Logarithmic mean and weighted sum of geometric and anti-harmonic means

Mira Anisiu
(original), paper

Abstract

We consider the problem of finding the optimal values\U{3b1}, \(\U{3b2} \in R\) for which the inequality \(\U{3b1} G(a,b)+(1-\U{3b1} )C(a,b)<L(a,b)<\U{3b2} G(a,b)+(1-\U{3b2} )C(a,b)\) holds for all \(a,\) \(b>0,a\neq b\), where \(G(a,b),L(a,b)\) and \(C(a,b)\) are respectivelythe geometric, logarithmic and anti-harmonic means of \(a\) and \(b\).

Authors

Mira-Cristiana Anisiu
T. Popoviciu Institute of Numerical Analysis, Cluj-Napoca, Romania

Valeriu Anisiu
Babes-Bolyai University, Faculty of Mathematics and Computer Science,  Cluj-Napoca, Romania

Keywords

Two-variable means; weighted arithmetic mean; inequalities; symbolic computer algebra.

Paper coordinates

M.-C. Anisiu, V. Anisiu, Logarithmic mean and weighted sum of geometric and anti-harmonic means, Rev. Anal. Numér. Théor. Approx. ,41 (2012) no. 2, pp. 95-98, https://doi.org/10.33993/jnaat412-971.

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https://ictp.acad.ro/jnaat/journal/article/view/2012-vol41-no2-art1/Anisiu-Anisiu-pdf

About this paper

Journal
Revue d’Analyse Numerique et de Theorie de l’Approximation
Publisher Name

Romanian Academy

DOI

https://doi.org/10.33993/jnaat412-971

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1222-9024

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[1] H. AlzerandS.-L. Qiu,Inequalities for means in two variables, Arch. Math. (Basel),80(2003), pp. 201–215.
[2] M.-C. AnisiuandV. Anisiu,Bilateral inequalities for means, communication, 9th Joint Conference on Mathematics and Computer Science (MACS 2012), Siofok, Hungary, Feb-ruary 9–12, Abstracts, p. 15, 2012.
[3] Y.-M.Chu, Y.-F. Qiu, M.-K. Wang and G.-D. Wang,The optimal convex combina-tion bounds of arithmetic and harmonic means for the Seiffert’s mean, J. Inequal. Appl.,2010, Article ID 436457, 2010, 7 pages, doi:10.1155/2010/436457.
[4] W.-F. XiaandY.-M. Chu, Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means, Rev. Anal. Numer. Theor. Approx.,39(2010) no. 2,pp. 176–183.Received by the editors: July 12, 2012.

jnaat2,+Journal+manager,+2012-2-Anisiu-Anisiu-15-10-19

LOGARITHMIC MEAN AND WEIGHTED SUM OF GEOMETRIC AND ANTI-HARMONIC MEANS

MIRA-CRISTIANA ANISIU* and VALERIU ANISIU † † ^(†){ }^{\dagger}†

Abstract

We consider the problem of finding the optimal values α , β ∈ R α , β ∈ R alpha,beta inR\alpha, \beta \in \mathbb{R}α,β∈R for which the inequality α G ( a , b ) + ( 1 − α ) C ( a , b ) < L ( a , b ) < β G ( a , b ) + ( 1 − β ) C ( a , b ) α G ( a , b ) + ( 1 − α ) C ( a , b ) < L ( a , b ) < β G ( a , b ) + ( 1 − β ) C ( a , b ) alpha G(a,b)+(1-alpha)C(a,b) < L(a,b) < beta G(a,b)+(1-beta)C(a,b)\alpha G(a, b)+(1-\alpha) C(a, b)<L(a, b)<\beta G(a, b)+(1-\beta) C(a, b)αG(a,b)+(1−α)C(a,b)<L(a,b)<βG(a,b)+(1−β)C(a,b) holds for all a , b > 0 , a ≠ b a , b > 0 , a ≠ b a,b > 0,a!=ba, b>0, a \neq ba,b>0,a≠b, where G ( a , b ) , L ( a , b ) G ( a , b ) , L ( a , b ) G(a,b),L(a,b)G(a, b), L(a, b)G(a,b),L(a,b) and C ( a , b ) C ( a , b ) C(a,b)C(a, b)C(a,b) are respectively the geometric, logarithmic and anti-harmonic means of a a aaa and b b bbb. MSC 2000. 26-04, 26D15, 26E60. Keywords. Two-variable means, weighted arithmetic mean, inequalities, symbolic computer algebra.

