# On Some Intermediate Mean Values

Authors: Slavko Simic
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## Abstract

We give a necessary and sufficient mean condition for the quotient of two Jensen functionals and define a new classΛf,g(a,b) of mean values where f,g are continuously differentiable convex functions satisfying the relation f″(t)=tg″(t),  t∈ℝ+. Then we asked for a characterization of f,g such that the inequalities H(a,b)≤Λf,g(a,b)≤A(a,b) or L(a,b)≤Λf,g(a,b)≤I(a,b) hold for each positive a,b, where H,A,L,I are the harmonic, arithmetic, logarithmic, and identric means, respectively. For a subclass of Λ with g″(t)=ts,s∈ℝ, this problem is thoroughly solved.


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## Body

## 1. Introduction

It is said that the meanP is intermediate relating to the means M and N,M≤N if the relation
(1)M(a,b)≤P(a,b)≤N(a,b)
holds for each two positive numbers a, b.It is also well known that(2)min{a,b}≤H(a,b)≤G(a,b)≤L(a,b)≤I(a,b)≤A(a,b)≤S(a,b)≤max{a,b},
where
(3)H=H(a,b):=2(1a+1b)-1;G=G(a,b):=ab;L=L(a,b):=b-alogb-loga;I=I(a,b):=(bb/aa)1/(b-a)e;A=A(a,b):=a+b2;S=S(a,b):=aa/(a+b)bb/(a+b)
are the harmonic, geometric, logarithmic, identric, arithmetic, and Gini mean, respectively.An easy task is to construct intermediate means related to two given meansM and N with M≤N. For instance, for an arbitrary mean P, we have that
(4)M(a,b)≤P(M(a,b),N(a,b))≤N(a,b).The problem is more difficult if we have to decide whether the given mean is intermediate or not. For example, the relation(5)L(a,b)≤Ss(a,b)≤I(a,b)
holds for each positive a and b if and only if 0≤s≤1, where the Stolarsky mean Ss is defined by (cf [1])
(6)Ss(a,b):=(bs-ass(b-a))1/(s-1).Also,(7)G(a,b)≤As(a,b)≤A(a,b)
holds if and only if 0≤s≤1, where the Hölder mean of order s is defined by
(8)As(a,b):=(as+bs2)1/s.An inverse problem is to find best possible approximation of a given meanP by elements of an ordered class of means S. A good example for this topic is comparison between the logarithmic mean and the class As of Hölder means of order s. Namely, since A0=lims→0As=G and A1=A, it follows from (2) that
(9)A0≤L≤A1.SinceAs is monotone increasing in s, an improving of the above is given by Carlson [2]:
(10)A0≤L≤A1/2.Finally, Lin showed in [3] that
(11)A0≤L≤A1/3
is the best possible approximation of the logarithmic mean by the means from the class As.Numerous similar results have been obtained recently. For example, an approximation of Seiffert’s mean by the classAs is given in [4, 5].In this paper we will give best possible approximations for a whole variety of elementary means (2) by the class λs defined below (see Theorem 5).Letf,g be twice continuously differentiable (strictly) convex functions on ℝ+. By definition (cf [6], page 5),
(12)f-(a,b):=f(a)+f(b)-2f(a+b2)>0,a≠b,f-(a,b)=0,
if and only if a=b.It turns out that the expression(13)Λf,g(a,b):=f-(a,b)g-(a,b)=f(a)+f(b)-2f((a+b)/2)g(a)+g(b)-2g((a+b)/2)
represents a mean of two positive numbers a, b; that is, the relation
(14)min{a,b}≤Λf,g(a,b)≤max{a,b}
holds for each a,b∈ℝ+, if and only if the relation
(15)f′′(t)=tg′′(t)
holds for each t∈ℝ+.Letf,g∈C∞(0,∞) and denote by Λ the set {(f,g)} of convex functions satisfying the relation (15). There is a natural question how to improve the bounds in (14); in this sense we come upon the following intermediate mean problem.Open Question. Under what additional conditions on f,g∈Λ, the inequalities
(16)H(a,b)≤Λf,g(a,b)≤A(a,b),
or, more tightly,
(17)L(a,b)≤Λf,g(a,b)≤I(a,b),
hold for each a,b∈ℝ+?As an illustration, consider the functionfs(t) defined to be
(18)fs(t)={ts-st+s-1s(s-1),s(s-1)≠0;t-logt-1,s=0;tlogt-t+1,s=1.Since(19)fs′(t)={ts-1-1s-1,s(s-1)≠0;1-lt,s=0;logt,s=1,fs′′(t)=ts-2,s∈ℝ,t>0,
it follows that fs(t) is a twice continuously differentiable convex function for s∈ℝ, t∈ℝ+.Moreover, it is evident that(fs+1,fs)∈Λ.We will give in the sequel a complete answer to the above question concerning the means(20)f-s+1(a,b)f-s(a,b):=λs(a,b)
defined by(21)λs(a,b)={s-1s+1as+1+bs+1-2((a+b)/2)s+1as+bs-2((a+b)/2)s,s∈ℝ/{-1,0,1};2log((a+b)/2)-loga-logb1/2a+1/2b-2/(a+b),s=-1;aloga+blogb-(a+b)log((a+b)/2)2log((a+b)/2)-loga-logb,s=0;(b-a)24(aloga+blogb-(a+b)log((a+b)/2)),s=1.Those means are obviously symmetric and homogeneous of order one.As a consequence we obtain some new intermediate mean values; for instance, we show that the inequalities(22)H(a,b)≤λ-1(a,b)≤G(a,b)≤λ0(a,b)≤L(a,b)≤λ1(a,b)≤I(a,b)
hold for arbitrary a,b∈ℝ+.  Note that
(23)λ-1=2G2log(A/G)A-H;λ0=Alog(S/A)log(A/G);λ1=12A-Hlog(S/A).


