Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 2, Article 42, 2004
AN EXTENSION OF THE HERMITE-HADAMARD INEQUALITY AND AN
APPLICATION FOR GINI AND STOLARSKY MEANS
PÉTER CZINDER AND ZSOLT PÁLES
B ERZE NAGY JÁNOS G RAMMAR S CHOOL
H-3200 G YÖNGYÖS
KOSSUTH STR . 33, H UNGARY.
pczinder@berze-nagy.sulinet.hu
I NSTITUTE OF M ATHEMATICS
U NIVERSITY OF D EBRECEN
H-4010 D EBRECEN
P F. 12, H UNGARY.
pales@math.klte.hu
Received 05 December, 2003; accepted 01 March, 2004
Communicated by S.S. Dragomir
A BSTRACT. In this paper we extend the Hermite-Hadamard inequality
Z q
p+q
1
f (p) + f (q)
f
≤
f (x)dx ≤
2
q−p p
2
for convex-concave symmetric functions. As consequences some new inequalities for Gini and
Stolarsky means are also derived.
Key words and phrases: Hadamard’s Inequality, Gini means, Stolarsky means.
2000 Mathematics Subject Classification. Primary 26D15, 26D07.
1. I NTRODUCTION
The so-called Hermite-Hadamard inequality [7] is one of the most investigated classical inequalities concerning convex functions. It reads as follows:
Theorem 1.1. Let I ⊂ R be an interval and f : I → R be a convex function. Then, for all
subintervals [p, q] ⊂ I,
Z q
p+q
1
f (p) + f (q)
(1.1)
f
≤
f (x)dx ≤
,
2
q−p p
2
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
The research was supported by the Hungarian National Research Fund (OTKA), Grant Nos. T-038072 and T-047373.
167-03
2
P. C ZINDER AND Z S . PÁLES
while in the case when f is concave all the inequalities are reversed, i.e.,
Z q
p+q
1
f (p) + f (q)
(1.2)
f
≥
f (x)dx ≥
.
2
q−p p
2
holds.
An account on the history of this inequality can be found in [9]. Surveys on various generalizations and developments can be found in [10] and [4]. The description of best possible
inequalities of Hadamard-Hermite type are due to Fink [5]. A generalization to higher-order
convex functions can be found in [1], while [2] offers a generalization for functions that are
Beckenbach-convex with respect to a two dimensional linear space of continuous functions.
In this form (1.1) and (1.2) are valid only for functions that are purely convex or concave
on their whole domain. In Section 2 we will see that under appropriate conditions the same
inequalities can be stated for a much larger family of functions. It will turn out that the results,
obtained for this situation, can be applied for the Gini and Stolarsky means. In this way, we will
obtain new inequalities for these classes of two variable homogeneous means.
2. T HE E XTENSION OF THE H ERMITE -H ADAMARD I NEQUALITY
Let I ⊂ R be an arbitrary real interval and m ∈ I. A function f : I → R is called symmetric
with respect to point m if the equation
(2.1)
f (m − t) + f (m + t) = 2f (m)
holds for all t ∈ (I − m) ∩ (m − I). Observe that when m is one of the endpoints of the interval
I, then (I − m) ∩ (m − I) is either empty or the singleton {m}, therefore (2.1) does not mean
any restriction on f .
Concerning symmetric functions, we have the following obvious statement.
Lemma 2.1. Let f : I → R be symmetric with respect to an element m ∈ I. Then
Z m+α
f (x)dx = 2αf (m)
m−α
for any positive α in (I − m) ∩ (m − I).
Proof. By splitting the integral at the point m and applying substitutions x = m − t and x =
m + t, respectively, we get that