1. INTRODUCTION

Given a , b > 0 , a ≠ b a , b > 0 , a ≠ b a,b > 0,a!=ba, b>0, a \neq ba,b>0,a≠b, the geometric, logarithmic and anti-harmonic means are defined by
G = a b , L = b − a ln ⁡ b − ln ⁡ a , C = a 2 + b 2 a + b G = a b , L = b − a ln ⁡ b − ln ⁡ a , C = a 2 + b 2 a + b G=sqrt(ab),quad L=(b-a)/(ln b-ln a),quad C=(a^(2)+b^(2))/(a+b)G=\sqrt{a b}, \quad L=\frac{b-a}{\ln b-\ln a}, \quad C=\frac{a^{2}+b^{2}}{a+b}G=ab,L=b−aln⁡b−ln⁡a,C=a2+b2a+b
It is well-known that
(1) G < L < C (1) G < L < C {:(1)G < L < C:}\begin{equation*} G<L<C \tag{1} \end{equation*}(1)G<L<C
In this paper we find the values of the parameters α , β ∈ R α , β ∈ R alpha,beta inR\alpha, \beta \in \mathbb{R}α,β∈R for which the inequality
α G ( a , b ) + ( 1 − α ) C ( a , b ) < L ( a , b ) < β G ( a , b ) + ( 1 − β ) C ( a , b ) α G ( a , b ) + ( 1 − α ) C ( a , b ) < L ( a , b ) < β G ( a , b ) + ( 1 − β ) C ( a , b ) alpha G(a,b)+(1-alpha)C(a,b) < L(a,b) < beta G(a,b)+(1-beta)C(a,b)\alpha G(a, b)+(1-\alpha) C(a, b)<L(a, b)<\beta G(a, b)+(1-\beta) C(a, b)αG(a,b)+(1−α)C(a,b)<L(a,b)<βG(a,b)+(1−β)C(a,b)
holds for all positive numbers a ≠ b a ≠ b a!=ba \neq ba≠b.
Recently, results of this type have been obtained for various triplets of means. Not being exhaustive, we mention Alzer and Qiu [1] for geometric, exponential (identric) and arithmetic means, Xia and Chu 4 for harmonic, logarithmic respectively identric and arithmetic means, and Chu et al. [3] for harmonic, Seiffert and arithmetic means. Several theorems concerning three means chosen from
(2) H < G < L < I < A < Q < S < C (2) H < G < L < I < A < Q < S < C {:(2)H < G < L < I < A < Q < S < C:}\begin{equation*} H<G<L<I<A<Q<S<C \tag{2} \end{equation*}(2)H<G<L<I<A<Q<S<C
are proved in [2]. For those means Symbolic Algebra Program Maple was used to find the interval where the parameters α α alpha\alphaα and β β beta\betaβ can vary, and then the proofs were given.
We can use Maple also to understand the expected degree of difficulty of the proof. Doing so, we found that the problem involving the means G , L G , L G,LG, LG,L and C C CCC is among the more difficult ones.

