volume 7, issue 1, article 14, 2006.
Received 28 June, 2005;
accepted 17 September, 2005.
Communicated by:A. Lupa¸s
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Journal of Inequalities in Pure and Applied Mathematics
A POTPOURRI OF SCHWARZ RELATED INEQUALITIES IN INNER PRODUCT SPACES (II)
S.S. DRAGOMIR
School of Computer Science and Mathematics Victoria University
PO Box 14428, Melbourne City Victoria 8001, Australia.
EMail:[email protected] URL:http://rgmia.vu.edu.au/dragomir
2000c Victoria University ISSN (electronic): 1443-5756 196-05
A Potpourri of Schwarz Related Inequalities in Inner Product
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Abstract
Further inequalities related to the Schwarz inequality in real or complex inner product spaces are given.
2000 Mathematics Subject Classification:46C05, 26D15.
Key words: Schwarz inequality, Inner product spaces, Reverse inequalities.
Contents
1 Introduction. . . 3
2 Quadratic Schwarz Related Inequalities. . . 6
3 Other Inequalities . . . 15
4 Applications for the Triangle Inequality. . . 22 References
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1. Introduction
Let(H;h·,·i)be an inner product space over the real or complex number field K. One of the most important inequalities in inner product spaces with numer- ous applications is the Schwarz inequality, that may be written in two forms:
(1.1) |hx, yi|2 ≤ kxk2kyk2, x, y ∈H (quadratic form) or, equivalently,
(1.2) |hx, yi| ≤ kxk kyk, x, y ∈H (simple form).
The case of equality holds in (1.1) or in (1.2) if and only if the vectorsxandy are linearly dependent.
In the previous paper [6], several results related to Schwarz inequalities have been established. We recall few of them below:
1. Ifx, y ∈H\ {0}andkxk ≥ kyk,then
(1.3) kxk kyk −Rehx, yi ≤
1
2r2kxk
kyk
r−1
kx−yk2 if r ≥1
1 2
kxk
kyk
1−r
kx−yk2 if r <1.
2. If(H;h·,·i)is complex, α ∈ C with Reα, Imα > 0and x, y ∈ H are such that
(1.4)
x− Imα Reα ·y
≤r
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then
(1.5) kxk kyk −Rehx, yi ≤ 1
2 · Reα Imαr2. 3. Ifα∈K\ {0},then for anyx, y ∈H
(1.6) kxk kyk −Re α2
|α|2 hx, yi
≤ 1
2 ·[|Reα| kx−yk+|Imα| kx+yk]2
|α|2 .
4. Ifp≥1,then for anyx, y ∈Hone has
(1.7) kxk kyk −Rehx, yi ≤ 1 2×
(kxk+kyk)2p− kx+yk2p1p , kx−yk2p− |kxk − kyk|2pp1
. 5. Ifα, γ >0andβ ∈Kwith|β|2 ≥αγ then forx, a∈Hwitha6= 0and
(1.8)
x− β αa
≤ |β|2−αγ12
α kak,
one has
kxk kak ≤ ReβRehx, ai+ ImβImhx, ai
√αγ (1.9)
≤ |β| |hx, ai|
√αγ
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and
(1.10) kxk2kak2− |hx, ai|2 ≤ |β|2 −αγ
αγ |hx, ai|2.
The aim of this paper is to provide other results related to the Schwarz in- equality. Applications for reversing the generalised triangle inequality are also given.
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2. Quadratic Schwarz Related Inequalities
The following result holds.
Theorem 2.1. Let (H;h·,·i)be a complex inner product space andx, y ∈ H, α ∈[0,1].Then
(2.1)
αkty−xk2+ (1−α)kity−xk2 kyk2
≥ kxk2kyk2−[(1−α) Imhx, yi+αRehx, yi]2 ≥0 for anyt ∈R.
Proof. Firstly, recall that for a quadratic polynomial P : R → R, P(t) = at2+ 2bt+c, a >0, we have that
(2.2) inf
t∈R
P (t) = P
−b a
= ac−b2 a . Now, consider the polynomialP :R→Rgiven by
(2.3) P(t) :=αkty−xk2+ (1−α)kity−xk2. Since
kty−xk2 =t2kyk2 −2tRehx, yi+kxk2 and
kity−xk2 =t2kyk2 −2tImhx, yi+kxk2, hence
P (t) = t2kyk2−2t[αRehx, yi+ (1−α) Imhx, yi] +kxk2.