## 2. Results

We prove firstly the followingTheorem 1.
Letf,g∈C2(I) with g′′>0. The expression Λf,g(a,b) represents a mean of arbitrary numbers a,b∈I if and only if the relation (15) holds for t∈I.Remark 2.
In the same way, for arbitraryp,q>0,p+q=1, it can be deduced that the quotient
(24)Λf,g(p,q;a,b):=pf(a)+qf(b)-f(pa+qb)pg(a)+qg(b)-g(pa+qb)
represents a mean value of numbers a, b if and only if (15) holds.A generalization of the above assertion is the next.Theorem 3.
Letf,g:I→ℝ be twice continuously differentiable functions with g′′>0 on I and let p={pi}, i=1,2,…,∑pi=1 be an arbitrary positive weight sequence. Then the quotient of two Jensen functionals
(25)Λf,g(p,x):=∑1npif(xi)-f(∑1npixi)∑1npig(xi)-g(∑1npixi),n≥2,
represents a mean of an arbitrary set of real numbers x1,x2,…,xn∈I if and only if the relation
(26)f′′(t)=tg′′(t)
holds for each t∈I.Remark 4.
It should be noted that the relationf′′(t)=tg′′(t) determines f in terms of g in an easy way. Precisely,
(27)f(t)=tg(t)-2G(t)+ct+d,
where G(t):=∫1t‍g(u)du and c and d are constants.Our results concerning the meansλs(a,b), s∈ℝ are included in the following.Theorem 5.
For the class of meansλs(a,b) defined above, the following assertions hold for each a,b∈ℝ+.(1)
The meansλs(a,b) are monotone increasing in s;(2)
λs(a,b)≤H(a,b) for each s≤-4;(3)
H(a,b)≤λs(a,b)≤G(a,b) for -3≤s≤-1;(4)
G(a,b)≤λs(a,b)≤L(a,b) for -1/2≤s≤0;(5)
there is a numbers0∈(1/12,1/11) such that  L(a,b)≤λs(a,b)≤I(a,b) for s0≤s≤1;(6)
there is a numbers1∈(1.03,1.04) such that  I(a,b)≤λs(a,b)≤A(a,b) for s1≤s≤2;(7)
A(a,b)≤λs(a,b)≤S(a,b) for each 2≤s≤5;(8)
there is no finites such that the inequality S(a,b)≤λs(a,b) holds for each a,b∈ℝ+.
The above estimations are best possible.


