = min{xk : 1 ≤ k ≤ n}. Then
rα x k ≤ √
xk .
s 2 k =1
k =1
n
n
(2.13)
The equality holds if and only if
n
xk = rα
k =1
n
(1 + i) xk a.
2s k=1
(2.14)
Proof. Apply Theorem 2.2 with r = r , s = s , p = r, q = s. Note that αr,s = rα/2s = βr ,s .
Theorem 2.4. Let a be a unit vector in (H; ·, ·). Suppose that the vectors xk ∈ H, k ∈
{1,...,n} satisfy
1/2
max rxk − sa : 1 ≤ k ≤ n ≤ p < (rα)2 + s2 ,
(2.15)
where r > 0, s > 0 and
α = min xk .
(2.16)
1≤k≤n
Let
2
r 2 xk − p2 + s2
:1≤k≤n .
αr,s = min
2rsxk (2.17)
Then
αr,s
n
n
xk ≤ x
.
k
k =1
(2.18)
k =1
Moreover, the equality holds if and only if
n
k =1
xk = αr,s
n
xk a.
k =1
(2.19)
A. H. Ansari and M. S. Moslehian 2887
Proof. Proof is similar to that of Theorem 2.2 in which we use Theorem 2.1 with m = 1,
ρ1 = 0.
Theorem 2.5. Let a be a unit vector in (H; ·, ·). Suppose that r,s > 0, and vectors xk ∈
H − {0}, k ∈ {1,...,n} satisfy
n
xk = 0.
(2.20)
k =1
Then
r 2 α2 + s2 ≤ max rxk − sa : 1 ≤ n ,
(2.21)
where
α = min xk .
(2.22)
1≤k≤n
√
Proof. Let p = max{rxk − sa : 1 ≤ k ≤ n}. If p < r 2 α2 + s2 , then, using Theorem 2.4,
we get
n
n
= 0.
xk ≤ x
αr,s
k
k =1
(2.23)
k =1
Hence αr,s = 0. On the other hand, (p2 − s2 )/r 2 < α2 , so
2
r 2 xk − p2 + s2
αr,s = min
:1≤k≤n >0
2rsxk (2.24)
holds a contradiction.
Theorem 2.6. Let a1 ,...,am be orthonormal vectors in the complex inner product space
(H; ·, ·), Mt ≥ mt > 0, Lt ≥ t > 0, 1 ≤ t ≤ m, and xk ∈ H − {0}, k ∈ {1,...,n} such that
Re Mt at − xk ,xk − mt at ≥ 0,
Re Lt iat − xk ,xk − t iat ≥ 0,
(2.25)
or equivalently
m t + Mt Mt − mt
x k −
≤
at ,
2
2
Lt + t Lt − t
xk −
≤
iat ,
2
2
(2.26)
for all 1 ≤ k ≤ n and 1 ≤ t ≤ m. Let
αmt ,Mt
αt ,Lt
2
x k + mt M t
= min
:1≤k≤n ,
m t + Mt x k 2
xk + t Lt
= min
:1≤k≤n ,
m t + Mt x k 1 ≤ t ≤ m,
(2.27)
1 ≤ t ≤ m,
2888
Reverse triangle inequality
then
m
t =1
1/2
α2mt ,Mt
+ α2t ,Lt
n
n
x k ≤ xk .
k =1
(2.28)
k =1
The equality holds if and only if
n
xk =
k =1
m
n
x k αmt ,Mt + iαt ,Lt at .
(2.29)
t =1
k =1
Proof. Given 1 ≤ t ≤ m and all 1 ≤ k ≤ n, it follows from xk − ((mt + Mt )/2)at ≤ (Mt −
mt )/2 that
2
xk + mt Mt ≤ mt + Mt Re xk ,at .
(2.30)
2
x + mt M t k
xk ≤ Re xk ,at ,
m t + Mt x k (2.31)
Then
and so
αmt ,Mt xk ≤ Re xk ,at .
(2.32)
Similarly, from the second inequality, we deduce that
αt ,Lt xk ≤ Im xk ,at .