Z m+α
Z 0
Z α
f (x)dx =
−f (m − t)dt +
f (m + t)dt
m−α
α
0
Z α
=
f (m − t) + f (m + t) dt.
0
Due to the symmetry of f , the integrand equals 2f (m), which completes the proof.
Theorem 2.2. Let f : I → R be symmetric with respect to an element m ∈ I, furthermore,
suppose that f is convex over the interval I ∩ (−∞, m] and concave over I ∩ [m, ∞). Then, for
any interval [p, q] ⊂ I
Z q
p+q
≥ 1
≥ f (p) + f (q)
(2.2)
f
f (x)dx
2
(≤) q − p p
(≤)
2
≥
holds if p+q
m.
2 (≤)
(In (2.2) the reversed inequalities are valid if f is concave over the interval I ∩ (−∞, m] and
convex over I ∩ [m, ∞)).
J. Inequal. Pure and Appl. Math., 5(2) Art. 42, 2004
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A N E XTENSION OF THE H ERMITE -H ADAMARD I NEQUALITY FOR G INI AND S TOLARSKY M EANS
3
Proof. Suppose first that p+q
≥ m. The case p, q ≥ m has no interest, since then Theorem 1.1
2
could be applied. Therefore, we may assume that p < m < q.
First we show the left hand side inequality in (2.2)
Z q
p+q
1
f
≥
f (x)dx.
2
q−p p
For this purpose, we split the integral into two parts:
Z q
Z 2m−p
Z
f (x)dx =
f (x)dx +
p
p
q
f (x)dx.
2m−p
Applying Lemma 2.1 with α = m − p, the first integral is equal to 2(m − p)f (m). Moreover,
due to the assumptions p < m < q and p+q
≥ m, we have that m < 2m − p ≤ q. Therefore
2
the function f is concave over the interval [2m − p, q], thus, by Theorem 1.1, the value of the
). That is, we have shown that
second integral is less than or equal to (q − 2m + p)f ( q+2m−p
2
Z q
q + 2m − p
f (x)dx ≤ 2(m − p)f (m) + (q − 2m + p)f
.
2
p
Using the concavity of f over the interval [m, q+2m−p
], we obtain
2
m−p
q − 2m + p
q + 2m − p
2
f (m) +
f
q−p
q−p
2
m−p
q − 2m + p q + 2m − p
p+q
≤f 2
·m+
·
=f
.
q−p
q−p
2
2
This inequality combined with previous one, immediately yields (2.2) and thus proof of the first
part is complete.
Now we prove the right hand side inequality in (2.2). Using the symmetry of f and the
concavity over the interval [2m − p, q], Lemma 2.1 and Theorem 1.1 yield
Z q
Z 2m−p
Z q
f (x)dx =
f (x)dx +
f (x)dx
p
p
2m−p
≥ 2(m − p)f (m) + (q − 2m + p)
f (2m − p) + f (q)
.
2
To complete the proof of (2.2), it is enough to show that
f (2m − p) + f (q)
f (p) + f (q)
(2.3)
(2m − 2p)f (m) + (q − 2m + p)
≥ (q − p)
.
2
2
For, we use, again, the concavity of f over the interval [m, q]. Thus,
q − 2m + p
m−p
(2.4)
f (2m − p) = f
·m+
·q
q−m
q−m
q − 2m + p
m−p
≥
f (m) +
f (q).
q−m
q−m
Substituting f (p) by 2f (m) − f (2m − p) in (2.3), one can easily check that (2.4) and (2.3) are
equivalent inequalities. Consequently, (2.3) follows from (2.4).
An analogous argument leads also to the result in the case p+q
≤ m. Finally, if f is concave
2
over the interval I ∩ (−∞, m] and convex over I ∩ [m, ∞) then, applying what we have already
proven for −f , the statement follows.
Remark 2.3. Theorem 1.1 can be considered as a special case of Theorem 2.2. For, one has to
take m to be one of the endpoints of I.
J. Inequal. Pure and Appl. Math., 5(2) Art. 42, 2004
http://jipam.vu.edu.au/
4
P. C ZINDER AND Z S . PÁLES
3. A N A PPLICATION FOR G INI AND S TOLARSKY M EANS
Given two real parameters a, b, if x, y are positive numbers, then their Gini mean Ga,b (cf.
[6]) is defined by:
 1
a−b
a +y a