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By the definition of P (see (2.3)), we observe thatP(t) ≥ 0for every t ∈ R, therefore 14∆≤0,i.e.,
[(1−α) Imhx, yi+αRehx, yi]2 − kxk2kyk2 ≤0, proving the second inequality in (2.1).
The first inequality follows by (2.2) and the theorem is proved.
The following particular cases are of interest.
Corollary 2.2. For anyx, y ∈H one has the inequalities:
kty−xk2kyk2 ≥ kαk2kyk2−[Rehx, yi]2 ≥0;
(2.4)
kity−xk2kyk2 ≥ kαk2kyk2−[Imhx, yi]2 ≥0;
(2.5) and (2.6) 1
2
kty−xk2+kity−xk2 kyk2
≥ kxk2kyk2 −
Imhx, yi+ Rehx, yi 2
2
≥0, for anyt ∈R.
The following corollary may be stated as well:
Corollary 2.3. Letx, y ∈H andMi, mi ∈R,i∈ {1,2}such thatMi ≥mi >
0, i ∈ {1,2}.If either
(2.7) RehM1y−x, x−m1yi ≥0 and RehM2iy−x, x−im2yi ≥0,
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or, equivalently,
x−M1 +m1
2 y
≤ 1
2(M1−m1)kyk and (2.8)
x− M2 +m2 2 iy
≤ 1
2(M2−m2)kyk hold, then
(0≤)kxk2kyk2−[(1−α) Imhx, yi+αRehx, yi]2 (2.9)
≤ 1 4kyk4
α(M1−m1)2+ (1−α) (M2−m2)2 for anyα∈[0,1].
Proof. It is easy to see that, if x, z, Z ∈ H,then the following statements are equivalent:
(i) RehZ −x, x−zi ≥0, (ii)
x−z+Z2
≤ 12kZ−zk.
Utilising this property one may simply realize that the statements (2.7) and (2.8) are equivalent.
Now, on making use of (2.8) and (2.1), one may deduce the desired inequal- ity (2.9).
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Remark 1. If one assumes thatM1 =M2 =M, m1 =m2 =min either (2.7) or (2.8), then
(0≤)kxk2kyk2−[(1−α) Imhx, yi+αRehx, yi]2 (2.10)
≤ 1
4kyk4(M −m)2 for eachα∈[0,1].
Remark 2. Corollary 2.3 may be seen as a potential source of some reverse results for the Schwarz inequality. For instance, ifx, y ∈ H andM ≥ m > 0 are such that either
(2.11) RehM y−x, x−myi ≥0 or
x− M +m
2 y
≤ 1
2(M −m)kyk hold, then
(2.12) (0≤)kxk2kyk2−[Rehx, yi]2 ≤ 1
4(M −m)2kyk4. Ifx, y ∈H andN ≥n >0are such that either
(2.13) RehN iy −x, x−niyi ≥0 or
x− N +n 2 iy
≤ 1
2(N −n)kyk hold, then
(2.14) (0≤)kxk2kyk2−[Imhx, yi]2 ≤ 1
4(N −n)2kyk4.
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We notice that (2.12) is an improvement of the inequality (0≤)kxk2kyk2− |hx, yi|2 ≤ 1
4(M−m)2kyk4
that has been established in [4] under the same condition (2.11) given above.
The following result may be stated as well.
Theorem 2.4. Let (H;h·,·i) be a real or complex inner product space and x, y ∈H, α ∈[0,1].Then
(2.15)
αkty−xk2+ (1−α)ky−txk2 αkyk2+ (1−α)kxk2
≥
(1−α)kxk2+αkyk2 αkxk2+ (1−α)kyk2
−[Rehx, yi]2 ≥0 for anyt ∈R.
