Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 4, Article 76, 2003
CERTAIN BOUNDS FOR THE DIFFERENCES OF MEANS
PENG GAO
D EPARTMENT OF M ATHEMATICS ,
U NIVERSITY OF M ICHIGAN ,
A NN A RBOR , MI 48109, USA.
penggao@umich.edu
Received 17 August, 2002; accepted 3 June, 2003
Communicated by K.B. Stolarsky
A BSTRACT. Let Pn,r (x) be the generalized weighted power means. We consider bounds for
the differences of means in the following form:
)
(
)
(
α
α
Pn,u
− Pn,v
Cu,v,β Cu,v,β
Cu,v,β Cu,v,β
max
, 2β−α σn,w0 ,β ≥
≥ min
, 2β−α σn,w,β .
α
x2β−α
xn
x2β−α
xn
1
1
Pn
β
Here β 6= 0, σn,t,β ( x) = i=1 ωi [xβi −Pn,t
(x)]2 and Cu,v,β = u−v
2β 2 . Some similar inequalities
are also considered. The results are applied to inequalities of Ky Fan’s type.
Key words and phrases: Ky Fan’s inequality, Levinson’s inequality, Generalized weighted power means, Mean value theorem.
2000 Mathematics Subject Classification. Primary 26D15, 26D20.
1. I NTRODUCTION
1
P
Let Pn,r (x) be the generalized
weighted power means: Pn,r (x) = ( ni=1 ωi xri ) r , where
Pn
ωi > 0, 1 ≤ i ≤ n with i=1 ωi = 1 and x = (x1 , x2 , . . . , xn ). Here Pn,0 (x) denotes the limit
of Pn,r (x) as r → 0+ . In this paper, we always assume that 0 < x1 ≤ x2 ≤ · · · ≤ xn . We write
σn,t,β (x) =
n
X
h
i2
β
ωi xβi − Pn,t
(x)
i=1
and denote σn,t as σn,t,1 .
We let
An (x) = Pn,1 (x), Gn (x) = Pn,0 (x), Hn (x) = Pn,−1 (x)
and we shall write Pn,r for Pn,r (x), An for An (x) and similarly for other means when there is
no risk of confusion.
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
The author thanks the referees for their many valuable comments and suggestions. These greatly improved the presentation of the paper.
089-02
2
P ENG G AO
We consider upper and lower bounds for the differences of the generalized weighted means
in the following forms (β 6= 0):
α
α
Pn,u
− Pn,v
Cu,v,β Cu,v,β
Cu,v,β Cu,v,β
, 2β−α σn,w,β ,
(1.1)
max
, 2β−α σn,w0 ,β ≥
≥ min
α
x2β−α
xn
x2β−α
xn
1
1
where Cu,v,β =
u−v
.
2β 2
If we set x1 = · · · = xn−1 6= xn , then we conclude from
α
α
Pn,u
− Pn,v
u−v
=
x1 →xn ασn,w,β
2β 2 x2β−α
n
0
0
that Cu,v,β is best possible. Here we define (Pn,u − Pn,v )/0 = ln(Pn,u /Pn,v ), the limit of
α
α
(Pn,u
− Pn,v
)/α as α → 0.
In what follows we will refer to (1.1) as (u, v, α, β, w, w0 ). D.I. Cartwright and M.J. Field [8]
first proved the case (1, 0, 1, 1, 1, 1). H. Alzer [4] proved (1, 0, 1, 1, 1, 0) and [5] (1, 0, α, 1, 1, 1)
with α ≤ 1. A.McD. Mercer [13] proved the right-hand side inequality with smaller constants
for α = β = u = 1, v = −1, w = ±1.
There is a close relationship between (1.1) and the following Ky Fan inequality, first published in the monograph
Q Inequalities by Beckenbach and Bellman [7].(In this section, we set
A0n = 1 − An , G0n = ni=1 (1 − xi )ωi . For general definitions, see the beginning of Section 3.)
Theorem 1.1. For xi ∈ 0, 12 ,
Lim
An
A0n
≤
0
Gn
Gn
with equality holding if and only if x1 = · · · = xn .
