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(x−yx+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+t≤f(t)≤t1−t. |
According to the results of Witkowski [2] we introduce the mean Mf of the form
Mf(x,y)={|x−y|2f(|x−y|x+y)x≠y,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(1−t,1+t), | (1.2) |
where
t=|x−y|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 x≠y, 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 x≠y. Then there is a fact that the inequality fM1(t)>fM2(t) holds if and only if M1<M2. That is to say,
1fM1<1fM2⟺M1<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)={|x−y|2sin(|x−y|x+y)x≠yxx=y, | (1.5) |
and the other called hyperbolic tangent mean
Mtanh(x,y)={|x−y|2tanh(|x−y|x+y)x≠yxx=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
(1−α)A+αCe<Msin<(1−β)A+βCe |
holds if and only if α≤ 1/2 and β≥(3/sin1)−3≈0.5652.
Proposition 1.2. The double inequality
(1−α)A+αCe<Mtanh<(1−β)A+βCe |
holds if and only if α≤ (3/tanh1)−3≈0.9391 and β≥1.
Proposition 1.3. The double inequality
(1−α)C−1e+αA−1<M−1sin<(1−β)C−1e+βA−1 |
holds if and only if α≤ 4sin1−3≈0.3659 and β≥1/2.
Proposition 1.4. The double inequality
(1−α)C−1e+αA−1<M−1tanh<(1−β)C−1e+βA−1 |
holds if and only if α≤ 0 and β≥4tanh1−3≈0.0464.
Proposition 1.5. The double inequality
(1−α)A2+αC2e<M2sin<(1−β)A2+βC2e |
holds if and only if α≤ 1/2 and β≥(9cot21)/7≈0.5301.
Proposition 1.6. The double inequality
(1−α)A2+αC2e<M2tanh<(1−β)A2+βC2e |
holds if and only if α≤ (9(coth21−1))/7≈0.9309 and β≥1.
Proposition 1.7. The double inequality
(1−α)C−2e+αA−2<M−2sin<(1−β)C−2e+βA−2 |
holds if and only if α≤(16sin21−9)/7≈0.3327 and β≥1/2.
Proposition 1.8. The double inequality
(1−α)C−2e+αA−2<M−2tanh<(1−β)C−2e+βA−2 |
holds if and only if α≤0 and β≥(16tanh21−9)/7≈0.0401.
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 p∈R
(1−αp)Ap+αpCep<Mpsin<(1−βp)Ap+βpCep, | (1.7) |
(1−λp)Ap+λpCep<Mptanh<(1−μp)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, x≠y, p≠0 and
p♣=3cos2+sin2+13sin2−cos2−3≈4.588. |
Then the following are considered.
(i) If p≥p♣, the double inequality
(1−αp)Ap+αpCep<Mpsin<(1−βp)Ap+βpCep | (1.9) |
holds if and only if αp≤3p(1−sinp1)/[(sinp1)(4p−3p)] and βp≥1/2.
(ii) If 0<p≤12/5, the double inequality
(1−αp)Ap+αpCep<Mpsin<(1−βp)Ap+βpCep | (1.10) |
holds if and only if αp≤1/2 and βp≥3p(1−sinp1)/[(sinp1)(4p−3p)].
(iii) If p<0, the double inequality
(1−βp)Ap+βpCep<Mpsin<(1−αp)Ap+αpCep | (1.11) |
holds if and only if αp≤1/2 and βp≥3p(1−sinp1)/[(sinp1)(4p−3p)].
Theorem 1.2. Let x,y>0, x≠y, p≠0 and
p∗=−16cosh2−3cosh4+4sinh2+3cosh4−12sinh2+15≈−3.4776. |
Then the following are considered:
(i) If p>0, the double inequality
(1−λp)Ap+λpCep<Mptanh<(1−μp)Ap+μpCep | (1.12) |
holds if and only if λp≤((coth1)p−1)/((4/3)p−1) and μp≥1.
(ii) If p∗≤p<0,
(1−μp)Ap+μpλpCep<Mptanh<(1−λp)Ap+λpCep | (1.13) |
holds if and only if λp≤((coth1)p−1)/((4/3)p−1) and μp≥1.
