Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 130, 2006
GENERALIZATIONS OF THE KY FAN INEQUALITY
AI-JUN LI, XUE-MIN WANG, AND CHAO-PING CHEN
J IAOZUO U NIVERSITY
J IAOZUO C ITY, H ENAN P ROVINCE
454000, C HINA
liaijun72@163.com
wangxm881@163.com
C OLLEGE OF M ATHEMATICS AND I NFORMATICS
R ESEARCH I NSTITUTE OF A PPLIED M ATHEMATICS
H ENAN P OLYTECHNIC U NIVERSITY
J IAOZUO C ITY, H ENAN 454010, C HINA
chenchaoping@hpu.edu.cn
Received 25 November, 2005; accepted 06 October, 2006
Communicated by D. Ştefǎnescu
A BSTRACT. In this paper, we extend the Ky Fan inequality to several general integral forms,
Ls (a,b)
and obtain the monotonic properties of the function Ls (α−a,α−b)
with α, a, b ∈ (0, +∞) and
s ∈ R.
Key words and phrases: Generalized logarithmic mean, Monotonicity, Ky Fan inequality.
2000 Mathematics Subject Classification. 26A48, 26D20.
1. I NTRODUCTION
The following inequality proposed by Ky Fan was recorded in [1, p. 5] : If 0 < xi ≤
i = 1, 2, . . . , n, then
Pn
Qn
n1
x
xi
i
i=1
Qn
(1.1)
≤ Pn i=1
,
i=1 (1 − xi )
i=1 (1 − xi )
unless x1 = x2 = · · · = xn .
With the notation
 P
1
 1 n xr r , r 6= 0;
i=1 i
n
(1.2)
Mr (x) =
(Qn x ) n1 ,
r = 0,
i=1 i
1
2
for
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
The authors were supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan
Province, China.
346-05
2
A.-J. L I , X.-M. WANG , AND C H .-P. C HEN
where Mr (x) denotes the r-order power mean of xi > 0 for i = 1, 2, . . . , n, the inequality (1.1)
can be written as
M0 (x)
M1 (x)
(1.3)
≤
.
M0 (1 − x)
M1 (1 − x)
In 1996, Zh. Wang, J. Chen and X. Li [12] found the necessary and sufficient condition for
Mr (x)
Ms (x)
(1.4)
≤
Mr (1 − x)
Ms (1 − x)
Lr (a,b)
when r < s. Recently, Ch.-P. Chen proved that the function Lr (1−a,1−b)
is strictly increasing for
1
1
0 < a < b ≤ 2 and strictly decreasing for 2 ≤ a < b < 1, where r ∈ (−∞, ∞) and Lr (a, b)
is the generalized logarithmic mean of two positive numbers a, b, which is a special case of the
extended means E(r, s; x, y) defined by Stolarsky [10] in 1975. For more information about the
extended means please refer to [4, 6, 8, 11] and references therein.
Moreover, we have,
Lr (a, b) = a,
a = b;
r+1
r1
b
− ar+1
Lr (a, b) =
,
(r + 1)(b − a)
b−a
L−1 (a, b) =
= L(a, b);
ln b − ln a
1
1 bb b−a
L0 (a, b) =
= I(a, b),
e aa
a 6= b, r 6= −1, 0;
where L(a, b) and I(a, b) are respectively the logarithmic mean and the exponential mean of
two positive numbers a and b. When a 6= b, Lr (a, b) is a strictly increasing function of r. In
particular,
lim Lr (a, b) = min{a, b},
r→−∞
L1 (a, b) = A(a, b),
lim Lr (a, b) = max{a, b},
r→+∞
L−2 (a, b) = G(a, b),
where A(a, b) and G(a, b) are the arithmetic and the geometric means, respectively. For a 6= b,
the following well known inequality holds:
G(a, b) < L(a, b) < I(a, b) < A(a, b).
(1.5)
In this paper, motivated by inequality (1.4), we will extend the inequality (1.4) to general
integral forms. Some monotonic properties of several related functions will be obtained.
Theorem 1.1. Let
Rb
fα (s) =
a
Rb
a
xs d x
(α − x)s d x
! 1s
=
Ls (a, b)
,
Ls (α − a, α − b)
s ∈ (−∞, +∞) and α be a positive number. Then fα (s) is a strictly increasing function for
[a, b] ⊆ (0, α2 ], and is a strictly decreasing function for [a, b] ⊆ [ α2 , α).
Corollary 1.2. If [a, b] ⊆ (0, α2 ] and α is a positive number, then
(1.6)
a
G(a, b)
L(a, b)
<
<
α−b
G(α − a, α − b)
L(α − a, α − b)
I(a, b)
A(a, b)
b
<
<
<
.
I(α − a, α − b)
A(α − a, α − b)
α−a
J. Inequal. Pure and Appl. Math., 7(4) Art. 130, 2006
http://jipam.vu.edu.au/
G ENERALIZATIONS OF THE K Y FAN I NEQUALITY
3
If [a, b] ⊆ [ α2 , α), the inequalities (1.6) is reversed.
R
1s
b s
x
d
x
a
Corollary 1.3. Let hα (s) = R α−a
, s ∈ (−∞, +∞) and α be a positive number. Then
xs d x
α−b
hα (s) is a strictly increasing function for [a, b] ⊆ (0, α2 ], or a strictly decreasing function for
[a, b] ⊆ [ α2 , α).
In [13], Feng Qi has proved that the function
! r1
Rb r
1
x
d
x
Lr (a, b)
b−a a
r 7→
=
R
b+δ
1
Lr (a, b + δ)
xr d x
b+δ−a a
is strictly decreasing with r ∈ (−∞, +∞). Now, we will extend the conclusion in the following
theorem.
Theorem 1.4. Let
f (s) =
1
b−a
1
d−c
Rb
a
xs d x
! 1s
Rd
=
Ls (a, b)
,
Ls (c, d)
=
Ls (a, b)
,
Ls (a, d)
xs d x
s ∈ (−∞, +∞) and a, b, c, d be positive numbers. Then f (s) is a strictly increasing function
for ad < bc, or a strictly decreasing function for ad > bc.
c
Corollary 1.5. Let
h(s) =
1
b−a
1
d−a
Rb
a
Rd
xs d x
! 1s
xs d x
s ∈ (−∞, +∞) and a, b, d are positive numbers. Then h(s) is a strictly increasing function for
d < b, or a strictly decreasing function for d > b.
a
2. P ROOFS OF T HEOREMS
In order to prove Theorem 1.1, we make use of the following elementary lemma which can
be found in [3, p. 395].
Lemma 2.1 ([3, p. 395]). Let the second derivative of φ(x) be continuous with x ∈ (−∞, ∞)
and φ(0) = 0. Define

