Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 64, 2005
REFINEMENTS OF REVERSE TRIANGLE INEQUALITIES IN INNER PRODUCT
SPACES
ARSALAN HOJJAT ANSARI AND MOHAMMAD SAL MOSLEHIAN
D EPARTMENT OF M ATHEMATICS
F ERDOWSI U NIVERSITY
P.O. B OX 1159, M ASHHAD 91775, I RAN .
msalm@math.um.ac.ir
URL: http://www.um.ac.ir/∼moslehian/
Received 01 February, 2005; accepted 20 May, 2005
Communicated by S.S. Dragomir
A BSTRACT. Refining some results of S.S. Dragomir, several new reverses of the triangle inequality in inner product spaces are obtained.
Key words and phrases: Triangle inequality, Reverse inequality, Diaz-Metkalf inequality, Inner product space.
2000 Mathematics Subject Classification. Primary 46C05; Secondary 26D15.
1. I NTRODUCTION
It is interesting to know under which conditions the triangle inequality reverses in a normed
space
words, we would like to know if there is a constant c with the property that
P X; in other P
c nk=1 kxk k ≤ k nk=1 xk k for some finite set x1 , . . . , xn ∈ X. M. Nakai and T. Tada [7]
proved that the normed spaces with this property for any finite set x1 , . . . , xn ∈ X are only
those of finite dimension.
The first authors to investigate the reverse of the triangle inequality in inner product spaces
were J. B. Diaz and F. T. Metcalf [2]. They did so by establishing the following result as an
extension of an inequality given by M. Petrovich [8] for complex numbers:
Theorem 1.1 (Diaz-Metcalf Theorem). Let a be a unit vector in the inner product space
(H; h·, ·i). Suppose the vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
0≤r≤
Rehxk , ai
,
kxk k
Then
r
n
X
k=1
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
026-05
k ∈ {1, . . . , n}.
n
X
kxk k ≤ xk ,
k=1
2
A RSALAN H OJJAT A NSARI AND M OHAMMAD S AL M OSLEHIAN
where equality holds if and only if
n
X
xk = r
k=1
n
X
kxk ka.
k=1
Inequalities related to the triangle inequality are of special interest; cf. Chapter XVII of [6]
and may be applied to obtain inequalities in complex numbers or to study vector-valued integral
inequalities [3], [4].
Using several ideas and the notation of [3], [4] we modify or refine some results of S.S.
Dragomir to procure some new reverses of the triangle inequality (see also [1]).
We use repeatedly the Cauchy-Schwarz inequality without mentioning it. The reader is referred to [9], [5] for the terminology of inner product spaces.
2. M AIN R ESULTS
The following theorem is an improvement of Theorem 2.1 of [4] in which the real numbers
r1 , r2 are not neccesarily nonnegative. The proof seems to be different as well.
Theorem 2.1. Let a be a unit vector in the complex inner product space (H; h·, ·i). Suppose
that the vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
(2.1)
0 ≤ r12 kxk k ≤ Rehxk , r1 ai,
0 ≤ r22 kxk k ≤ Imhxk , r2 ai
for some r1 , r2 ∈ [−1, 1]. Then we have the inequality
1
(r12 + r22 ) 2
(2.2)
n
X
k=1
n
X
kxk k ≤ xk .
k=1
The equality holds in (2.2) if and only if
n
X
(2.3)
xk = (r1 + ir2 )
k=1
n
X
kxk ka.
k=1
Proof. If r12 + r22 = 0, the theorem is trivial. Assume that r12 + r22 6= 0. Summing inequalities
(2.1) over k from 1 to n, we have
* n
+
* n
+
n
X
X
X
(r12 + r22 )
kxk k ≤ Re
xk , r1 a + Im
xk , r2 a
k=1
k=1
= Re
* n
X
k=1
+
xk , (r1 + ir2 )a
* n k=1
+
X
xk , (r1 + ir2 )a ≤
k=1 n
X
≤
xk k(r1 + ir2 )ak
k=1
n
X
2
2 21 = (r1 + r2 ) xk .
k=1
Hence (2.2) holds.