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,p∈R, and p≠0. 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(1−t,1+t)=λtfM(t), | (2.3) |
where
{u=λ2xx+yv=λ2yx+y, 0<x<y, λ>0. |
and 0<t<1,
t=y−xx+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=1x−∞∑n=122n(2n)!|B2n|x2n−1, 0<|x|<π, | (2.4) |
1sin2x=csc2x=−(cotx)′=1x2+∞∑n=122n(2n−1)(2n)!|B2n|x2n−2, 0<|x|<π. | (2.5) |
Lemma 2.5. ([16,17,18,19,20]) Let B2n the even-indexed Bernoulli numbers, n=1,2,…. Then
22n−1−122n+1−1(2n+2)(2n+1)π2<|B2n+2||B2n|<22n−122n+2−1(2n+2)(2n+1)π2. |
Lemma 2.6. Let l1(t) be defined by
l1(t)=s1(t)r1(t), |
where
s1(t)=6t2+2t4−12sin2t−2t3costsint+6tcostsintsin2t,r1(t)=8t2sin2t+2t4sin2t−6t2−2t4−6sin2t+12tcostsintsin2t. |
Then the double inequality
125<l1(t)<p♣=3cos2+sin2+13sin2−cos2−3≈4.588 | (2.6) |
holds for all t∈(0,1), where the constants 12/5 and (3cos2+sin2+1)/(3sin2−cos2−3)≈4.588 are the best possible in (2.6).
Proof. Since
1l1(t)=r1(t)s1(t), |
and
r1(t)=8t2sin2t+2t4sin2t−6t2−2t4−6sin2t+12tcostsintsin2t=8t2−2t41sin2t−6t21sin2t+2t4+12tcostsint−6=8t2−2t4[1t2+∞∑n=122n(2n−1)(2n)!|B2n|t2n−2]−6t2[1t2+∞∑n=122n(2n−1)(2n)!|B2n|t2n−2]+2t4+12t[1t−∞∑n=122n(2n)!|B2n|t2n−1]−6=23t4−∞∑n=3[22n−1(2n−3)(2n−2)!|B2n−2|+6⋅22n(2n+1)(2n)!|B2n|]t2n=:∞∑n=2ant2n, |
where
a2=23,an=−[22n−1(2n−3)(2n−2)!|B2n−2|+6⋅22n(2n+1)(2n)!|B2n|], n=3,4,…, |
s1(t)=6t2+2t4−12sin2t−2t3costsint+6tcostsintsin2t=6t21sin2t+2t41sin2t+6tcostsint−2t3costsint−12=6t2[1t2+∞∑n=122n(2n−1)(2n)!|B2n|t2n−2]+2t4[1t2+∞∑n=122n(2n−1)(2n)!|B2n|t2n−2]+6t[1t−∞∑n=122n(2n)!|B2n|t2n−1]−2t3[1t−∞∑n=122n(2n)!|B2n|t2n−1]−12=∞∑n=212⋅22n(n−1)(2n)!|B2n|t2n+∞∑n=14n⋅22n(2n)!|B2n|t2n+2=∞∑n=212⋅22n(n−1)(2n)!|B2n|t2n+∞∑n=2(n−1)⋅22n(2n−2)!|B2n−2|t2n=∞∑n=2[12⋅22n(n−1)(2n)!|B2n|+(n−1)⋅22n(2n−2)!|B2n−2|]t2n=85t4+∞∑n=3[12⋅22n(n−1)(2n)!|B2n|+(n−1)⋅22n(2n−2)!|B2n−2|]t2n=:∞∑n=2bnt2n, |
where
b2=85,bn=12⋅22n(n−1)(2n)!|B2n|+(n−1)⋅22n(2n−2)!|B2n−2|>0, n=3,4,…. |
Setting
qn=anbn, n=2,3,…, |
we have
q2=512=0.41667,qn=−22n−1(2n−3)(2n−2)!|B2n−2|+6⋅22n(2n+1)(2n)!|B2n|12⋅22n(n−1)(2n)!|B2n|+(n−1)⋅22n(2n−2)!|B2n−2|, n=3,4,…. |
Here we prove that the sequence {qn}n≥2 decreases monotonously. Obviously, q2>0>q3. We shall prove that for n≥3,
qn>qn+1⟺−22n−1(2n−3)(2n−2)!|B2n−2|+6⋅22n(2n+1)(2n)!|B2n|12⋅22n(n−1)(2n)!|B2n|+(n−1)⋅22n(2n−2)!|B2n−2|>−22n+1(2n−1)(2n)!|B2n|+6⋅22n+2(2n+3)(2n+2)!|B2n+2|12⋅22n+2n(2n+2)!|B2n+2|+n⋅22n+2(2n)!|B2n|⟺22n−1(2n−3)(2n−2)!|B2n−2|+6⋅22n(2n+1)(2n)!|B2n|12⋅22n(n−1)(2n)!|B2n|+(n−1)⋅22n(2n−2)!|B2n−2|<22n+1(2n−1)(2n)!|B2n|+6⋅22n+2(2n+3)(2n+2)!|B2n+2|12⋅22n+2n(2n+2)!|B2n+2|+n⋅22n+2(2n)!|B2n|, |
that is,
2(2n)!(2n−2)!|B2n−2||B2n|+24(4n−3)(2n−2)!(2n+2)!|B2n+2||B2n−2||B2n||B2n|>24(4n−1)((2n)!)2+864(2n)!(2n+2)!|B2n+2||B2n|. | (2.7) |
By Lemma 2.5 we have
