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Research article

A new reverse Mulholland-type inequality with multi-parameters

  • Received: 01 February 2021 Accepted: 22 June 2021 Published: 05 July 2021
  • MSC : 26D15, 26D10, 26A42

  • In this paper, we present a new reverse Mulholland-type inequality with multi-parameters and deal with its equivalent forms. Based on the obtained inequalities, the equivalent statements of the best possible constant factor related to several parameters are discussed. As an application, some interesting inequalities for double series are derived from the special cases of our main results.

    Citation: Bicheng Yang, Shanhe Wu, Aizhen Wang. A new reverse Mulholland-type inequality with multi-parameters[J]. AIMS Mathematics, 2021, 6(9): 9939-9954. doi: 10.3934/math.2021578

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  • In this paper, we present a new reverse Mulholland-type inequality with multi-parameters and deal with its equivalent forms. Based on the obtained inequalities, the equivalent statements of the best possible constant factor related to several parameters are discussed. As an application, some interesting inequalities for double series are derived from the special cases of our main results.



    In 1925, Hardy [1] gave a generalization of the Hilbert's inequality by introducing one pair of conjugated exponents (p,q) which satisfies 1p+1q=1 and p>1, as follows:

    If am0,bn0, 0<m=1apm< and 0<n=1bqn<, then

    m=1n=1ambnm+n<πsin(π/p)(m=1apm)1/p  (n=1bqn)1/q, (1)

    where the constant factor πsin(π/p) is the best possible. The inequality (1) is called the Hardy-Hilbert inequality. In particular, when p=q=2, the Hardy-Hilbert inequality reduces to the Hilbert's inequality (see [2]). As is known to us, the Hardy-Hilbert inequality plays an important role in analysis number theory, real analysis and divergent series theory (see [3]).

    For the continuous case, the integral version of Hardy-Hilbert inequality can be stated as follows (see [3], Theorem 316):

    If f(x)0,g(y)0,0<0fp(x)dx< and 0<0gq(y)dy<, then

    00f(x)g(y)x+ydxdy<πsin(π/p)(0fp(x)dx)1/p  (0gq(y)dy)1/q, (2)

    where the constant factor πsin(π/p) is the best possible.

    The Hardy-Hilbert inequalities (1) and (2) has been studied extensively, numerous variants, generalizations, and extensions can be found in the literatures (see [4,5,6,7,8]).

    Motivated by the Hardy-Hilbert inequality, in 1929, Mulholland [9] proposed a similar version of inequality (1), which contains the same best possible constant factor πsin(π/p) as in (1), i.e.,

    m=2n=2ambnmnlnmn<πsin(π/p)(m=21mapm)1/p  (n=21nbqn)1/q. (3)

    Obviously, the Mulholland's inequality (3) can be rewritten in an equivalent form as:

    m=2n=2ambnlnmn<πsin(π/p)(m=21m1papm)1/p  (n=21n1qbqn)1/q. (4)

    In recent years, the Mulholland inequality has been generalized by various methods of constructing parameters, see [10,11,12,13,14,15] and the references cited therein.

    Let us recall some results which are connected with the current investigation. Hong and Wen [16] and Hong [17] studied the Hilbert type series inequalities and the Hilbert type integral inequalities with homogeneous kernel, respectively. They established the necessary and sufficient condition for which the inequalities hold under the best constant factor. Subsequently, by using the quasi-homogeneous kernels instead of the homogeneous kernel in the Hilbert type integral inequalities, Hong, He and Yang [18] established the necessary and sufficient condition for which the inequalities hold under the best constant factor. Recently, with the help of the Euler-Maclaurin summation formula, Yang, Wu and Liao [19] and gave the extension of Hardy-Hilbert's inequality and its equivalent forms. In [20], Yang, Wu and Chen investigated the generalization of Hardy-Littlewood-Polya's inequality and its equivalent forms.

    In this paper, following the ideas of [16,17,18,19,20], we will study the Mulholland-type inequalities. The present research objects are structurally different from Hilbert type inequalities, and this will involve new techniques in dealing with the inequalities. Specifically, we will establish a reverse Mulholland-type inequality with multi-parameters. And then, we discuss the equivalent statements of the best possible constant factor associated with several parameters. Finally, we illustrate that some new inequalities of Mulholland-type can be derived from the equivalent expressions of the reverse Mulholland-type inequality.

