# Sharp Bounds for the Weighted Geometric Mean of the First Seiffert and Logarithmic Means in terms of Weighted Generalized Heronian Mean

Authors: Ladislav Matejíčka
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Optimal bounds for the weighted geometric mean of the first Seiffert and logarithmic means by weighted generalized Heronian mean are proved. We answer the question: forα∈(0,1), what the greatest value p(α) and the least value q(α) such that the double inequality, Hp(α)(a,b)<Pα(a,b)L1-α(a,b)<Hq(α)(a,b), holds for all a,b>0 with a≠b are. Here, P(a,b),L(a,b), and Hω(a,b) denote the first Seiffert, logarithmic, and weighted generalized Heronian means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Recently, means has been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert, logarithmic, and Heronian mean can be found in the literature [1–11]. In the paper [1], authors proved the following optimal inequalities:Leta>0, b>0, a≠b then
(1)Hδ(a,b)<P(a,b)<Hβ(a,b)forδ≥π-2,β≤1,δ=π-2,β=1arethebestconstants.Hγ(a,b)<L(a,b)<Hτ(a,b)forγ=+∞,τ≤4,γ=+∞,τ=4arethebestconstants.P(a,b) is the first Seiffert mean, which was introduced by Seiffert in [9]
(2)P(a,b)=a-b4arctan(a/b)-π=a-b2arcsin((a-b)/(a+b))fora,b>0,a≠b.
In [9], Seiffert proved that L(a,b)<P(a,b)<I(a,b), where I(a,b) is the identric mean
(3)I(a,b)=1e(aabb)1/(b-a)ifa≠b,I(a,a)=a.L(a,b) is the logarithmic mean
(4)L(a,b)=a-bloga-logbfora,b>0,a≠b.Gα(a,b) is the weighted geometric mean
(5)Gα(a,b)=aαb1-αfora,b>0,0≤α≤1.Hω(a,b) is the weighted generalized Heronian mean introduced by Janous [7]
(6)Hω(a,b)=a+ωab+bω+2for0≤ω<+∞=abforω=+∞.
It is well known, that Hω(a,b) is a strictly decreasing continuous function of the argument ω. From this and from results of [1], it is natural to assume that there exist optimal functions p(α), q(α), 0≤α≤1 such that
(7)Hp(α)(a,b)<Pα(a,b)L1-α(a,b)<Hq(α)(a,b).
The purpose of this paper is to find the optimal functions. For some other details about means, see [1–11] and the related references cited there in.


## 2. Main Results

The main result of this paper is the following theorem.Theorem 1.
Leta,b>0, a≠b, α∈(0,1). Then
(8)Hp(a,b)<Pα(a,b)L1-α(a,b)<Hq(a,b),forp=+∞,q≤q(α),
where p=p(α)=+∞, q(α)=2(2-α)/(1+α) are the best possible functions.Proof.
First, we prove the left inequality of (8). The inequalities (1) imply that
(9)H+∞(a,b)<Pα(a,b)L1-α(a,b)fora,b>0,a≠b,0<α<1.
From limt→0+G(t,α)=+∞ for α∈(0,1) (see (14)) we obtain that p(α)=+∞ is the optimal function.
Without loss of generality, we assume that0<a<b. Let t=a/b; then 0<t<1. The right inequality of (8) can be rewritten as
(10)1bαPα(a,b)1b1-αL1-α(a,b)<1bHq(a,b)fora,b>0,a≠b,0<α<1.
Simple computations lead to
(11)1-t2(π-4arctant)α(-2lnt)1-α-t2+qt+1q+2<0for0<t<1,0<α<1.
Then the inequality (11) is equivalent to
(12)q(1-t2-t(π-4arctant)α(-2lnt)1-α)<(1+t2)(π-4arctant)α(-2lnt)1-α-2(1-t2).