(1.2)
P. Mercer [15] observed that the validity of (1, 0, 1, 1, 1, 1) leads to the following refinement
of the additive Ky Fan inequality:
Theorem 1.2. Let 0 < a ≤ xi ≤ b < 1 (1 ≤ i ≤ n, n ≥ 2). For a 6= b we have
A0 − G0n
b
a
< n
<
.
(1.3)
1−a
An − G n
1−b
Thus, by a result of P. Gao [9], it yields the following refinement of Ky Fan’s inequality, first
proved by Alzer [6]:
2
2
a
b
( 1−a
( 1−b
)
)
An
A0n
An
≤ 0 ≤
.
Gn
Gn
Gn
For an account of Ky Fan’s inequality, we refer the reader to the survey article [2] and the
references therein.
The additive Ky Fan’s inequality for generalized weighted means is a consequence of (1.1).
Since it does not always hold (see [9]), it follows that (1.1) does not hold for arbitrary
(u, v, α, β, w, w0 ).
Our main result is a theorem that shows the validity of (1.1) for some α, β, u, v, w, w0 . We
apply it in Section 3 to obtain further refinements and generalizations of inequalities of Ky Fan’s
type.
One can obtain further refinements of (1.1). Recently, A.McD. Mercer proved the following
theorem [14]:
Theorem 1.3. If x1 6= xn , n ≥ 2, then
Gn − x 1
xn − Gn
(1.4)
σn,1 > An − Gn >
σn,1 .
2x1 (An − x1 )
2xn (xn − An )
We generalize this in Section 2.
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C ERTAIN B OUNDS FOR THE D IFFERENCES OF M EANS
3
2. T HE M AIN T HEOREM
Theorem 2.1. 1, rs , 1, γr , rt ,
t0
r
, r 6= s, r 6= 0, γ 6= 0 holds for the following three cases:
t0
(1) γs ≤ γr ≤ 2, 1 ≥ γt , γ ≥ γs ≥ γr − 1;
0
(2) γr ≥ 2, γr − 1 ≥ γs ≥ γt , tγ ≥ 1;
0
(3) γr ≤ γs ≤ γt , tγ ≤ 1,
with equality holding if and only if x1 = · · · = xn for all the cases.
Proof. Let γ = 1 and r 6= s. We will show that (1.1) holds for the following three cases:
(1) s ≤ r ≤ 2, 1 ≥ t, t0 ≥ s ≥ r − 1;
(2) r ≥ 2, r − 1 ≥ s ≥ t, t0 ≥ 1;
(3) r ≤ s ≤ t, t0 ≤ 1.
For case (1), consider the right-hand side inequality of (1.1) and let
n
1 2
r(r − s) X r1
r
s
(2.1)
Dn (x) = An − Pn, r −
ωi xi − Pn, t .
2
−1
r
2xnr
i=1
We want to show that Dn ≥ 0 here. We can assume that x1 < x2 < · · · < xn and use
induction. The case n = 1 is clear, so assume that the inequality holds for n − 1 variables. Then
(2.2)
1 ∂Dn
·
=1−
ωn ∂xn
"
Pn, rs
xn
r1 #r−s
− (r − s) 1 −
Pn, rt
r1 !
+ S,
xn
where
1−t
n
(2 − r)(r − s) X
1
r
1
r
2
Pn,rt 1
r
1
r
ωi xi − Pn, t + (r − s)
Pn, 1 − Pn, t .
2
−2
r
r
r
2ωn xnr
xn
i=1
Thus, when s ≤ r ≤ 2, t ≤ 1, S ≥ 0.
Now by the mean value theorem
"
1 #r−s
s r1 !
s r1 !
Pn, rs r
P
Pn, r
n, r
1−
= (r − s)η r−s−1 1 −
≥ (r − s) 1 −
xn
xn
xn
S=
r
2−t
r
for r ≥ s ≥ r − 1 with
( ( 1 )
1 )
Pn, rs r
Pn, rs r
min 1,
≤ η ≤ max 1,
.
xn
xn
This implies
"
"
1 #r−s
t r1 !