    In this section, we present some preliminary results which are essential for establishing our main results in subsequent sections. We begin with introducing the notations k(η)λ(x,y) and kλ,η(γ) with their associated formulas.

    (i) In view of the following expression (see [21])

    cotx=1x+k=1(1xπk+1x+πk)(x(0,π)),

    for b(0,1), we have

    Ab:=0ub1  1udu=10ub1  1udu+1ub1  1udu=10ub1  1udu10vb1vdv=10ub1  ub1udu=10k=0(uk+b1  ukb)du=k=010(uk+b1  ukb)du=k=0(1k+b1k+1b)=π[1πb+k=1(1πbπk+1πb+πk)]=πcotπbR:=(,+).

    Moreover, it is easy to observe that Ab>0 for b(0,12); Ab<0 for b(12,1); Ab=0 for b=12.

    (ii) For λ,η>0, we set the homogeneous function of order λ as follows:

    k(η)λ(x,y):=xηyηxλ+ηyλ+η(x,y>0),

    which satisfies k(η)λ(ux,uy)=uλk(η)λ(x,y)(u,x,y>0), and k(η)λ(v,v):=η(λ+η)vλ(v>0).

    It follows that k(η)λ(x,y) is a positive and continuous function with respect to x,y>0. For xy, we obtain

    xk(η)λ(x,y)=xη1  (xλ+ηyλ+η)2φ(x,y),

    where φ(x,y):=λxλ+η(λ+η)yηxλ+ηyλ+η(x,y>0).

    We find that for 0<x<y,

    xφ(x,y)=λ(λ+η)xλ1  (xηyη)<0;

    for x>y, xφ(x,y)>0. It follows that φ(x,y) is strictly decreasing (resp. increasing) with respect to x<y (resp. x>y). Since φ(y,y)=minx>0φ(x,y)=0(y>0), then φ(x,y)>0(xy), namely, xk(η)λ(x,y)<0 (xy). Note that k(η)λ(x,y) is continuous at x=y, we conform that k(η)λ(x,y) is strictly decreasing with respect to x>0. In the same way, we can show that k(η)λ(x,y) is also strictly decreasing with respect to y>0.

    (iii) For γ(0,λ), since k(η)λ(x,y)>0, by (i), we obtain (see [22])

    kλ,η(γ):=0k(η)λ(u,1)uγ1  du=0uη1  uλ+η1  uγ1  duv=uλ+η=1λ+η(0vγλ+η1  1vdv0vγ+ηλ+η1  1vdv)=πλ+η[cot(πγλ+η)cot(π(γ+η)λ+η)]=πλ+η[cot(πγλ+η)+cot(π(λγ)λ+η)]R+:=(0,).

    In what follows, we suppose that

    p<0(0<q<1),1p+1q=1,α,β,η>0,0<λ1,λ2<λ,λ11α, λ21β,k(η)λ(lnαm,lnβn)=lnαηmlnβηnlnα(λ+η)mlnβ(λ+η)n(m,nN{1}:={2,3,});

    for γ=λ1,λλ2((0,λ)),

    kλ,η(γ)=πλ+η[cot(πγλ+η)+cot(π(λγ)λ+η)]. (5)

    Also, we assume am,bn0(m,nN{1}) such that

    0<m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm<  and  0<n=2lnq[1β(λ2p+λλ1q)]1  nn1pbqn<.