Denote
(13)s(t,α)=1-t2-t(π-4arctant)α(-2lnt)1-α,r(t)=π-4arctant+2lnt,v(t)=t2-2tlnt-1.
From r(1)=0 and r′(t)=(2-4t+2t2)/(t+t3)>0 we have r(t)<0 for t∈(0,1). It implies ((π-4arctant)/(-2lnt))α<1. From v(1)=0, v′(1)=0, v′′(t)=2-2/t<0 we obtain v′(t)>0 and so v(t)<0. It implies that s(t,α)>0 for t, α∈(0,1). This leads to
(14)q<G(t,α)=(1+t2)(π-4arctant)α(-2lnt)1-α-2(1-t2)1-t2-t(π-4arctant)α(-2lnt)1-α.
If we show Gt′(t,α)<0 for t,α∈(0,1), then q(α)=limt→1-G(t,α) will be the best function in (8). Simple computations lead to Gt′(t,α)<0 which is equivalent to
(15)H(t,α)=2(1-t)lnt+4α(1+t)(1-t)2lntπ-4arctant-(1-α)(1+t)(1-t)2t+2(1+t)ln2t(π-4arctant-2lnt)α<0.
Using the inequality tα<1-α(1-t) for t,α∈(0,1) it suffices to show that
(16)R(t,α)=2(1-t)lnt+4α(1+t)(1-t)2lntπ-4arctant-(1-α)(1+t)(1-t)2t+(2lnt+α(-2lnt-π+4arctant))(1+t)lnt<0.
It will be done, if we show R(t,0)<0 and R(t,1)<0. It follows from R(t,α) being a linear continuous function in the argument α(17)R(t,0)=2(1-t)lnt-(1+t)(1-t)2t+2(1+t)ln2t<0
is equivalent to
(18)s(t)=2(1-t)lnt1+t-(1-t)2t+2ln2t<0.
From s(1)=0 it suffices to show that s′(t)>0 which is equivalent to
(19)v(t)=lnt+(1+3t-3t2-t3)(1+t)4t(1+t+t2)>0.
It follows from v(1)=0 and
(20)v′(t)=w(t)4t2(1+t+t2)2<0,
where w(t)=-(1-t)2(1-t2)2.
Next, we show that(21)R(t,1)=4(1-t)lnt+8(1+t)(1-t)2lntπ-4arctant-2(π-4arctant)(1+t)lnt<0.
The inequality (21) is equivalent to
(22)(π-4arctant-1-t1+t)2<(1-t1+t)2-4(1-t)2.
So, it suffices to show that
(23)g(t)=-π+4arctant+1-t1+t(1+1+4(1+t)2)>0.
It is easy to see that
(24)1+1+4(1+t)2>1+2(1+t)+15(1+t)=16+25t+10t25(1+t)for0≤t≤1.
Because of
(25)g(t)>g1(t)=-π+4arctant+1-t1+t(16+25t+10t25(1+t))>0,
it suffices to prove g1(t)>0 for 0<t<1. From
(26)g1(t)=15g2(t)=15(-5π+20arctant+(1-t)(16+25t+10t2)(1+t)2),arctan(t)>t-t3/3, for t∈(0,1), g2(1)=0 we have done it, if we show
(27)g3(t)=-5π+20(t-t33)+(1-t)(16+25t+10t2)(1+t)2>0,
on (0,0.67〉 and g2′(t)<0 on 〈0.67,1).
Simple computation gives(28)g2′(t)=201+t2-(23+39t+30t2+10t3(1+t)3).
The inequality g2′(t)<0 is equivalent to
(29)ch(t)=-3+21t+7t2-29t3-30t4-10t5<0.
From ch′′′(t)<0 we get ch′′(t) is a decreasing function. ch′′(0.67)=-324.3366 implies ch′′(t)<0 on 〈0.67,1). So, we obtain ch′(t) is a decreasing function. From ch′(0.67)=-54.8414 we have ch′(t)<0 on 〈0.67,1). It implies that ch(t) is a decreasing function. From ch(0.67)=-1.9053 we get ch(t)<0 on 〈0.67,1). So g2′(t)<0 on 〈0.67,1).
