Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 2, Issue 2, Article 19, 2001
MONOTONIC REFINEMENTS OF A KY FAN INEQUALITY
K. K. CHONG
D EPARTMENT OF A PPLIED M ATHEMATICS
T HE H ONG KONG P OLYTECHNIC U NIVERSITY
H UNG H OM , KOWLOON
H ONG KONG
T HE P EOPLE ’ S R EPUBLIC OF C HINA .
makkchon@inet.polyu.edu.hk
Received 3 October, 2000; accepted 2 February, 2001
Communicated by Feng Qi
A BSTRACT. It is well-known that inequalities between means play a very important role in many
branches of mathematics. Please refer to [1, 3, 7], etc. The main aims of the present article are:
(i) to show that there are monotonic and continuous functions H(t), K(t), P (t) and Q(t) on
[0, 1] such that for all t ∈ [0, 1],
Hn ≤ H(t) ≤ Gn ≤ K(t) ≤ An and
Hn /(1 − Hn ) ≤ P (t) ≤ Gn /G0n ≤ Q(t) ≤ An /A0n ,
where An , Gn and Hn are respectively the weighted arithmetic, geometric and harmonic
means of the positive numbers x1 , x2 , ..., xn in (0, 1/2], with positive weights α1 , α2 , ..., αn ;
while A0n and G0n are respectively the weighted arithmetic and geometric means of the
numbers 1 − x1 , 1 − x2 , ..., 1 − xn with the same positive weights α1 , α2 , ..., αn ;
(ii) to present more general monotonic refinements for the Ky Fan inequality as well as some
inequalities involving means; and
(iii) to present some generalized and new inequalities in this connection.
Key words and phrases: Ky Fan inequality, monotonic refinements of inequalities, arithmetic, geometric and harmonic means.
2000 Mathematics Subject Classification. 26D15, 26A48.
1. I NTRODUCTION
Let n be a positive integer. To two given sequences of positive numbers x1 , x2 , . . . , xn and
α1 , α2 , . . . , αn such that α1 + α2 + · · · + αn = 1, we denote by An , Gn and Hn respectively
the weighted arithmetic, geometric and harmonic means, that is,
An =
n
X
αi xi ,
i=1
ISSN (electronic): 1443-5756
c 2001 Victoria University. All rights reserved.
The author would like to thank the referees for their invaluable comments and suggestions.
035-00
2
K. K. C HONG
Gn =
n
Y
xαi i ,
i=1
n
X
Hn =
!−1
αi /xi
.
i=1
We use the symbols an , gn and hn to denote the corresponding unweighted arithmetic, geometric and harmonic means of the n positive numbers x1 , x2 , . . . , xn . The following well-known
inequality has been proved, using many different methods: (Please refer to [3].)
H n ≤ G n ≤ An
(1.1)
Let the real numbers xi be such that 0 < xi ≤ 1/2, for all i = 1, 2, . . . , n. We denote
by A0n , G0n and Hn0 the weighted arithmetic, geometric and harmonic means of the numbers
1 − x1 , 1 − x2 , . . . , 1 − xn , namely,
n
X
0
An =
αi (1 − xi ) ,
i=1
G0n
=
n
Y
(1 − xi )αi ,
i=1
Hn0 =
n
X
!−1
αi / (1 − xi )
.
i=1
Also, let a0n , gn0 and h0n denote the corresponding unweighted arithmetic, geometric and harmonic means of the numbers 1 − x1 , 1 − x2 , . . . , 1 − xn respectively. In recent years many
interesting inequalities involving these mean values have been published, in particular, the following well-known Ky Fan and Wang-Wang inequalities :
Hn
Gn
An
(1.2)
0 ≤
0 ≤
Hn
Gn
A0n
with equality holding if and only if x1 = · · · = xn . Please refer to the following papers by
H. Alzer [1] – [2] and Wang-Wang [10] or [7], etc. The right-hand inequality of (1.2) is the
famous Ky Fan inequality; the left-hand inequality for the unweighted case was first discovered
by Wang-Wang in 1984 [10]. The main purpose of this paper is to present some new monotonic
continuous functions H(λ), K(λ), P (λ) and Q(λ) on [0, 1] such that
Hn ≤ H(λ) ≤ Gn ≤ K(λ) ≤ An
and
Hn /(1 − Hn ) ≤ P (λ) ≤ Gn /G0n ≤ Q(λ) ≤ An /A0n .
