INEQUALITIES OF JENSEN-PEČARIĆ-SVRTAN-FAN
TYPE
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
CHAOBANG GAO AND JIAJIN WEN
College Of Mathematics And Information Science
Chengdu University
Chengdu, 610106, P.R. China
EMail: kobren427@163.com
wenjiajin623@163.com
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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Received:
22 January, 2007
Accepted:
06 July, 2008
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
26D15, 26E60.
Key words:
Jensen’s inequality, Pečarić-Svrtan’s inequality, Fan’s inequality, Theory of majorization, Hermite matrix.
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Abstract:
By using the theory of majorization, the following inequalities of JensenPečarić-Svrtan-Fan type are established: Let I be an interval, f : I → R
and t ∈ I, x, a, b ∈ I n . If a1 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b1 , a1 + b1 ≤ · · · ≤
an + bn ; f (t) > 0, f 0 (t) > 0, f 00 (t) > 0, f 000 (t) < 0 for any t ∈ I, then
f (A(a))
fn,n (a)
fk+1,n (a)
fk,n (a)
=
≤ ··· ≤
≤
f (A(b))
fn,n (b)
fk+1,n (b)
fk,n (b)
f1,n (a)
A(f (a))
≤ ··· ≤
=
,
f1,n (b)
A(f (b))
the inequalities are reversed for f 00 (t) < 0, f 000 (t) > 0, ∀t ∈ I, where A(·) is
the arithmetic mean and
X
1
xi1 + · · · + xik
fk,n (x) := n
f
, k = 1, . . . , n.
k
k
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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1≤i1 <···<ik ≤n
Acknowledgements:
This work is supported by the Natural Science Foundation of China
(10671136) and the Natural Science Foundation of Sichuan Province Education Department (07ZA207).
The authors are deeply indebted to anonymous referees for many useful comments and keen observations which led to the present improved version of the
paper as it stands, and also are grateful to their friend Professor Wan-lan Wang
for numerous discussions and helpful suggestions in preparation of this paper.
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Contents
1
Introduction
4
2
Proof of Theorem 1.1
6
3
Corollary of Theorem 1.1
12
4
A Matrix Variant
14
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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1.
Introduction
In what follows, we shall use the following symbols:
x := (x1 , . . . , xn );
f (x) := (f (x1 ), . . . , f (xn ));
G(x) := (x1 x2 · · · xn )1/n ;
x1 + x2 + · · · + xn
; Rn+ := [0, +∞)n ; Rn++ := (0, +∞)n ;
n
I n := {x|xi ∈ I, i = 1, . . . , n, I is an interval};
X
1
xi1 + · · · + xik
fk,n (x) := n
f
, k = 1, . . . , n.
k
k 1≤i <···<i ≤n
A(x) :=
1
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
k
Jensen’s inequality states that: Let f : I → R be a convex function and x ∈ I n .
Then
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f (A(x)) ≤ A(f (x)).
(1.1)
This well-known inequality has a great number of generalizations in the literature
(see [1] – [6]). An interesting generalization of (1.1) due to Pečarić and Svrtan [5]
is:
f (A(x)) = fn,n (x) ≤ · · · ≤ fk+1,n (x) ≤ fk,n (x)
≤ · · · ≤ f1,n (x) = A(f (x)).
(1.2)
In 2003, Tang and Wen [6] obtained the following generalization of (1.2):
(1.3)
fr,s,n ≥ · · · ≥ fr,s,i ≥ · · · ≥ fr,s,s ≥ · · · ≥ fr,j,j ≥ · · · ≥ fr,r,r = 0,
where
fr,s,n
n n
:=
(fr,n − fs,n ),
r
s
fk,n := fk,n (x),
1 ≤ r ≤ s ≤ n.
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Ky Fan’s arithmetic-geometric mean inequality is (see [7]): Let x ∈ (0, 1/2]n . Then
A(x)
G(x)
≥
.
