Key words and phrases: Hilbert’s inequality, Hölder’s inequality, Jensen’s inequality, Power mean inequality

http://jipam.vu.edu.au/

Volume 7, Issue 4, Article 154, 2006

ON SEVERAL NEW INEQUALITIES CLOSE TO HILBERT-PACHPATTE’S INEQUALITY

BING HE AND YONGJIN LI DEPARTMENT OFMATHEMATICS

SUNYAT-SENUNIVERSITY

GUANGZHOU, 510275 P. R. CHINA

hzs314@163.com stslyj@mail.sysu.edu.cn

Received 27 April, 2006; accepted 09 October, 2006 Communicated by B. Yang

ABSTRACT. In this paper we establish several new inequalities similar to Hilbert-Pachpatte’s inequality. Moreover, some new generalizations of Hilbert-Pachpatte’s inequality are presented.

Key words and phrases: Hilbert’s inequality, Hölder’s inequality, Jensen’s inequality, Power mean inequality.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

The well known Hardy-Hilbert’s inequality is (see [1]):

Theorem 1.1. Ifp >1, p⁰ =p/(p−1)andP∞

m=1a^p_m ≤A,P∞

n=1b^p_n⁰ ≤B, then (1.1)

∞

X

n=1

∞

X

m=1

a_mb_n

m+n < π sin

π p

(A)^1p(B)^p10, unless the sequence{a_m}or{b_n}is null.

The integral analogue can be stated as follows:

Theorem 1.2. Ifp >1, p⁰ =p/(p−1)andR∞

0 f^p(x)dx≤F,R∞

0 g^p0(y)dy≤G, then (1.2)

Z ∞

0

Z ∞

0

f(x)g(y)

x+y dxdy < π sin

π p

F^1pG^p10

unlessf ≡0org ≡0.

The following two theorems were studied by Pachpatte (see [2])

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

The work was partially supported by the Foundation of Sun Yat-sen University Advanced Research Centre.

124-06

Theorem 1.3. Let p ≥ 1, q ≥ 1 andf(σ) ≥ 0, g(τ) ≥ 0 forσ ∈ (0, x), τ ∈ (0, y), where x, y are positive real numbers and defineF(s) =Rs

0 f(σ)dσ, G(t) =Rt

0 g(τ)dτ, fors∈(0, x), t∈(0, y). Then

Z x

0

Z y

0

F^p(s)G^q(t)

s+t dsdt≤D(p, q, x, y) Z x

0

(x−s)(F^p−1(s)f(s))²ds ¹₂

× Z y

0

(y−t)(G^q−1(t)g(t))²dt ¹₂

,

unlessf ≡0org ≡0, where

D(p, q, x, y) = 1 2pq√

xy.

Theorem 1.4. Let f, g, F, G be as in the above theorem, let p(σ) and q(τ) be two positive functions defined for σ ∈ (0, x), τ ∈ (0, y)and defineP(s) = Rs

0 p(σ)dσ, Q(t) = Rt

0 q(τ)dτ for s ∈ (0, x), t ∈ (0, y), where x, y are positive real numbers. Letφ andψ be real-valued, nonnegative, convex, and sub-multiplicative functions defined onR⁺= [0,∞). Then

(1.3) Z x

0

Z y

0

φ(F(s))ψ(G(t))

s+t dsdt≤L(x, y) (Z x

0

(x−s)

p(s)φ

f(s) p(s)

2

ds )¹₂

× (Z y

0

(y−t)

q(t)ψ g(t)

q(t) 2

dt )¹₂

where

L(x, y) = 1 2

Z x

0

φ(P(s)) P(s)

2

ds

!¹₂ Z y

0

ψ(Q(t)) Q(t)

2

dt

!¹₂ .

The inequalities in these theorems were studied extensively and numerous variants, general- izations, and extensions have appeared in the literature (see [3] – [9]).

The main purpose of the present article is to establish some new inequalities similar to the Hilbert-Pachpatte inequalities.

2. MAINRESULTS

Lemma 2.1. Supposep > 1,¹_p +¹_q = 1, r >1,¹_r + _w¹ = 1, λ > 0, define the weight functions ωe1(w, p, x),eω2(w, p, x),ωe3(w, p, x)as

(2.1) ωe1(w, p, x) :=

Z ∞

0

1

x^λ+y^λ · x^{(p−1)(1−λr)}

y^1−wλ dy, x∈(0,∞),

(2.2) ωe₂(w, p, x) :=

Z ∞

0

1

max{x^λ, y^λ} · x^{(p−1)(1−λr)}

y^1−λw dy, x∈(0,∞),

(2.3) ωe₃(w, p, x) :=

Z ∞

0

lnx/y

x^λ−y^λ · x^{(p−1)(1−λr)}

y^1−wλ dy, x∈(0,∞).