Proof. Consider the polynomialP :R→Rgiven by
(2.16) P(t) :=αkty−xk2 + (1−α)ky−txk2. Since
kty−xk2 =t2kyk2 −2tRehx, yi+kxk2 and
ky−txk2 =t2kxk2−2tRehx, yi+kyk2, hence
P(t) =
αkyk2+ (1−α)kxk2
t2−2tRehx, yi+
αkxk2+ (1−α)kyk2
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for anyt∈R.
By the definition of P (see (2.16)), we observe that P (t) ≥ 0 for every t ∈R, therefore 14∆≤0,i.e., the second inequality in (2.15) holds true.
The first inequality follows by (2.2) and the theorem is proved.
Remark 3. We observe that if eitherα= 0orα= 1,then (2.15) will generate the same reverse of the Schwarz inequality as (2.4) does.
Corollary 2.5. Ifx, y ∈H,then
(2.17) kty−xk2+ky−txk2
2 · kxk2 +kyk2 2
≥ kxk2+kyk2 2
!2
−[Rehx, yi]2 ≥0
for anyt ∈Rand (2.18) kx±yk2
αkyk2+ (1−α)kxk2
≥
(1−α)kxk2+αkyk2 αkxk2+ (1−α)kyk2
−[Rehx, yi]2 ≥0 for anyα∈[0,1].
In particular, we have
(2.19) kx±yk2· kxk2+kyk2 2
!
≥ kxk2+kyk2 2
!2
−[Rehx, yi]2 ≥0.
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In [7, p. 210], C.S. Lin has proved the following reverse of the Schwarz inequality in real or complex inner product spaces(H;h·,·i) :
(0≤)kxk2kyk2− |hx, yi|2 ≤ 1
r2 kxk2kx−ryk2 for anyr∈R,r 6= 0andx, y ∈H.
The following slightly more general result may be stated:
Theorem 2.6. Let (H;h·,·i)be a real or complex inner product space. Then for anyx, y ∈H andα∈K(C,R)withα6= 0we have
(2.20) (0≤)kxk2kyk2− |hx, yi|2 ≤ 1
|α|2 kxk2kx−αyk2. The case of equality holds in (2.20) if and only if
(2.21) Rehx, αyi=kxk2.
Proof. Observe that
I(α) := kxk2kx−αyk2− |α|2
kxk2kyk2− |hx, yi|2
=kxk2
kxk2−2 Re [ ¯αhx, yi] +|α|2kyk2
− |α|2kxk2kyk2+|α|2|hx, yi|2
=kxk4−2kxk2Re [ ¯αhx, yi] +|α|2|hx, yi|2. Since
(2.22) Re [ ¯αhx, yi]≤ |α| |hx, yi|,
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hence
I(α)≥ kxk4−2kxk2|α| |hx, yi|+|α|2|hx, yi|2 (2.23)
= kxk2− |α| |hx, yi|2
≥0.
Conversely, if (2.21) holds true, thenI(α) = 0,showing that the equality case holds in (2.20).
Now, if the equality case holds in (2.20), then we must have equality in (2.22) and in (2.23) implying that
Re [hx, αyi] =|α| |hx, yi| and |α| |hx, yi|=kxk2 which imply (2.21).
The following result may be stated.
Corollary 2.7. Let(H;h·,·i)be as above andx, y ∈H, α∈K\ {0}andr >0 such that|α| ≥r.If
(2.24) kx−αyk ≤rkyk,
then
(2.25)
q
|α|2−r2
|α| ≤ |hx, yi|
kxk kyk(≤1). Proof. From (2.24) and (2.20) we have
kxk2kyk2 − |hx, yi|2 ≤ r2
|α|2 kxk2kyk2,
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that is,
|α|2−r2
|α|2 kxk2kyk2 ≤ |hx, yi|2, which is clearly equivalent to (2.25).
Remark 4. Since forΓ, γ ∈Kthe following statements are equivalent (i) RehΓy−x, x−γyi ≥0,
(ii)
x−γ+Γ2 ·y
≤ 12|Γ−γ| kyk, hence by Corollary2.7we deduce
(2.26) 2 [Re (Γ¯γ)]12
|Γ +γ| ≤ |hx, yi|
kxk kyk,
providedRe (Γ¯γ) >0,an inequality that has been obtained in a different way in [3].