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R EFINEMENTS OF R EVERSE T RIANGLE I NEQUALITIES IN I NNER P RODUCT S PACES
3
If (2.3) holds, then
n
n
n
X
X
X
2 12
2
xk = (r1 + ir2 )
kxk ka = (r1 + r2 )
kxk k.
k=1
k=1
k=1
Conversely, if the equality holds in (2.2), we have
n
n
X X
1 (r12 + r22 ) 2 xk = (r12 + r22 )
kxk k
k=1
k=1
* n
+
X
≤ Re
xk , (r1 + ir2 )a
* k=1
+
n
X
≤
xk , (r1 + ir2 )a k=1
n
X
2
2 12 ≤ (r1 + r2 ) xk .
k=1
From this we deduce
* n
+ n
X
X xk , (r1 + ir2 )a = xk k(r1 + ir2 )ak.
k=1
k=1
Consequently there exists η ≥ 0 such that
n
X
xk = η(r1 + ir2 )a.
k=1
From this we have
n
n
X X
2
2 12
2
2
xk = (r1 + r2 )
kxk k.
(r1 + r2 ) η = kη(r1 + ir2 )ak = 1
2
k=1
Hence
η=
n
X
k=1
kxk k.
k=1
pThe next theorem
p is a refinement of Corollary 1 of [4] since, in the notation of Theorem 2.1,
2 − p21 − p22 ≤ α12 + α22 .
Theorem 2.2. Let a be a unit vector in the complex inner product space (H; h·, ·i). Suppose the
vectors xk ∈ H − {0}, k ∈ {1, . . . , n}, are such that
√
(2.4)
kxk − ak ≤ p1 , kxk − iak ≤ p2 , p1 , p2 ∈ 0, α2 + 1 ,
where α = min1≤k≤n kxk k. Let
kxk k2 − p21 + 1
α1 = min
:1≤k≤n ,
2kxk k
kxk k2 − p22 + 1
α2 = min
:1≤k≤n .
2kxk k
J. Inequal. Pure and Appl. Math., 6(3) Art. 64, 2005
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4
A RSALAN H OJJAT A NSARI AND M OHAMMAD S AL M OSLEHIAN
Then we have the inequality
(α12 + α22 )
1
2
n
X
k=1
n
X
kxk k ≤ xk ,
k=1
where the equality holds if and only if
n
X
xk = (α1 + iα2 )
k=1
n
X
kxk ka.
k=1
Proof. From the first inequality in (2.4) we have
hxk − a, xk − ai ≤ p21 ,
kxk k2 + 1 − p21 ≤ 2 Rehxk , ai,
k = 1, . . . , n,
and
kxk k2 − p21 + 1
kxk k ≤ Rehxk , ai.
2kxk k
Consequently,
α1 kxk k ≤ Rehxk , ai.
Similarly from the second inequality we obtain
α2 kxk k ≤ Rehxk , iai = Imhxk , ai.
Now apply Theorem 2.1 for r1 = α1 , r2 = α2 .
Corollary 2.3. Let a be a unit vector in the complex inner product space (H; h·, ·i). Suppose
that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n} such that
kxk − ak ≤ 1,
Then
kxk − iak ≤ 1.
n
n
X
α X
√
kxk k ≤ xk ,
2 k=1
k=1
in which α = min1≤k≤n kxk k. The equality holds if and only if
n
n
X
(1 + i) X
kxk ka.
xk = α
2
k=1
k=1
Proof. Apply Theorem 2.2 for α1 =
α
2
= α2 .
Theorem 2.4. Let a be a unit vector in the inner product space (H; h·, ·i) over the real or
complex number field. Suppose that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n} satisfy
√
kxk − ak ≤ p, p ∈ 0, α2 + 1 , α = min kxk k.
1≤k≤n
Then we have the inequality
α1
n
X
k=1
n
X
kxk k ≤ xk ,
k=1
where
α1 = min
J. Inequal. Pure and Appl. Math., 6(3) Art. 64, 2005
kxk k2 − p2 + 1
:1≤k≤n .
2kxk k
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R EFINEMENTS OF R EVERSE T RIANGLE I NEQUALITIES IN I NNER P RODUCT S PACES
5
The equality holds if and only if
n
X
n
X
xk = α1
k=1
kxk ka.
k=1
Proof. The proof is similar to Theorem 2.2 in which we use Theorem 2.1 with r2 = 0.
The next theorem is a generalization of Theorem 2.1. It is a modification of Theorem 3 of
[4], however our proof is apparently different.
Theorem 2.5. Let a1 , . . . , am be orthonormal vectors in the complex inner product space (H; h·, ·i).
Suppose that for 1 ≤ t ≤ m, rt , ρt ∈ R and that the vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
(2.5)
0 ≤ ρ2t kxk k ≤ Imhxk , ρt at i,
0 ≤ rt2 kxk k ≤ Rehxk , rt at i,
t ∈ {1, . . . , m}.