    Definition 1. We define the following weight functions:

    ωλ,η(λ2,m):=lnα(λλ2)mn=2k(η)λ(lnαm,lnβn)lnβλ21  nn(mN{1}), (6)
    ϖλ,η(λ1,n):=lnβ(λλ1)nm=2k(η)λ(lnαm,lnβn)lnαλ11  mm(nN{1}). (7)

    Lemma 1. For βλ21, we have

    ωλ,η(λ2,m)<1βkλ,η(λλ2)R+(mN{1}); (8)

    For αλ11, we have

    0<1αkλ,η(λ1)(1θλ,η(λ1,n))<ϖλ,η(λ1,n)<1αkλ,η(λ1)(nN{1}), (9)

    where θλ,η(λ1,n)(>0) is indicated as

    θλ,η(λ1,n):=1kλ,η(λ1)lnα2lnβn0k(η)λ(u,1)uλ11  du=O(1lnβλ1n). (10)

    Proof. For fixed mN{1},βλ210,k(η)λ(lnαm,lnβt)lnβ21tt is a strictly decreasing function with respect to t>1. By the decreasing property of series, setting u=lnαmlnβt, we find that

    ωλ,η(λ2,m)<lnα(λλ2)m1k(η)λ(lnαm,lnβt)lnβλ21  ttdt=1β0k(η)λ(u,1)u(λλ2)1  du=1βkλ,η(λλ2).

    Hence, we get the inequality (8).

    For αλ110, it is evident that k(η)λ(lnαt,lnβn)lnαλ11  tt is a strictly decreasing function with respect to t>1. By the decreasing property of series, setting u=lnαtlnβn, we find that

    ϖλ,η(λ1,n)<lnβ(λλ1)n1k(η)λ(lnαt,lnβn)lnαλ11  ttdt=1α0k(η)λ(u,1)uλ11  du=1αkλ,η(λ1).

    By the decreasing property of series and (10), we obtain

    ϖλ,η(λ1,n)>lnβ(λλ1)n2k(η)λ(lnαt,lnβn)lnαλ11  ttdt=1αlnα2lnβnk(η)λ(u,1)uλ11  du=1αkλ,η(λ1)(1θλ,η(λ1,n))>0,
    0<θλ,η(λ1,n)1kλ,η(λ1)lnα2lnβn0uλ11  du=1λ1kλ,η(λ1)(lnα2lnβn)λ1.

    Hence, we deduce the inequality (9). The Lemma 1 is proved.

    Lemma 2. We have the following inequality:

    I:=n=2m=2k(η)λ(lnαm,lnβn)ambn
    >1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1){m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm}1p
    ×{n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1q  . (11)

    Proof. By the reverse Hӧlder inequality [23], we obtain

    I=n=2m=2k(η)λ(lnαm,lnβn)[ln(βλ21)pnn1/p  ln(1αλ1)/qmm1/qam]
    ×[ln(αλ11)/qmm1/qln(1βλ2)/pnn1/pbn]{m=2[lnαλ1mn=2k(η)λ(lnαm,lnβn)lnβλ21  nn]lnp(1αλ1)1  mm1papm}1p
    ×{n=2[lnβλ2nm=2k(η)λ(lnαm,lnβn)lnαλ11  mm]lnq(1βλ2)1  nn1qbqn}1q  
    ={m=2ωλ,η(λ2,m)lnp[1α(λλ2p+λ1q)]1  mm1papm}1p{n=2ϖλ,η(λ1,n)lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1q  .

    By (8) and (9), we get the inequality (11). This completes the proof of Lemma 2.

    Remark 1. By (11), for λ1+λ2=λ, we find

    0<m=2lnp(1αλ1)1  mm1p  apm<,0<n=2lnq(1βλ2)1  nn1pbqn<,

    and obtain the following inequality:

    n=2m=2k(η)λ(lnαm,lnβn)ambn
    >1β1/p  α1/qkλ,η(λ1)[m=2lnp(1αλ1)1  mm1p  apm]1p[n=2(1θλ,η(λ1,n))lnq(1βλ2)1  nn1qbqn]1q  . (12)

    In particular, for α=β=1, we have

    ˜θλ,η(λ1,n):=1kλ,η(λ1)ln2lnn0k(η)λ(u,1)uλ11  du=O(1lnλ1n),
    0<m=2lnp(1λ1)1  mm1p  apm<,0<n=2lnq(1λ2)1  nn1pbqn<,

    and the following inequality:

    n=2m=2lnηmlnηnlnλ+ηmlnλ+ηnambn
    >kλ,η(λ1)[m=2lnp(1λ1)1  mm1p  apm]1p[n=2(1˜θλ,η(λ1,n))lnq(1λ2)1  nn1qbqn]1q  . (13)

    Hence, inequality (12) is an extension of inequality (13).