## 3. Proofs

### 3.1. Proof of Theorem1

We prove firstly the necessity of the condition (15).SinceΛf,g(a,b) is a mean value for arbitrary a,b∈I; a≠b, we have
(28)min{a,b}≤Λf,g(a,b)≤max{a,b}.
Hence
(29)limb→aΛf,g(a,b)=a.From the other hand, due to l’Hospital’s rule we obtain(30)limb→aΛf,g(a,b)=limb→a(f′(b)-f′((a+b)/2)g′(b)-g′((a+b)/2))=limb→a(2f′′(b)-f′′((a+b)/2)2g′′(b)-g′′((a+b)/2))=f′′(a)g′′(a).
Comparing (29) and (30) the desired result follows.Suppose now that (15) holds and let a<b. Since g′′(t)>0t∈[a,b] by the Cauchy mean value theorem there exists ξ∈((a+t)/2,t) such that
(31)f′(t)-f′((a+t)/2)g′(t)-g′((a+t)/2)=f′′(ξ)g′′(ξ)=ξ.But,(32)a≤a+t2<ξ<t≤b,
and, since g′ is strictly increasing,  g′(t)-g′((a+t)/2)>0,t∈[a,b].Therefore, by (31) we get
(33)a(g′(t)-g′(a+t2))≤f′(t)-f′(a+t2)≤b(g′(t)-g′(a+t2)).
Finally, integrating (33) over t∈[a,b] we obtain the assertion from Theorem 1.


### 3.2. Proof of Theorem3

We will give a proof of this assertion by induction onn.By Remark2, it holds for n=2.Next, it is not difficult to check the identity(34)∑1n‍pif(xi)-f(∑1n‍pixi)=(1-pn)(∑1n-1‍pi′f(xi)-f(∑1n-1‍pi′xi))+[(1-pn)f(T)+pnf(xn)-f((1-pn)T+pnxn)],
where
(35)T≔∑1n-1‍pi′xi;pi′≔pi(1-pn),i=1,2,…,n-1;∑1n-1‍pi′=1.Therefore, by induction hypothesis and Remark2, we get
(36)∑1n‍pif(xi)-f(∑1n‍pixi)≤max{x1,x2,…xn-1}(1-pn)×(∑1n-1‍pi′g(xi)-g(∑1n-1‍pi′xi))+max{T,xn}[(1-pn)g(T)+png(xn)+max{T,xn}-g((1-pn)T+pnxn)]≤max{x1,x2,…,xn}×((1-pn)(∑1n-1‍pi′g(xi)-g(∑1n-1‍pi′xi))+[(1-pn)g(T)+png(xn)-g((1-pn)T+pnxn)](1-pn)(∑1n-1‍p′ig(xi)-g(∑1n-1‍p′ixi)))=max{x1,x2,…,xn}(∑1n‍pig(xi)-g(∑1n‍pixi)).The inequality(37)min{x1,x2,…,xn}≤Λf,g(p,x)
can be proved analogously.For the proof of necessity, putx2=x3=⋯=xn and proceed as in Theorem 1.Remark 6.
It is evident from (15) that if I⊆ℝ+ then f has to be also convex on I. Otherwise, it shouldn't be the case. For example, the conditions of Theorem 3 are satisfied with f(t)=t3/3, g(t)=t2,t∈ℝ. Hence, for an arbitrary sequence {xi}1n of real numbers, we obtain
(38)min{x1,x2,…,xn}≤∑1npixi3-(∑1npixi)33(∑1npixi2-(∑1npixi)2)≤max{x1,x2,…,xn}.Because the above inequality does not depend onn, a probabilistic interpretation of the above result is contained in the following.Theorem 7.
For an arbitrary probability lawF of random variable X with support on (-∞,+∞), one has
(39)(EX)3+3(minX)σX2≤EX3≤(EX)3+3(maxX)σX2.