Then we have the inequality
m
X
(2.6)
! 21
n
X
rt2 + ρ2t
t=1
k=1
n
X kxk k ≤ xk .
k=1
The equality holds in (2.7) if and only if
n
n
m
X
X
X
(2.7)
xk =
kxk k
(rt + iρt )at .
k=1
t=1
k=1
Pm
P
2
2
Proof. If
= 0, the theorem is trivial. Assume that m
t=1 (rt + ρt ) 6= 0. Summing
inequalities (2.6) over k from 1 to n and again over t from 1 to m, we get
* n
+
* n
+
m
n
m
m
X
X
X
X
X
X
kxk k ≤ Re
xk ,
rt at + Im
(rt2 + ρ2t )
xk ,
ρ t at
2
2
t=1 (rt + ρt )
t=1
k=1
t=1
* n
X
m
X
k=1
= Re
= Re
xk ,
k=1
t=1
* n
X
m
X
xk ,
k=1
+
r t at
+ Re
* n
X
k=1
t=1
xk , i
m
X
+
ρ t at
t=1
+
(rt + iρt )at
t=1
* n k=1 m
+
X
X
≤
xk ,
(rt + iρt )at t=1
nk=1 m
X X
xk (rt + iρt )at ≤
t=1
k=1
n
m
! 12
X X
=
xk (rt2 + ρ2t )
.
t=1
k=1
Then
m
X
(2.8)
! 12
(rt2 + ρ2t )
t=1
n
X
k=1
n
X
kxk k ≤ xk .
k=1
If (2.8) holds, then
! 12
n
n
m
n
m
X
X
X
X
X
.
xk = kxk k
(rt + iρt )at =
kxk k
(rt2 + ρ2t )
t=1
t=1
k=1
k=1
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k=1
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6
A RSALAN H OJJAT A NSARI AND M OHAMMAD S AL M OSLEHIAN
Conversely, if the equality holds in (2.7), we obtain from (2.6) that
! 21 n
m
m
n
X X
X
X
2
2
2
2
(rt + ρt )
xk =
kxk k
(rt + ρt )
t=1
t=1
k=1
k=1
* n
+
m
X
X
≤ Re
xk ,
(rt + iρt )at
t=1
* k=1
+
m
n
X
X
xk ,
(rt + iρt )at ≤
t=1
k=1 n
m
X
X
≤
xk (rt + iρt )at t=1
k=1
! 12
n
m
X
X
.
=
xk (r2 + ρ2t )
t=1 t
k=1
Thus we have
*
+ n
n
m
m
X
X X
X
xk ,
(rt + iρt )at = xk (rt + iρt )at .
k=1
t=1
t=1
k=1
Consequently there exists η ≥ 0 such that
m
n
X
X
xk = η
(rt + iρt )at
t=1
k=1
from which we have
η
m
X
! 12
(rt2 + ρ2t )
t=1
m
X
= η
(rt + iρt )at nt=1 X =
xk k=1
=
n
X
kxk k
η=
n
X
! 12
(rt2 + ρ2t )
.
t=1
k=1
Hence
m
X
kxk k.
k=1
Corollary 2.6. Let a1 , . . . , am be orthornormal vectors in the inner product space (H; h·, ·i)
over the real or complex number field. Suppose for 1 ≤ t ≤ m that the vectors xk ∈ H,
k ∈ {1, . . . , n} satisfy
0 ≤ rt2 kxk k ≤ Rehxk , rt at i.
Then we have the inequality
! 21 n
n
m
X
X
X
2
kxk k ≤ xk .
rt
t=1
J. Inequal. Pure and Appl. Math., 6(3) Art. 64, 2005
k=1
k=1
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R EFINEMENTS OF R EVERSE T RIANGLE I NEQUALITIES IN I NNER P RODUCT S PACES
7
The equality holds if and only if
n
X
xk =
k=1
n
X
kxk k
m
X
rt at .
t=1
k=1
Proof. Apply Theorem 2.5 for ρt = 0.
Theorem 2.7. Let a1 , . . . , am be orthornormal vectors in the complex inner product space
(H; h·, ·i). Suppose that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n} satisfy
√
kxk − at k ≤ pt , kxk − iat k ≤ qt , pt , qt ∈ 0, α2 + 1 , 1 ≤ t ≤ m,
where α = min1≤k≤n kxk k. Let
kxk k2 − p2t + 1
αt = min
:1≤k≤n ,
2kxk k
kxk k2 − qt2 + 1
βt = min
:1≤k≤n .