    Lemma 3. For any ε>0, we have

    L:=n=2O(1nlnβ(λ1+ε)+1n)=O(1). (14)

    Proof. There exist constants m>0,M>0 such that

    0<mn=21nlnβ(λ1+ε)+1nLM[12lnβ(λ1+ε)+1  2+n=31nlnβ(λ1+ε)+1  n].

    By the decreasing property of series, it follows that

    0<LM[12lnβ(λ1+ε)+12+21ylnβ(λ1+ε)+1ydy]=M[12lnβ(λ1+ε)+12+1β(λ1+ε)lnβ(λ1+ε)2]M(12lnβλ1+12+1βλ1lnβλ12)<.

    Hence, inequality (14) follows. The proof of Lemma 3 is complete.

    Lemma 4. For δ0:=12mini=1,2{λi,λλi},0<ε<qδ0, we have

    kλ,η(λ1+εq)kλ,η(λ1)(ε0+). (15)

    Proof. By Levi Theorem (see [24]), it follows that

    10k(η)λ(u,1)uλ1+εq1  du10k(η)λ(u,1)uλ11  du(ε0+).

    For 0<ε<qδ0, we have 0<εq<δ0, and then λ1+δ0<λ,λλ1+δ0<λ,

    01k(η)λ(u,1)uλ1+εq1  du1k(η)λ(u,1)uλ1+δ01  dukλ,η(λ1+δ0)<.

    By using Lebesgue control convergence theorem [24], we have

    1k(η)λ(u,1)uλ1+εq1  du1k(η)λ(u,1)uλ11  du(ε0+).

    Then we find

    kλ,η(λ1+εq)=10k(η)λ(u,1)uλ1+εq1  du+1k(η)λ(u,1)uλ1+εq1  du10k(η)λ(u,1)uλ11  du+1k(η)λ(u,1)uλ11  du=kλ,η(λ1)(ε0+).

    The Lemma 4 is proved.

    Lemma 5. If λ1+λ2=λ, then the constant factor kλ,η(λ1) in (12) is the best possible.

    Proof. For 0<ε<qδ0, we set

    ˜am:=lnα(λ1εp)1  mm,˜bn:=lnβ(λ2εq)1  nn(m,nN{1}).

    If there exists a constant M1β1/p  α1/qkλ,η(λ1) such that the inequalities (12) is valid when replacing 1β1/p  α1/qkλ,η(λ1) by M, then, in particular, we have

    ˜I:=n=2m=2k(η)λ(lnαm,lnβn)˜am˜bn
    >M[m=2lnp(1αλ1)1  mm1p˜  apm]1p[n=2(1θλ,η(λ1,n))lnq(1βλ2)1  nn1q˜bqn]1q  .

    In view of (14), we obtain

    ˜I>M[m=2lnp(1αλ1)1  mm1plnpαλ1αεpmmp]1p
    ×[n=2(1θλ,η(λ1,n))lnq(1βλ2)1  nn1qlnqβλ2βεqnnq]1q  =M(lnαε1  22+m=3lnαε1  mm)1p(n=2lnβε1  nnn=2O(1lnβλ1n)lnβε1  nn)1q  >M(lnαε1  22+2lnαε1  xxdx)1p(2lnβε1  yydyn=2O(1nlnβ(λ1+ε)+1n))1q  =Mε(εlnαε1  22+1αlnαε2)1p(1βlnβε2εO(1))1q  .

    By (8), for ˆλ2=λ2εq((0,1β))(ˆλ1=λ1+εq<λ), we find

    ˜I=m=2[lnα(λ1+εq)mn=2k(η)λ(lnαm,lnβn)lnβ(λ2εq)1  nn]lnαε1  mm=m=2ωλ,η(ˆλ2,m)lnαε1  mm <1βkλ,η(ˆλ1)(ln1αε22+m=3ln1αεmm)
    1βkλ,η(ˆλ1)(ln1αε22+2ln1αεxxdx)=1εβαkλ,η(ˆλ1)(εαln1αε22+lnαε2).

