Refinements of Reverse Triangle Inequalities in Inner Product Spaces

volume 6, issue 3, article 64,2005.

Received 01 February, 2005;accepted 20 May, 2005.

Communicated by: S.S. Dragomir

Abstract

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Journal of Inequalities in Pure andApplied Mathematics

REFINEMENTS OF REVERSE TRIANGLE INEQUALITIES ININNER PRODUCT SPACES

ARSALAN HOJJAT ANSARI AND MOHAMMAD SAL MOSLEHIANDepartment of MathematicsFerdowsi UniversityP.O. Box 1159, Mashhad 91775, Iran.

EMail : [email protected]: http://www.um.ac.ir/∼moslehian/

c©2000Victoria UniversityISSN (electronic): 1443-5756026-05

Please quote this number (026-05) in correspondence regarding this paper with the Editorial Office.

mailto:[email protected]

http://jipam.vu.edu.au/


http://www.um.ac.ir/~moslehian/

http://www.vu.edu.au/

Refinements of ReverseTriangle Inequalities in Inner

Product Spaces

Arsalan Hojjat Ansari andMohammad Sal Moslehian

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J. Ineq. Pure and Appl. Math. 6(3) Art. 64, 2005

http://jipam.vu.edu.au

Abstract

Refining some results of S.S. Dragomir, several new reverses of the triangleinequality in inner product spaces are obtained.

2000 Mathematics Subject Classification: Primary 46C05; Secondary 26D15.Key words: Triangle inequality, Reverse inequality, Diaz-Metkalf inequality, Inner

product space.

Contents1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Main Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5References





http://www.ams.org/msc/


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1. IntroductionIt is interesting to know under which conditions the triangle inequality reversesin a normed spaceX; in other words, we would like to know if there is a con-stantc with the property thatc

∑nk=1 ‖xk‖ ≤ ‖

∑nk=1 xk‖ for some finite set

x1, . . . , xn ∈ X. M. Nakai and T. Tada [7] proved that the normed spaces withthis property for any finite setx1, . . . , xn ∈ X are only those of finite dimen-sion.

The first authors to investigate the reverse of the triangle inequality in innerproduct spaces were J. B. Diaz and F. T. Metcalf [2]. They did so by establishingthe 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 innerproduct space(H; 〈·, ·〉). Suppose the vectorsxk ∈ H, k ∈ {1, . . . , n} satisfy

0 ≤ r ≤ Re〈xk, a〉‖xk‖

, k ∈ {1, . . . , n}.

Then

rn∑

k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ,

where equality holds if and only if

n∑k=1

xk = rn∑

k=1

‖xk‖a.






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Inequalities related to the triangle inequality are of special interest; cf. Chap-ter XVII of [ 6] and may be applied to obtain inequalities in complex numbersor to study vector-valued integral inequalities [3], [4].

Using several ideas and the notation of [3], [4] we modify or refine someresults 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.






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2. Main ResultsThe following theorem is an improvement of Theorem2.1 of [4] in which thereal numbersr1, r2 are not neccesarily nonnegative. The proof seems to bedifferent as well.

Theorem 2.1.Leta be a unit vector in the complex inner product space(H; 〈·, ·〉).Suppose that the vectorsxk ∈ H, k ∈ {1, . . . , n} satisfy

(2.1) 0 ≤ r21‖xk‖ ≤ Re〈xk, r1a〉, 0 ≤ r2

2‖xk‖ ≤ Im〈xk, r2a〉

for somer1, r2 ∈ [−1, 1]. Then we have the inequality

(2.2) (r21 + r2

2)12

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .

The equality holds in (2.2) if and only if

(2.3)n∑

k=1

xk = (r1 + ir2)n∑

k=1

‖xk‖a.