A(1 − x)
G(1 − x)
(1.4)
In this paper, we shall establish further extensions of (1.2) and (1.4) as follows:
Theorem 1.1. Let I be an interval. If f : I → R, a, b ∈ I n (n ≥ 2) and
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
(i) a1 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b1 , a1 + b1 ≤ · · · ≤ an + bn ;
(ii) f (t) > 0, f 0 (t) > 0, f 00 (t) > 0, f 000 (t) < 0 for any t ∈ I,
then
(1.5)
fn,n (a)
fk+1,n (a)
fk,n (a)
f (A(a))
=
≤ ··· ≤
≤
f (A(b))
fn,n (b)
fk+1,n (b)
fk,n (b)
f1,n (a)
A(f (a))
≤ ··· ≤
=
.
f1,n (b)
A(f (b))
vol. 9, iss. 3, art. 74, 2008
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The inequalities are reversed for f 00 (t) < 0, f 000 (t) > 0, ∀t ∈ I. The equality
signs hold if and only if a1 = · · · = an and b1 = · · · = bn .
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In Section 3, several interesting results of Ky Fan shall be deduced. In Section 4,
the matrix variant of (1.5) will be established.
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2.
Proof of Theorem 1.1
Lemma 2.1. Let f : I → R be a function whose second derivative exists and x ∈ I n ,
α ∈ Ωn = {α ∈ Rn+ : α1 + · · · + αn = 1}.
Writing
S(α, x) :=
1 X
f (α1 xi1 + · · · + αn xin ),
n! i ···i
1
where
P
i1 ···in
n
denotes summation over all permutations of {1, 2, . . . , n},
S(α, a)
F (α) := log
, a, b ∈ I n ,
S(α, b)
ui (x) := α1 xi1 + α2 xi2 +
vi (x) := α1 xi2 + α2 xi1 +
n
X
j=3
n
X
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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αj xij ,
αj xij ,
i = (i1 , i2 , . . . , in ).
j=3
Then there exist ξi (a) between ui (a) and vi (a), and ξi (b) between ui (b) and vi (b)
such that
∂F
∂F
(2.1) (α1 − α2 )
−
∂α1 ∂α2
X f 00 (ξi (a))(ui (a) − vi (a))2 f 00 (ξi (b))(ui (b) − vi (b))2 1 X
−
,
=
n! i ···i 1≤i <i ≤i
S(α, a)
S(α, b)
n
n
3
1
2
P
where i3 ···in denotes the summation over all permutations of {1, 2, . . . , n}\{i1 , i2 }.
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Proof. Note the following identities:
1 X X
f (α1 xi1 + · · · + αn xin )
n! i ···i 1≤i 6=i ≤n
n
3
1
2
1 X X
=
[f (ui (x)) + f (vi (x))];
n! i ···i 1≤i <i ≤n
S(α, x) =
n
3
1
2
∂
∂
[f (ui ) + f (vi )] −
[f (ui ) + f (vi )] = [f 0 (ui ) − f 0 (vi )](xi1 − xi2 );
∂α1
∂α2
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
(2.2)
∂S
∂S
−
(α1 − α2 )
∂α1 ∂α2
1 X X
=
[f 0 (ui ) − f 0 (vi )](α1 − α2 )(xi1 − xi2 )
n! i ···i 1≤i <i ≤n
n
3
1
2
1 X X
=
[f 0 (ui ) − f 0 (vi )](ui − vi ).
n! i ···i 1≤i <i ≤n
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
3
n
1
2
By F (α) = log S(α, a) − log S(α, b) and (2.2), we have
∂F
∂F
(α1 − α2 )
−
∂α1 ∂α2
∂S(α, a)
−1 ∂S(α, a)
= (α1 − α2 ) [S(α, a)]
−
∂α1
∂α2
∂S(α, b)
−1 ∂S(α, b)
−[S(α, b)]
−
∂α1
∂α2
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[f 0 (ui (a)) − f 0 (vi (a))][ui (a) − vi (a)]
S(α, a)
n 1≤i1 <i2 ≤n
3
[f 0 (ui (b)) − f 0 (vi (b))][ui (b) − vi (b)]
−
S(α, b)
00
X
X
f (ξi (a))[ui (a) − vi (a)]2 f 00 (ξi (b))[ui (b) − vi (b)]2
1
=
−
.
n! i ···i 1≤i <i ≤n
S(α, a)
S(α, b)
1 X
=
n! i ···i
3
n
X
1
2
Here we used the Mean Value Theorem for f 0 (t). This completes the proof.