    Then we have

    1βαkλ,η(λ1+εq)(εαln1αε22+lnαε2)
    ε˜I>M(εlnαε1  22+1αlnαε2)1p(1βlnε2εO(1))1q  .

    For ε0+, by (15), we get

    1βαkλ,η(λ1)M(1α)1p(1β)1q  ,

    namely, 1β1/p  α1/qkλ,η(λ1)M. Hence, M= 1β1/p  α1/qkλ,η(λ1) is the best possible constant factor of (12). The proof of Lemma 5 is completed.

    Remark 2. Setting ˜λ1:=λλ2p+λ1q=λ1+λλ1λ2p,˜λ2:=λλ1q+λ2p, we find

    ˜λ1+˜λ2=λλ2p+λ1q+λλ1q+λ2p=λp+λq=λ.

    For |λλ1λ2|<|p|δ0, it means that δ0<λλ1λ2p<δ0, 0<˜λ1<λ,0<˜λ2=λ˜λ1<λ, and then kλ,η(˜λ1)R+. We reduce (11) as follows:

    I>1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1)[m=2lnp(1α˜λ1)1  mm1p  apm]1p
    ×[n=2(1θλ,η(λ1,n))lnq(1β˜λ2)1  nn1qbqn]1q  . (16)

    Lemma 6. If |λλ1λ2|<|p|δ0, and the constant factor 1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1) in (11) is the best possible, then we have λ=λ1+λ2.

    Proof. If the constant factor k1pλ,η(λλ2)k1qλ,η(λ1) in (11) (or (16)) is the best possible, then by (12) (for λi=˜λi(i=1,2)), we have

    1β1/p  α1/q k1pλ,η(λλ2)k1qλ,η  (λ1) 1β1/p  α1/qkλ,η(˜λ1)(R+),

    namely, kλ,η(˜λ1) k1pλ,η(λλ2)k1qλ,η  (λ1).

    By the reverse Hӧlder inequality, we find

    kλ,η(˜λ1)=kλ,η(λλ2p+λ1q)=0k(η)λ(u,1)uλλ2p+λ1q1  du=0k(η)λ(u,1)(uλλ21  p)(uλ11  q)du(0k(η)λ(u,1)uλλ21  du)1p(0k(η)λ(u,1)uλ11  du)1q  
    =k1pλ,η(λλ2)k1qλ,η  (λ1). (17)

    It follows that (17) keeps the form of equality.

    We observe that (17) keeps the form of equality if and only if there exist constants A and B such that they are not both zero and

    Auλλ21=Buλ11  a.e.  in R+=(0,).

    Assuming that A0, it follows that uλλ2λ1=ABa.e. in R+, and then λλ2λ1=0. Namely, λ=λ1+λ2. This completes the proof of Lemma 6.

    Theorem 1. Inequality (11) is equivalent to the following inequalities:

    J:=[n=2lnpβ(λλ1q+λ2p)1  n(1θλ,η(λ1,n))p1  n(m=2k(η)λ(lnαm,lnβn)am)p]1p
    >1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1){m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm}1p, (18)
    J1:=[m=2lnqα(λλ2p+λ1q)1  mm(n=2k(η)λ(lnαm,lnβn)bn)q]1q  
    >1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1){n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1q  . (19)

    Proof. Suppose that (18) is valid. By the reverse Hӧlder inequality, we have

    I=n=2[ln1  p+β(λλ1q+λ2p)n(1θλ,η(λ1,n))1/qn1/p  m=2k(η)λ(lnαm,lnβn)am]×[(1θλ,η(λ1,n))1q  ln1pβ(λλ1q+λ2p)nn1/pbn]
    J{n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1q  . (20)

    Then by (18), we obtain (11). On the other hand, assuming that (11) is valid, we set

    bn:=lnpβ(λλ1q+λ2p)1  n(1θλ,η(λ1,n))p1  n(m=2k(η)λ(lnαm,lnβn)am)p1  ,nN{1}.