2. MAIN RESULT

Theorem 1. The inequality
(3) α G ( a , b ) + ( 1 − α ) C ( a , b ) < L ( a , b ) < β G ( a , b ) + ( 1 − β ) C ( a , b ) (3) α G ( a , b ) + ( 1 − α ) C ( a , b ) < L ( a , b ) < β G ( a , b ) + ( 1 − β ) C ( a , b ) {:(3)alpha G(a","b)+(1-alpha)C(a","b) < L(a","b) < beta G(a","b)+(1-beta)C(a","b):}\begin{equation*} \alpha G(a, b)+(1-\alpha) C(a, b)<L(a, b)<\beta G(a, b)+(1-\beta) C(a, b) \tag{3} \end{equation*}(3)αG(a,b)+(1−α)C(a,b)<L(a,b)<βG(a,b)+(1−β)C(a,b)
holds for all positive numbers a ≠ b a ≠ b a!=ba \neq ba≠b if and only if α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1 and β < β 0 β < β 0 beta < beta_(0)\beta<\beta_{0}β<β0, where β 0 = g ( x 0 ) = 0.87002762 … β 0 = g x 0 = 0.87002762 … beta_(0)=g(x_(0))=0.87002762 dots\beta_{0}=g\left(x_{0}\right)=0.87002762 \ldotsβ0=g(x0)=0.87002762…, with x 0 x 0 x_(0)x_{0}x0 the unique root of (7) which is greater than 1 , and with g g ggg defined in (6).
Proof. The double inequality (3) is equivalent to
(4) β < C ( a , b ) − L ( a , b ) C ( a , b ) − G ( a , b ) < α (4) β < C ( a , b ) − L ( a , b ) C ( a , b ) − G ( a , b ) < α {:(4)beta < (C(a,b)-L(a,b))/(C(a,b)-G(a,b)) < alpha:}\begin{equation*} \beta<\frac{C(a, b)-L(a, b)}{C(a, b)-G(a, b)}<\alpha \tag{4} \end{equation*}(4)β<C(a,b)−L(a,b)C(a,b)−G(a,b)<α
Without loss of generality, we can consider 0 < a < b 0 < a < b 0 < a < b0<a<b0<a<b. Denoting by t = b / a t = b / a t=b//at=b / at=b/a, t > 1 t > 1 t > 1t>1t>1, due to the homogeneity of the means, the problem reduces to find inf f inf f i n f f\inf finff and sup f sup f s u p f\sup fsupf, where
(5) f ( t ) = C ( 1 , t ) − L ( 1 , t ) C ( 1 , t ) − G ( 1 , t ) = ( t 2 + 1 ) ln ⁡ t − t 2 + 1 ( t − 1 ) 2 ( t + t + 1 ) ln ⁡ t (5) f ( t ) = C ( 1 , t ) − L ( 1 , t ) C ( 1 , t ) − G ( 1 , t ) = t 2 + 1 ln ⁡ t − t 2 + 1 ( t − 1 ) 2 ( t + t + 1 ) ln ⁡ t {:(5)f(t)=(C(1,t)-L(1,t))/(C(1,t)-G(1,t))=((t^(2)+1)ln t-t^(2)+1)/((sqrtt-1)^(2)(t+sqrtt+1)ln t):}\begin{equation*} f(t)=\frac{C(1, t)-L(1, t)}{C(1, t)-G(1, t)}=\frac{\left(t^{2}+1\right) \ln t-t^{2}+1}{(\sqrt{t}-1)^{2}(t+\sqrt{t}+1) \ln t} \tag{5} \end{equation*}(5)f(t)=C(1,t)−L(1,t)C(1,t)−G(1,t)=(t2+1)ln⁡t−t2+1(t−1)2(t+t+1)ln⁡t
The function f f fff is obviously bounded, 0 ≤ f ( t ) ≤ 1 0 ≤ f ( t ) ≤ 1 0 <= f(t) <= 10 \leq f(t) \leq 10≤f(t)≤1. We shall find inf f inf f i n f f\inf finff and sup f sup f s u p f\sup fsupf for t > 1 t > 1 t > 1t>1t>1.
Define