Lemma 2.2. Under the hypotheses of Theorem 1.1, F is a Schur-convex function or
a Schur-concave function on Ωn , where F is defined by Lemma 2.1.
Proof. It is easy to see that Ωn is a symmetric convex set and F is a differentiable
symmetric function on Ωn . To prove that F is a Schur-convex function on Ωn , it is
enough from [8, p .57] to prove that
∂F
∂F
(2.3)
(α1 − α2 )
−
≥ 0, ∀α ∈ Ωn .
∂α1 ∂α2
To prove (2.3), it is enough from Lemma 2.1 to prove
(2.4)
f 00 (ξi (a))[ui (a) − vi (a)]2
f 00 (ξi (b))[ui (b) − vi (b)]2
≥
.
S(α, a)
S(α, b)
Using the given conditions a1 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b1 , f (t) > 0 and f 0 (t) > 0,
we obtain that aj ≤ bj (j = 1, 2, . . . , n) and the inequalities:
(2.5)
1
1
≥
> 0.
S(α, a)
S(α, b)
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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By the given condition (i) of Theorem 1.1 and 1 ≤ i1 < i2 ≤ n, we have
ai2 − ai1 ≥ bi1 − bi2 ≥ 0
and
(2.6)
[ui (a) − vi (a)]2 ≥ [ui (b) − vi (b)]2 ≥ 0.
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
From (2.5) and (2.6), we get
(2.7)
[ui (b) − vi (b)]2
[ui (a) − vi (a)]2
≥
≥ 0.
S(α, a)
S(α, b)
Note that a, b ∈ I n , ui (a), vi (a), ui (b), vi (b) ∈ I, and
min{ui (a), vi (a)} ≤ ξi (a)
≤ max{ui (a), vi (a)}
≤ min{ui (b), vi (b)}
≤ ξi (b)
≤ max{ui (b), vi (b)}.
ξi (a) ≤ ξi (b)
(ξi (a), ξi (b) ∈ I).
If f 00 (t) > 0, f 000 (t) < 0 for any t ∈ I, from these and (2.8) we get
(2.9)
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It follows that
(2.8)
Chaobang Gao and Jiajin Wen
f 00 (ξi (a)) ≥ f 00 (ξi (b)) > 0.
Combining with (2.7) and (2.9), we have proven that (2.4) holds, hence, F is a
Schur-convex function on Ωn .
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Similarly, if f 00 (t) < 0, f 000 (t) > 0 for any t ∈ I, we obtain
(2.10)
−f 00 (ξi (a)) ≥ −f 00 (ξi (b)) > 0.
Combining with (2.7) and (2.10), we know that the inequalities are reversed in (2.4)
and (2.3). Therefore, F is a Schur-concave function on Ωn . This ends the proof of
Lemma 2.2.
Remark 1. When α1 6= α2 , there is equality in (2.3) if a1 = · · · = an and b1 = · · · =
bn . In fact, there is equality in (2.3) if and only if there is equality in (2.5), (2.8),
(2.9) and the first inequality in (2.6) or all the equality signs hold in (2.6). For the
first case, by a1 ≤ · · · ≤ an ≤ bn ≤ · · · ≤ b1 , we get a1 = · · · = an , b1 = · · · = bn .
For the second case, we have ai1 − ai2 = 0 = bi1 − bi2 . Since 1 ≤ i1 < i2 ≤ n and
i1 , i2 are arbitrary, we get a1 = · · · = an , b1 = · · · = bn . Clearly, if a1 = · · · = an ,
b1 = · · · = bn , then (2.3) reduces to an equality.
Proof of Theorem 1.1. First we note that if


α = αk := k −1 , k −1 , . . . , k −1 , 0, . . . , 0 ,
|
{z
}
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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k
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we obtain that
S(αk , x) = fk,n (x)
and
(2.11)
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F (αk ) = log
fk,n (a)
.
fk,n (b)
By Lemma 2.2, we observe that F (α) is a Schur-convex(concave) function on Ωn .