Proof. If r21 +r2

2 = 0, the theorem is trivial. Assume thatr21 +r2

2 6= 0. Summinginequalities (2.1) overk from 1 to n, we have

(r21 + r2

2)n∑

k=1

‖xk‖ ≤ Re

⟨n∑

k=1

xk, r1a

⟩+ Im

⟨n∑

k=1

xk, r2a

⟩






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= Re

⟨n∑

k=1

xk, (r1 + ir2)a

⟩

≤

∣∣∣∣∣⟨

n∑k=1

xk, (r1 + ir2)a

⟩∣∣∣∣∣≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ‖(r1 + ir2)a‖

= (r21 + r2

2)12

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .

Hence (2.2) holds.If (2.3) holds, then∥∥∥∥∥

n∑k=1

xk

∥∥∥∥∥ =

∥∥∥∥∥(r1 + ir2)n∑

k=1

‖xk‖a

∥∥∥∥∥ = (r21 + r2

2)12

n∑k=1

‖xk‖.

Conversely, if the equality holds in (2.2), we have

(r21 + r2

2)12

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ = (r21 + r2

2)n∑

k=1

‖xk‖

≤ Re

⟨n∑

k=1

xk, (r1 + ir2)a

⟩

≤

∣∣∣∣∣⟨

n∑k=1

xk, (r1 + ir2)a

⟩∣∣∣∣∣ ≤ (r21 + r2

2)12

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .






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From this we deduce∣∣∣∣∣⟨

n∑k=1

xk, (r1 + ir2)a

⟩∣∣∣∣∣ =

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ‖(r1 + ir2)a‖.

Consequently there existsη ≥ 0 such that

n∑k=1

xk = η(r1 + ir2)a.

From this we have

(r21 + r2

2)12 η = ‖η(r1 + ir2)a‖ =

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ = (r21 + r2

2)12

n∑k=1

‖xk‖.

Hence

η =n∑

k=1

‖xk‖.

The next theorem is a refinement of Corollary 1 of [4] since, in the notationof Theorem2.1,

√2 − p2

1 − p22 ≤

√α2

1 + α22.

Theorem 2.2.Leta be a unit vector in the complex inner product space(H; 〈·, ·〉).Suppose the vectorsxk ∈ H − {0}, k ∈ {1, . . . , n}, are such that

(2.4) ‖xk − a‖ ≤ p1, ‖xk − ia‖ ≤ p2, p1, p2 ∈(0,√

α2 + 1)

,






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whereα = min1≤k≤n‖xk‖. Let

α1 = min

{‖xk‖2 − p2

1 + 1

2‖xk‖: 1 ≤ k ≤ n

},

α2 = min

{‖xk‖2 − p2

2 + 1

2‖xk‖: 1 ≤ k ≤ n

}.

Then we have the inequality

(α21 + α2

2)12

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ,

where the equality holds if and only if

n∑k=1

xk = (α1 + iα2)n∑

k=1

‖xk‖a.

Proof. From the first inequality in (2.4) we have

〈xk − a, xk − a〉 ≤ p21,

‖xk‖2 + 1 − p21 ≤ 2 Re〈xk, a〉, k = 1, . . . , n,

and‖xk‖2 − p2

1 + 1

2‖xk‖‖xk‖ ≤ Re〈xk, a〉.

Consequently,α1‖xk‖ ≤ Re〈xk, a〉.






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Similarly from the second inequality we obtain

α2‖xk‖ ≤ Re〈xk, ia〉 = Im〈xk, a〉.

Now apply Theorem2.1for r1 = α1, r2 = α2.

Corollary 2.3. Leta be a unit vector in the complex inner product space(H; 〈·, ·〉).Suppose that the vectorsxk ∈ H − {0}, k ∈ {1, . . . , n} such that

‖xk − a‖ ≤ 1, ‖xk − ia‖ ≤ 1.

Thenα√2

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ,

in whichα = min1≤k≤n‖xk‖. The equality holds if and only if

n∑k=1

xk = α(1 + i)

2

n∑k=1

‖xk‖a.

Proof. Apply Theorem2.2for α1 = α2

= α2.