x
 xb +yb
if a 6= b,
Ga,b (x, y) =

 exp xa log x+ya log y
if a = b,
a
a
x +y
while their Stolarsky mean Sa,b (cf. [14], [15]) is the following:
 1

b(xa −y a ) a−b

if (a − b)ab 6= 0, x 6= y,


a(xb −y b )



a
a log y



exp − a1 + x logxx−y
if a = b 6= 0, x 6= y,
a −y a


a1
Sa,b (x, y) =
xa −y a

if a 6= 0, b = 0, x 6= y,

a(log x−log y)



√


xy
if a = b = 0,





x,
if x = y.
These definitions create a continuous, moreover, infinitely many times differentiable function
(a, b, x, y) 7→ Ma,b (x, y)
on the domain R2 × R2+ , where Ma,b (x, y) can stand for either Ga,b (x, y) or Sa,b (x, y).
Nevertheless the cases in the definitions seem quite different, we will see that they all can
be derived from the case of equal parameters, which – in a sense – plays a central role in our
treatment. The following lemma is true:
Lemma 3.1. Let the positive numbers x and y be fixed. Then for any real numbers a, b (a 6= b)
the following formula holds:
Z a
1
log Mt,t (x, y)dt.
(3.1)
log Ma,b (x, y) =
a−b b
Proof. For Gini means, we have
Z a
Z a t
1
1
x ln x + y t ln y
ln Gt,t (x, y)dt =
dt
a−b b
a−b b
xt + y t
ia
1 h
=
ln xt + y t
a−b
b
a
a
1
x +y
=
ln b
= ln Ga,b (x, y).
a − b x + yb
In the Stolarsky case we will assume that x > y and a > b. If 0 < b < a or b < a < 0 then
Z a
Z a
1
1
1 xt log x − y t log y
ln St,t (x, y)dt =
− +
dt
a−b b
a−b b
t
xt − y t
t
a
1
x − yt
=
ln
a−b
t
b
1
=
ln
a−b
J. Inequal. Pure and Appl. Math., 5(2) Art. 42, 2004
xa −y a
a
xb −y b
b
= ln Sa,b (x, y).
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A N E XTENSION OF THE H ERMITE -H ADAMARD I NEQUALITY FOR G INI AND S TOLARSKY M EANS
5
If 0 = b < a or b < a = 0 then we can apply the continuity of the integral as the function of its
limits. For example,
Z
Z a
1 a
1
1 xt log x − y t log y
dt
ln St,t (x, y)dt = lim
− +
b→0+
a 0
a−b b
t
xt − y t
t
a
1
x − yt
= lim log
a b→0+
t
b
a
a
1
x −y
xb − y b
=
log
− lim log
b→0+
a
a
b
a
a
1
x −y
=
log
− log(log x − log y)
a
a
= log Sa,0 (x, y).
Finally, in the case b < 0 < a
Z 0
Z a
Z a
1
1
ln St,t (x, y)dt =
log St,t (x, y)dt +
ln St,t (x, y)dt
a−b b
a−b
b
0
1
1
xa − y a
=
a
log
− log(log x − log y)
a−b
a
a
1
xb − y b
−b
log
− log(log x − log y)
b
b
= log Sa,b (x, y).
In the sequel, the following results will prove to be useful.
Lemma 3.2. For any positive x 6= 1,
x(x + 1)
<
2
(3.2)
x−1
log x
3
.
Proof. By Karamata’s classical inequality (see [8, p. 272]), we have that
x + x1/3
x−1
<
.
1/3
1+x
log x
(3.3)
Thus, it suffices to show that
x(x + 1)
<
2
(3.4)
x + x1/3
1 + x1/3
3
.
Dividing both sides by x, then multiplying them by 2(1 + x1/3 )3 , finally, collecting the terms
on the right side, one can easily check that (3.4) becomes
0 < (x2/3 + x1/3 + 1)(x1/3 − 1)4 ,
which is obviously true for all positive x 6= 1.
The inequality stated in the above lemma can be translated to an inequality concerning the
geometric, arithmetic and logarithmic means.
Corollary 3.3. For all x, y > 0,
(3.5)
2
3
(x, y).
S0,0
(x, y) · S2,1 (x, y) ≤ S1,0
J. Inequal. Pure and Appl. Math., 5(2) Art. 42, 2004
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6
P. C ZINDER AND Z S . PÁLES
Proof. If x = y, then (3.5) is obvious. If x 6= 1 and y = 1, then (3.5) is literally the same
as (3.2), hence (3.5) holds in this case, too. Now replacing x by x/y in (3.2), and using the
homogeneity of the Stolarsky means, we get that (3.5) is valid for all positive x 6= y.
Remark 3.4. Arguing in the same way as in the proof of Corollary 3.4, one can deduce that the