Then

ωe₁(w, p, x) = π

λsin ^π_rx^{p(1−λr)−1},

ωe₂(w, p, x) = rw

λ x^{p(1−λr)−1},

ωe₃(w, p, x) =

"

π λsin ^π_r

#2

x^{p(1−λr)−1}.

Proof. For fixedx, letu=y^λ/x^λ, then (2.1) turns into

ωe₁(w, p, x) = 1

λx^{p(1−λr)−1} Z ∞

0

1

1 +uu^−1+w1du

= 1

λx^{p(1−λr)−1}B 1

w,1 r

= π

λsin

π p

x^{p(1−λr)−1}

and similarly, we can prove the others.

Theorem 2.2. Letm ≥ 1, n ≥ 1, pi > 1, _p¹

i + _q¹

i = 1 for i = 0,1,2,3,4, p0 = p, p3 = k, p₄ = r;q₀ = q, q₃ = l, q₄ = w and f(σ) ≥ 0, g(τ) ≥ 0. Define F(s) = Rs

0 f(σ)dσ and G(t) =Rt

0 g(τ)dτ such that 0<

Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds < ∞, 0<

Z ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt <∞ forσ, τ, s, t∈(0,∞). Then

(2.4) Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p1 +kt^l/p2)(s^λ +t^λ)dsdt

≤E₁(m, n, k, r, λ) Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_pZ ∞

0

t^{q(1−λw)−1}G^q_g(t)dt ¹_q

unlessf ≡0org ≡0, whereE₁(m, n, k, r, λ) = _λkl^πmn_sin(π/r),

Ff(s) = Z s

0

(F^m−1(σ)f(σ))^q1dσ _q¹

1

and

G_g(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ _q¹

2

.

Proof. From the hypotheses, we can easily observe that

(2.5) F^m(s) =m

Z s

0

F^m−1(σ)f(σ)dσ, s∈(0,∞),

(2.6) Gⁿ(t) = n

Z t

0

Gⁿ⁻¹(τ)g(τ)dτ, t∈(0,∞).

From (2.5) and (2.6), applying Hölder’s inequality, equality (2.1) and Young’s inequality:

ab≤ a^p p + b^q

q,

wherea≥0, b≥0, p >1, ¹_p +¹_q = 1,we obtain F^m(s)Gⁿ(t)

=mn Z s

0

F^m−1(σ)f(σ)dσ

Z t

0

Gⁿ⁻¹(τ)g(τ)dτ

≤mns^1/p1t^1/p2 Z s

0

(F^m−1(σ)f(σ))^q1dσ) _q¹

1 Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ) _q¹

2

≤mn

s^k/p1

k +t^l/p2 l

Z s

0

(F^m−1(σ)f(σ))^q1dσ) _q¹

1 Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ) _q¹

2

.

Note that F_f(s) =

Z s

0

(F^m−1(σ)f(σ))^q1dσ _q¹

1 , G_g(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ _q¹

2

.

Hence Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p1 +kt^l/p2)(s^λ+t^λ)dsdt

≤ mn kl

Z ∞

0

Z ∞

0

Rs

0(F^m−1(σ)f(σ))^q1dσ)

1 q1

nRt

0(Gⁿ⁻¹(τ)g(τ))^q2dτ)o_q¹

2

s^λ+t^λ dsdt

= mn kl

Z ∞

0

Z ∞

0

1 s^λ+t^λ

"

F_f(s)s^(1−λr)/q t^(1−wλ)/p

# "

G_g(t)t^(1−wλ)/p s^(1−λr)/q

# dsdt

≤ mn kl

(Z ∞

0

Z ∞

0

F_f^p(s)

s^λ +t^λ · s^{(p−1)(1−λr)} t^(1−wλ) dsdt

)¹_p( Z ∞

0

Z ∞

0

G^q_g(t)

s^λ +t^λ ·t^{(q−1)(1−wλ)} s^(1−λr) dsdt

)¹_q

= mn kl

Z ∞

0

ωe1(w, p, s)F_f^p(s)ds

¹_pZ ∞

0

ωe1(r, q, t)G^q_g(t)dt ¹_q

= πmn

λklsin(π/r) Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_pZ ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt ¹_q

=E₁(m, n, k, r, λ) Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_pZ ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt ¹_q

.

This completes the proof.