Next, we showg3(t)>0 on (0,0.67〉.
Simple computation gives(30)g3(t)=(10316-5π+t(29-10π)+t2(25-5π)+t3103-t4403-t5203)((1+t)2)-1.
The inequality g3(t)>0 is equivalent to
(31)h(t)=16-5π+t(29-10π)+t2(25-5π)+t3103-t4403-t5203>0,
on (0,0.67〉. From h(0)=16-5π>0, h(0.1)=0.1453, h(0.15)=0.1427, h(0.67)=0.2602, h′(0.15)=0.3998 it suffices to show that h′(t)<0 on (0,0.1〉; h(t)>0 on 〈0.1,0.15〉 and h′(t) has only one root in (0.15,0.67).
First, we showh(t)>0 on 〈0.1,0.15〉. From t3>0.1t2, t4<0.152t2, t5<0.153t2 we have
(32)h(t)>16-5π+t(29-10π)+t2(25-5π+0.1103-0.152403-0.153203)>l(t),
where
(33)l(t)=16-5π+t(29-10π)+t29.3.
It is easy to see that l′(t)=0 for t=(10π-29)/18.6=0.1299. From l′′(t)>0 on 〈0.1,0.15〉 and l(0.1299)=0.1351 we have l(t)>0. It implies h(t)>0 on 〈0.1,0.15〉.
Next, we showh′(t)<0 on (0,0.1〉. Simple computation gives
(34)h′(t)=(29-10π)+(50-10π)t+10t2-1603t3-1003t4<j(t),
where
(35)j(t)=(29-10π)+(50-10π)t+10t2,j(0)=29-10π<0, j′′(t)>0, j(0.1)=-0.45750 imply j(t)<0 so h′(t)<0 on (0,0.1〉.
Finally, we show thath′(t) has only one root on (0.15,0.67). From h′′′′(t)<0 we obtain h′′′(t) is a decreasing function. Because of h′′′(0.15)=-37 we have h′′′(t)<0 on (0.15,0.67) so h′(t) is a concave function. From h′(0.15)=0.3998 and h′(0.67)=-8.2333 we have that h′(t) has only one root on (0.15,0.67). It implies h(t)>0 on 〈0.15,0.67〉. So, the proof of decreasing of G(t,α) is complete.
In what follows, we find the representation of the functionq(α).
It is easy to see that(36)q(α)=limt→1-((-2lnt)1-α)-1((1+t2)(π-4arctant)α×(-2lnt)1-α-2(1-t2))×(1-t2-t(π-4arctant)α×(-2lnt)1-α)-1((1+t2)(π-4arctant)α).
Equation (36) can be rewritten as
(37)q(α)=limt→1-(1+t2)Y(t,α)U(t,α)-2(1+t)1+t-tY(t,α)U(t,α),
where
(38)Y(t,α)=(π-4arctant1-t)α,U(t,α)=(-2lnt1-t)1-α.
Simple computations give
(39)Y(t,α)=2α(1+1-t2+(1-t)26+y(t)(1-t)3),
where y(t) is a suitable function. Similarly we have
(40)U(t,α)=21-α(1+1-t2+(1-t)23+u(t)(1-t)3),
where u(t) is a suitable function. Denote S(t,α)=Y(t,α)U(t,α). Then
(41)S(t,α)=2(1+1-t2+(2-α)(1-t)26+s(t)(1-t)3),
where s(t) is a suitable function. Using the L’Hospital's rule we obtain
(42)q(α)=limt→1-(1+t2)S(t,α)-2(1+t)1+t-tS(t,α)=limt→1-2S(t,α)+4tSt′(t,α)+(1+t2)Stt′′(t,α)-2St′(t,α)-tStt′′(t,α)=2(2-α)1+α.
The proof is complete.



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