2(2n)!(2n−2)!|B2n−2||B2n|+24(4n−3)(2n−2)!(2n+2)!|B2n+2||B2n−2||B2n||B2n|>2(2n)!(2n−2)!22n−122n−2−1π2(2n)(2n−1)+24(4n−3)(2n−2)!(2n+2)!22n−1−122n+1−1(2n+2)(2n+1)π222n−122n−2−1π2(2n)(2n−1)=2π2(2n)!(2n)!22n−122n−2−1+24(4n−3)(2n)!(2n)!22n−1−122n+1−122n−122n−2−1, |
and
24(4n−1)(2n)!2+864(2n)!(2n+2)!|B2n+2||B2n|<24(4n−1)(2n)!2+864(2n)!(2n+2)!22n−122n+2−1(2n+2)(2n+1)π2=24(4n−1)(2n)!2+864(2n)!(2n)!22n−122n+2−11π2. |
So we can complete the prove (2.7) when proving
2π2(2n)!(2n)!22n−122n−2−1+24(4n−3)(2n)!(2n)!22n−1−122n+1−122n−122n−2−1>24(4n−1)((2n)!)2+864(2n)!(2n)!22n−122n+2−11π2 |
or
2π2(22n−1)22n−2−1+24(4n−3)22n−1−122n+1−122n−122n−2−1>24(4n−1)+22n−122n+2−1864π2. |
In fact,
2π2(22n−1)22n−2−1+24(4n−3)22n−1−122n+1−122n−122n−2−1−[24(4n−1)+22n−122n+2−1864π2]=:8H(n)π2(22n+2−1)(22n−4)(22n+1−1), |
where
H(n)=8⋅26n(π+3)(π−3)(π2+3)+2⋅24n(72π2n+60π2−7π4+594)−22n(36π2n+123π2−7π4+1404)+(24π2−π4+432)>0 |
for all n≥3.
So the sequence {qn}n≥2 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
limt→0+l1(t)=125 and limt→1−l1(t)=p♣=3cos2+sin2+13sin2−cos2−3≈4.588, |
the proof of this lemma is complete.
Lemma 2.7. Let l2(t) be defined by
l2(t)=2⋅3cosh4t−12t2cosh2t−4t4cosh2t+2t3sinh2t−6tsinh2t−3t2cosh4t−3cosh4t+24tsinh2t−25t2−8t4+3=:2B(t)A(t), 0<t<∞, |
where
A(t)=t2cosh4t−3cosh4t+24tsinh2t−25t2−8t4+3,B(t)=3cosh4t−12t2cosh2t−4t4cosh2t+2t3sinh2t−6tsinh2t−3. |
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+2n2−23)22n+48(2n+2)(2n+2)!]22n, n=3,4,…,dn=[48⋅22n−8(n+1)(5n−n2+2n3+6)(2n+2)!]22n, n=2,3,…, |
Setting
kn=cndn=48(n+1)+22n(3n+2n2−23)4(6⋅22n−11n−4n2−n3−2n4−6), n=2,3,…, |
Here we prove that the sequence {kn}n≥2 decreases monotonously. Obviously, k2=0<k3. For n≥3,
kn<kn+1⟺48(n+1)+22n(3n+2n2−23)4(6⋅22n−11n−4n2−n3−2n4−6)<48(n+2)+22n+2(3(n+1)+2(n+1)2−23)4(6⋅22n+2−11(n+1)−4(n+1)2−(n+1)3−2(n+1)4−6)⟺48(n+1)+22n(3n+2n2−23)6⋅22n−11n−4n2−n3−2n4−6<48n+96+22n+2(7n+2n2−18)6⋅22n+2−30n−19n2−9n3−2n4−24 |
follows from Δ(n)>0 for all n≥2, where
Δ(n)=(48n+96+22n+2(7n+2n2−18))(6⋅22n−11n−4n2−n3−2n4−6)−(48(n+1)+22n(3n+2n2−23))(6⋅22n+2−30n−19n2−9n3−2n4−24)=24⋅24n(4n+5)−22n(858n+367n2+218n3−103n4+40n5+12n6+696)+1248n+1440n2+1056n3+288n4+576=:22n[j(n)22n−i(n)]+w(n) |
with
j(n)=24(4n+5),i(n)=858n+367n2+218n3−103n4+40n5+12n6+696,w(n)=1248n+1440n2+1056n3+288n4+576>0. |
We have that Δ(2)=5376>0 and shall prove that
j(n)22n−i(n)>0⟺22n>i(n)j(n) | (2.8) |
holds for all n≥3. 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)=941/17≈55.353, 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)=4⋅22m>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+90720m6−60000m5−97176m4+1169232m3+2266104m2+3581136m+2154816=146337408+234401616(m−3)+189746328(m−3)2+92580720(m−3)3+27579624(m−3)4+4838880(m−3)5+453600(m−3)6+17280(m−3)7>0 |
