Research article

Optimal bounds for the sine and hyperbolic tangent means by arithmetic and centroidal means in exponential type

  • Received: 27 March 2021 Accepted: 07 September 2021 Published: 14 September 2021
  • MSC : 26D15

  • In this paper, optimal bounds for the sine and hyperbolic tangent means by arithmetic and centroidal means in exponential type are established using the monotone form of L'Hospital's rule and the criterion for the monotonicity of the quotient of power series.

    Citation: Ling Zhu, Branko Malešević. Optimal bounds for the sine and hyperbolic tangent means by arithmetic and centroidal means in exponential type[J]. AIMS Mathematics, 2021, 6(12): 13024-13040. doi: 10.3934/math.2021753

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  • In this paper, optimal bounds for the sine and hyperbolic tangent means by arithmetic and centroidal means in exponential type are established using the monotone form of L'Hospital's rule and the criterion for the monotonicity of the quotient of power series.



    In 1998, Kahlig and Matkowski [1] proved in particular that every homogeneous bivariable mean M in (0,) can be represented in the form

    M(x,y)=A(x,y)fM,A(xyx+y),

    where A is the arithmetic mean and fM,A: (1,1)(0,2) is a unique single variable function (with the graph laying in a set of a butterfly shape), called an A-index of M.

    In this paper we consider Seiffert function f:(0,1)R which fulfils the following condition

    t1+tf(t)t1t.

    According to the results of Witkowski [2] we introduce the mean Mf of the form

    Mf(x,y)={|xy|2f(|xy|x+y)xy,xx=y. (1.1)

    In this paper a mean Mf:R2+R is the function that is symmetric, positively homogeneous and internal in sense [2]. Basic result of Witkowski is correspondence between a mean Mf and Seiffert function f of Mf is given by the following formula

    f(t)=tMf(1t,1+t),  (1.2)

    where

    t=|xy|x+y. (1.3)

    Therefore, f and Mf form a one-to-one correspondence via (1.1) and (1.2). For this reason, in the following we can rewrite f=:fM.

    Throughout this article, we say xy, that is, t(0,1). For convenience, we note that M1<M2 means M1(x,y)<M2(x,y) holds for two means M1 and M2 with xy. Then there is a fact that the inequality fM1(t)>fM2(t) holds if and only if M1<M2. That is to say,

    1fM1<1fM2M1<M2. (1.4)

    The above relationship (1.4) inspires us to ask a question: Can we transform the means inequality problem into the reciprocal inequality problem of the corresponding Seiffert functions? Witkowski [2] answers this question from the perspective of one-to-one correspondence. We find that these two kinds of inequalities are equivalent in similar linear inequalities. We describe this result in Lemma 2.1 as a support of this paper.

    As we know, the study of inequalities for mean values has always been a hot topic in the field of inequalities. For example, two common means can be used to define some new means. The recent success in this respect can be seen in references [3,4,5,6,7,8]. In [2], Witkowski introduced the following two new means, one called sine mean

    Msin(x,y)={|xy|2sin(|xy|x+y)xyxx=y, (1.5)

    and the other called hyperbolic tangent mean

    Mtanh(x,y)={|xy|2tanh(|xy|x+y)xyxx=y. (1.6)

    Recently, Nowicka and Witkowski [9] determined various optimal bounds for the Msin(x,y) and Mtanh(x,y) by the arithmetic mean A(x,y)=(x+y)/2 and centroidal mean

    Ce(x,y)=23x2+xy+y2x+y

    as follows:

    Proposition 1.1. The double inequality

    (1α)A+αCe<Msin<(1β)A+βCe

    holds if and only if α 1/2 and β(3/sin1)30.5652.

    Proposition 1.2. The double inequality

    (1α)A+αCe<Mtanh<(1β)A+βCe

    holds if and only if α (3/tanh1)30.9391 and β1.

    Proposition 1.3. The double inequality

    (1α)C1e+αA1<M1sin<(1β)C1e+βA1

    holds if and only if α 4sin130.3659 and β1/2.

    Proposition 1.4. The double inequality

    (1α)C1e+αA1<M1tanh<(1β)C1e+βA1

    holds if and only if α 0 and β4tanh130.0464.