(6) g ( x ) = f ( x 2 ) = 2 ( x 4 + 1 ) ln ⁡ x − x 4 + 1 2 ( x − 1 ) 2 ( x 2 + x + 1 ) ln ⁡ x , x > 1 . (6) g ( x ) = f x 2 = 2 x 4 + 1 ln ⁡ x − x 4 + 1 2 ( x − 1 ) 2 x 2 + x + 1 ln ⁡ x , x > 1 . {:(6)g(x)=f(x^(2))=(2(x^(4)+1)ln x-x^(4)+1)/(2(x-1)^(2)(x^(2)+x+1)ln x)","quad x > 1.:}\begin{equation*} g(x)=f\left(x^{2}\right)=\frac{2\left(x^{4}+1\right) \ln x-x^{4}+1}{2(x-1)^{2}\left(x^{2}+x+1\right) \ln x}, \quad x>1 . \tag{6} \end{equation*}(6)g(x)=f(x2)=2(x4+1)ln⁡x−x4+12(x−1)2(x2+x+1)ln⁡x,x>1.
In order to find inf f = inf g inf f = inf g i n f f=i n f g\inf f=\inf ginff=infg and sup f = sup g sup f = sup g s u p f=s u p g\sup f=\sup gsupf=supg we shall show first that
(*) g ′ has a unique root in ( 1 , ∞ ) . (*) g ′  has a unique root in  ( 1 , ∞ ) . {:(*)g^(')" has a unique root in "(1","oo).:}\begin{equation*} g^{\prime} \text { has a unique root in }(1, \infty) . \tag{*} \end{equation*}(*)g′ has a unique root in (1,∞).
Suppose for a moment that this is true and denote by x 0 x 0 x_(0)x_{0}x0 this root. We have lim x → 1 g ( x ) = 8 / 9 , lim x → ∞ g ( x ) = 1 , g ( 7 ) = 0.87003995 … < 8 / 9 lim x → 1   g ( x ) = 8 / 9 , lim x → ∞   g ( x ) = 1 , g ( 7 ) = 0.87003995 … < 8 / 9 lim_(x rarr1)g(x)=8//9,lim_(x rarr oo)g(x)=1,g(7)=0.87003995 dots < 8//9\lim _{x \rightarrow 1} g(x)=8 / 9, \lim _{x \rightarrow \infty} g(x)=1, g(7)=0.87003995 \ldots<8 / 9limx→1g(x)=8/9,limx→∞g(x)=1,g(7)=0.87003995…<8/9. It follows that g g ggg has a minimal point in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ), so this point must be x 0 x 0 x_(0)x_{0}x0. Furthermore, g g ggg must be monotonic in ( 1 , x 0 ) 1 , x 0 (1,x_(0))\left(1, x_{0}\right)(1,x0) and ( x 0 , ∞ ) x 0 , ∞ (x_(0),oo)\left(x_{0}, \infty\right)(x0,∞) and so β 0 = inf g = g ( x 0 ) , sup g = max ( 1 , 8 / 9 ) = 1 β 0 = inf g = g x 0 , sup g = max ( 1 , 8 / 9 ) = 1 beta_(0)=i n f g=g(x_(0)),s u p g=max(1,8//9)=1\beta_{0}=\inf g=g\left(x_{0}\right), \sup g= \max (1,8 / 9)=1β0=infg=g(x0),supg=max(1,8/9)=1.
So, it remains to prove (*).
The derivative of g g ggg is given by
g ′ ( x ) = h ( x ) 2 x ( x − 1 ) 3 ( x 2 + x + 1 ) 2 ( ln ⁡ x ) 2 g ′ ( x ) = h ( x ) 2 x ( x − 1 ) 3 x 2 + x + 1 2 ( ln ⁡ x ) 2 g^(')(x)=(h(x))/(2x(x-1)^(3)(x^(2)+x+1)^(2)(ln x)^(2))g^{\prime}(x)=\frac{h(x)}{2 x(x-1)^{3}\left(x^{2}+x+1\right)^{2}(\ln x)^{2}}g′(x)=h(x)2x(x−1)3(x2+x+1)2(ln⁡x)2
where
h ( x ) = − 2 x ( x + 1 ) ( x 4 + 4 x 2 + 1 ) ( ln ⁡ x ) 2 (7) + x ( x − 1 ) ( x 4 + 2 x 3 + 6 x 2 + 2 x + 1 ) ln ⁡ x + ( x + 1 ) ( x 2 + x + 1 ) ( x 2 + 1 ) ( x − 1 ) 2 h ( x ) = − 2 x ( x + 1 ) x 4 + 4 x 2 + 1 ( ln ⁡ x ) 2 (7) + x ( x − 1 ) x 4 + 2 x 3 + 6 x 2 + 2 x + 1 ln ⁡ x + ( x + 1 ) x 2 + x + 1 x 2 + 1 ( x − 1 ) 2 {:[h(x)=-2x(x+1)(x^(4)+4x^(2)+1)(ln x)^(2)],[(7)+x(x-1)(x^(4)+2x^(3)+6x^(2)+2x+1)ln x],[+(x+1)(x^(2)+x+1)(x^(2)+1)(x-1)^(2)]:}\begin{align*} h(x)= & -2 x(x+1)\left(x^{4}+4 x^{2}+1\right)(\ln x)^{2} \\ & +x(x-1)\left(x^{4}+2 x^{3}+6 x^{2}+2 x+1\right) \ln x \tag{7}\\ & +(x+1)\left(x^{2}+x+1\right)\left(x^{2}+1\right)(x-1)^{2} \end{align*}h(x)=−2x(x+1)(x4+4x2+1)(ln⁡x)2(7)+x(x−1)(x4+2x3+6x2+2x+1)ln⁡x+(x+1)(x2+x+1)(x2+1)(x−1)2