Theorem 2.4. Let a be a unit vector in the inner product space(H; 〈·, ·〉) overthe real or complex number field. Suppose that the vectorsxk ∈ H − {0},k ∈ {1, . . . , n} satisfy

‖xk − a‖ ≤ p, p ∈(0,√

α2 + 1)

, α = min1≤k≤n

‖xk‖.






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α1

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ,

whereα1 = min

{‖xk‖2 − p2 + 1

2‖xk‖: 1 ≤ k ≤ n

}.

The equality holds if and only ifn∑

k=1

xk = α1

n∑k=1

‖xk‖a.

Proof. The proof is similar to Theorem2.2 in which we use Theorem2.1withr2 = 0.

The next theorem is a generalization of Theorem2.1. It is a modification ofTheorem 3 of [4], however our proof is apparently different.

Theorem 2.5. Let a1, . . . , am be orthonormal vectors in the complex innerproduct space(H; 〈·, ·〉). Suppose that for1 ≤ t ≤ m, rt, ρt ∈ R and thatthe vectorsxk ∈ H, k ∈ {1, . . . , n} satisfy

0 ≤ r2t ‖xk‖ ≤ Re〈xk, rtat〉,(2.5)

0 ≤ ρ2t‖xk‖ ≤ Im〈xk, ρtat〉, t ∈ {1, . . . ,m}.


(2.6)

(m∑

t=1

r2t + ρ2

t

) 12 n∑

k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .






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The equality holds in (2.7) if and only if

(2.7)n∑

k=1

xk =n∑

k=1

‖xk‖m∑

t=1

(rt + iρt)at.

Proof. If∑m

t=1(r2t + ρ2

t ) = 0, the theorem is trivial. Assume that∑m

t=1(r2t +

ρ2t ) 6= 0. Summing inequalities (2.6) overk from 1 to n and again overt from 1

to m, we get

m∑t=1

(r2t + ρ2

t )n∑

k=1

‖xk‖ ≤ Re

⟨n∑

k=1

xk,m∑

t=1

rtat

⟩+ Im

⟨n∑

k=1

xk,m∑

t=1

ρtat

⟩

= Re

⟨n∑

k=1

xk,m∑

t=1

rtat

⟩+ Re

⟨n∑

k=1

xk, im∑

t=1

ρtat

⟩

= Re

⟨n∑

k=1

xk,m∑

t=1

(rt + iρt)at

⟩

≤

∣∣∣∣∣⟨

n∑k=1

xk,m∑

t=1

(rt + iρt)at

⟩∣∣∣∣∣≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥∥∥∥∥∥

m∑t=1

(rt + iρt)at

∥∥∥∥∥=

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥(

m∑t=1

(r2t + ρ2

t )

) 12

.






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Then

(2.8)

(m∑

t=1

(r2t + ρ2

t )

) 12 n∑

k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .

If (2.8) holds, then∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ =

∥∥∥∥∥n∑

k=1

‖xk‖m∑

t=1

(rt + iρt)at

∥∥∥∥∥ =n∑

k=1

‖xk‖

(m∑

t=1

(r2t + ρ2

t )

) 12

.

Conversely, if the equality holds in(2.7), we obtain from (2.6) that(m∑

t=1

(r2t + ρ2

t )

) 12∥∥∥∥∥

n∑k=1

xk

∥∥∥∥∥ =m∑

t=1

(r2t + ρ2

t )n∑

k=1

‖xk‖

≤ Re

⟨n∑

k=1

xk,m∑

t=1

(rt + iρt)at

⟩

≤

∣∣∣∣∣⟨

n∑k=1

xk,m∑

t=1

(rt + iρt)at

⟩∣∣∣∣∣≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥∥∥∥∥∥

m∑t=1

(rt + iρt)at

∥∥∥∥∥=

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥(

m∑t=1

(r2t + ρ2

t )

) 12

.






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Thus we have∣∣∣∣∣⟨

n∑k=1

xk,m∑

t=1

(rt + iρt)at

⟩∣∣∣∣∣ =

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥∥∥∥∥∥

m∑t=1

(rt + iρt)at

∥∥∥∥∥ .