    Proposition 1.5. The double inequality

    (1α)A2+αC2e<M2sin<(1β)A2+βC2e

    holds if and only if α 1/2 and β(9cot21)/70.5301.

    Proposition 1.6. The double inequality

    (1α)A2+αC2e<M2tanh<(1β)A2+βC2e

    holds if and only if α (9(coth211))/70.9309 and β1.

    Proposition 1.7. The double inequality

    (1α)C2e+αA2<M2sin<(1β)C2e+βA2

    holds if and only if α(16sin219)/70.3327 and β1/2.

    Proposition 1.8. The double inequality

    (1α)C2e+αA2<M2tanh<(1β)C2e+βA2

    holds if and only if α0 and β(16tanh219)/70.0401.

    In essence, the above results are how the two new means Msin and Mtanh are expressed linearly, harmoniously, squarely, and harmoniously in square by the two classical means Ce(x,y) and A(x,y). In this paper, we study the following two-sided inequalities in exponential form for nonzero number pR

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep, (1.7)
    (1λp)Ap+λpCep<Mptanh<(1μp)Ap+μpCep (1.8)

    in order to reach a broader conclusion including all the above properties. The main conclusions of this paper are as follows:

    Theorem 1.1. Let x,y>0, xy, p0 and

    p=3cos2+sin2+13sin2cos234.588.

    Then the following are considered.

    (i) If pp, the double inequality

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep (1.9)

    holds if and only if αp3p(1sinp1)/[(sinp1)(4p3p)] and βp1/2.

    (ii) If 0<p12/5, the double inequality

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep (1.10)

    holds if and only if αp1/2 and βp3p(1sinp1)/[(sinp1)(4p3p)].

    (iii) If p<0, the double inequality

    (1βp)Ap+βpCep<Mpsin<(1αp)Ap+αpCep (1.11)

    holds if and only if αp1/2 and βp3p(1sinp1)/[(sinp1)(4p3p)].

    Theorem 1.2. Let x,y>0, xy, p0 and

    p=16cosh23cosh4+4sinh2+3cosh412sinh2+153.4776.

    Then the following are considered:

    (i) If p>0, the double inequality

    (1λp)Ap+λpCep<Mptanh<(1μp)Ap+μpCep (1.12)

    holds if and only if λp((coth1)p1)/((4/3)p1) and μp1.

    (ii) If pp<0,

    (1μp)Ap+μpλpCep<Mptanh<(1λp)Ap+λpCep (1.13)

    holds if and only if λp((coth1)p1)/((4/3)p1) and μp1.

    We first introduce a theoretical support of this paper.

    Lemma 2.1. ([10]) Let K(x,y),R(x,y), and N(x,y) be three means with two positive distinct parameters x and y; fK(t), fR(t), and fN(t) be the corresponding Seiffert functions of the former, ϑ1,ϑ2,θ1,θ2,pR, and p0. Then

    ϑ1Kp(x,y)+ϑ2Np(x,y)Rp(x,y)θ1Kp(x,y)+θ2Np(x,y) (2.1)
    ϑ1fpK(t)+ϑ2fpN(t)1fpR(t)θ1fpK(t)+θ2fpN(t). (2.2)

    It must be mentioned that the key steps to prove the above results are following:

    Mf(u,v)=Mf(λ2xx+y,λ2yx+y)=λMf(2xx+y,2yx+y)=λMf(1t,1+t)=λtfM(t), (2.3)

    where

    {u=λ2xx+yv=λ2yx+y, 0<x<y, λ>0.

    and 0<t<1,

    t=yxx+y.

    In order to prove the main conclusions, we shall introduce some very suitable methods which are called the monotone form of L'Hospital's rule (see Lemma 2.2) and the criterion for the monotonicity of the quotient of power series (see Lemma 2.3).

    Lemma 2.2. ([11,12]) For <a<b<, let f,g:[a,b]R be continuous functions that are differentiable on (a,b), with f(a)=g(a)=0 or f(b)=g(b)=0. Assume that g(t)0 for each x in (a,b). If f/g is increasing (decreasing) on (a,b), then so is f/g.