 φ(x) , x 6= 0;
x
(2.1)
g(x) =
 0
φ (0), x = 0.
Then φ(x) is strictly convex (concave) if and only if g(x) is strictly increasing (decreasing) with
x ∈ (−∞, ∞).
Remark 2.2. A general conclusion was given in [7, p. 18]: A function φ is convex on [a, b] if
0)
and only if φ(x)−φ(x
is nondecreasing on [a, b] for every point x0 ∈ [a, b].
x−x0
Proof of Theorem 1.1. It is obvious that
! 1s Rb s
1s
s+1
s+1
x
d
x
b
−
a
fα (s) = R b a
=
(α − a)s+1 − (α − b)s+1
(α − x)s d x
a
Ls (a, b)
=
.
Ls (α − a, α − b)
J. Inequal. Pure and Appl. Math., 7(4) Art. 130, 2006
http://jipam.vu.edu.au/
4
A.-J. L I , X.-M. WANG , AND C H .-P. C HEN
Define for s ∈ (−∞, ∞),
 s+1
s+1
b
−
a


ln
, s 6= −1;

(α − a)s+1 − (α − b)s+1
ϕ(s) =

ln(b/a)

ln
,
s = −1.

ln[(α − a)/(α − b)]
(2.2)
Then

 ϕ(s) , s 6= 0;
s
ln fα (s) =
 0
ϕ (0), s = 0.
(2.3)
In order to prove that ln fα is strictly increasing (decreasing), it suffices to show that ϕ is
strictly convex (concave) on (−∞, ∞). A simple calculation reveals that
(α − a)(α − b)
,
ab
which implies that ϕ00 (−1 − s) = ϕ00 (−1 + s), and ϕ has the same convexity (concavity) on
both (−∞, −1) and (−1, ∞). Hence, it is sufficient to prove that ϕ is strictly convex (concave)
on (−1, ∞).
A computation yields
ϕ(−1 − s) = ϕ(−1 + s) + s ln
(2.4)
bs+1 ln b − as+1 ln a (α − b)s+1 ln(α − b) − (α − a)s+1 ln(α − a)
−
,
bs+1 − as+1
(α − b)s+1 − (α − a)s+1
"
#
α−b 2
)
as+1 bs+1 (ln ab )2 (α − a)s+1 (α − b)s+1 (ln α−a
2 00
2
(s + 1) ϕ (s) = (s + 1) − s+1
+
(b − as+1 )2
[(α − a)s+1 − (α − b)s+1 ]2
ϕ0 (s) =
=−
α−b s+1
α−b s+1 2
) [ln( α−a
) ]
( ab )s+1 [ln( ab )s+1 ]2 ( α−a
+
.
a s+1 2
α−b s+1 2
[1 − ( b ) ]
[1 − ( α−a
) ]
Define for 0 < t < 1,
ω(t) =
(2.5)
t(ln t)2
.
(1 − t)2
Differentiation yields
(2.6)
∞
X
ω 0 (t)
n−1
(1 − t)t ln t
= (1 + t) ln t + 2(1 − t) = −
(1 − t)n+1 < 0,
ω(t)
n(n
+
1)
n=2
which implies that ω 0 (t) > 0 for 0 < t < 1. It is easy to see that
a s+1 α − b s+1
αi
(2.7)
0<
<
< 1 for [a, b] ⊆ 0,
, s > −1,
b
α−a
2
(2.8)
0<
α−b
α−a
s+1
<
a s+1
b
< 1 for [a, b] ⊆
hα
2
, α , s > −1,
and therefore ϕ00 (s) > 0 for [a, b] ⊆ (0, α2 ] and s > −1, ϕ00 (s) < 0 for [a, b] ⊆ [ α2 , α) and