The equation g ′ ( x ) = 0 g ′ ( x ) = 0 g^(')(x)=0g^{\prime}(x)=0g′(x)=0 is equivalent to h ( x ) = 0 h ( x ) = 0 h(x)=0h(x)=0h(x)=0, hence to
(8) ln ⁡ x − ( x 5 + 2 x 4 + 6 x 3 + 2 x 2 + x + p ) ( x − 1 ) 4 x ( x 5 + x 4 + 4 x 3 + 4 x 2 + x + 1 ) = 0 (8) ln ⁡ x − x 5 + 2 x 4 + 6 x 3 + 2 x 2 + x + p ( x − 1 ) 4 x x 5 + x 4 + 4 x 3 + 4 x 2 + x + 1 = 0 {:(8)ln x-((x^(5)+2x^(4)+6x^(3)+2x^(2)+x+sqrtp)(x-1))/(4x(x^(5)+x^(4)+4x^(3)+4x^(2)+x+1))=0:}\begin{equation*} \ln x-\frac{\left(x^{5}+2 x^{4}+6 x^{3}+2 x^{2}+x+\sqrt{p}\right)(x-1)}{4 x\left(x^{5}+x^{4}+4 x^{3}+4 x^{2}+x+1\right)}=0 \tag{8} \end{equation*}(8)ln⁡x−(x5+2x4+6x3+2x2+x+p)(x−1)4x(x5+x4+4x3+4x2+x+1)=0
where
p = 8 x 11 + 25 x 10 + 76 x 9 + 160 x 8 + 236 x 7 + 286 x 6 (9) + 236 x 5 + 160 x 4 + 76 x 3 + 25 x 2 + 8 x . p = 8 x 11 + 25 x 10 + 76 x 9 + 160 x 8 + 236 x 7 + 286 x 6 (9) + 236 x 5 + 160 x 4 + 76 x 3 + 25 x 2 + 8 x . {:[p=8x^(11)+25x^(10)+76x^(9)+160x^(8)+236x^(7)+286x^(6)],[(9)+236x^(5)+160x^(4)+76x^(3)+25x^(2)+8x.]:}\begin{align*} p= & 8 x^{11}+25 x^{10}+76 x^{9}+160 x^{8}+236 x^{7}+286 x^{6} \\ & +236 x^{5}+160 x^{4}+76 x^{3}+25 x^{2}+8 x . \tag{9} \end{align*}p=8x11+25x10+76x9+160x8+236x7+286x6(9)+236x5+160x4+76x3+25x2+8x.
We have considered in (8) the positive root of the quadratic in ln ⁡ x ln ⁡ x ln x\ln xln⁡x equation h ( x ) = 0 h ( x ) = 0 h(x)=0h(x)=0h(x)=0. Let us denote the left hand side of (8) by k ( x ) k ( x ) k(x)k(x)k(x).
We have to show that k k kkk has a unique root in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ). To this aim we compute k ′ ( x ) k ′ ( x ) k^(')(x)k^{\prime}(x)k′(x). The Computer Algebra System Maple will help us to do and organize the computations.
We are interested in the numerator of k ′ ( x ) k ′ ( x ) k^(')(x)k^{\prime}(x)k′(x) expressed in terms of d = p d = p d=sqrtpd=\sqrt{p}d=p, where p p ppp is the polynomial given in (9), i.e.
> p := 8 ∗ x ∧ 11 + 25 ∗ x ∧ 10 + 76 ∗ x ∧ 9 + 160 ∗ x ∧ 8 + 236 ∗ x ∧ 7 + 286 ∗ x ∧ 6 + 236 ∗ x ∧ 5 + 160 ∗ x ∧ 4 + 76 ∗ x ∧ 3 + 25 ∗ x ∧ 2 + 8 ∗ x ; > p := 8 ∗ x ∧ 11 + 25 ∗ x ∧ 10 + 76 ∗ x ∧ 9 + 160 ∗ x ∧ 8 + 236 ∗ x ∧ 7 + 286 ∗ x ∧ 6 + 236 ∗ x ∧ 5 + 160 ∗ x ∧ 4 + 76 ∗ x ∧ 3 + 25 ∗ x ∧ 2 + 8 ∗ x ; {:[ > p:=8**x^11+25**x^10+76**x^9+160**x^8+236**x^7+286**x^6+236**x^5],[+160**x^4+76**x^3+25**x^2+8**x;]:}\begin{aligned} & >p:=8 * x^{\wedge} 11+25 * x^{\wedge} 10+76 * x^{\wedge} 9+160 * x^{\wedge} 8+236 * x^{\wedge} 7+286 * x^{\wedge} 6+236 * x^{\wedge} 5 \\ & +160 * x^{\wedge} 4+76 * x^{\wedge} 3+25 * x^{\wedge} 2+8 * x ; \end{aligned}>p:=8∗x∧11+25∗x∧10+76∗x∧9+160∗x∧8+236∗x∧7+286∗x∧6+236∗x∧5+160∗x∧4+76∗x∧3+25∗x∧2+8∗x;