    Lemma 2.3. ([13]) Let an and bn (n=0,1,2,) be real numbers, and let the power series A(x)=n=0anxn and B(x)=n=0bnxn be convergent for |x|<R (R+). If bn>0 for n=0,1,2,, and if εn=an/bn is strictly increasing (or decreasing) for n=0,1,2,, then the function A(x)/B(x) is strictly increasing (or decreasing) on (0,R) (R+).

    Lemma 2.4. ([14,15]) Let B2n be the even-indexed Bernoulli numbers. Then we have the following power series expansions

    cotx=1xn=122n(2n)!|B2n|x2n1, 0<|x|<π, (2.4)
    1sin2x=csc2x=(cotx)=1x2+n=122n(2n1)(2n)!|B2n|x2n2, 0<|x|<π.  (2.5)

    Lemma 2.5. ([16,17,18,19,20]) Let B2n the even-indexed Bernoulli numbers, n=1,2,. Then

    22n1122n+11(2n+2)(2n+1)π2<|B2n+2||B2n|<22n122n+21(2n+2)(2n+1)π2.

    Lemma 2.6. Let l1(t) be defined by

    l1(t)=s1(t)r1(t),

    where

    s1(t)=6t2+2t412sin2t2t3costsint+6tcostsintsin2t,r1(t)=8t2sin2t+2t4sin2t6t22t46sin2t+12tcostsintsin2t.

    Then the double inequality

    125<l1(t)<p=3cos2+sin2+13sin2cos234.588 (2.6)

    holds for all t(0,1), where the constants 12/5 and (3cos2+sin2+1)/(3sin2cos23)4.588 are the best possible in (2.6).

    Proof. Since

    1l1(t)=r1(t)s1(t),

    and

    r1(t)=8t2sin2t+2t4sin2t6t22t46sin2t+12tcostsintsin2t=8t22t41sin2t6t21sin2t+2t4+12tcostsint6=8t22t4[1t2+n=122n(2n1)(2n)!|B2n|t2n2]6t2[1t2+n=122n(2n1)(2n)!|B2n|t2n2]+2t4+12t[1tn=122n(2n)!|B2n|t2n1]6=23t4n=3[22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|]t2n=:n=2ant2n,

    where

    a2=23,an=[22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|], n=3,4,,
    s1(t)=6t2+2t412sin2t2t3costsint+6tcostsintsin2t=6t21sin2t+2t41sin2t+6tcostsint2t3costsint12=6t2[1t2+n=122n(2n1)(2n)!|B2n|t2n2]+2t4[1t2+n=122n(2n1)(2n)!|B2n|t2n2]+6t[1tn=122n(2n)!|B2n|t2n1]2t3[1tn=122n(2n)!|B2n|t2n1]12=n=21222n(n1)(2n)!|B2n|t2n+n=14n22n(2n)!|B2n|t2n+2=n=21222n(n1)(2n)!|B2n|t2n+n=2(n1)22n(2n2)!|B2n2|t2n=n=2[1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|]t2n=85t4+n=3[1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|]t2n=:n=2bnt2n,

    where

    b2=85,bn=1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|>0, n=3,4,.

    Setting

    qn=anbn, n=2,3,,

    we have

    q2=512=0.41667,qn=22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|, n=3,4,.

    Here we prove that the sequence {qn}n2 decreases monotonously. Obviously, q2>0>q3. We shall prove that for n3,

    qn>qn+122n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|>22n+1(2n1)(2n)!|B2n|+622n+2(2n+3)(2n+2)!|B2n+2|1222n+2n(2n+2)!|B2n+2|+n22n+2(2n)!|B2n|22n1(2n3)(2n2)!|B2n2|+622n(2n+1)(2n)!|B2n|1222n(n1)(2n)!|B2n|+(n1)22n(2n2)!|B2n2|<22n+1(2n1)(2n)!|B2n|+622n+2(2n+3)(2n+2)!|B2n+2|1222n+2n(2n+2)!|B2n+2|+n22n+2(2n)!|B2n|,

    that is,

    2(2n)!(2n2)!|B2n2||B2n|+24(4n3)(2n2)!(2n+2)!|B2n+2||B2n2||B2n||B2n|>24(4n1)((2n)!)2+864(2n)!(2n+2)!|B2n+2||B2n|. (2.7)