In fact, our theorems here generalize results of Wang and Yang in [9] and a theorem of K.M.
Chong [6]. In Section 2, we shall generalize refinements of inequalities between means. In
Section 3, we shall present generalizations to refinements of the Ky Fan inequality, with some
new inequalities deduced. Finally, in Section 4, we shall show that Theorem 3.1 can be used to
deduce many other established refinements of the Ky Fan inequality.
In a recent paper of Wang and Yang [9], the following two interesting theorems were put
forward. In fact, they are refinements of inequalities (1.1) and (1.2) in Section 1, for the discrete
unweighted case. They are restated here without proof. For the details of the proof, please refer
to [9].
Theorem 1.1. Given a sequence { x1 , x2 , . . . , xn } of positive numbers, which are not all equal:
J. Inequal. Pure and Appl. Math., 2(2) Art. 19, 2001
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M ONOTONIC R EFINEMENTS OF A K Y FAN I NEQUALITY
3
(a) For any t in [0, 1/n] , let
(1.3)
h (t) =
"
n
Y
i=1
n X
1
1
1
+t
−
xi
x
xi
j
j=1
#−1/n
Then, h(t) is continuous, strictly decreasing and hn = h(1/n) ≤ h(t) ≤ h(0) = gn on
[0, 1/n] .
(b) For any t in [0, 1/n] , let
"
#1/n
n
n
X
Y
xi + t
(xj − xi )
(1.4)
k (t) =
i=1
j=1
Then, k(t) is continuous, strictly increasing and gn = k(0) ≤ k(t) ≤ k(1/n) = an on
[0, 1/n] .
Theorem 1.2. Given a sequence { x1 , x2 , . . . , xn } with xi in (0, 1/2], i = 1, 2, . . . , n, which
are not all equal:
(a) For any t in [0, 1/n] , let
#−1/n
"
n n
X
Y
1
1
1
(1.5)
p (t) =
+t
−
−1
xi
xj
xi
j=1
i=1
Then, p(t) is continuous, strictly decreasing, and hn /(1 − hn ) = p(1/n) ≤ p(t) ≤
p(0) = gn /gn0 on [0, 1/n] .
(b) For any t in [0, 1/n] , let
"
#1/n
n
n
Q
P
xi + t
(xj − xi )
(1.6)
i=1
q (t) =
j=1
"
n
Q
1 − xi − t
i=1
n
P
#1/n
(xj − xi )
j=1
Then, q(t) is continuous, strictly increasing and gn /gn0 = q(0) ≤ q(t) ≤ q(1/n) =
an /a0n on [0, 1/n] .
2. S OME G ENERALIZATIONS
In this section, we are going to present and prove a generalization of Theorem 1.1 and Theorem 1.2(a), in particular, to the case for weighted means. Its statement runs as follows:
Theorem 2.1. Let a1 , a2 ,P
. . . , an , and α1 , α2 , . . . , αn be two sequences of positive numbers,
n
with aiP
not all equal and
i=1 αi = 1. Let a be any positive number such that An ≤ a, where
An = ni=1 αi ai , and k is a constant such that k < ai , for all i = 1, 2, . . . , n. Let
n
Y
(2.1)
F (λ) =
[λa + (1 − λ) ai − k]αi
i=1
(2.2)
G (λ) =
n
Y
[λa + (1 − λ) ai − k]−αi
i=1
Then,
(i) F (λ) is continuous and strictly increasing on [0, 1] ;
(ii) G(λ) is continuous and strictly decreasing on [0, 1] .
J. Inequal. Pure and Appl. Math., 2(2) Art. 19, 2001
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4
K. K. C HONG
Proof.