1 s r1 #
Pn, r
Pn, rt r
Pn, rs r
Pn, r
−
,
1−
− (r − s) 1 −
≥ (r − s)
xn
xn
xn
xn
which is positive if s ≤ t.
n
Thus for s ≤ r ≤ 2, 1 ≥ t ≥ s ≥ r − 1, we have ∂D
≥ 0. By letting xn tend to xn−1 , we
∂xn
have Dn ≥ Dn−1 (with weights ω1 , . . . , ωn−2 , ωn−1 + ωn ) and thus the right-hand side inequality
of (1.1) holds by induction. It is also easy to see that equality holds if and only if x1 = · · · = xn .
Now consider the left-hand side inequality of (1.1) and write
2
n
1
1
r(r − s) X
ωi xir − P r t0 .
(2.3)
En (x) = An − Pn, rs −
2
−1
n, r
r
2x1
i=1
J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
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4
P ENG G AO
n
has an expression similar to (2.2) with xn ↔ x1 , ωn ↔ ω1 , t ↔ t0 . It is then easy
Now ω11 ∂E
∂x1
n
to see under the same condition, ∂E
≥ 0. Thus the left-hand side inequality of (1.1) holds by
∂x1
a similar induction process with the equality holding if and only if x1 = · · · = xn .
Similarly, we can show Dn (x) ≤ 0, En (x) ≥ 0 for cases (2) and (3) with equality holding if
and only if x1 = · · · = xn for all the cases.
Now for an arbitrary γ, a change of variables y → y/γ for y = r, s, t, t0 in the above cases
leads to the desired conclusion.
In what follows our results often include the cases r = 0 or s = 0 and we will leave the
proofs of these special cases to the reader since they are similar to what we give in the paper.
Corollary 2.2. For r > s, min{1, r − 1} ≤ s ≤ max{1, r − 1} and min{1, s} ≤ t, t0 ≤
max{1, s}, (r, s, r, 1, t, t0 ) holds. For s ≤ r ≤ t, t0 ≤ 1, (r, s, s, 1, t, t0 ) holds, with equality
holding if and only if x1 = · · · = xn for all the cases.
Proof. This follows from taking γ = 1 in Theorem 2.1 and another change of variables: x1 →
min{xr1 , xrn }, xn → max{xr1 , xrn } and xi = xri for 2 ≤ i ≤ n − 1 if n ≥ 3 and exchanging r
and s for the case s > r.
We remark here since σn,t0 = σn,t + (2An − Pn,t − Pn,t0 )(Pn,t − Pn,t0 ), we have σn,1 ≤ σn,t for
t 6= 1 and σn,t ≤ σn,t0 for t0 ≤ t ≤ 1, σn,t ≥ σn,t0 for t ≥ t0 ≥ 1. Thus the optimal choices for
the set {t, t0 } will be {1, s} for the case (r, s, r, 1, t, t0 ) and {1, r} for the case (r, s, s, 1, t, t0 ).
Our next two propositions give relations between differences of means with different powers:
Proposition 2.3. For l − r ≥ t − s ≥ 0, l 6= t, xi ∈ [a, b], a > 0,
(P r − P r )/r (r − s) 1
(r − s) 1
n,r
n,s
(2.4)
(l − t) al−r ≥ (P l − P l )/l ≥ (l − t) bl−r .
n,t
n,l
Except for the trivial cases r = s or (l, t) = (r, s), the equality holds if and only if x1 = · · · =
xn , where we define 0/0 = xr−l
for any i.
i
Proof. This is a generalization of a result A.McD. Mercer [12]. We may assume that x1 =
a, xn = b and consider
r
r
D(x) = Pn,r
− Pn,s
−
r(r − s)
l
l
(Pn,l
− Pn,t
),
l(l − t)xl−r
n
r(r − s)
l
l
(Pn,l
− Pn,t
).
l−r
l(l − t)x1
We will show that Dn · En ≤ 0. Suppose r − s ≥ 0 here; the case r − s ≤ 0 is similar. We have

l−t 
"
r−s
r−s # r−s
1−r
xn
∂Dn
Pn,s
r−s
Pn,t
 + S,
·
=1−
−
1−
rωn ∂xn
xn
l−t
xn
r
r
E(x) = Pn,r
− Pn,s
−
where
S=
(r − s)(l − r) l
l
(Pn,l − Pn,t
) ≥ 0.
l(l − t)xl−2
ω
n
n
Now by the mean value theorem
l−t
"
r−s # r−s
Pn,t
l − t l−t−r+s
1−
=
η
xn
r−s
J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
1−
Pn,t
xn
r−s !