    It follows that

    Jp=n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn=I. (21)

    If J=, then (18) is naturally valid; if J=0, then it is impossible that makes (18) valid, namely, J>0. Suppose that 0<J<. By (11), we find

    n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn=Jp=I
    >k1pλ,η(λλ2)k1qλ,η  (λ1){m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm}1pJp1  ,
    J={n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1p>k1pλ,η(λλ2)k1qλ,η  (λ1){m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm}1p,

    thus, (18) follows, which is equivalent to (11).

    Suppose that (19) is valid. By the reverse Hӧlder inequality, we have

    I=m=2[ln1qα(λλ2p+λ1q)mm1/qam][ln1  q+α(λλ2p+λ1q)mm1/qn=2k(η)λ(lnαm,lnβn)bn]
    {m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm}1pJ1. (22)

    Then by (19), we obtain (11). On the other hand, assuming that (11) is valid, we set

    am:=lnqα(λλ2p+λ1q)1  mm(n=2k(η)λ(lnαm,lnβn)bn)q1  ,mN{1}.

    It follows that

    Jq1=m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm=I. (23)

    If J1=, then (19) is naturally valid; if J1=0, then it is impossible that makes (19) valid, namely, J1>0. Suppose that 0<J1<. By (11), we have

    m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm=Jq1=I
    >k1pλ,η(λλ2)k1qλ,η  (λ1)Jq11{n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1q  ,
    J1={m=2lnp[1α(λλ2p+λ1q)]1  mm1p  apm}1q  
    >k1pλ,η(λλ2)k1qλ,η  (λ1){n=2(1θλ,η(λ1,n))lnq[1β(λλ1q+λ2p)]1  nn1qbqn}1q  ,

    thereby, (19) follows, which is equivalent to (11). Hence, inequalities (11), (18) and (19) are equivalent.

    This completes the proof of Theorem 1.

    Theorem 2. Suppose that |λλ1λ2|<|p|δ0. The following statements (i), (ii), (iii), (iv) and (v) are equivalent:

    (i) Both k1pλ,η(λλ2)k1qλ,η  (λ1) and kλ,η(λλ2p+λ1q) are independent of p,q;

    (ii) k1pλ,η(λλ2)k1qλ,η  (λ1) =kλ,η(λλ2p+λ1q);

    (iii) λ=λ1+λ2;

    (iv) 1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1) is the best possible constant factor of (11);

    (v) 1β1/p  α1/qk1pλ,η(λλ2)k1qλ,η  (λ1) is the best possible constant factor of (18) (resp. (19)).

    Moreover, if the statement (iii) follows, namely, λ=λ1+λ2, then we have (12) and the following equivalent inequalities with the best possible constant factor 1β1/p  α1/q kλ,η(λ1) :

    [n=2lnpβλ21  n(1θλ,η(λ1,n))p1  n(m=2k(η)λ(lnαm,lnβn)am)p]1p
    >1β1/p  α1/qkλ,η(λ1)[m=2lnp(1αλ1)1  mm1p  apm]1p, (24)
    [m=2lnqαλ11  mm(n=2k(η)λ(lnαm,lnβn)bn)q]1q  
    >1β1/p  α1/qkλ,η(λ1)[n=2(1θλ,η(λ1,n))lnq(1βλ2)1  nn1qbqn]1q  . (25)

    Proof. (i)(ii). By (i), we have

    k1pλ,η(λλ2)k1qλ,η  (λ1)=limplimq1k1pλ,η(λλ2)k1qλ,η  (λ1)=kλ,η(λ1).

    Then by (i), (15) and the above result, we find

    kλ,η(λλ2p+λ1q)=limpkλ,η(λ1+λλ1λ2p)=kλ,η(λ1)= k1pλ,η(λλ2)k1qλ,η  (λ1).

    (ii) (iii). by (ii), (17) keeps the form of equality. In view of the proof of Lemma 6, it follows that λ=λ1+λ2.

    (iii) (i). By (iii), for λ=λ1+λ2, both k1pλ,η(λλ2)k1qλ,η  (λ1) and kλ,η(λλ2p+λ1q) are equal to kλ,η(λ1), which are independent of p,q. Hence, we have (i) (ii) (iii).

    (iii) (iv). By Lemma 5 and Lemma 6, we have the conclusions.