Using αk+1 ≺ αk for αk , αk+1 ∈ Ωn and the definition of Schur-convex(concave)
functions, we have [8]
(2.12)
F (αk+1 ) ≤ (≥)F (αk ),
k = 1, . . . , n − 1.
It follows from (2.11) and (2.12) that (1.5) holds. Since αk+1 6= αk , combining this
fact with Remark 1, we observe that the equality signs hold in (1.5) if and only if
a1 = · · · = an , b1 = · · · = bn . This completes the proof of Theorem 1.1.
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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3.
Corollary of Theorem 1.1
Corollary 3.1. Let 0 < r < 1, s ≥ 1, 0 < ai ≤ 2−1/s , bi = (1 − asi )1/s , i = 1, . . . , n,
f (t) = tr , t ∈ (0, 1). Then the inequalities in (1.5) are reversed.
Proof. Without loss of generality, we can assume that 0 < a1 ≤ · · · ≤ an . By
bi = (1 − asi )1/s and 0 < ai ≤ 2−1/s (i = 1, . . . , n), we have
0 < a1 ≤ · · · ≤ an ≤ 2−1/s ≤ bn ≤ · · · ≤ b1 < 1.
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Now we take g(t) := t + (1 − ts )1/s (0 < t ≤ 2−1/s ), so g 0 (t) = 1 − (1 −
ts )(1/s)−1 ts−1 ≥ 0, i.e., g is an increasing function. Thus
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
a1 + b 1 ≤ · · · ≤ an + b n .
It is easy to see that f (t) = tr > 0, f 0 (t) = rtr−1 > 0, f 00 (t) = r(r − 1)tr−2 < 0,
f 000 (t) = r(r − 1)(r − 2)tr−3 > 0 for any t ∈ (0, 1). By Theorem 1.1, Corollary 3.1
can be deduced. This completes the proof.
Corollary 3.2. Let a ∈ (0, 1/2]n . Writing
[AG; x]k,n :=
Y
1≤i1 <···<ik
xi1 + · · · + xik
k
≤n
!
1
n
k
()
,
we have
(3.1)
[AG; a]n,n
A(a)
=
A(1 − a)
[AG; 1 − a]n,n
[AG; a]k,n
[AG; a]k+1,n
≥ ··· ≥
≥
[AG; 1 − a]k+1,n
[AG; 1 − a]k,n
G(a)
[AG; a]1,n
=
.
≥ ··· ≥
[AG; 1 − a]1,n
G(1 − a)
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Equalities hold throughout if and only if a1 = · · · = an . (Compare (3.1) with
[7, 10, 11])
Proof. We choose s = 1 in Corollary 3.1. Raising each term to the power of 1/r and
letting r → 0 in (1.5), (3.1) can be deduced. This ends the proof.
Corollary 3.3. Let f : I → R be such that f (t) > 0, f 0 (t) > 0, f 00 (t) > 0,
f 000 (t) < 0 for any t ∈ I. Let Φ : I0 → I be increasing and Ψ : I0 → I be
decreasing, and suppose that Φ + Ψ is increasing and sup Φ ≤ inf Ψ. Then
R
R
f |I0 |−1 I0 Φdt
f (Φ)dt
≤ RI0
(3.2)
,
R
f
(Ψ)dt
−1
f |I0 |
Ψdt
I0
I0
where |I0 | is the length of the interval I0 . The inequality is reversed for f 00 (t) < 0,
f 000 (t) > 0, ∀t ∈ I.
f (A(a))
f (A(b))
A(f (a))
,
A(f (b))
In fact, since (3.2) is an integral version of the inequality
≤
therefore (3.2) holds by Theorem 1.1.
According to Theorem 1.1, (1.5) implies inequalities (1.1), (1.2) and (3.1), and
the implication (3.1) to (1.4) is obvious. Consequently, Theorem 1.1 is a generalization of Jensen’s inequality (1.1), Pečarić-Svrtan’s inequalities (1.2) and Fan’s
inequality (1.4). Note that Theorem 1.1 contains a great number of inequalities as
special cases. To save space we omit the details.
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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4.