### 3.3. Proof of Theorem5, Part (1)

We will prove a general assertion of this type. Namely, for an arbitrary positive sequencex={xi} and an associated weight sequence p={pi}, i=1,2,…, denote
(40)χs(p,x):={∑pixis-(∑pixi)ss(s-1),s∈ℝ/{0,1};log(∑‍pixi)-∑pilogxi,s=0;∑‍pixilogxi-(∑‍pixi)log(∑‍pixi),s=1.Fors∈ℝ, r>0 we have
(41)χs(p,x)χs+r+1(p,x)≥χs+1(p,x)χs+r(p,x),
which is equivalent toTheorem 8.
The sequence
{
χ
s
+
1
(
p
,
x
)
/
χ
s
(
p
,
x
)
}  is monotone increasing ins, s∈ℝ.
This assertion follows applying the result from [7, Theorem  2] which states the following.Lemma 9.
For-∞<a<b<c<+∞, the inequality
(42)(χb(p,x))c-a≤(χa(p,x))c-b(χc(p,x))b-a
holds for arbitrary sequences p, x.Putting therea=s, b=s+1, c=s+r+1 and a=s, b=s+r, c=s+r+1, we successively obtain
(43)(χs+1(p,x))r+1≤(χs(p,x))rχs+r+1(p,x),(χs+r(p,x))r+1≤χs(p,x)(χs+r+1(p,x))r.Sincer>0, multiplying those inequalities we get the relation (41), that is, the proof of Theorem 8.The part (1) of Theorem5 follows for p1=p2=1/2.A general way to prove the rest of Theorem5 is to use an easy-checkable identity
(44)λs(a,b)A(a,b)=λs(1+t,1-t),
with t:=(b-a)/(b+a).Since0<a<b, we get 0<t<1. Also,
(45)H(a,b)A(a,b)=1-t2;G(a,b)A(a,b)=1-t2;L(a,b)A(a,b)=2tlog(1+t)-log(1-t);I(a,b)A(a,b)=exp((1+t)log(1+t)-(1-t)log(1-t)2t-1);S(a,b)A(a,b)=exp(12((1+t)log(1+t)+(1-t)log(1-t))).Therefore, we have to compare some one-variable inequalities and to check their validness for eacht∈(0,1).For example, we will prove that the inequality(46)λs(a,b)≤L(a,b)
holds for each positive a,b if and only if s≤0.Sinceλs(a,b) is monotone increasing in s, it is enough to prove that
(47)λ0(a,b)L(a,b)≤1.By the above formulae, this is equivalent to the assertion that the inequality(48)ϕ(t)≤0
holds for each t∈(0,1), with
(49)ϕ(t):=log(1+t)-log(1-t)2t×((1+t)log(1+t)+(1-t)log(1-t))+log(1+t)+log(1-t).We will prove that the power series expansion ofϕ(t) have non-positive coefficients. Thus the relation (48) will be proved.Since(50)log(1+t)-log(1-t)2t=∑0∞‍t2k2k+1;log(1+t)+log(1-t)=-t2∑0∞‍t2kk+1;(1+t)log(1+t)+(1-t)log(1-t)=t2∑0∞‍t2k(k+1)(2k+1),
we get
(51)ϕ(t)t2=∑n=0∞‍(-1n+1+∑k=0n‍1(2n-2k+1)(k+1)(2k+1))t2n=∑0∞‍cnt2n.Hence,(52)c0=c1=0;c2=-190,
and, after some calculation, we get
(53)cn=2(n+1)(2n+3)((n+2)∑1n‍12k+1-(n+1)∑1n‍12k),n>1.Now, one can easily prove (by induction, e.g.) that(54)dn:=(n+2)∑1n‍12k+1-(n+1)∑1n12k
is a negative real number for n≥2. Therefore cn≤0, and the proof of the first part is done.  For 0<s<1 we have
(55)λs(a,b)L(a,b)-1=(1-s)((1+t)s+1+(1-t)s+1-2)log((1+t)/(1-t))2t(1+s)(2-(1+t)s-(1-t)s)-1=16st2+O(t4)(t→0).Therefore,λs(a,b)>L(a,b)  for s>0 and sufficiently small t:=(b-a)/(b+a).Similarly, we will prove that the inequality(56)λs(a,b)≤I(a,b)