2kxk k
Then we have the inequality
m
X
! 12
αt2 + βt2
t=1
n
X
k=1
n
X
kxk k ≤ xk ,
k=1
where equality holds if and only if
n
n
m
X
X
X
xk =
kxk k
(αt + iβt )at .
k=1
k=1
t=1
Proof. For 1 ≤ t ≤ m, 1 ≤ k ≤ n it follows from kxk − at k ≤ pt that
hxk − at i, xk − at i ≤ p2t ,
kxk k2 − p2t + 1
kxk k ≤ Rehxk , at i,
2kxk k
αt kxk k ≤ Rehxk , at i,
and similarly
βt kxk k ≤ Rehxk , iat i = Imhxk , at i.
Now applying Theorem 2.4 with rt = αt , ρt = βt we deduce the desired inequality.
Corollary 2.8. Let a1 , . . . , am be orthornormal vectors in the complex inner product space
(H; h·, ·i). Suppose that the vectors xk ∈ H, k ∈ {1, . . . , n} satisfy
kxk − at k ≤ 1,
Then
kxk − iat k ≤ 1,
1 ≤ t ≤ m.
n
n
X α √ X
√ m
kxk k ≤ xk .
2
k=1
k=1
The equality holds if and only if
n
X
n
m
X
(1 + i) X
xk = α
kxk k
at .
2 k=1
t=1
k=1
Proof. Apply Theorem 2.7 for αt =
J. Inequal. Pure and Appl. Math., 6(3) Art. 64, 2005
α
2
= βt .
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8
A RSALAN H OJJAT A NSARI AND M OHAMMAD S AL M OSLEHIAN
Remark 2.9. It is interesting to note that
P
k nk=1 xk k
α √
√ m ≤ Pn
≤ 1,
2
k=1 kxk k
where
r
α≤
2
.
m
Corollary 2.10. Let a be a unit vector in the complex inner product space (H; h·, ·i). Suppose
that the vectors xk ∈ H − {0}, k ∈ {1, . . . , n} satisfy
kxk − ak ≤ p1 ,
kxk − iak ≤ p2 ,
p1 , p2 ∈ (0, 1].
Let
kxk k2 − p21 + 1
α1 = min
:1≤k≤n ,
2kxk k
kxk k2 − p22 + 1
α2 = min
:1≤k≤n .
2kxk k
1
1
If α1 6= (1 − p21 ) 2 , or α2 6= (1 − p22 ) 2 , then we have the following strictly inequality
n
n
X X
1
(2 − p21 − p22 ) 2
kxk k < xk .
k=1
k=1
Proof. If equality holds, then by Theorem 2.2 we have
n
n
n
X X
X
1
2
2 2
2
2 12
(α1 + α2 )
kxk k ≤ xk = (2 − p1 − p2 )
kxk k
k=1
k=1
k=1
and so
1
1
(α12 + α22 ) 2 ≤ (2 − p21 − p22 ) 2 .
On the other hand for 1 ≤ k ≤ n,
1
kxk k2 − p21 + 1
≥ (1 − p21 ) 2
2kxk k
and so
1
α1 ≥ (1 − p21 ) 2 .
Similarly
1
α2 ≥ (1 − p22 ) 2 .
Hence
1
1
(2 − p21 − p22 ) 2 ≤ (α12 + α22 ) 2 .
Thus
q
1
α12 + α22 = (2 − p21 − p22 ) 2 .
Therefore
1
α1 = (1 − p21 ) 2
and
1
α2 = (1 − p22 ) 2 ,
a contradiction.
The following result looks like Corollary 2 of [4].
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R EFINEMENTS OF R EVERSE T RIANGLE I NEQUALITIES IN I NNER P RODUCT S PACES
9
Theorem 2.11. Let a be a unit vector in the complex inner product space (H; h·, ·i), M ≥ m >
0, L ≥ ` > 0 and xk ∈ H − {0}, k ∈ {1, . . . , n} such that
RehM a − xk , xk − mai ≥ 0,
RehLia − xk , xk − `iai ≥ 0,
or equivalently,
kxk −
m+M
M −m
ak ≤
,
2
2
Let
αm,M = min
and
α`,L = min
kxk −
L+`
L−`
iak ≤
.
2
2
kxk k2 + mM
:1≤k≤n
(m + M )kxk k
kxk k2 + `L
:1≤k≤n .