    (iv) (v). If the constant factor in (11) is the best possible, then so is the constant factor in (18) (resp. (19)). Otherwise, by (20) (resp. (22)), we would reach a contradiction that the constant factor in (11) is not the best possible. On the other hand, if the constant factor in (18) (resp. (19)) is the best possible, then so is the constant factor in (11). Otherwise, by (21) (resp. (23)), we would reach a contradiction that the constant factor in (18) (resp. (19)) is not the best possible.

    Hence, the statements (i), (ii), (iii), (iv) and (v) are equivalent. The proof of Theorem 2 is complete.

    In this section, we illustrate that some novel inequalities of Mulholland type can be derived from our main results as special cases.

    Remark 3. Taking α=β=1 in (24) and (25) respectively, we have the following inequalities which is equivalent to (13) with the best possible constant factor kλ,η(λ1) :

    [n=2lnpλ21  n(1˜θλ,η(λ1,n))p1  n(m=2lnηmlnηnlnλ+ηmlnλ+ηnam)p]1p
    >kλ,η(λ1)[m=2lnp(1λ1)1  mm1p  apm]1p, (26)
    [m=2lnqλ11  mm(n=2lnηmlnηnlnλ+ηmlnλ+ηnbn)q]1q  
    >kλ,η(λ1)[n=2(1˜θλ,η(λ1,n))lnq(1λ2)1  nn1qbqn]1q  . (27)

    In particular, putting λ=1,λ1=λ2=12 in (13), (26) and (27) respectively, we have the following equivalent reverse inequalities with the best possible constant factor 2π1+ηcotπ2(1+η) :

    n=2m=2lnηmlnηnln1+ηmln1+ηnambn
    >2π1+ηcotπ2(1+η)(m=2lnp21  mm1p  apm)1p[n=2(1˜θ1,η(12,n))lnq21  nn1qbqn]1q  , (28)
    [n=2lnp21  n(1˜θ1,η(12,n))p1  n(m=2lnηmlnηnln1+ηmln1+ηnam)p]1p
    >2π1+ηcotπ2(1+η)
    (m=2lnp21  mm1p  apm)1p, (29)
    [m=2lnq21  mm(n=2lnηmlnηnln1+ηmln1+ηnbn)q]1q  
    >2π1+ηcotπ2(1+η)
    [n=2(1˜θ1,η(12,n))lnq21  nn1qbqn]1q  , (30)

    where ˜θ1,η(12,n):=1+η2πtanπ2(1+η)ln2lnn0uη1  u1+η1  u12du=O(1ln1/2n).

    Choosing η=1 in (28), (29) and (30) respectively, we have the reverse of inequality (3) and the equivalent forms with the best possible constant factor π :

    n=2m=21lnmnambn
    >π(m=2lnp21  mm1p  apm)1p[n=2(1˜θ(n))lnq21  nn1qbqn]1q  , (31)
    [n=2lnp21  n(1˜θ(n))p1  n(m=21lnmnam)p]1p >π (m=2lnp21  mm1p  apm)1p, (32)
    [m=2lnq21  mm(n=21lnmnbn)q]1q   >π [n=2(1˜θ(n))lnq21nn1qbqn]1q  , (33)

    where ˜θ(n):=˜θ1,1(12,n)=1πln2lnn01u+1  u12du=O(1ln1/2n)(0,1).

    In this paper, by the use of the weight coefficients and the idea of introducing parameters, a new reverse Mulholland-type inequality with multi-parameters and the equivalent forms were given in Lemma 2 and Theorem 1. The equivalent statements of the best possible constant factor related to several parameters were obtained in Theorem 2. Some other inequalities associated with reverse Mulholland-type inequality were established in Remarks 1 and 3. The lemmas and theorems presented in this paper provided an extensive account of this type of inequalities.

    This work was supported by the National Natural Science Foundation (No. 61772140), the Characteristic Innovation Project of Guangdong Provincial Colleges and Universities in 2020 (No. 2020KTSCX088), and the Natural Science Foundation of Fujian Province of China (No. 2020J01365). The authors would like to express sincere appreciation to the anonymous reviewers for their helpful comments and suggestions. All authors contributed equally to the manuscript and read and approved the final manuscript.

    The authors declare that they have no competing interests.



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