holds for each a, b; 0<a<b if and only if s≤1.As before, it is enough to consider the expression(57)I(a,b)λ1(a,b)=eμ(t)ν(t):=ψ(t),
with
(58)μ(t)=(1+t)log(1+t)-(1-t)log(1-t)2t-1;ν(t)=(1+t)log(1+t)+(1-t)log(1-t)t2.It is not difficult to check the identity(59)ψ′(t)=-eμ(t)ϕ(t)t3.Hence by (48), we get ψ′(t)>0, that is, ψ(t) is monotone increasing for t∈(0,1).Therefore(60)I(a,b)λ1(a,b)≥limt→0+ψ(t)=1.By monotonicity it follows thatλs(a,b)≤I(a,b) for s≤1.Fors>1,(b-a)/(b+a)=t, we have
(61)λs(a,b)-I(a,b)=(16(s-1)t2+O(t4))A(a,b)(t→0+).Hence,λs(a,b)>I(a,b) for s>1 and t sufficiently small.From the other hand,(62)limt→1-[λs(a,b)I(a,b)-1]=e(s-1)(2s+1-2)2(s+1)(2s-2)-1:=τ(s).Examining the functionτ(s), we find out that it has the only real zero at s0≈1.0376 and is negative for s∈(1,s0).Remark 10.
Sinceψ(t) is monotone increasing, we also get
(63)I(a,b)λ1(a,b)≤limt→1-ψ(t)=4log2e.
Hence
(64)1≤I(a,b)λ1(a,b)≤4log2e.
A calculation gives4log2/e≈1.0200.Note also that(65)λ2(a,b)≡A(a,b).Therefore, applying the assertion from the part 1, we get(66)λs(a,b)≤A(a,b),s≤2;λs(a,b)≥A(a,b),s≥2.Finally, we give a detailed proof of the part 7.We have to prove thatλs(a,b)≤S(a,b) for s≤5. Since λs(a,b) is monotone increasing in s, it is sufficient to prove that the inequality
(67)λ5(a,b)≤S(a,b)
holds for each a,b∈ℝ+.Therefore, by the transformation given above, we get(68)logλ5A=log[23(1+t)6+(1-t)6-2(1+t)5+(1-t)5-2]=log[21515+15t2+t42+t2]≤log[1+t2+t4/41+t2/2]=log(1+t22)=t22-t48+t624-⋯≤t22+t412+t630+⋯=12((1+t)log(1+t)+(1-t)log(1-t))=logSA,
and the proof is done.Further, we have to show thatλs(a,b)>S(a,b) for some positive a, b whenever s>5.Indeed, since(69)(1+t)s+(1-t)s-2=(s2)t2+(s4)t4+O(t6),
for s>5 and sufficiently small t, we get
(70)λsA=s-1s+1(s+12)t2+(s+14)t4+O(t6)(s2)t2+(s4)t4+O(t6)=1+(s-1)(s-2)t2/12+O(t4)1+(s-2)(s-3)t2/12+O(t4)=1+(s6-13)t2+O(t4).
Similarly,
(71)SA=exp(12((1+t)log(1+t)+(1-t)log(1-t)))=exp(t22+O(t4))=1+t22+O(t4).
Hence,
(72)1A(λs-S)=16(s-5)t2+O(t4),
and this expression is positive for s>5 and t sufficiently small, that is, a sufficiently close to b.As for the part 8, applying the above transformation we obtain(73)λs(a,b)S(a,b)=s-1s+1(1+t)s+1+(1-t)s+1-2(1+t)s+(1-t)s-2×exp(-12((1+t)log(1+t)+(1-t)log(1-t))),
where 0<a<b, t=(b-a)/(b+a).Since fors>5,
(74)limt→1-λsS=s-1s+12s-12s-2,
and the last expression is less than one, it follows that the inequality S(a,b)<λs(a,b) cannot hold whenever b/a is sufficiently large.The rest of the proof is straightforward.