s > −1. Then ϕ is strictly convex (concave) on (−1, ∞) for [a, b] ⊆ (0, α2 ] ([a, b] ⊆ [ α2 , α))
respectively. By Lemma 2.1 above, Theorem 1.1 holds.
J. Inequal. Pure and Appl. Math., 7(4) Art. 130, 2006
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G ENERALIZATIONS OF THE K Y FAN I NEQUALITY
5
Since fα (s) is a strictly increasing (decreasing) function for [a, b] ⊆ (0, α2 ] ([a, b] ⊆ [ α2 , α)),
put s = −2, −1, 0, 1 respectively. The inequalities (1.6) are deduced.
Rb s
1s
a x dx
Then, let (α − x) = t and apply it to the function R b (α−x)s d x . We get Corollary 1.3.
a
Proof of Theorem 1.4. Using an analogous method of proof to that of Theorem 1.1, we get
f (s) =
1
b−a
1
d−c
Rb
a
Rd
c
xs d x
xs
" bs+1 −as+1 # 1s
! 1s
(s+1)(b−a)
ds+1 −cs+1
(s+1)(d−c)
=
dx
(d − c) (bs+1 − as+1 )
=
(b − a) (ds+1 − cs+1 )
Let M =
(2.9)
(d−c)
,
(b−a)
1s
=
Ls (a, b)
.
Ls (c, d)
and define for s ∈ (−∞, ∞),
 bs+1 − as+1



ln M ds+1 − cs+1 , s 6= −1;
ϕ(s) =

ln(b/a)


,
s = −1.
ln M
ln(d/c)
Then

 ϕ(s) , s 6= 0;
s
ln f (s) =
 0
ϕ (0), s = 0,
(2.10)
and ϕ has the same convexity (concavity) on both (−∞, −1) and (−1, ∞).
A computation yields
( ab )s+1 [ln( ab )s+1 ]2 ( dc )s+1 [ln( dc )s+1 ]2
(s + 1) ϕ (s) = −
+
.
[1 − ( ab )s+1 ]2
[1 − ( dc )s+1 ]2
00
2
Define for 0 < t < 1,
ω(t) =
(2.11)
t(ln t)2
.
(1 − t)2
Differentiation yields ω 0 (t) > 0 for 0 < t < 1. It is easy to see that
a s+1 c s+1
(2.12)
0<
<
< 1 for ad < bc, s > −1,
b
d
(2.13)
0<
c s+1
d
<
a s+1
b
< 1 for
ad > bc, s > −1,
and therefore ϕ00 (s) > 0 for ad < bc and s > −1, ϕ00 (s) < 0 for ad > bc and s > −1 Then
ϕ is strictly convex (concave) on (−1, ∞) for ad < bc (ad > bc) respectively. The proof is
complete.
In Theorem 1.4, let a = c. Then f (s) is a strictly increasing function for d < b, or a strictly
decreasing function for d > b. Thus Corollary 1.5 holds.
J. Inequal. Pure and Appl. Math., 7(4) Art. 130, 2006
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6
A.-J. L I , X.-M. WANG , AND C H .-P. C HEN
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J. Inequal. Pure and Appl. Math., 7(4) Art. 130, 2006
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