,
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C ERTAIN B OUNDS FOR THE D IFFERENCES OF M EANS
where
Pn,t
xn
5
< η < 1 and
∂Dn
x1−r
n
≥1−
rωn ∂xn
Pn,s
xn
r−s
−
1−
Pn,t
xn
r−s !
≥0
since t ≥ s.
x1−r
∂En
1
Similarly, we have rω
≥ 0 and by a similar induction process as the one in the proof of
1 ∂x1
Theorem 2.1, we have Dn · En ≤ 0. This completes the proof.
By taking l = 2, t = 0, r = 1, s = −1 in the proposition, we get the following inequality:
1
1
2
(2.5)
(Pn,2
− G2n ) ≥ An − Hn ≥
(P 2 − G2n )
2x1
2xn n,2
and the right-hand side inequality above gives a refinement of a result of A.McD. Mercer [13].
Proposition 2.4. For r > s, α > β,
(2.6)
β−α
xβ−α
≥ Pn,s
≥
1
β
β
(Pn,r
− Pn,s
)/β
β−α
≥ Pn,r
≥ xβ−α
n
α
α
(Pn,r − Pn,s )/α
with equality holding if and only if x1 = · · · = xn , where we define 0/0 = xβ−α
for any i.
i
Proof. By the mean value theorem,
β
β
α β/α
α β/α
Pn,r
− Pn,s
= (Pn,r
)
− (Pn,s
)
=
β β−α α
α
η
(Pn,r − Pn,s
),
α
where Pn,s < η < Pn,r and (2.6) follows.
We apply (2.6) to the case (1, 0, 1, 1, 1, 1) to see that (1, 0, α, 1, 1, 1) holds with α ≤ 1, a
result of Alzer [5]. We end this section with a generalization of (1.4) and leave the formulation
of similar refinements to the reader.
Theorem 2.5. If x1 6= xn , n ≥ 2, then for 1 > s ≥ 0
(2.7)
1−s
− x1−s
Pn,s
1
2x1−s
1 (An − x1 )
σn,1 > An − Pn,s >
1−s
x1−s
− Pn,s
n
σn,1 .
2x1−s
n (xn − An )
Proof. We prove the right-hand inequality; the left-hand side inequality is similar. Let
1−s
x1−s
− Pn,s
n
Dn (x) = (xn − An )(An − Pn,s ) −
σn,1 .
2x1−s
n
We show by induction that Dn ≥ 0. We have
1−s s ∂Dn
1 − s Pn,s
xn
= (1 − ωn )(An − Pn,s ) −
1−
ωn σn,1
∂xn
2xn
xn
Pn,s
1−s
≥ (1 − ωn ) An − Pn,s −
σn,1 ≥ 0,
2xn
where the last inequality holds by Theorem 2.1. By an induction process similar to the one in
the proof of Theorem 2.1, we have Dn ≥ 0. Since not all the xi ’s are equal, we get the desired
result.
Corollary 2.6. For 1 > s ≥ 0,
1 − s Pn,s
1−s
(2.8)
·
σn,1 ≥ An − Pn,s ≥
σn,s ,
2x1
An
2xn
with equality holding if and only if x1 = · · · = xn .
J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
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6
P ENG G AO
Proof. By Theorem 2.5, we only need to show
by using the mean value theorem.