Theorem 2.3. Letm ≥ 1, n ≥ 1, p_i > 1, _p¹

i + _q¹

i = 1 for i = 0,1,2,3,4, p₀ = p, p₃ = k, p₄ = r;q₀ = q, q₃ = l, q₄ = w and f(σ) ≥ 0, g(τ) ≥ 0. Define F(s) = Rs

0 f(σ)dσ and G(t) = Rt

0 g(τ)dτ such that0 < R∞

0 s^{p(1−λr)−1}F_f^p(s)ds < ∞,0 < R∞

0 t^{q(1−wλ)−1}G^q_g(t)dt < ∞ forσ, τ, s, t∈(0,∞). Then

Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p1 +kt^l/p2) max{s^λ, t^λ}dsdt

≤E₂(m, n, k, r, λ) Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_p Z ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt ¹_q

,

unlessf ≡0org ≡0, whereE₂(m, n, k, r, λ) = ^mnrw_λkl and

F_f(s) = Z s

0

(F^m−1(σ)f(σ))^q1dσ _q¹

1

,

Gg(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ _q¹

2

.

Proof. By (2.5) and (2.6), using Hölder’s inequality, equality (2.2) and Young’s inequality:

ab≤ ^a_p^p + ^b_q^q,wherea≥0, b≥0, p >1, ¹_p + ¹_q = 1.We obtain F^m(s)Gⁿ(t)

=mn Z s

0

F^m−1(σ)f(σ)dσ

Z t

0

Gⁿ⁻¹(τ)g(τ)dτ

≤mns^1/p1t^1/p2 Z s

0

(F^m−1(σ)f(σ))^q1dσ) _q¹

1 Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ) _q¹

2

≤mn

s^k/p1

k +t^l/p2 l

Z s

0

(F^m−1(σ)f(σ))^q1dσ) _q¹

1 Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ) _q¹

2

.

Note

F_f(s) = Z s

0

(F^m−1(σ)f(σ))^q1dσ _q¹

1 ,

G_g(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ _q¹₂

.

Hence Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p1 +kt^l/p2) max{s^λ, t^λ}dsdt

≤ mn kl

Z ∞

0

Z ∞

0

Rs

0(F^m−1(σ)f(σ))^q1dσ)

1 q1

nRt

0(Gⁿ⁻¹(τ)g(τ))^q2dτ)o_q¹

2

max{s^λ, t^λ} dsdt

= mn kl

Z ∞

0

Z ∞

0

1 max{s^λ, t^λ}

"

F_f(s)s^(1−λr)/q t^(1−wλ)/p

# "

G_g(t)t^(1−wλ)/p s^(1−λr)/q

# dsdt

≤ mn kl

(Z ∞

0

Z ∞

0

F_f^p(s) max{s^λ, t^λ}

s^{(p−1)(1−λr)} t^(1−wλ) dsdt

)¹_p( Z ∞

0

Z ∞

0

G^q_g(t) max{s^λ, t^λ}

t^{(q−1)(1−wλ)} s^(1−λr) dsdt

)¹_q

= mn kl

Z ∞

0 ωe₂(w, p, s)F_f^p(s)ds

¹_pZ ∞

0 ωe₂(r, q, t)G^q_g(t)dt ¹_q

= mnrw λkl

Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_pZ ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt ¹_q

=E₂(m, n, k, r, λ) Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_p Z ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt ¹_q

.

This completes the proof.

Theorem 2.4. Letm ≥ 1, n ≥ 1, p_i > 1, _p¹

i + _q¹

i = 1 for i = 0,1,2,3,4, p₀ = p, p₃ = k, p₄ = r; q₀ = q, q₃ = l, q₄ = w and f(σ) ≥ 0, g(τ) ≥ 0. Define F(s) = Rs

0 f(σ)dσ and G(t) =Rt

0 g(τ)dτ such that

0<

Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds <∞, 0<

Z ∞

0

t^{q(1−λw)−1}G^q_g(t)dt <∞

forσ, τ, s, t∈(0,∞). Then Z ∞

0

Z ∞

0

ln(s/t)

(s^λ−t^λ)(ls^k/p1 +kt^l/p2)F^m(s)Gⁿ(t)dsdt≤E₃(m, n, k, r, λ)

× Z ∞

0

s^{p(1−λr)−1}F_f^p(s)ds

¹_p Z ∞

0

t^{q(1−wλ)−1}G^q_g(t)dt ¹_q

,

unlessf ≡0org ≡0, whereE3(m, n, k, r, λ) = _λ2kl(sin(π/r))^{π2mn 2},

Ff(s) = Z s

0

(F^m−1(σ)f(σ))^q1dσ _q¹

1

,

G_g(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q2dτ _q¹

2

.