## 3.1. Proof of Theorem1

We prove firstly the necessity of the condition (15).SinceΛf,g(a,b) is a mean value for arbitrary a,b∈I; a≠b, we have
(28)min{a,b}≤Λf,g(a,b)≤max{a,b}.
Hence
(29)limb→aΛf,g(a,b)=a.From the other hand, due to l’Hospital’s rule we obtain(30)limb→aΛf,g(a,b)=limb→a(f′(b)-f′((a+b)/2)g′(b)-g′((a+b)/2))=limb→a(2f′′(b)-f′′((a+b)/2)2g′′(b)-g′′((a+b)/2))=f′′(a)g′′(a).
Comparing (29) and (30) the desired result follows.Suppose now that (15) holds and let a<b. Since g′′(t)>0t∈[a,b] by the Cauchy mean value theorem there exists ξ∈((a+t)/2,t) such that
(31)f′(t)-f′((a+t)/2)g′(t)-g′((a+t)/2)=f′′(ξ)g′′(ξ)=ξ.But,(32)a≤a+t2<ξ<t≤b,
and, since g′ is strictly increasing,  g′(t)-g′((a+t)/2)>0,t∈[a,b].Therefore, by (31) we get
(33)a(g′(t)-g′(a+t2))≤f′(t)-f′(a+t2)≤b(g′(t)-g′(a+t2)).
Finally, integrating (33) over t∈[a,b] we obtain the assertion from Theorem 1.


## 3.2. Proof of Theorem3

We will give a proof of this assertion by induction onn.By Remark2, it holds for n=2.Next, it is not difficult to check the identity(34)∑1n‍pif(xi)-f(∑1n‍pixi)=(1-pn)(∑1n-1‍pi′f(xi)-f(∑1n-1‍pi′xi))+[(1-pn)f(T)+pnf(xn)-f((1-pn)T+pnxn)],
where
(35)T≔∑1n-1‍pi′xi;pi′≔pi(1-pn),i=1,2,…,n-1;∑1n-1‍pi′=1.Therefore, by induction hypothesis and Remark2, we get
(36)∑1n‍pif(xi)-f(∑1n‍pixi)≤max{x1,x2,…xn-1}(1-pn)×(∑1n-1‍pi′g(xi)-g(∑1n-1‍pi′xi))+max{T,xn}[(1-pn)g(T)+png(xn)+max{T,xn}-g((1-pn)T+pnxn)]≤max{x1,x2,…,xn}×((1-pn)(∑1n-1‍pi′g(xi)-g(∑1n-1‍pi′xi))+[(1-pn)g(T)+png(xn)-g((1-pn)T+pnxn)](1-pn)(∑1n-1‍p′ig(xi)-g(∑1n-1‍p′ixi)))=max{x1,x2,…,xn}(∑1n‍pig(xi)-g(∑1n‍pixi)).The inequality(37)min{x1,x2,…,xn}≤Λf,g(p,x)
can be proved analogously.For the proof of necessity, putx2=x3=⋯=xn and proceed as in Theorem 1.Remark 6.
It is evident from (15) that if I⊆ℝ+ then f has to be also convex on I. Otherwise, it shouldn't be the case. For example, the conditions of Theorem 3 are satisfied with f(t)=t3/3, g(t)=t2,t∈ℝ. Hence, for an arbitrary sequence {xi}1n of real numbers, we obtain
(38)min{x1,x2,…,xn}≤∑1npixi3-(∑1npixi)33(∑1npixi2-(∑1npixi)2)≤max{x1,x2,…,xn}.Because the above inequality does not depend onn, a probabilistic interpretation of the above result is contained in the following.Theorem 7.
For an arbitrary probability lawF of random variable X with support on (-∞,+∞), one has
(39)(EX)3+3(minX)σX2≤EX3≤(EX)3+3(maxX)σX2.