(i) Taking the logarithm of F (λ), we have,
ln F (λ) =
n
X
αi ln [λa + (1 − λ) ai − k]
i=1
Differentiating the last expression with respect to λ, we have:
n
F 0 (λ) X
αi (a − ai )
=
F (λ)
[λa + (1 − λ) ai − k]
i=1
(2.3)
Differentiating again, we obtain:
0
0
n
X
F (λ)
αi (a − ai )2
=−
<0
F (λ)
[λa + (1 − λ) ai − k]2
i=1
for all λ in [0, 1] , as the ai are not all equal. Hence, F 0 (λ)/F (λ) is strictly decreasing
on [0, 1] . Also, as An ≤ a and k < a, we have :
n
(2.4)
F 0 (1) X αi (a − ai )
a − An
=
=
≥0
F (1)
a−k
a−k
i=1
Therefore, F 0 (λ)/F (λ) > 0, for all λ in [0, 1). As F (λ) is positive for all λ in [0, 1] ,
F 0 (λ) > 0 for λ in [0, 1). Hence, F (λ) is strictly increasing on [0, 1] . The continuity of
F (λ) on [0, 1] is obvious.
(ii) As F (λ) is positive for all λ in [0, 1] and G(λ) = 1/F (λ), G(λ) is continuous and
strictly decreasing on [0, 1]. Hence, the proof of Theorem 2.1 is complete.
Now, we use Theorem 2.1 to deduce some established theorems.
Remark 2.2.
(i) From Theorem 2.1, we have, for all λ ∈ (0, 1),
F (0) < F (λ) < F (1),
which yields for not all equal ai ,
n
Y
(2.5)
(ai − k)αi < a − k.
i=1
In particular, if a = An , for not all equal ai we have,
(2.6)
n
Y
(ai − k)αi < An − k,
i=1
which is a generalization of the weighted arithmetic-geometric
means inequality.
P
(ii) Again, in Theorem 2.1, we let k = 0, a = ni=1 αi ai = An . Then, F (λ) will reduce to,
say
(2.7)
K(λ) =
n
Y
[λAn + (1 − λ)ai ]αi
i=1
It is clear that K(λ) is continuous and strictly increasing on [0, 1] , and for all λ ∈ (0, 1) ,
(2.8)
K(0) = Gn < K(λ) < K(1) = An .
This is a refinement of the weighted arithmetic-geometric means inequality.
J. Inequal. Pure and Appl. Math., 2(2) Art. 19, 2001
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M ONOTONIC R EFINEMENTS OF A K Y FAN I NEQUALITY
(iii) Furthermore, if we put λ = nt, αi =
t ∈ [0, 1/n] ,
K(nt) =
n
Y
1
,i
n
5
= 1, 2, . . . , n, into K(λ), we obtain for all
[ntAn + (1 − nt)ai ]1/n =
i=1
n
Y
"
ai + t
i=1
n
X
#1/n
(aj − ai )
.
j=1
The last expression is in fact the function k(t) of Theorem 1.1(b). Hence, we have
shown that Theorem 1.1(b) is a particular case of Theorem 2.1.
Remark 2.3. If, in Theorem 2.1, we let k = 0, ai = x1i , i = 1, 2, . . . , n, a = H1n =
α1
+ · · · + αxnn , then G(λ) will reduce to,
x1
−αi
n Y
λ
1
(2.9)
H(λ) =
+ (1 − λ)
.
Hn
xi
i=1
Then, H(λ) is continuous and strictly decreasing on [0, 1] , and for all λ ∈ (0, 1),
(2.10)
H(1) = Hn < H(λ) < H(0) = Gn .
(2.10) is a refinement of the weighted means inequality. Furthermore, if we put λ = nt, αi = n1 ,
ai = 1/xi , i = 1, 2, . . . , n, into H(λ), we obtain for all t in [0, 1/n] ,
"
−1/n Y
#−1/n
n n
n Y
X
1
1
1
1
H (nt) =
nta + (1 − nt)
=
+t
−
x
x
x
xi
i
i
j
i=1
i=1
j=1
This is the function h(t) in Theorem 1.1(a). Hence, we have deduced Theorem 1.1(a) as a
particular case of Theorem 2.1.
Theorem 2.4. Let x1 , x2 , . . . , xn be n positive numbers, not all equal, with xi ∈ (0, 1/2] for all
i = 1, 2, .., n. Let α1 , α2 , . . . , αn be the corresponding weights, i.e., αi > 0, i = 1, 2, . . . , n and
α1 + · · · + αn = 1. Let γ be a constant such that γ < x1i , for all i = 1, 2, . . . , n. We define P (λ)
as :
−αi
n Y
λ
1
(2.11)
P (λ) =
+ (1 − λ) − γ
Hn
xi
i=1
Then,
(i) P (λ) is continuous and strictly decreasing on [0, 1] ;
(ii) for all λ ∈ (0, 1), we have,
n
(2.12)
Y
Hn
P (1) =
< P (λ) < P (0) =
1 − γHn
i=1
xi
1 − γxi
α i
.