1−s
Pn,s
−x1−s
1
2x1−s
(A
n −x1 )
1
≤
1−s Pn,s
2x1 An
and this is easily verified
3. A PPLICATIONS TO I NEQUALITIES OF K Y FAN ’ S T YPE
Let f (x, y) be a real function. We regard y as an implicit function defined by f (x, y) = 0 and
0
0
0
for y = (y1 , . . . , yn ), let f (xi , yi ) = 0, 1 ≤ i ≤ n. We write Pn,r
= Pn,r (y) with An = Pn,1
,
0
0
0
0
Gn = Pn,0 , Hn = Pn,−1 . Furthermore, we write x1 = a > 0 and xn = b so that xi ∈ [a, b] with
yi ∈ [a0 , b0 ], a0 > 0 and require that fx0 , fy0 exist for xi ∈ [a, b], yi ∈ [a0 , b0 ].
To simplify expressions, we define:
∆r,s,α =
(3.1)
α
α
Pn,r
(y) − Pn,s
(y)
α
α
Pn,r (x) − Pn,s (x)
. (y)
Pn,r (x)
with ∆r,s,0 = ln PPn,r
ln
and, in order to include the case of equality for various
Pn,s (x)
n,s (y)
inequalities in our discussion, we define 0/0 = 1 from now on.
In this section, we apply our results above to inequalities of Ky Fan’s type. Let f (x, y) be
any function satisfying the conditions in the first paragraph of this section. We now show how
to get inequalities of Ky Fan’s type in general.
0
Suppose (1.1) holds for some α > 0, r > s, β = 1 and t = t0 = 1 , write σn,1 (y) = σn,1
,
apply (1.1) to sequences x, y and then take their quotients to get
0
0
aσn,1
bσn,1
≤
∆
≤
.
r,s,α
b0 σn,1
a0 σn,1
0
Since σn,1
=
Pn
i=1
P
wi ( nk=1 wk (yi − yk ))2 , the mean value theorem yields
yi − yk = −
fx0
(ξ, y(ξ))(xi − xk )
fy0
for some ξ ∈ (a, b). Thus
0 2
0 2
fx f 0
min 0 σn,1 ≤ σn,1 ≤ max x0 σn,1 ,
a≤x≤b fy
a≤x≤b fy
which implies
0 2
0 2
fx a
b
≤ ∆r,s,α ≤ max fx .
min
b0 a≤x≤b fy0 a0 a≤x≤b fy0 We next apply the above argument to a special case.
1
Corollary 3.1. Let f (x, y) = cxp + dy p − 1, 0 < c ≤ d, p ≥ 1, xi ∈ [0, (c + d)− p ]. For
s ∈ [0, 2] and α = max{s, 1} we have
(3.2)
∆1,s,α ≤ 1
with equality holding if and only if x1 = · · · = xn .
Proof. This follows from Corollary 2.2 by the appropriate choice of r and s.
From now on we will concentrate on the case f (x, y) = x + y − 1. Extensions to the case of
general functions f (x, y) are left to the reader.
J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
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C ERTAIN B OUNDS FOR THE D IFFERENCES OF M EANS
7
Corollary 3.2. Let f (x, y) = x + y − 1, 0 < a < b < 1 and xi ∈ [a, b] (i = 1, . . . , n), n ≥ 2.
Then for r > s, min{1, r − 1} ≤ s ≤ max{1, r − 1}
(
(
2−r 2−r )
2−r 2−r )
b
a
b
a
(3.3) max
,
> ∆r,s,r > min
,
.
1−b
1−a
1−b
1−a
For s < r ≤ 1,
(
(
2−s 2−s )
2−s 2−s )
b
a
b
a
(3.4) max
,
> ∆r,s,s > min
,
.
1−b
1−a
1−b
1−a
Proof. Apply Corollary 2.2 to sequences x, y with t = t0 = 1 and take their quotients, by
noticing σn,1 (x) = σn,1 (y).
As a special case of the above corollary, by taking r = 0, s = −1, we get the following
refinement of the Wang-Wang inequality [17]:
2
2
a
b
( 1−a
( 1−b
)
)
Gn
G0n
Gn
(3.5)
≤ 0 ≤
.