    By Lemma 2.5 we have

    2(2n)!(2n2)!|B2n2||B2n|+24(4n3)(2n2)!(2n+2)!|B2n+2||B2n2||B2n||B2n|>2(2n)!(2n2)!22n122n21π2(2n)(2n1)+24(4n3)(2n2)!(2n+2)!22n1122n+11(2n+2)(2n+1)π222n122n21π2(2n)(2n1)=2π2(2n)!(2n)!22n122n21+24(4n3)(2n)!(2n)!22n1122n+1122n122n21,

    and

    24(4n1)(2n)!2+864(2n)!(2n+2)!|B2n+2||B2n|<24(4n1)(2n)!2+864(2n)!(2n+2)!22n122n+21(2n+2)(2n+1)π2=24(4n1)(2n)!2+864(2n)!(2n)!22n122n+211π2.

    So we can complete the prove (2.7) when proving

    2π2(2n)!(2n)!22n122n21+24(4n3)(2n)!(2n)!22n1122n+1122n122n21>24(4n1)((2n)!)2+864(2n)!(2n)!22n122n+211π2

    or

    2π2(22n1)22n21+24(4n3)22n1122n+1122n122n21>24(4n1)+22n122n+21864π2.

    In fact,

    2π2(22n1)22n21+24(4n3)22n1122n+1122n122n21[24(4n1)+22n122n+21864π2]=:8H(n)π2(22n+21)(22n4)(22n+11),

    where

    H(n)=826n(π+3)(π3)(π2+3)+224n(72π2n+60π27π4+594)22n(36π2n+123π27π4+1404)+(24π2π4+432)>0

    for all n3.

    So the sequence {qn}n2 decreases monotonously. By Lemma 2.3 we obtain that r1(t)/s1(t) is decreasing on (0,1), which means that the function l1(t) is increasing on (0,1). In view of

    limt0+l1(t)=125 and limt1l1(t)=p=3cos2+sin2+13sin2cos234.588,

    the proof of this lemma is complete.

    Lemma 2.7. Let l2(t) be defined by

    l2(t)=23cosh4t12t2cosh2t4t4cosh2t+2t3sinh2t6tsinh2t3t2cosh4t3cosh4t+24tsinh2t25t28t4+3=:2B(t)A(t), 0<t<,

    where

    A(t)=t2cosh4t3cosh4t+24tsinh2t25t28t4+3,B(t)=3cosh4t12t2cosh2t4t4cosh2t+2t3sinh2t6tsinh2t3.

    Then l2(t) is strictly decreasing on (0,).

    Proof. Let's take the power series expansions

    sinhkt=n=0k2n+1(2n+1)!t2n+1, coshkt=n=0k2n(2n)!t2n

    into A(t) and B(t), and get

    A(t)=n=2cnt2n+2, B(t)=n=2dnt2n+2,

    where

    c2=0,cn=[2(3n+2n223)22n+48(2n+2)(2n+2)!]22n, n=3,4,,dn=[4822n8(n+1)(5nn2+2n3+6)(2n+2)!]22n, n=2,3,,

    Setting

    kn=cndn=48(n+1)+22n(3n+2n223)4(622n11n4n2n32n46), n=2,3,,

    Here we prove that the sequence {kn}n2 decreases monotonously. Obviously, k2=0<k3. For n3,

    kn<kn+148(n+1)+22n(3n+2n223)4(622n11n4n2n32n46)<48(n+2)+22n+2(3(n+1)+2(n+1)223)4(622n+211(n+1)4(n+1)2(n+1)32(n+1)46)48(n+1)+22n(3n+2n223)622n11n4n2n32n46<48n+96+22n+2(7n+2n218)622n+230n19n29n32n424

    follows from Δ(n)>0 for all n2, where

    Δ(n)=(48n+96+22n+2(7n+2n218))(622n11n4n2n32n46)(48(n+1)+22n(3n+2n223))(622n+230n19n29n32n424)=2424n(4n+5)22n(858n+367n2+218n3103n4+40n5+12n6+696)+1248n+1440n2+1056n3+288n4+576=:22n[j(n)22ni(n)]+w(n)

    with

    j(n)=24(4n+5),i(n)=858n+367n2+218n3103n4+40n5+12n6+696,w(n)=1248n+1440n2+1056n3+288n4+576>0.