Consequently there existsη ≥ 0 such that

n∑k=1

xk = ηm∑

t=1

(rt + iρt)at

from which we have

η

(m∑

t=1

(r2t + ρ2

t )

) 12

=

∥∥∥∥∥ηm∑

t=1

(rt + iρt)at

∥∥∥∥∥=

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥=

n∑k=1

‖xk‖

(m∑

t=1

(r2t + ρ2

t )

) 12

.

Hence

η =n∑

k=1

‖xk‖.






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Corollary 2.6. Let a1, . . . , am be orthornormal vectors in the inner productspace(H; 〈·, ·〉) over the real or complex number field. Suppose for1 ≤ t ≤ mthat the vectorsxk ∈ H, k ∈ {1, . . . , n} satisfy

0 ≤ r2t ‖xk‖ ≤ Re〈xk, rtat〉.

Then we have the inequality(m∑

t=1

r2t

) 12 n∑

k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .


k=1

xk =n∑

k=1

‖xk‖m∑

t=1

rtat.

Proof. Apply Theorem2.5for ρt = 0.

Theorem 2.7. Let a1, . . . , am be orthornormal vectors in the complex innerproduct space(H; 〈·, ·〉). Suppose that the vectorsxk ∈ H−{0}, k ∈ {1, . . . , n}satisfy

‖xk − at‖ ≤ pt, ‖xk − iat‖ ≤ qt, pt, qt ∈(0,√

α2 + 1)

, 1 ≤ t ≤ m,

whereα = min1≤k≤n‖xk‖. Let

αt = min

{‖xk‖2 − p2

t + 1

2‖xk‖: 1 ≤ k ≤ n

},

βt = min

{‖xk‖2 − q2

t + 1

2‖xk‖: 1 ≤ k ≤ n

}.






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Then we have the inequality(m∑

t=1

α2t + β2

t

) 12 n∑

k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ ,

where equality holds if and only if

n∑k=1

xk =n∑

k=1

‖xk‖m∑

t=1

(αt + iβt)at.

Proof. For1 ≤ t ≤ m, 1 ≤ k ≤ n it follows from ‖xk − at‖ ≤ pt that

〈xk − at〉, xk − at〉 ≤ p2t ,

‖xk‖2 − p2t + 1

2‖xk‖‖xk‖ ≤ Re〈xk, at〉,

αt‖xk‖ ≤ Re〈xk, at〉,

and similarlyβt‖xk‖ ≤ Re〈xk, iat〉 = Im〈xk, at〉.

Now applying Theorem2.4 with rt = αt, ρt = βt we deduce the desired in-equality.

Corollary 2.8. Let a1, . . . , am be orthornormal vectors in the complex innerproduct space(H; 〈·, ·〉). Suppose that the vectorsxk ∈ H, k ∈ {1, . . . , n}satisfy

‖xk − at‖ ≤ 1, ‖xk − iat‖ ≤ 1, 1 ≤ t ≤ m.






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Thenα√2

√m

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .


k=1

xk = α(1 + i)

2

n∑k=1

‖xk‖m∑

t=1

at.

Proof. Apply Theorem2.7for αt = α2

= βt.

Remark 1. It is interesting to note that

α√2

√m ≤ ‖

∑nk=1 xk‖∑n

k=1 ‖xk‖≤ 1,

where

α ≤√

2

m.

Corollary 2.9. Leta be a unit vector in the complex inner product space(H; 〈·, ·〉).Suppose that the vectorsxk ∈ H − {0}, k ∈ {1, . . . , n} satisfy

‖xk − a‖ ≤ p1, ‖xk − ia‖ ≤ p2, p1, p2 ∈ (0, 1].

Let

α1 = min

{‖xk‖2 − p2

1 + 1

2‖xk‖: 1 ≤ k ≤ n

},

α2 = min

{‖xk‖2 − p2

2 + 1

2‖xk‖: 1 ≤ k ≤ n

}.