Corollary 2.8. Ifx, y ∈ H, α ∈ K\ {0} andρ > 0such thatkx−αyk ≤ ρ, then
(2.27) (0≤)kxk2kyk2− |hx, yi|2 ≤ ρ2
|α|2 kxk2.
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3. Other Inequalities
The following result holds.
Proposition 3.1. Letx, y ∈H\ {0}andε ∈(0,12]. If (3.1) (0≤) 1−ε−√
1−2ε≤ kxk
kyk ≤1−ε+√
1−2ε, then
(3.2) (0≤)kxk kyk −Rehx, yi ≤εkx−yk2. Proof. Ifx=y,then (3.2) is trivial.
Supposex6=y.Utilising the inequality (2.5) from [6], we can state that kxk kyk −Rehx, yi
kx−yk2 ≤ 2kxk kyk (kxk+kyk)2 for anyx, y ∈H\ {0}, x6=y.
Now, if we assume that
2kxk kyk
(kxk+kyk)2 ≤ε, then, after some manipulation, we get that
εkxk2+ 2 (ε−1)kxk kyk+εkyk2 ≥0, which, forε∈(0,12]andy6= 0,is clearly equivalent to (3.1).
The proof is complete.
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The following result may be stated:
Proposition 3.2. Let(H;h·,·i)be a real or complex inner product space. Then for anyx, y ∈H andϕ∈Rone has:
kxk kyk −[cos 2ϕ·Rehx, yi+ sin 2ϕ·Imhx, yi]
≤ 1
2[|cosϕ| kx−yk+|sinϕ| kx+yk]2.
Proof. For ϕ ∈ R, consider the complex number α = cosϕ −isinϕ. Then α2 = cos 2ϕ −isin 2ϕ, |α| = 1 and by the inequality (1.6) we deduce the desired result.
From a different perspective, we may consider the following results as well:
Theorem 3.3. Let(H;h·,·i)be a real or complex inner product space, α ∈K with|α−1|= 1.Then for anye∈Hwithkek= 1andx, y ∈H,we have (3.3) |hx, yi −αhx, ei he, yi| ≤ kxk kyk.
The equality holds in (3.3) if and only if there exists aλ ∈Ksuch that
(3.4) αhx, eie=x+λy.
Proof. It is known that foru, v ∈H,we have equality in the Schwarz inequality
(3.5) |hu, vi| ≤ kuk kvk
if and only if there exists aλ∈Ksuch thatu=λv.
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If we apply (3.5) foru=αhx, eie−x, v =y,we get (3.6) |hαhx, eie−x, yi| ≤ kαhx, eie−xk kyk with equality iff there exists aλ∈Ksuch that
αhx, eie=x+λy.
Since
kαhx, eie−xk2 =|α|2|hx, ei|2−2 Re [α]|hx, ei|2+kxk2
= |α|2−2 Re [α]
|hx, ei|2+kxk2
= |α−1|2−1
|hx, ei|2+kxk2
=kxk2 and
hαhx, eie−x, yi=αhx, ei he, yi − hx, yi hence by (3.6) we deduce the desired inequality (3.3).
Remark 5. Ifα= 0in (3.3), then it reduces to the Schwarz inequality.
Remark 6. Ifα6= 0,then (3.3) is equivalent to
(3.7)
hx, ei he, yi − 1 α hx, yi
≤ 1
|α|kxk kyk.
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Utilising the continuity property of modulus for complex numbers, i.e.,
|z−w| ≥ ||z| − |w||we then obtain
|hx, ei he, yi| − 1
|α||hx, yi|
≤ 1
|α|kxk kyk, which implies that
(3.8) |hx, ei he, yi| ≤ 1
|α|[|hx, yi|+kxk kyk]. Fore= kzkz , z6= 0,we get from (3.8) that
(3.9) |hx, zi hz, yi| ≤ 1
|α|[|hx, yi|+kxk kyk]kzk2 for anyα∈K\ {0}with|α−1|= 1andx, y, z ∈H.
Forα= 2,we get from (3.9) the Buzano inequality [1]
(3.10) |hx, zi hz, yi| ≤ 1
2[|hx, yi|+kxk kyk]kzk2 for anyx, y, z ∈H.