    We have that Δ(2)=5376>0 and shall prove that

    j(n)22ni(n)>022n>i(n)j(n) (2.8)

    holds for all n3. Now we use mathematical induction to prove (2.8). When n=3, the left-hand side and right-hand side of (2.8) are 26=64 and i(3)/j(3)=941/1755.353, which implies (2.8) holds for n=3. Assuming that (2.8) holds for n=m, that is,

    22m>i(m)j(m). (2.9)

    Next, we prove that (2.8) is valid for n=m+1. By (2.9) we have

    22(m+1)=422m>4i(m)j(m),

    in order to complete the proof of (2.8) it suffices to show that

    4i(m)j(m)>i(m+1)j(m+1)4i(m)j(m+1)i(m+1)j(m)>0.

    In fact,

    4i(m)j(m+1)i(m+1)j(m)=17280m7+90720m660000m597176m4+1169232m3+2266104m2+3581136m+2154816=146337408+234401616(m3)+189746328(m3)2+92580720(m3)3+27579624(m3)4+4838880(m3)5+453600(m3)6+17280(m3)7>0

    for m3 due to the coefficients of the power square of (m1) are positive.

    By Lemma 2.3 we get that A(t)/B(t) is strictly increasing on (0,). So the function l2(x) is strictly decreasing on (0,).

    The proof of Lemma 2.7 is complete.

    Via (1.3) and (1.2) we can obtain

    fA(t)=t,fCe(t)=3t3+t2,fMsin(t)=sint,fMtanh(t)=tanht.

    Then by Lemma 2.1 and (2.3) we have

    αp<MpsinApCepAp<βpαp<(1sint)p(1t)p(3+t23t)p(1t)p<βp,λp<MptanhApCepAp<μpλp<(1tanht)p(1t)p(3+t23t)p(1t)p<μp.

    So we turn to the proof of the following two theorems.

    Theorem 3.1. Let t(0,1) and

    p=3cos2+sin2+13sin2cos234.588.

    Then,

    (i) if pp, the double inequality

    αp<(1sint)p(1t)p(3+t23t)p(1t)p<βp (3.1)

    holds if and only if αp3p(1sinp1)/[(sinp1)(4p3p)] and βp1/2;

    (ii) if 0p12/5=2.4 and p0 the double inequality

    βp<(1sint)p(1t)p(3+t23t)p(1t)p<αp (3.2)

    holds if and only if αp1/2 and β3p(1sinp1)/[(sinp1)(4p3p)].

    Theorem 3.2. Let t(0,1) and

    p=16cosh23cosh4+4sinh2+3cosh412sinh2+153.4776.

    If 0p3.4776, the double inequality

    λp<(1tanht)p(1t)p(3+t23t)p(1t)p<μp (3.3)

    holds if and only if λp((coth1)p1)/((4/3)p1) and μp1.

    Let

    F(t)=(1sint)p(1t)p(3+t23t)p(1t)p=(tsint)p1(3+t23)p1=:f(t)g(t)=f(t)f(0+)g(t)g(0+),

    where

    f(t)=(tsint)p1,g(t)=(3+t23)p1.

    Then

    f(t)=psin2t(sinttcost)(tsint)p1,g(t)=23(13)p1pt(t2+3)p1,
    f(t)g(t)=3p21tsin2t(sinttcost)(t(t2+3)sint)p1,

    and

    (f(t)g(t))=14(3t(sint)(t2+3))pr1(t)t3sin2t[s1(t)r1(t)p]=:14(3t(sint)(t2+3))pr1(t)t3sin2t[l1(t)p],

    where the three functions s1(t), r1(t), and l1(t) are shown in Lemma 2.6.

    By Lemma 2.6 we can obtain the following results:

    (a) When pmaxt(0,1)l1(t)=:p=(3cos2+sin2+1)/(3sin2cos23)4.588,

    (f(t)g(t))0f(t)g(t)  is decreasing on  (0,1)

    this leads to F(t)=f(t)/g(t) is decreasing on (0,1) by Lemma 2.1. In view of

    F(0+)=12, F(1)=3p(1sinp1)(sinp1)(4p3p), (3.4)

    we have that (3.1) holds.