for m≥3 due to the coefficients of the power square of (m−1) 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<Mpsin−ApCep−Ap<βp⟺αp<(1sint)p−(1t)p(3+t23t)p−(1t)p<βp,λp<Mptanh−ApCep−Ap<μp⟺λp<(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+13sin2−cos2−3≈4.588. |
Then,
(i) if p≥p♣, the double inequality
αp<(1sint)p−(1t)p(3+t23t)p−(1t)p<βp | (3.1) |
holds if and only if αp≤3p(1−sinp1)/[(sinp1)(4p−3p)] and βp≥1/2;
(ii) if 0≠p≤12/5=2.4 and p≠0 the double inequality
βp<(1sint)p−(1t)p(3+t23t)p−(1t)p<αp | (3.2) |
holds if and only if αp≤1/2 and β≥3p(1−sinp1)/[(sinp1)(4p−3p)].
Theorem 3.2. Let t∈(0,1) and
p∗=−16cosh2−3cosh4+4sinh2+3cosh4−12sinh2+15≈−3.4776. |
If 0≠p≥−3.4776, the double inequality
λp<(1tanht)p−(1t)p(3+t23t)p−(1t)p<μp | (3.3) |
holds if and only if λp≤((coth1)p−1)/((4/3)p−1) and μp≥1.
Let
F(t)=(1sint)p−(1t)p(3+t23t)p−(1t)p=(tsint)p−1(3+t23)p−1=:f(t)g(t)=f(t)−f(0+)g(t)−g(0+), |
where
f(t)=(tsint)p−1,g(t)=(3+t23)p−1. |
Then
f′(t)=psin2t(sint−tcost)(tsint)p−1,g′(t)=23(13)p−1pt(t2+3)p−1, |
f′(t)g′(t)=3p21tsin2t(sint−tcost)(t(t2+3)sint)p−1, |
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 p≥maxt∈(0,1)l1(t)=:p♣=(3cos2+sin2+1)/(3sin2−cos2−3)≈4.588,
(f′(t)g′(t))′≤0⟹f′(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(1−sinp1)(sinp1)(4p−3p), | (3.4) |
we have that (3.1) holds.
(b) When 0≠p≤12/5=mint∈(0,1)l1(t),
(f′(t)g′(t))′≥0⟹f′(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)p−1(3+t23)p−1=:u(t)v(t)=u(t)−u(0+)v(t)−v(0+). |
Then
u′(t)=ptanh2t(ttanht)p−1(ttanh2t+tanht−t),v′(t)=23pt(t2+33)p−1, |
u′(t)v′(t)=32ttanh2t+tanht−tttanh2t[3t(t2+3)tanht]p−1, |
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
limt→0+l2(t)=∞,limt→1−l2(t)=16cosh2−3cosh4+4sinh2+3cosh4−12sinh2+15=:p#≈3.4776, |
we obtain the following result:
When p≥maxt∈(0,1){−l2(t)}=−p#=:p∗≈−3.4776,
(u′(t)v′(t))′≤0⟹u′(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)p−1(43)p−1, |
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, x≠y, and
p♣=3cos2+sin2+13sin2−cos2−3≈4.588,α=3p♣(1−sinp♣1)(sinp♣1)(4p♣−3p♣)≈0.44025,β=12. |
Then the double inequality
(1−α)Ap♣+αCep♣<Mp♣sin<(1−β)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, x≠y, and
θ=12,ϑ=312/5(1−sin12/51)(sin12/51)(412/5−312/5)≈0.51603. |
Then the double inequality
(1−θ)A12/5+θCe12/5<M12/5sin<(1−ϑ)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, x≠y, and
p∗=−16cosh2−3cosh4+4sinh2+3cosh4−12sinh2+15≈−3.4776,λ=(coth1)p∗−1(4/3)p∗−1≈0.96813,μ=1. |
Then the double inequality
(1−μ)Ap∗+μCep∗<Mp∗tanh<(1−λ)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 p∈R:
(1−αp)Ap+αpCep<Mpsin<(1−βp)Ap+βpCep,(1−λp)Ap+λpCep<Mptanh<(1−μp)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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