(i) P (λ) is continuous and strictly decreasing on [0, 1] , as we get P (λ) from the
continuous and strictly decreasing function G(λ), by putting k = γ, ai = 1/xi with
xi ∈ (0, 1/2], i = 1, 2, . . . , n, a = 1/Hn = α1 /x1 + · · · + αn /xn into G(λ) of Theorem
2.1.
Q n xi α i
(ii) We have : P (0) = G(0) = i=1 1−γxi
,
Proof.
P (1) = G(1) = Hn /(1 − γHn ).
Hence, for all λ ∈ (0, 1),
Hn /(1 − γHn ) < P (λ) <
n Y
i=1
J. Inequal. Pure and Appl. Math., 2(2) Art. 19, 2001
xi
1 − γxi
α i
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6
K. K. C HONG
This completes the proof of Theorem 2.4.
Remark 2.5. If we put λ = nt, αi = 1/n, i = 1, 2, . . . , n, γ = 1 into P (λ) of Theorem 2.4,
we obtain for any t in [0, 1/n] ,
"
#−1/n
−1/n Y
n n
n Y
X
nt
1
1
1
P (nt) =
+ (1 − nt) ai − 1
=
+t
−
−1
hn
xi
xj
xi
i=1
i=1
j=1
This is the function p(t) of Theorem 1.2(a), and we have deduced Theorem 1.2(a) as a particular
case of Theorem 2.1.
The only part in Section 1, which is not yet dealt with, is Theorem 1.2(b). Its proof is postponed to the next section, with some additional theorems. We end this section by considering
another similar theorem. In [6], K.M. Chong presented the following theorem:
Theorem 2.6. Let a1 , a2 , . . . , an be positive
Pnnumbers and α1 , α2 , . . . , αn be their corresponding
weights, i.e. αi > 0, i = 1, 2, . . . , n and i=1 αi = 1. Let f (λ) be defined as:
" n
#αi
n
Y
X
(2.13)
f (λ) =
λ
αj aj + (1 − λ) ai
i=1
j=1
Then f (λ) is a strictly increasing function of λ for λ ∈ [0, 1], unless a1 = a2 = · · · = an ; in
which case f (0) = Gn = An = f (1) .
Proof. It is obvious that when k = 0 and a = An in Theorem 2.1 we obtain K.M. Chong’s
theorem at once.
3. M ONOTONIC R EFINEMENTS OF THE K Y FAN I NEQUALITY
In the previous section, we have seen that Theorem 2.1 is a generalization of various theorems. In this section, we shall present a refinement of the well-known Ky Fan inequality, which
is a generalization of Theorem 1.2(b).
Theorem 3.1. Let x1 , x2 , . . . , xn be n positive numbers, not all equal, such that xi ∈ (0, 1/2]
for all i = 1, 2,P
. . . , n and let α1 , α2 , . . . , αn be their corresponding weights i.e. αi > 0, i =
1, 2, . . . , n and ni=1 αi = 1. Let β and δ be two constants such that δ ≤ β and β < xi , for all
i = 1, 2, . . . , n. Let r(λ) be defined as :
n
Q
[λz + (1 − λ)xi − β]αi
i=1
(3.1)
r(λ) = Q
n
[λ(1 − z) + (1 − λ)(1 − xi ) − δ]αi
i=1
P
for any λ ∈ [0, 1], and any real number z such that ni=1 αi xi ≤ z ≤ 1/2.
Then,
(i) r(λ) is continuous and strictly increasing on [0, 1] ; and
(ii) when β = δ = 0, we have, for all λ ∈ (0, 1) ,
z
0
(3.2)
Gn /Gn = r(0) < r(λ) < r(1) =
.
1−z
Proof.