## 3.3. Proof of Theorem5, Part (1)

We will prove a general assertion of this type. Namely, for an arbitrary positive sequencex={xi} and an associated weight sequence p={pi}, i=1,2,…, denote
(40)χs(p,x):={∑pixis-(∑pixi)ss(s-1),s∈ℝ/{0,1};log(∑‍pixi)-∑pilogxi,s=0;∑‍pixilogxi-(∑‍pixi)log(∑‍pixi),s=1.Fors∈ℝ, r>0 we have
(41)χs(p,x)χs+r+1(p,x)≥χs+1(p,x)χs+r(p,x),
which is equivalent toTheorem 8.
The sequence
{
χ
s
+
1
(
p
,
x
)
/
χ
s
(
p
,
x
)
}  is monotone increasing ins, s∈ℝ.
This assertion follows applying the result from [7, Theorem  2] which states the following.Lemma 9.
For-∞<a<b<c<+∞, the inequality
(42)(χb(p,x))c-a≤(χa(p,x))c-b(χc(p,x))b-a
holds for arbitrary sequences p, x.Putting therea=s, b=s+1, c=s+r+1 and a=s, b=s+r, c=s+r+1, we successively obtain
(43)(χs+1(p,x))r+1≤(χs(p,x))rχs+r+1(p,x),(χs+r(p,x))r+1≤χs(p,x)(χs+r+1(p,x))r.Sincer>0, multiplying those inequalities we get the relation (41), that is, the proof of Theorem 8.The part (1) of Theorem5 follows for p1=p2=1/2.A general way to prove the rest of Theorem5 is to use an easy-checkable identity
(44)λs(a,b)A(a,b)=λs(1+t,1-t),
with t:=(b-a)/(b+a).Since0<a<b, we get 0<t<1. Also,
(45)H(a,b)A(a,b)=1-t2;G(a,b)A(a,b)=1-t2;L(a,b)A(a,b)=2tlog(1+t)-log(1-t);I(a,b)A(a,b)=exp((1+t)log(1+t)-(1-t)log(1-t)2t-1);S(a,b)A(a,b)=exp(12((1+t)log(1+t)+(1-t)log(1-t))).Therefore, we have to compare some one-variable inequalities and to check their validness for eacht∈(0,1).For example, we will prove that the inequality(46)λs(a,b)≤L(a,b)
holds for each positive a,b if and only if s≤0.Sinceλs(a,b) is monotone increasing in s, it is enough to prove that
(47)λ0(a,b)L(a,b)≤1.By the above formulae, this is equivalent to the assertion that the inequality(48)ϕ(t)≤0
holds for each t∈(0,1), with
(49)ϕ(t):=log(1+t)-log(1-t)2t×((1+t)log(1+t)+(1-t)log(1-t))+log(1+t)+log(1-t).We will prove that the power series expansion ofϕ(t) have non-positive coefficients. Thus the relation (48) will be proved.Since(50)log(1+t)-log(1-t)2t=∑0∞‍t2k2k+1;log(1+t)+log(1-t)=-t2∑0∞‍t2kk+1;(1+t)log(1+t)+(1-t)log(1-t)=t2∑0∞‍t2k(k+1)(2k+1),
we get
(51)ϕ(t)t2=∑n=0∞‍(-1n+1+∑k=0n‍1(2n-2k+1)(k+1)(2k+1))t2n=∑0∞‍cnt2n.Hence,(52)c0=c1=0;c2=-190,
and, after some calculation, we get
(53)cn=2(n+1)(2n+3)((n+2)∑1n‍12k+1-(n+1)∑1n‍12k),n>1.Now, one can easily prove (by induction, e.g.) that(54)dn:=(n+2)∑1n‍12k+1-(n+1)∑1n12k
is a negative real number for n≥2. Therefore cn≤0, and the proof of the first part is done.  For 0<s<1 we have
(55)λs(a,b)L(a,b)-1=(1-s)((1+t)s+1+(1-t)s+1-2)log((1+t)/(1-t))2t(1+s)(2-(1+t)s-(1-t)s)-1=16st2+O(t4)(t→0).Therefore,λs(a,b)>L(a,b)  for s>0 and sufficiently small t:=(b-a)/(b+a).Similarly, we will prove that the inequality(56)λs(a,b)≤I(a,b)