inequalities (3.3) and (3.4) are equivalent to
2
3
S0,0
(x, y) · G 2 , 1 (x, y) ≤ S1,0
(x, y)
3 3
and
S2,1 (x, y) = G0,1 (x, y) ≤ G 2 , 1 (x, y)
3 3
respectively. The latter inequality can also be derived from the comparison theorem of two
variable Gini means (cf. [12], [13], [3]).
Our aim is to apply the results in Theorem 2.2 for Gini and Stolarsky means. For this purpose
we will show that, for fixed positive x, y, the function
(3.6)
µx,y : R → R, t 7→ log Mt,t (x, y)
satisfies the assumptions of Theorem 2.2.
Lemma 3.5. Let x, y be arbitrary positive numbers. Then the function µx,y defined in (3.6) has
the following properties:
(i) µx,y (t) + µx,y (−t) = 2µx,y (0) (t ∈ R),
(ii) µx,y is convex over R− and concave over R+ .
Proof.
(i) For Gini means:
xt log x + y t log y x−t log x + y −t log y
+
xt + y t
x−t + y −t
xt log x + y t log y y t log x + xt log y
+
=
xt + y t
y t + xt
xt log(xy) + y t log(xy)
=
xt + y t
= log(xy) = 2µx,y (0),
µx,y (t) + µx,y (−t) =
while for Stolarsky means – assuming that t 6= 0 –
1 xt log x − y t log y 1 x−t log x − y −t log y
µx,y (t) + µx,y (−t) = − +
+ +
t
xt − y t
t
x−t − y −t
xt log x − y t log y y t log x − xt log y
=
+
xt − y t
y t − xt
xt log(xy) − y t log(xy)
=
xt − y t
= log(xy) = 2µx,y (0).
(ii) If x = y, then µx,y (t) = x for all t ∈ R, hence µx,y is convex-concave everywhere.
Therefore, we may assume that x 6= y.
In the case of Gini means,
xt y t (log xt − log y t )3 (xt − y t )
3 00
t µx,y (t) = −
.
(xt + y t )3
Since the sign of xt − y t is the same as that of log xt − log y t , therefore, t3 µ00x,y (t) ≥ 0
for all t ∈ R. Thus, µx,y is convex over R− and concave over R+ .
J. Inequal. Pure and Appl. Math., 5(2) Art. 42, 2004
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A N E XTENSION OF THE H ERMITE -H ADAMARD I NEQUALITY FOR G INI AND S TOLARSKY M EANS
7
In the setting of Stolarsky means, we have that
xt y t (log xt − log y t )3 (xt + y t )
t3 µ00x,y (t) = −2 +
(xt − y t )3
2
S0,0
(xt , y t )S2,1 (xt , y t )
= −2 1 −
.
3
S1,0
(xt , y t )
In view of Corollary 3.4, it follows that t3 µ00x,y (t) ≥ 0 for all t ∈ R. Therefore, µx,y is
convex over R− and concave over R+ in this case, too.
As a consequence of Lemma 3.5 and Theoremp
2.2, we can provide a lower and an upper
estimate for Ma,b in terms of the means M a+b and Ma,a · Mb,b .
2
≥
Theorem 3.6. Let a, b be real numbers so that a + b (≤)
0. Then
q
≥
≥
G a+b , a+b (x, y)
Ga,b (x, y)
Ga,a (x, y)Gb,b (x, y)
2
2
(≤)
(≤)
and
q
≥
≥
S a+b , a+b (x, y)
Sa,a (x, y)Sb,b (x, y)
Sa,b (x, y)
2
2
(≤)
(≤)
hold for any positive numbers x, y.
Proof. Let x, y be fixed positive numbers. By Lemma 3.5, the function µx,y is symmetric with
respect to m = 0 and is convex (concave) on R− (on R+ ). Therefore, Theorem 2.2 can be
applied to f := µx,y . Then, by (2.2),
Z a
a+b ≥ 1
≥ µx,y (a) + µx,y (b)
µx,y
µx,y (t)dt
2
(≤) a − b b
(≤)
2
≥
if a+b
0. Thus, by the definition of µx,y and in view of Lemma 3.1, the following inequality
2 (≤)
holds:
≥ log Ma,a (x, y) + log Mb,b (x, y)
≥
log M a+b , a+b (x, y)
log Ma,b (x, y)
2
2
(≤)
(≤)
2
≥
if a + b (≤)
0. Applying the exponential function to this inequality, we get that
q
≥
≥
M a+b , a+b (x, y)
Ma,b (x, y)
Ma,a (x, y)Mb,b (x, y)
2
2
(≤)
(≤)
≥
if a + b (≤)
0. Hence the stated inequalities follow in the Gini and Stolarsky means setting,
respectively.
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J. Inequal. Pure and Appl. Math., 5(2) Art. 42, 2004
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