Hn
Hn
Hn
We can use Corollary 2.6 to get further refinements of inequalities of Ky Fan’s type. Since
σn,s = σn,1 + (An − Pn,s )2 , we can rewrite the right-hand side inequality in (2.8) as
1−s
1−s
(3.6)
(Pn,1 (x) − Pn,s (x)) 1 −
(Pn,1 (x) − Pn,s (x)) ≥
σn,1 .
2b
2b
Apply (2.8) to y and taking the quotient with (3.6), we get
0
0
0
bσn,1
Pn,s
Pn,1 (y) − Pn,s (y)
b Pn,s
≤
=
.
a0 σn,1 A0n
a0 A0n
(Pn,1 (x) − Pn,s (x)) 1 − 1−s
(Pn,1 (x) − Pn,s (x))
2b
Similarly,
(Pn,1 (y) − Pn,s (y)) 1 − 1−s
(Pn,1 (y) − Pn,s (y))
a An
2a0
≥ 0
.
Pn,1 (x) − Pn,s (x)
b Pn,s
Combining these with a result in [9], we obtain the following refinement of Ky Fan’s inequality:
Corollary 3.3. Let 0 < a < b < 1 and xi ∈ [a, b] (i = 1, . . . , n), n ≥ 2. Then for α ≤ 1,
0≤s<1
2−α 0
2−α
Pn,s
b
a
An
(3.7)
B > ∆1,s,α >
A,
0
1−b
An
1−a
Pn,s
where
−1
1−s
A= 1−
(Pn,1 (y) − Pn,s (y))
,
2a0
1−s
B =1−
(Pn,1 (x) − Pn,s (x)).
2b
We note here when α = 1, s = 0, b ≤ 12 , the left-hand side inequality of (3.7) yields
(3.8)
A0n − G0n
b G0n 0
<
(A + Gn )
An − G n
1 − b A0n n
a refinement of the following two results of H. Alzer [1]: A0n /G0n ≤ (1 − Gn )/(1 − An ), which
is equivalent to (A0n − G0n )/(An − Gn ) < G0n /A0n and [3]: A0n − G0n ≤ (An − Gn )(A0n + Gn ).
Next, we give a result related to Levinson’s generalization of Ky Fan’s inequality. We first
generalize a lemma of A.McD. Mercer [12].
J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
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8
P ENG G AO
Lemma 3.4. Let J(x) be the smallest closed interval that contains all of xi and let y ∈ J(x)
and f (x), g(x) ∈ C 2 (J(x)) be two twice continuously differentiable functions. Then
Pn
P
ωi f (xi ) − f (y) − ( ni=1 ωi xi − y)f 0 (y)
f 00 (ξ)
i=1
Pn
Pn
(3.9)
=
0
g 00 (ξ)
i=1 ωi g(xi ) − g(y) − (
i=1 ωi xi − y)g (y)
for some ξ ∈ J(x), provided that the denominator of the left-hand side is nonzero.
Proof. The proof is very similar to the one given in [12]. Write
n
X
(Qf )(t) =
wi f (txi + (1 − t)y) − f (y) − t(A − y)f 0 (y)
i=1
and consider W (t) = (Qf )(t) − K(Qg)(t), where K is the left-hand side expression in (3.9).
The lemma then follows by the same argument as in [12].
By taking g(x) = x2 , y = Pn,t in the lemma, we get:
Corollary 3.5. Let f (x) ∈ C 2 [a, b] with m = min f 00 (x), M = max f 00 (x). Then
a≤x≤b
a≤x≤b
!
n
n
X
X
m
M
σn,t ≥
ωi f (xi ) − f
ωi xi − (An − Pn,t )f 0 (Pn,t ) ≥ σn,t .
(3.10)
2
2
i=1
i=1
Moreover, if f 000 (x) exists for x ∈ [a, b] with f 000 (x) > 0 or f 000 (x) < 0 for x ∈ [a, b] then the
equality holds if and only if x1 = · · · = xn .