(i) Taking the logarithm, we have,
ln{r(λ)} =
n
X
αi ln[λz + (1 − λ)xi − β] −
i=1
J. Inequal. Pure and Appl. Math., 2(2) Art. 19, 2001
n
X
αi ln[λ(1 − z) + (1 − λ)(1 − xi ) − δ]
i=1
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M ONOTONIC R EFINEMENTS OF A K Y FAN I NEQUALITY
7
Differentiating with respect to λ, we have :
n
n
X
X
r0 (λ)
z − xi
(1 − z) − (1 − xi )
=
αi
−
αi
r(λ)
λz + (1 − λ)xi − β
λ(1 − z) + (1 − λ)(1 − xi ) − δ
i=1
i=1
n
X
=
n
αi
i=1
X
z − xi
z − xi
+
αi
.
λz + (1 − λ)xi − β
λ(1
−
z)
+
(1
−
λ)(1
−
x
)
−
δ
i
i=1
r0 (λ)
.
r(λ)
We are going to show that ln{r(λ)} and hence r(λ) are both strictly increasing by
showing that u(λ) > 0 for all λ ∈ [0, 1).
Differentiating u(λ) with respect to λ, we have :
Let u(λ) =
0
u (λ) = −
n
X
i=1
n
X
αi (z − xi )2
αi (z − xi )2
+
<0
[λz + (1 − λ)xi − β]2 i=1 [λ(1 − z) + (1 − λ)(1 − xi ) − δ]2
as
1
1
>
[λz + (1 − λ)xi − β]2
[λ(1 − z) + (1 − λ)(1 − xi ) − δ]2
i = 1, 2, . . . , n, unless z = x1 = x2 = . . . = xn = 1/2, and β = δ.
Hence u(λ) is strictly decreasing on [0, 1] .
u(1) =
n
X
i=1
n
z − xi
z − xi X
+
αi
αi
z−β
1−z−δ
i=1
n
n
X
1 X
1
αi (z − xi ) +
αi (z − xi )
z − β i=1
1 − z − δ i=1
!
n
n
X
X
1−β−δ
=
z−
αi xi
≥ 0,
for
αi xi ≤ z ≤ 1/2.
(z − β)(1 − z − δ)
i=1
i=1
=
0
(λ)
Hence, u(λ) = rr(λ)
> 0, for all λ ∈ [0, 1).
As r(λ) is always positive, we have r0 (λ) > 0 for all λ ∈ [0, 1) and r(λ) is strictly
increasing on [0, 1] .
0
z
and r(0) < r(λ) <
(ii) It is easy to see that when β = δ = 0, r(0) = Gn /Gn , r(1) = 1−z
r(1) on (0, 1).
Pn
It is remarked, that if β = δ = 0, z =
i=1 αi xi in Theorem 3.1, then the chain of
inequalities in (3.2), with r(λ) replaced by Q(λ), will become : for any λ ∈ (0, 1),
(3.3)
Gn
z
An
= 0
0 = Q(0) < Q(λ) < Q(1) =
Gn
1−z
An
This is a refinement of the Ky Fan inequality.
Remark 3.2. (3.3) is a refinement of the weighted Ky Fan inequality and we have r(0) <
r(λ) < r(1), unless x1 = x2 = . . . = xn . In general, (3.2) yields a generalization of the Ky Fan
inequality as follows :
J. Inequal. Pure and Appl. Math., 2(2) Art. 19, 2001
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8
K. K. C HONG
For An ≤ z ≤ 1/2 and δ ≤ β < xi ∈ (0, 1/2], for all i = 1, 2, . . . , n, we have,
n
Q
[xi − β]αi
z−β
i=1
r(0) = Q
≤
= r(1),
n
1−z−δ
αi
[1 − xi − δ]
(3.4)
i=1
with equality if and only if x1 = x2 = · · · = xn .
If in (3.4) we let z = An the weighted arithmetic mean of x1 , x2 , . . . , xn , we obtain a generalization of the weighted Ky Fan inequality :
n
Q
[xi − β]αi
i=1
(3.5)
n
Q
≤
[1 − xi − δ]αi
An − β
,
A0n − δ
i=1
with equality if and only if x1 = x2 = · · · = xn .
P
Remark 3.3. If we put αi = 1/n, i = 1, 2, . . . , n, z = ni=1 αi xi , β = δ = 0 and λ = nt into
(3.1), we then obtain after simplification,
"
#1/n
n
n 1
Q
P
nt
xj + (1 − nt)xi
i=1
j=1 n
r(nt) =
"
!