(` + L)kxk k
Then we have the inequlity
2
2
(αm,M
+ α`,L
)
1
2
n
X
k=1
The equality holds if and only if
n
X
n
X
kxk k ≤ xk .
k=1
xk = (αm,M + iα`,L )
k=1
n
X
kxk ka.
k=1
Proof. For each 1 ≤ k ≤ n, it follows from
M −m
m+M
ak ≤
kxk −
2
2
that
2
m+M
M −m
m+M
a, xk −
≤
.
xk −
2
2
2
Hence
kxk k2 + mM ≤ (m + M ) Rehxk , ai.
Then
kxk k2 + mM
kxk k ≤ Rehxk , ai,
(m + M )kxk k
and consequently
αm,M kxk k ≤ Rehxk , ai.
Similarly from the second inequality we deduce
α`,L kxk k ≤ Imhxk , ai.
Applying Theorem 2.1 for r1 = αm,M , r2 = α`,L , we infer the desired inequality.
Theorem 2.12. Let a be a unit vector in the complex inner product space (H; h·, ·i), M ≥ m >
0, L ≥ ` > 0 and xk ∈ H − {0}, k ∈ {1, . . . , n} such that
RehM a − xk , xk − mai ≥ 0,
RehLia − xk , xk − `iai ≥ 0,
xk − m + M a ≤ M − m ,
2
2
xk − L + ` ia ≤ L − ` .
2
2
or equivalently
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10
A RSALAN H OJJAT A NSARI AND M OHAMMAD S AL M OSLEHIAN
Let
αm,M = min
kxk k2 + mM
:1≤k≤n
(m + M )kxk k
and
α`,L = min
kxk k2 + `L
:1≤k≤n .
(` + L)kxk k
√
√
mM
`L
If αm,M 6= 2 m+M
, or α`,L 6= 2 `+L
, then we have
2
`L
mM
+
2
(m + M )
(` + L)2
12 X
n
k=1
n
X
kxk k < xk .
k=1
Proof. If
2
mM
`L
+
2
(m + M )
(` + L)2
12 X
n
k=1
n
X
kxk k = xk k=1
then by Theorem 2.11 we have
2
(αm,M
n
X 2
+ α`,L
)
kxk k ≤ xk k=1
k=1
12 X
n
`L
mM
=2
+
kxk k.
(m + M )2 (` + L)2
k=1
1
2
n
X
Consequently
2
(αm,M
+
1
2
)2
α`,L
≤2
`L
mM
+
2
(m + M )
(` + L)2
12
.
On the other hand for 1 ≤ k ≤ n,
√
kxk k2 + mM
mM
≥2
,
(m + M )kxk k
m+M
and
√
kxk k2 + `L
`L
≥2
,
(` + L)kxk k
`+L
so
2
(αm,M
+
1
2
α`,L
)2
+
1
2
α`,L
)2
≥2
mM
`L
+
2
(m + M )
(` + L)2
12
mM
`L
+
2
(m + M )
(` + L)2
12
.
Then
2
(αm,M
=2
Hence
.
√
αm,M = 2
mM
m+M
and
√
α`,L = 2
`L
`+L
a contradiction.
Finally we mention two applications of our results to complex numbers.
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R EFINEMENTS OF R EVERSE T RIANGLE I NEQUALITIES IN I NNER P RODUCT S PACES
11
Corollary 2.13. Let a ∈ C with |a| = 1. Suppose that zk ∈ C, k ∈ {1, . . . , n} such that
√
2
|zk − a| ≤ p1 , |zk − ia| ≤ p2 , p1 , p2 ∈ 0, α + 1 ,
where
α = min{|zk | : 1 ≤ k ≤ n}.
Let
|zk |2 − p21 + 1
α1 = min
:1≤k≤n ,
2|zk |
|zk |2 − p22 + 1
α2 = min
:1≤k≤n .
2|zk |
Then we have the inequality
n
n
q
X
X
2
2
α1 + α2
|zk | ≤ zk .
k=1
The equality holds if and only if
n
X
k=1
zk = (α1 + iα2 )
k=1
n
X
!
|zk | a.
k=1
Proof. Apply Theorem 2.2 for H = C.
Corollary 2.14. Let a ∈ C with |a| = 1. Suppose that zk ∈ C, k ∈ {1, . . . , n} such that
|zk − a| ≤ 1,
|zk − ia| ≤ 1.
If α = min{|zk | : 1 ≤ k ≤ n}. Then we have the inequality
n
n
X
α X
√
|zk | ≤ zk 2 k=1
k=1
the equality holds if and only if
n
X
(1 + i)
zk = α
2
k=1
n
X
!
|zk | a.
k=1
Proof. Apply Corollary 2.3 for H = C.
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