The case t = 1 in the above corollary was treated by A.McD. Mercer [11]. Note for an
arbitrary f (x), equality can hold even if the condition x1 = · · · = xn is not satisfied, for
example, for f (x) = x2 , we have the following identity:
!2
!2
n
n
n
n
X
X
X
X
ωi x2i −
ωi xi
=
ωi xi −
ωk xk .
i=1
i=1
i=1
k=1
Corollary 3.5 can be regarded as a refinement of Jensen’s inequality and it leads to the following well-known Levinson’s inequality for 3-convex functions [10]:
Corollary 3.6. Let xi ∈ (0, a]. If f 000 (x) ≥ 0 in (0, 2a), then
!
n
n
n
X
X
X
(3.11)
ωi f (xi ) − f
ωi xi ≤
ωi f (2a − xi ) − f
i=1
i=1
i=1
n
X
!
ωi (2a − xi ) .
i=1
If f 000 (x) > 0 on (0, 2a) then equality holds if and only if x1 = · · · = xn .
Proof. Take t = 1 in (3.10) and apply Corollary 3.5 to (x1 , . . . , xn ) and (2a − x1 , . . . , 2a − xn ).
Since f 000 (x) ≥ 0 in (0, 2a), it follows that max f 00 (x) ≤ min f 00 (x) and the corollary is
0≤x≤a
a≤x≤2a
proved.
Now we establish an inequality relating different ∆r,s,α ’s:
Corollary 3.7. For l − r ≥ t − s ≥ 0, l 6= t, r 6= s, (l, t) 6= (r, s), xi ∈ [a, b], yi ∈ [a, b], n ≥ 2,
(3.12)
l−r ∆r,s,r a l−r
b
>
> .
a0
∆l,t,l b0
Proof. Apply (2.4) to both x and y and take their quotients.
For another proof of inequality (3.5), use this corollary with l = 1, t = 0, s = −1 and r = 0.
J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
http://jipam.vu.edu.au/
C ERTAIN B OUNDS FOR THE D IFFERENCES OF M EANS
9
4. A F EW C OMMENTS
A variant of (1.1) is the following conjecture by A.McD. Mercer [13] (r > s, t, t0 = r, s):
Pn,r − Pn,s
r−s r−s
r−s r−s
(4.1)
max
, 2−r σn,t0 ≥
≥ min
, 2−r σn,t .
1−r
2xn
Pn,r
2xn
2x2−r
2x2−r
1
1
The conjecture presented here has been reformulated (one can compare it with the original
one in [13]), since here (r − s)/2 is the best possible constant by the same argument as above.
Note when r = 1, (4.1) coincides with (1.1) and thus the conjecture in general is false.
There are many other kinds of expressions for the bounds of the difference between the
arithmetic and geometric means. See Chapter II of the book Classical and New Inequalities in
Analysis [16].
In [12], A.McD. Mercer showed
2
2
Pn,2
− G2n
Pn,2
− G2n
(4.2)
≥ An − G n ≥
.
4x1
4xn
He also pointed out that the above inequality is not comparable to either of the inequalities in
(1.1) with α = β = u = 1, v = 0, t = t0 = 0, 1. We note that (4.2) can be obtained from (1.1)
by averaging the case α = β = u = t = t0 = 1, v = 0 with the following trivial bound:
A2 − G2n
A2n − G2n
≥ An − G n ≥ n
.
2x1
2xn
Thus the incomparability of (4.2) and (4.1) with r = 1, s = 0, t = 1 reflects the fact that
2
− A2n and A2n − G2n are in general not comparable.
Pn,2
We also note when replacing Cu,v,β by a smaller constant, that we sometimes get a trivial
bound. For example, for s ≤ 21 , the following inequality holds:
n
n
2
1 X 1/2
1 X
1/2
ωk xk − An
≥
ωk (xk − An )2 .
An − Pn,s ≥
2 k=1
8xn k=1
1/2
1/2
The first inequality is equivalent to Pn,1/2 An ≥ Pn,s . For the second, simply apply the mean
value theorem to
2
2 1
1
1/2
−1/2
1/2
x k − An
=
ξk (xk − An ) ≥
(xk − An )2 ,
2
4xn
with ξk in between xk and An .
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J. Inequal. Pure and Appl. Math., 4(4) Art. 76, 2003
http://jipam.vu.edu.au/