(2.33)
Applying Theorem 2.5 for rt = αmt ,Mt and ρt = αt ,Lt , we obtain the required inequality.
We will need [4, Theorem 7]. We mention it for the sake of completeness.
Theorem 2.7. Let a be a unit vector in (H; ·, ·), and xk ∈ H − {0}, k ∈ {1,...,n}. If
rk ≥ 0, k ∈ {1,...,n} such that
xk − Re xk ,a ≤ rk ,
then
n
n
n
x k − ≤
x
rk .
k
k =1
k =1
(2.34)
(2.35)
k =1
The equality holds if and only if
n
n
x k ≥
rk ,
k =1
n
k =1
xk =
k =1
n
k =1
n
x k −
rk a.
k =1
(2.36)
A. H. Ansari and M. S. Moslehian 2889
Theorem 2.8. Let a be a unit vector in (H; ·, ·), and xk ∈ H − {0}, k ∈ {1,...,n}. Let
α = min xk : 1 ≤ k ≤ n ,
max xk − a : 1 ≤ k ≤ n ≤ p,
p ∈ 0, α2 + 1 ,
2
x k − p 2 + 1
:1≤k≤n .
β = min
2xk (2.37)
Then
n
n
n
1−β
x k − ≤
xk xk ,a .
Re
β
k =1
k =1
k =1
(2.38)
The equality holds if and only if
n
1−β
x k ≥
Re
β
k =1
n
xk =
k =1
n
xk ,a ,
k =1
n
1−β
xk −
Re
β
k =1
n
xk ,a
(2.39)
a.
k =1
Proof. Since max{xk − a : 1 ≤ k ≤ n} ≤ p, we have
2
xk + 1 − p2 ≤ 2Re xk ,a ,
xk − a,xk − a ≤ p2 ,
2
xk − p2 + 1 xk ≤ Re xk ,a ,
2 xk
βxk ≤ Re xk ,a ,
(2.40)
1
xk ≤ Re xk ,a ,
β
for all k ∈ {1,...,n}. Then
xk − Re xk ,a ≤ 1 − β Re xk ,a ,
β
k ∈ {1,...,n}.
(2.41)
Applying Theorem 2.7 for rk = ((1 − β)/β)Rexk ,a, k ∈ {1,...,n}, we deduce the desired
inequality.
As a corollary, we obtain a result similar to [4, Theorem 9].
Corollary 2.9. Let a be a unit vector in (H; ·, ·), and xk ∈ H − {0}, k ∈ {1,...,n}. Let
max xk − a : 1 ≤ k ≤ n ≤ 1,
α = min xk : 1 ≤ k ≤ n .
(2.42)
n
n
n
2−α
x k − ≤
xk xk ,a .
Re
α
k =1
k =1
k =1
(2.43)
Then
2890
Reverse triangle inequality
The equality holds if and only if
n
2−α
x k ≥
Re
α
k =1
n
xk =
k =1
n
n
xk ,a ,
k =1
2−α
x k −
Re
α
k =1
n
xk ,a
(2.44)
a.
k =1
Proof. Apply Theorem 2.8 with β = α/2.
Theorem 2.10. Let a be a unit vector in (H; ·, ·), M ≥ m > 0, and xk ∈ H − {0}, k ∈
{1,...,n} such that
Re Ma − xk ,xk − ma ≥ 0
(2.45)
x k − m + M a ≤ M − m .
2
2
(2.46)
or equivalently
Let
αm,M
Then
2
xk + mM
= min
:1≤k≤n .
(m + M)xk (2.47)
n
n
n
1 − αm,M
x k − ≤
x
Re
x
,a
.
k
k
αm,M
k =1
k =1
k =1
(2.48)
The equality holds if and only if
n
1 − αm,M
x k ≥
Re
αm,M
k =1
n
xk =
k =1
n
xk ,a ,
k =1
n
1 − αm,M
xk −
Re
k =1
αm,M
n
xk ,a
(2.49)
a.
k =1
Proof. For each 1 ≤ k ≤ n, it follows from the inequality
x k − m + M a ≤ M − m
2
2
(2.50)
that
m+M
m+M
M −m
xk −
a,xk −
a ≤
2
2
2
2
.