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If α1 6= (1 − p21)

12 , or α2 6= (1 − p2

2)12 , then we have the following strictly

inequality

(2 − p21 − p2

2)12

n∑k=1

‖xk‖ <

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .

Proof. If equality holds, then by Theorem2.2we have

(α21 + α2

2)12

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ = (2 − p21 − p2

2)12

n∑k=1

‖xk‖

and so(α2

1 + α22)

12 ≤ (2 − p2

1 − p22)

12 .

On the other hand for1 ≤ k ≤ n,

‖xk‖2 − p21 + 1

2‖xk‖≥ (1 − p2

1)12

and soα1 ≥ (1 − p2

1)12 .

Similarlyα2 ≥ (1 − p2

2)12 .

Hence(2 − p2

1 − p22)

12 ≤ (α2

1 + α22)

12 .

Thus √α2

1 + α22 = (2 − p2

1 − p22)

12 .






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Thereforeα1 = (1 − p2

1)12 and α2 = (1 − p2

2)12 ,

a contradiction.

The following result looks like Corollary 2 of [4].

Theorem 2.10. Let a be a unit vector in the complex inner product space(H; 〈·, ·〉), M ≥ m > 0, L ≥ ` > 0 and xk ∈ H − {0}, k ∈ {1, . . . , n}such that

Re〈Ma− xk, xk −ma〉 ≥ 0, Re〈Lia − xk, xk − `ia〉 ≥ 0,

or equivalently,

‖xk −m + M

2a‖ ≤ M −m

2, ‖xk −

L + `

2ia‖ ≤ L − `

2.

Let

αm,M = min

{‖xk‖2 + mM

(m + M)‖xk‖: 1 ≤ k ≤ n

}and

α`,L = min

{‖xk‖2 + `L

(` + L)‖xk‖: 1 ≤ k ≤ n

}.

Then we have the inequlity

(α2m,M + α2

`,L)12

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .






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The equality holds if and only if

n∑k=1

xk = (αm,M + iα`,L)n∑

k=1

‖xk‖a.

Proof. For each1 ≤ k ≤ n, it follows from

‖xk −m + M

2a‖ ≤ M −m

2

that ⟨xk −

m + M

2a, xk −

m + M

2

⟩≤(

M −m

2

)2

.

Hence‖xk‖2 + mM ≤ (m + M) Re〈xk, a〉.

Then‖xk‖2 + mM

(m + M)‖xk‖‖xk‖ ≤ Re〈xk, a〉,

and consequentlyαm,M‖xk‖ ≤ Re〈xk, a〉.

Similarly from the second inequality we deduce

α`,L‖xk‖ ≤ Im〈xk, a〉.

Applying Theorem2.1 for r1 = αm,M , r2 = α`,L, we infer the desired inequal-ity.






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Theorem 2.11. Let a be a unit vector in the complex inner product space(H; 〈·, ·〉), M ≥ m > 0, L ≥ ` > 0 and xk ∈ H − {0}, k ∈ {1, . . . , n}such that

Re〈Ma− xk, xk −ma〉 ≥ 0, Re〈Lia − xk, xk − `ia〉 ≥ 0,

or equivalently∥∥∥∥xk −m + M

2a

∥∥∥∥ ≤ M −m

2,

∥∥∥∥xk −L + `

2ia

∥∥∥∥ ≤ L − `

2.

Let

αm,M = min

{‖xk‖2 + mM

(m + M)‖xk‖: 1 ≤ k ≤ n

}and

α`,L = min

{‖xk‖2 + `L

(` + L)‖xk‖: 1 ≤ k ≤ n

}.

If αm,M 6= 2√

mMm+M

, or α`,L 6= 2√

`L`+L

, then we have

2

(mM

(m + M)2+

`L

(` + L)2

) 12

n∑k=1

‖xk‖ <

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥ .

Proof. If

2

(mM

(m + M)2+

`L

(` + L)2

) 12

n∑k=1

‖xk‖ =

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥






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then by Theorem2.10we have

(α2m,M + α2

`,L)12

n∑k=1

‖xk‖ ≤

∥∥∥∥∥n∑

k=1

xk

∥∥∥∥∥= 2

(mM

(m + M)2+

`L

(` + L)2

) 12

n∑k=1

‖xk‖.