holds for each a, b; 0<a<b if and only if s≤1.As before, it is enough to consider the expression(57)I(a,b)λ1(a,b)=eμ(t)ν(t):=ψ(t),
with
(58)μ(t)=(1+t)log(1+t)-(1-t)log(1-t)2t-1;ν(t)=(1+t)log(1+t)+(1-t)log(1-t)t2.It is not difficult to check the identity(59)ψ′(t)=-eμ(t)ϕ(t)t3.Hence by (48), we get ψ′(t)>0, that is, ψ(t) is monotone increasing for t∈(0,1).Therefore(60)I(a,b)λ1(a,b)≥limt→0+ψ(t)=1.By monotonicity it follows thatλs(a,b)≤I(a,b) for s≤1.Fors>1,(b-a)/(b+a)=t, we have
(61)λs(a,b)-I(a,b)=(16(s-1)t2+O(t4))A(a,b)(t→0+).Hence,λs(a,b)>I(a,b) for s>1 and t sufficiently small.From the other hand,(62)limt→1-[λs(a,b)I(a,b)-1]=e(s-1)(2s+1-2)2(s+1)(2s-2)-1:=τ(s).Examining the functionτ(s), we find out that it has the only real zero at s0≈1.0376 and is negative for s∈(1,s0).Remark 10.
Sinceψ(t) is monotone increasing, we also get
(63)I(a,b)λ1(a,b)≤limt→1-ψ(t)=4log2e.
Hence
(64)1≤I(a,b)λ1(a,b)≤4log2e.
A calculation gives4log2/e≈1.0200.Note also that(65)λ2(a,b)≡A(a,b).Therefore, applying the assertion from the part 1, we get(66)λs(a,b)≤A(a,b),s≤2;λs(a,b)≥A(a,b),s≥2.Finally, we give a detailed proof of the part 7.We have to prove thatλs(a,b)≤S(a,b) for s≤5. Since λs(a,b) is monotone increasing in s, it is sufficient to prove that the inequality
(67)λ5(a,b)≤S(a,b)
holds for each a,b∈ℝ+.Therefore, by the transformation given above, we get(68)logλ5A=log[23(1+t)6+(1-t)6-2(1+t)5+(1-t)5-2]=log[21515+15t2+t42+t2]≤log[1+t2+t4/41+t2/2]=log(1+t22)=t22-t48+t624-⋯≤t22+t412+t630+⋯=12((1+t)log(1+t)+(1-t)log(1-t))=logSA,
and the proof is done.Further, we have to show thatλs(a,b)>S(a,b) for some positive a, b whenever s>5.Indeed, since(69)(1+t)s+(1-t)s-2=(s2)t2+(s4)t4+O(t6),
for s>5 and sufficiently small t, we get
(70)λsA=s-1s+1(s+12)t2+(s+14)t4+O(t6)(s2)t2+(s4)t4+O(t6)=1+(s-1)(s-2)t2/12+O(t4)1+(s-2)(s-3)t2/12+O(t4)=1+(s6-13)t2+O(t4).
Similarly,
(71)SA=exp(12((1+t)log(1+t)+(1-t)log(1-t)))=exp(t22+O(t4))=1+t22+O(t4).
Hence,
(72)1A(λs-S)=16(s-5)t2+O(t4),
and this expression is positive for s>5 and t sufficiently small, that is, a sufficiently close to b.As for the part 8, applying the above transformation we obtain(73)λs(a,b)S(a,b)=s-1s+1(1+t)s+1+(1-t)s+1-2(1+t)s+(1-t)s-2×exp(-12((1+t)log(1+t)+(1-t)log(1-t))),
where 0<a<b, t=(b-a)/(b+a).Since fors>5,
(74)limt→1-λsS=s-1s+12s-12s-2,
and the last expression is less than one, it follows that the inequality S(a,b)<λs(a,b) cannot hold whenever b/a is sufficiently large.The rest of the proof is straightforward.



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*Source: 283127-2013-02-12.xml*