Proof. From (2.5) and (2.6), by Hölder’s inequality, equality (2.3) and Young’s inequality:ab≤

a^p

p +^b_q^q, wherea ≥0, b≥0, p >1,¹_p + ¹_q = 1, and using a similar method of proof to that of

Theorem 2.2, the result can be clearly seen.

Theorem 2.5. Let m_i ≥ 1, p_i > 1, Pn i=1

1

pi = 1, _p¹

i + _q¹

i = 1, and f_i(σ_i) ≥ 0 for σ_i ∈ (0, xi), where xi are positive real numbers and defineFi(ti) = Rti

0 fi(σi)dσi, forti ∈ (0, xi), i= 1,2, . . . , n. Then

(2.7) Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i^mi(t_i) Pn

i=1 ti

pi

dt₁dt₂· · ·dt_n

≤

n

Y

i=1

D(m_i, x_i, p_i) Z xi

0

(x_i−t_i)

Fe_i(t_i)qi

dt_i ¹

qi ,

unlessf ≡0org ≡0, whereD(m_i, x_i, p_i) =m_ix

1 pi

i ,Fe_i(t_i) =F^mi−1(t_i)f_i(t_i).

Proof. From the hypotheses, we know that

(2.8) F_i^mi(ti) = mi

Z ti

0

F_i^(mi−1)(σi)fi(σi)dσi, ti ∈(0, xi).

Then (2.9)

n

Y

i=1

F_i^mi(t_i) =

n

Y

i=1

m_i Z ti

0

F_i^mi−1(σ_i)f_i(σ_i)dσ_i.

Using Hölder’s inequality, we have Z ti

0

F_i^mi−1(σ_i)dσ_i ≤t

1 pi

i

Z ti

0

(F_i^mi−1(σ_i)f_i(σ_i))^qidσ_{i qi}¹ (2.10)

,t

1 pi

i

Z ti

0

Fe_i(σ_i)qi

dσ_{i qi}¹

.

By (2.9) and (2.10) it follows that

n

Y

i=1

F_i^mi(t_i)≤

n

Y

i=1

m_it

1 pi

i

Z ti

0

Fe_i(σ_i)qi

dσ_{i qi}¹

.

Applying Young’s inequality

n

Y

k=1

|a_k| ≤

n

X

k=1

1

p_k|a_k|^pk,

where1< p_k<∞,Pn k=1

1

pk = 1,we observe that Qn

i=1F_i^mi(t_i) Pn

i=1 ti

pi

≤

n

Y

i=1

m_i Z ti

0

Fe_i(σ_i)qi

dσ_{i qi}¹

.

Integrating overt_i from0tox_i whereiruns from1ton, and using the Hölder’s inequality, we get

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i^mi(t_i) Pn

i=1 ti

pi

dt₁dt₂· · ·dt_n

≤

n

Y

i=1

m_i

"

Z xi

0

Z ti

0

Fe_i(σ_i)qi

dσ_{i qi}¹

dt_i

#

≤

n

Y

i=1

m_ix

1 pi

i

Z xi

0

Z ti

0

Fe_i(σ_i)qi

dσ_i

dt_{i qi}¹

=

n

Y

i=1

D(mi, xi, pi) Z xi

0

(xi−ti)

Fei(ti) qi

dti

_qi¹

This completes the proof.

Remark 2.6. In the special case whenp₁ =p₂ =· · ·=p_n=n, the inequality (2.7) reduces to the following inequality,

(2.11)

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i^mi(t_i) Pn

i=1t_i dt₁dt₂· · ·dt_n

≤ 1 n

n

Y

i=1

D(m_i, x_i) Z xi

0

(x_i−t_i)

Fe_i(t_i)_n−1ⁿ dt_i

ⁿ⁻¹_n

where D(m_i, x_i) = m_ix

1 n

i . Moreover, (i) (2.11) reduces to Theorem 1.3 which belongs to Pachpatte forn = 2; (ii) whenm₁ =m₂ =· · ·=m_n = 1, (2.11) turns into

(2.12)

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i(t_i) Pn

i=1t_i dt₁dt₂· · ·dt_n

≤ 1 n

n

Y

i=1

D(m_i, x_i) Z xi

0

(x_i−t_i)(f_i(t_i))ⁿ⁻¹ⁿ dt_i ⁿ⁻¹_n

Theorem 2.7. Letf_i, F_i be as in the above theorem,p > 1, _p¹

i + _q¹

i = 1, letp_i(σ_i)be positive function defined for σ_i ∈ (0, x_i), and define P_i(t_i) = Rti