Consequently

(α2m,M + α2

`,L)12 ≤ 2

(mM

(m + M)2+

`L

(` + L)2

) 12

.

On the other hand for1 ≤ k ≤ n,

‖xk‖2 + mM

(m + M)‖xk‖≥ 2

√mM

m + M, and

‖xk‖2 + `L

(` + L)‖xk‖≥ 2

√`L

` + L,

so

(α2m,M + α2

`,L)12 ≥ 2

(mM

(m + M)2+

`L

(` + L)2

) 12

.

Then

(α2m,M + α2

`,L)12 = 2

(mM

(m + M)2+

`L

(` + L)2

) 12

.

Hence

αm,M = 2

√mM

m + M






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and

α`,L = 2

√`L

` + La contradiction.

Finally we mention two applications of our results to complex numbers.

Corollary 2.12. Let a ∈ C with |a| = 1. Suppose thatzk ∈ C, k ∈ {1, . . . , n}such that

|zk − a| ≤ p1, |zk − ia| ≤ p2, p1, p2 ∈(0,√

α2 + 1)

,

whereα = min{|zk| : 1 ≤ k ≤ n}.

Let

α1 = min

{|zk|2 − p2

1 + 1

2|zk|: 1 ≤ k ≤ n

},

α2 = min

{|zk|2 − p2

2 + 1

2|zk|: 1 ≤ k ≤ n

}.

Then we have the inequality√α2

1 + α22

n∑k=1

|zk| ≤

∣∣∣∣∣n∑

k=1

zk

∣∣∣∣∣ .The equality holds if and only if

n∑k=1

zk = (α1 + iα2)

(n∑

k=1

|zk|

)a.






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Proof. Apply Theorem2.2for H = C.

Corollary 2.13. Let a ∈ C with |a| = 1. Suppose thatzk ∈ C, k ∈ {1, . . . , n}such that

|zk − a| ≤ 1, |zk − ia| ≤ 1.

If α = min{|zk| : 1 ≤ k ≤ n}. Then we have the inequality

α√2

n∑k=1

|zk| ≤

∣∣∣∣∣n∑

k=1

zk

∣∣∣∣∣the equality holds if and only if

n∑k=1

zk = α(1 + i)

2

(n∑

k=1

|zk|

)a.

Proof. Apply Corollary2.3for H = C.






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References[1] A.H. ANSARI AND M.S. MOSLEHIAN, More on reverse triangle inequal-

ity in inner product spaces, arXiv:math.FA/0506198.

[2] J.B. DIAZ AND F.T. METCALF, A complementary triangle inequality inHilbert and Banach spaces,Proc. Amer. Math. Soc., 17(1) (1966), 88–97.

[3] S.S. DRAGOMIR, Reverses of the triangle inequality in inner productspaces, arXiv:math.FA/0405495.

[4] S.S. DRAGOMIR, Some reverses of the generalized triangle inequality incomplex inner product spaces, arXiv:math.FA/0405497.

[5] S.S. DRAGOMIR, Discrete Inequalities of the Cauchy-Bunyakovsky-Schwarz Type, Nova Science Publishers, Inc., Hauppauge, NY, 2004.

[6] D.S. MITRINOVIC, J.E. PECARIC AND A.M. FINK, Classical andNew Inequalities in Analysis, Kluwer Academic Publishers, Dor-drecht/Boston/London, 1993.

[7] M. NAKAI AND T. TADA, The reverse triangle inequality in normedspaces,New Zealand J. Math., 25(2) (1996), 181–193.

[8] M. PETROVICH, Module d’une somme,L’ Ensignement Mathématique,19 (1917), 53–56.

[9] Th.M. RASSIAS,Inner Product Spaces and Applications, Chapman-Hall,1997.





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Refinements of Reverse Triangle Inequalities in Inner Product Spaces