Remark 7. In the case of real spaces the condition|α−1|= 1is equivalent to eitherα= 0orα = 2.Forα= 2we deduce from (3.7) that
(3.11) 1
2[hx, yi − kxk kyk]≤ hx, ei he, yi ≤ 1
2[hx, yi+kxk kyk]
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for anyx, y ∈H ande ∈H withkek = 1,which implies Richard’s inequality [8]:
(3.12) 1
2[hx, yi − kxk kyk]kzk2 ≤ hx, zi hz, yi ≤ 1
2[hx, yi+kxk kyk]kzk2, for anyx, y, z ∈H.
The following result concerning a generalisation for orthornormal families of the inequality (3.3) may be stated.
Theorem 3.4. Let(H;h·,·i)be a real or complex inner product space,{ei}i∈F a finite orthonormal family, i.e., hei, eji = δij for i, j ∈ F, where δij is Kro- necker’s delta and αi ∈ K, i ∈ F such that |αi−1| = 1 for each i ∈ F.
Then
(3.13)
hx, yi −X
i∈F
αihx, eii hei, yi
≤ kxk kyk.
The equality holds in (3.13) if and only if there exists a constantλ ∈ K such that
(3.14) X
i∈F
αihx, eiiei =x+λy.
Proof. As above, by Schwarz’s inequality, we have
(3.15)
* X
i∈F
αihx, eiiei−x, y +
≤
X
i∈F
αihx, eiiei−x
kyk
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with equality if and only if there exists aλ ∈ Ksuch thatP
i∈Fαihx, eiiei = x+λy.
Since
X
i∈F
αihx, eiiei−x
2
=kxk2−2 Re
* x,X
i∈F
αihx, eiiei +
+
X
i∈F
αihx, eiiei
2
=kxk2−2X
i∈F
αihx, eii hx, eii+X
i∈F
|αi|2|hx, eii|2
=kxk2+X
i∈F
|hx, eii|2 |αi|2−2 Reαi
=kxk2+X
i∈F
|hx, eii|2 |αi−1|2−1
=kxk2,
hence the inequality (3.13) is obtained.
Remark 8. If the space is real, then the nontrivial case one can get from (3.13) is for allαi = 2,obtaining the inequality
(3.16) 1
2[hx, yi − kxk kyk]≤X
i∈F
hx, eii hei, yi ≤ 1
2[hx, yi+kxk kyk]
that has been obtained in [5].
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Corollary 3.5. With the above assumptions, we have
X
i∈F
αihx, eii hei, yi
≤ |hx, yi|+
hx, yi −X
i∈F
αihx, eii hei, yi (3.17)
≤ |hx, yi|+kxk kyk, x, y ∈H,
where|αi−1|= 1for eachi∈F and{ei}i∈F is an orthonormal family inH.
A Potpourri of Schwarz Related Inequalities in Inner Product
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4. Applications for the Triangle Inequality
In 1966, Diaz and Metcalf [2] proved the following reverse of the triangle inequality:
(4.1)
n
X
i=1
xi
≥r
n
X
i=1
kxik,
provided the vectors xi in the inner product space (H;h·,·i) over the real or complex number fieldKare nonzero and
(4.2) 0≤r≤ Rehxi, ai
kxik for each i∈ {1, . . . , n}, wherea∈H,kak= 1.The equality holds in (4.2) if and only if (4.3)
n
X
i=1
xi =r
n
X
i=1
kxik
! a.
The following result may be stated:
Proposition 4.1. Let e ∈ H with kek = 1, ε ∈ 0,12
and xi ∈ H, i ∈ {1, . . . , n}with the property that
(4.4) (0≤) 1−ε−√
1−2ε≤ kxik ≤1−ε+√ 1−2ε for eachi∈ {1, . . . , n}.Then
(4.5)
n
X
i=1
kxik ≤
n
X
i=1
xi
+ε
n
X
i=1
kxi−ek2.
A Potpourri of Schwarz Related Inequalities in Inner Product
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Proof. Utilising Proposition3.1forx=xi andy =e,we can state that kxik −Rehxi, ei ≤εkxi−ek2
for eachi∈ {1, . . . , n}.Summing overifrom1ton,we deduce that (4.6)
n
X
i=1
kxik ≤Re
* n X
i=1
xi, e +
+ε
n
X
i=1
kxi−ek2.