    (b) When 0p12/5=mint(0,1)l1(t),

    (f(t)g(t))0f(t)g(t) is increasing on (0,1)

    this leads to F(t)=f(t)/g(t) is increasing on (0,1) by Lemma 2.2. In view of (3.4) we have that (3.2) holds.

    The proof of Theorem 3.1 is complete.

    Let

    G(t)=(1tanht)p(1t)p(3+t23t)p(1t)p=(ttanht)p1(3+t23)p1=:u(t)v(t)=u(t)u(0+)v(t)v(0+).

    Then

    u(t)=ptanh2t(ttanht)p1(ttanh2t+tanhtt),v(t)=23pt(t2+33)p1,
    u(t)v(t)=32ttanh2t+tanhttttanh2t[3t(t2+3)tanht]p1,

    and

    (u(t)v(t))=116[3tcosht(3+t2)sinht]pA(t)t3cosh2tsinh2t[p+2B(t)A(t)]=:116[3tcosht(3+t2)sinht]pA(t)t3cosh2tsinh2t[p+l2(t)],

    where the three functions A(t), B(t), and l2(t) are shown in Lemma 2.7. By Lemma 2.7 we see that l2(x) is strictly decreasing on (0,1). Since

    limt0+l2(t)=,limt1l2(t)=16cosh23cosh4+4sinh2+3cosh412sinh2+15=:p#3.4776,

    we obtain the following result:

    When pmaxt(0,1){l2(t)}=p#=:p3.4776,

    (u(t)v(t))0u(t)v(t)  is decreasing on  (0,1)

    this leads to G(t)=u(t)/v(t) is decreasing on (0,1) by Lemma 2.2. Since

    G(0+)=1,G(1)=(cosh1sinh1)p1(43)p1,

    we have

    G(1)<G(t)<G(0+),

    which completes the proof of Theorem 3.2.

    Remark 4.1. Letting p=1,1,2,2 in Theorems 1.1 and 1.2 respectively, one can obtain Propositions 1.1–1.8.

    From Theorems 1.1 and 1.2, we can also get the following important conclusions:

    Corollary 4.1. Let x,y>0, xy, and

    p=3cos2+sin2+13sin2cos234.588,α=3p(1sinp1)(sinp1)(4p3p)0.44025,β=12.

    Then the double inequality

    (1α)Ap+αCep<Mpsin<(1β)Ap+βCep (4.1)

    holds, where the constants α and β are the best possible in (4.1).

    Corollary 4.2. Let x,y>0, xy, and

    θ=12,ϑ=312/5(1sin12/51)(sin12/51)(412/5312/5)0.51603.

    Then the double inequality

    (1θ)A12/5+θCe12/5<M12/5sin<(1ϑ)A12/5+ϑCe12/5 (4.2)

    holds, where the constants θ and ϑ are the best possible in (4.2).

    Corollary 4.3. Let x,y>0, xy, and

    p=16cosh23cosh4+4sinh2+3cosh412sinh2+153.4776,λ=(coth1)p1(4/3)p10.96813,μ=1.

    Then the double inequality

    (1μ)Ap+μCep<Mptanh<(1λ)Ap+λCep (4.3)

    holds, where the constants λ and μ are the best possible in (4.3).

    In this paper, we have studied exponential type inequalities for Msin and Mtanh in term of A and Ce for nonzero number pR:

    (1αp)Ap+αpCep<Mpsin<(1βp)Ap+βpCep,(1λp)Ap+λpCep<Mptanh<(1μp)Ap+μpCep,

    obtained a lot of interesting conclusions which include the ones of the previous similar literature. In fact, we can consider similar inequalities for dual means of the two means Msin and Mtanh, and we can replace A and Ce by other famous means. Therefore, the content of this research is very extensive.

    The authors are grateful to editor and anonymous referees for their careful corrections to and valuable comments on the original version of this paper.

    The first author was supported by the National Natural Science Foundation of China (no. 61772025). The second author was supported in part by the Serbian Ministry of Education, Science and Technological Development, under projects ON 174032 and III 44006.

    The authors declare that they have no conflict of interest.



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