#1/n
n
n 1
Q
P
nt 1 −
xj + (1 − nt)(1 − xi )
i=1
j=1 n
"
#1/n
n
n
Q
P
xi + t (xj − xi )
=
i=1
n
Q
j=1
"
1 − xi − t
i=1
n
P
for t ∈ [0, 1/n].
#1/n = q(t),
(xj − xi )
j=1
This is the function q(t) in Theorem 1.2(b), showing that Theorem 3.1 is a generalization of
Theorem 1.2(b).
Remark 3.4. In [5], we have the following theorem, which can be easily seen to follow as a
particular case of Theorem 3.1, when β = 0 and δ = 0 :
Theorem 3.5. Let x1 , x2 , . . . , xn be n positive numbers, such that xi ∈ (0, 1/2], for all i =
1, 2, . . . , n, and let α1 , α2 , . . . , αn be their corresponding positive weights,
with α1 + α2 + · · · +
P
αn = 1. Let z be a constant such that An ≤ z ≤ 1/2, where An = ni=1 αi xi . We define w(λ),
for any λ ∈ [0, 1], to be the function :
n
Q
(3.6)
w(λ) = Q
n
[λz + (1 − λ)xi ]αi
i=1
.
αi
[λ(1 − z) + (1 − λ)(1 − xi )]
i=1
Then,
(i) w(λ) is continuous and strictly increasing on [0, 1] , unless x1 = x2 = · · · = xn ;
0
z
, for λ ∈ [0, 1] .
(ii) Gn /Gn = w(0) ≤ w(λ) ≤ w(1) = 1−z
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M ONOTONIC R EFINEMENTS OF A K Y FAN I NEQUALITY
9
4. S ECOND P ROOF OF T HEOREM 2.1
In this section, we shall show that Theorem 3.1 is not only a generalization of Theorem
1.2(b), but also it can be used to deduce some elementary theorems.
Second proof of Theorem 2.1. Suppose the n numbers P
x1 , x2 , . . . , xn are not all equal.
For xi ∈ (0, 1/2], i = 1, 2, . . . , n, z lying between ni=1 αi xi and 1/2, the function r(λ) of
Theorem 3.1 is strictly increasing on [0, 1] , where for λ ∈ [0, 1] , r(λ) is defined as :
n
Q
[λz + (1 − λ)xi − β]αi
i=1
(4.1)
r(λ) = Q
.
n
α
i
[λ(1 − z) + (1 − λ)(1 − xi ) − δ]
i=1
ai
,
l
Now, we put xi =
i = 1, 2, . . . , n, where l is a large positive number, and let z = al , β = kl
with δ = β. Then, the function r(λ) becomes :
n
1Q
[λa + (1 − λ)ai − k]αi
l i=1
(4.2)
r(λ) = Q
iαi .
n h
a
ai
λ(1 − ) + (1 − λ)(1 − ) − β
l
l
i=1
By Theorem 3.1,
n
Q
[λa + (1 − λ)ai − k]αi
α
(4.3)
v(λ) = n i=1
Q
a
ai
k i
λ(1 − ) + (1 − λ)(1 − ) −
l
l
l
i=1
is strictly increasing as λ increases from 0 to 1.
We let l tend to +∞, the denominator tends to 1 and we have shown that the function in
Theorem 2.1
n
Y
(4.4)
F (λ) =
[λa + (1 − λ)ai − k]αi
i=1
is an increasing function on [0, 1] .
Differentiation calculations as in Theorem 2.1 easily reveal that in fact F (λ) is strictly increasing on [0, 1] . This completes the proof of Theorem 2.1.
Remark 4.1. From the discussions in Section 2, it can be seen that Theorem 2.1 generalizes
Theorem 1.1, Theorem 1.2(a), Theorem 2.4 and Theorem 2.6. From the discussions of the last
two sections, it can be seen that Theorem 3.1 generalizes Theorem 1.2(b), Theorem 3.5 and
Theorem 2.1. As a whole, we have shown that Theorem 3.1 is a generalization of all other
refinements of inequalities (1.1) and (1.2), appearing in this paper.
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