By Schwarz’s inequality in(H;h·,·i),we also have Re
* n X
i=1
xi, e +
≤
Re
* n X
i=1
xi, e +
≤
* n X
i=1
xi, e +
(4.7)
≤
n
X
i=1
xi
kek=
n
X
i=1
xi .
Therefore, by (4.6) and (4.7) we deduce (4.5).
In the same spirit, we can prove the following result as well:
Proposition 4.2. Let (H;h·,·i) be a real or complex inner product space and e ∈H withkek= 1.Then for anyϕ ∈Rone has the inequality:
(4.8)
n
X
i=1
kxik ≤
n
X
i=1
xi
+1 2
n
X
i=1
[|cosϕ| kxi−ek+|sinϕ| kxi+ek]2.
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Proof. Applying Proposition3.2forx=xi andy =e,we have:
(4.9) kxik ≤cos 2ϕ·Rehxi, ei+ sin 2ϕ·Imhxi, ei + 1
2[|cosϕ| kxi−ek+|sinϕ| kxi+ek]2 for anyi∈ {1, . . . , n}.
Summing in (4.5) overifrom1ton,we have:
(4.10)
n
X
i=1
kxik ≤cos 2ϕ·Re
* n X
i=1
xi, e +
+ sin 2ϕ·Im
* n X
i=1
xi, e +
+1 2
n
X
i=1
[|cosϕ| kxi−ek+|sinϕ| kxi+ek]2. Now, by the elementary inequality for the real numbersm, p, M andP,
mM +pP ≤ m2+p212
M2 +P212 , we have
cos 2ϕ·Re
* n X
i=1
xi, e +
+ sin 2ϕ·Im
* n X
i=1
xi, e + (4.11)
≤ cos22ϕ+ sin22ϕ12
×
"
Re
* n X
i=1
xi, e +#2
+
"
Im
* n X
i=1
xi, e +#2
1 2
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=
* n X
i=1
xi, e +
≤
n
X
i=1
xi
kek=
n
X
i=1
xi ,
where for the last inequality we used Schwarz’s inequality in(H;h·,·i). Finally, by (4.10) and (4.11) we deduce the desired result (4.8).
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References
[1] M.L. BUZANO, Generalizzazione della disiguaglianza di Cauchy-Schwarz (Italian), Rend. Sem. Mat. Univ. e Politech. Torino, 31 (1971/73), 405-409 (1974).
[2] J.B. DIAZ AND F.T. METCALF, A complementary inequality in Hilbert and Banach spaces, Proc. Amer. Math. Soc., 17(1) (1966), 88-97.
[3] S.S. DRAGOMIR, Reverses of Schwarz, triangle and Bessel inequal- ities in inner product spaces, J. Inequal. Pure & Appl. Math., 5(3) (2004), Art. 76. [ONLINE http://jipam.vu.edu.au/article.
php?sid=432].
[4] S.S. DRAGOMIR, A counterpart of Schwarz’s inequality in inner product spaces, East Asian Math. J., 20(1) (2004), 1-10.
[5] S.S. DRAGOMIR, Inequalities for orthornormal families of vectors in in- ner product spaces related to Buzano’s, Richard’s and Kurepa’s results, RGMIA Res. Rep. Coll., 7(2004), Article 25. [ONLINEhttp://rgmia.
vu.edu.au/v7(E).html].
[6] S.S. DRAGOMIR, A potpourri of Schwarz related inequalities in inner product spaces (I), J. Inequal. Pure & Appl. Math., 6(3) (2004), Art.
59. [ONLINE http://jipam.vu.edu.au/article.php?sid=
532].
[7] C.S. LIN, Cauchy-Schwarz inequality and generalized Bernstein-type in- equalities, in Inequality Theory and Applications, Vol. 1, Eds. Y.J. Cho, J.K. Kim and S.S. Dragomir, Nova Science Publishers Inc., N.Y., 2001.
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[8] U. RICHARD, Sur des inégalités du type Wirtinger et leurs application aux équationes différentielles ordinaires, Colloquium of Analysis held in Rio de Janeiro, August, 1972, pp. 233-244.