(2.51)
Hence
2
xk + mM ≤ (m + M)Re xk ,a .
(2.52)
A. H. Ansari and M. S. Moslehian 2891
So that
αm,M xk ≤ Re xk ,a ,
(2.53)
xk − Re xk ,a ≤ 1 − αm,M Re xk ,a .
(2.54)
consequently
αm,M
Now apply Theorem 2.7 for rk = ((1 − αm,M )/αm,M )Rexk ,a, k ∈ {1,...,n}.
3. Reverses of Schwarz inequality
In this section, we provide some reverses of the Schwarz inequality. The first theorem is
an extension of [4, Proposition 5.1].
Theorem 3.1. Let a be a unit vector in (H; ·, ·). Suppose that r,s > 0, p ∈ (0,s], and
D = x ∈ H, rx − sa ≤ p .
(3.1)
If 0 = x1 ∈ D, 0 = x2 ∈ D, then
2
2
x1 x2 − Re x1 ,x2
r x1 − p2 + s2 2
1
≤
1−
2
x1 + x2 2
2rsx1 (3.2)
2
2
x1 x2 − Re x1 ,x2
r x2 − p2 + s2 2
1
≤
1−
.
2
x 1 + x 2 2
2rsx2 (3.3)
or
Proof. Put αr,s = min{(r 2 xk 2 − p2 + s2 )/2rsxk : 1 ≤ k ≤ 2}. By Theorem 2.4, we obtain
αr,s x1 + x2 ≤ x1 + x2 .
(3.4)
Then
2
2
2
2
α2r,s x1 + 2x1 x2 + x2 ≤ x1 + 2Re x1 ,x2 + x2 .
(3.5)
Set α2r,s = 1 − t 2 . Then
x1 x2 − Re x1 ,x2
1
≤ t2 ,
2
x 1 + x 2 2
(3.6)
x1 x2 − Re x1 ,x2
1
≤ 1 − α2r,s .
2
x 1 + x 2 2
(3.7)
namely,
2892
Reverse triangle inequality
Corollary 3.2. Let a be a unit vector in (H; ·, ·). Suppose that r,s > 0 and
D = x ∈ H, rx − sa ≤ s .
(3.8)
If x, y ∈ D and 0 < x < y , then
x y − Rex, y 1
r x 2
≤
1−
.
2
2
2s
x + y (3.9)
Proof. In the notation of the proof of Theorem 3.1, we get from p = s, x1 = x, x2 = y that
αr,s = r x/2s. Now apply Theorem 3.1.
Corollary 3.3. Let a be a unit vector in (H; ·, ·), M ≥ m > 0, and xk ∈ H − {0}, k = 1,2
such that
Re Ma − xk ,xk − ma ≥ 0
(3.10)
x k − m + M a ≤ M − m .
2
2
(3.11)
x1 2 + mM 2
x1 x2 − Re x1 ,x2
1
≤
1−
2
x 1 + x 2 2
(m + M)x1 (3.12)
x2 2 + mM 2
x1 x2 − Re x1 ,x2
1
≤
1−
.
2
x 1 + x 2 2
(m + M)x2 (3.13)
or equivalently
Then
or
Proof. Put r = 1, s = (m + M)/2, p = (M − m)/2, x = x1 , and y = x2 in Theorem 3.1.
Acknowledgment
The authors would like to thank the referees for their valuable suggestions.
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[4]
[5]
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A. H. Ansari and M. S. Moslehian 2893
[6]
[7]
[8]
[9]
D. S. Mitrinović, J. E. Pečarić, and A. M. Fink, Classical and New Inequalities in Analysis, Mathematics and Its Applications (East European Series), vol. 61, Kluwer Academic, Dordrecht,
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A. H. Ansari: Department of Mathematics, Ferdowsi University, P.O. Box 1159, Mashhad 91775,
Iran
E-mail address: arsalan h a@yahoo.com
M. S. Moslehian: Department of Mathematics, Ferdowsi University, P.O. Box 1159, Mashhad
91775, Iran
E-mail address: moslehian@ferdowsi.um.ac.ir