0 p_i(σ_i)dσ_i for t_i ∈ (0, x_i),wherex_i are positive real numbers. Let φ_i be real-valued, nonnegative, convex, and sub-multiplicative function defined onR+ = [0,∞),i= 1,2, . . . , n.Then

(2.13)

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1φ_i(F_i(t_i)) Pn

i=1 ti

pi

dt₁dt₂· · ·dt_n

≤

n

Y

i=1

L(xi, pi) Z xi

0

(xi−ti)

pi(ti)φi

f_i(t_i) p_i(t_i)

qi

dti

_qi¹ ,

where

L(x_i, p_i) = Z xi

0

φ_i(P_i(t_i)) P_i(t_i)

pi

dt_{i pi}¹

.

Proof. Applying Jensen’s inequality and Hölder’s inequality, it is clear to observe that

φ_i(F_i(t_i)) = φ_i Pi(ti)Rti

0 pi(σi)^f_p^i(σi)

i(σi)dσi

Rti

0 p_i(σ_i)dσ_i

!

≤ φ_i(P_i(t_i)) P_i(t_i)

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) p_i(σ_i)

dσ_i

≤ φi(Pi(ti)) P_i(t_i) t

1 pi

i

Z ti

0

p_i(σ_i)φ_i

fi(σi) p_i(σ_i)

qi

dσ_{i qi}¹

.

Using Young’s inequality, we obtain that

Qn

i=1φ_i(F_i(t_i)) Pn

i=1 ti

pi

≤

n

Y

i=1

φ_i(P_i(t_i)) Pi(ti)

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) pi(σi)

qi

dσ_{i qi}¹

.

Integrating both sides of the above inequality overt_i from 0tox_i withi running fromi ton, and applying the Hölder inequality, we get

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1φ_i(F_i(t_i)) Pn

i=1 ti

pi

dt₁dt₂· · ·dt_n

≤

n

Y

i=1

"

Z xi

0

φ_i(P_i(t_i)) P_i(t_i)

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) p_i(σ_i)

qi

dσ_{i qi}¹

dt_i

#

≤

n

Y

i=1

Z xi

0

φ_i(P_i(t_i)) P_i(t_i)

pi

dt_i

_pi¹ Z xi

0

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) p_i(σ_i)

qi

dσ_i

dt_{i qi}¹

=

n

Y

i=1

Z xi

0

φi(Pi(ti)) P_i(t_i)

pi

dt_i

_pi¹ Z xi

0

(x_i−t_i)

p_i(t_i)φ_i

fi(ti) p_i(t_i)

qi

dt_{i qi}¹

=

n

Y

i=1

L(x_i, p_i) Z xi

0

(x_i−t_i)

p_i(t_i)φ_i

f_i(t_i) p_i(t_i)

qi

dt_{i qi}¹

This completes the theorem.

REFERENCES

[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ. Press, Cam- bridge, 1952.

[2] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequality, J. Math. Anal. Appl., 226(1) (1998), 166–179.

[3] G.S. DAVIES AND G.M. PETERSEN, On an inequality of Hardy’s, (II), Quart. J. Math. Oxford Ser., 15(2) (1964), 35–40.

[4] D.S. MITRINOVI ´C, Analytic inequalities. In cooperation with P. M. Vasi´c. Die Grundlehren der mathematischen Wisenschaften, Band 1965, Springer-Verlag, New York-Berlin, 1970.

[5] D.S. MITRINOVI ´CAND J.E. PE ˇCARI ´C, On inequalities of Hilbert and Widder, Proc. Edinburgh Math. Soc. (2), 34(3) (1991), 411–414.

[6] Z. LÜ, Some new inequalities similar to Hilbert-Pachpatte type inequalities, J. Inequal. Pure Appl.

Math., 4(2) (2003), Art. 33. [ONLINE:http://jipam.vu.edu.au/article.php?sid=

271].

[7] Y.H. KIM, Some new inverse-type Hilbert-Pachpatte integral inequalities, Acta Math. Sin. (Engl.

Ser.), 20(1) (2004), 57–62.

[8] B. YANG, On an extension of Hilbert’s integral inequality with some parameters, Aust. J. Math.

Anal. Appl., 1(1) (2004), Art. 11. [ONLINE:http://ajmaa.org].

[9] E.F. BECKENBACH AND R. BELLMAN, Inequalities. Second revised printing. Ergebnisse der Mathematik und ihrer Grenzgebiete. Neue Folge, Band 30, Springer-Verlag, New York, Inc. 1965.