A Matrix Variant
P
Let A = (aij )n×n (n ≥ 2) be a Hermite matrix of order n. Then tr A = ni=1 aii
is the trace of A. As is well-known, there exists a unitary matrix U such that A =
U diag(λ1 , . . . , λn )U ∗ , where U ∗ is the transpose conjugate matrix of U and the
components of λ = (λ1 , . . . , λn ) are the eigenvalues of A. Thus tr A = λ1 +· · ·+λn .
Let λ ∈ I n . Then, for f : I → R, we define f (A) := U diag(f (λ1 ), . . . , f (λn ))U ∗
(see [9]). Note that diag(λ1 , . . . , λn ) = U ∗ AU . Based on the above, we may use
the following symbols: If, for A, we keep the elements on the cross points of the
i1 , . . . , ik th rows and the i1 , . . . , ik th columns; replacing the other elements by nulls,
then we denote this new matrix by Ai1 ···ik . Clearly, we have tr[U ∗ AU ]i1 ···ik = λi1 +
· · · + λik . Thus we also define that
X
1
λi1 + · · · + λik
fk,n (A) := n
f
k
k 1≤i1 <···<ik ≤n
X
1
1
∗
f
tr[U AU ]i1 ···ik .
= n
k
k 1≤i <···<i ≤n
1
k
In particular, we have
n
1X
1
f1,n (A) =
f (λi ) = tr(f (A));
n i=1
n
λ1 + · · · + λn
1
fn,n (A) = f
=f
trA ;
n
n
n
1X
trA − λi
1
E · trA − A
fn−1,n (A) =
f
= trf
,
n i=1
n−1
n
n−1
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
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where E is a unit matrix. In fact, from
trA − λ1
E · trA − A
trA − λn
∗
U
U = diag
,...,
,
n−1
n−1
n−1
we get
X
n
trA − λi
E · trA − A
trf
=
f
.
n−1
n
−
1
i=1
Based on the above facts and Theorem 1.1, we observe the following.
Chaobang Gao and Jiajin Wen
n
Theorem 4.1. Let I be an interval and let λ, µ ∈ I . Suppose the components of
λ,µ are the eigenvalues of Hermitian matrices A and B. If
(i) λ1 ≤ · · · ≤ λn ≤ µn ≤ · · · ≤ µ1 , λ1 + µ1 ≤ · · · ≤ λn + µn ;
0
00
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
vol. 9, iss. 3, art. 74, 2008
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(ii) the function f : I → R satisfies f (t) > 0, f (t) > 0, f (t) > 0, f (t) < 0 for
any t ∈ I, and we have
trf E·trA−A
f n1 trA
f
(A)
fk,n (A)
trf (A)
n−1
≤ · · · ≤ k+1,n
≤
≤
≤ ··· ≤
.
E·trB−B
1
f
(B)
f
(B)
trf
(B)
trf
f n trB
k+1,n
k,n
n−1
The inequalities are reversed for f 00 (t) < 0, f 000 (t) > 0, ∀t ∈ I. Equalities hold
throughout if and only if λ1 = · · · = λn and µ1 = · · · = µn .
Remark 2. If I = (0, 1/2], 0 < λ1 ≤ · · · ≤ λn ≤ 1/2, B = E − A, then the
precondition (i) of Theorem 4.1 can be satisfied.
Remark 3. Lemma 2.2 possesses a general and meaningful result that should be an
important theorem. Theorem 1.1 is only an application of Lemma 2.2.
Remark 4. If f (t) < 0, f 0 (t) < 0 for any t ∈ I, then we can apply Theorem 1.1 to
−f .
Remark 5. In [12, 13], several applications on Jensen’s inequalities are displayed.
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References
[1] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and
new inequalities in analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
[2] J.E. PEČARIĆ, Inverse of Jensen-Steffensen’s inequality, Glasnik, Math.,
16(3) (1981), 229–233.
Jensen-Pečarić-Svrtan-Fan Type
Inequalities
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Jensen-Pečarić-Svrtan-Fan Type
Inequalities
Chaobang Gao and Jiajin Wen
vol. 9, iss. 3, art. 74, 2008
Title Page
Contents
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