The numerator of k ′ ( x ) k ′ ( x ) k^(')(x)k^{\prime}(x)k′(x) is given by
> numer ( normal ( subs ( p = d ∧ 2 , normal ( diff ( k ( x ) , x ) ) ) ) ) assuming d > 0 : > ndk := collect ( % , d ) ; > numer normal subs p = d ∧ 2 , normal ( diff ( k ( x ) , x ) )  assuming  d > 0 : > ndk := collect ( % , d ) ; {:[ > numer(normal(subs(p=d^2,normal(diff(k(x),x)))))],[quad" assuming "d > 0:],[ > ndk:=collect(%","d);]:}\begin{aligned} & >\operatorname{numer}\left(\operatorname{normal}\left(\operatorname{subs}\left(\mathrm{p}=\mathrm{d}^{\wedge} 2, \operatorname{normal}(\operatorname{diff}(\mathrm{k}(\mathrm{x}), \mathrm{x}))\right)\right)\right) \\ & \quad \text { assuming } \mathrm{d}>0: \\ & >\operatorname{ndk}:=\operatorname{collect}(\%, \mathrm{~d}) ; \end{aligned}>numer(normal(subs(p=d∧2,normal(diff(k(x),x))))) assuming d>0:>ndk:=collect(%, d);
n d k := ( 9 x 2 + 10 x 3 + 20 x 4 + 2 x + 2 x 9 + 20 x 6 + 9 x 8 + 10 x 7 + x 10 + 24 x 5 + 1 ) d − 1 − 4 x − 658 x 8 − 306 x 5 − 586 x 7 − 478 x 6 − 478 x 10 − 586 x 9 − 306 x 11 − 154 x 12 − 64 x 13 − 64 x 3 − 22 x 2 − 154 x 4 − x 16 − 22 x 14 − 4 x 15 n d k := 9 x 2 + 10 x 3 + 20 x 4 + 2 x + 2 x 9 + 20 x 6 + 9 x 8 + 10 x 7 + x 10 + 24 x 5 + 1 ) d − 1 − 4 x − 658 x 8 − 306 x 5 − 586 x 7 − 478 x 6 − 478 x 10 − 586 x 9 − 306 x 11 − 154 x 12 − 64 x 13 − 64 x 3 − 22 x 2 − 154 x 4 − x 16 − 22 x 14 − 4 x 15 {:[ndk:=(9x^(2)+10x^(3)+20x^(4)+2x+2x^(9)+20x^(6)+9x^(8)+10x^(7)+x^(10)+24x^(5):}],[+1)d-1-4x-658x^(8)-306x^(5)-586x^(7)-478x^(6)-478x^(10)-586x^(9)],[-306x^(11)-154x^(12)-64x^(13)-64x^(3)-22x^(2)-154x^(4)-x^(16)-22x^(14)-4x^(15)]:}\begin{aligned} n d k:= & \left(9 x^{2}+10 x^{3}+20 x^{4}+2 x+2 x^{9}+20 x^{6}+9 x^{8}+10 x^{7}+x^{10}+24 x^{5}\right. \\ & +1) d-1-4 x-658 x^{8}-306 x^{5}-586 x^{7}-478 x^{6}-478 x^{10}-586 x^{9} \\ & -306 x^{11}-154 x^{12}-64 x^{13}-64 x^{3}-22 x^{2}-154 x^{4}-x^{16}-22 x^{14}-4 x^{15} \end{aligned}ndk:=(9x2+10x3+20x4+2x+2x9+20x6+9x8+10x7+x10+24x5+1)d−1−4x−658x8−306x5−586x7−478x6−478x10−586x9−306x11−154x12−64x13−64x3−22x2−154x4−x16−22x14−4x15
Therefore the numerator n d k n d k ndkn d kndk of k ′ ( x ) k ′ ( x ) k^(')(x)k^{\prime}(x)k′(x) is of the form p 1 d + p 0 p 1 d + p 0 p_(1)d+p_(0)p_{1} d+p_{0}p1d+p0 ( p 0 p 0 p_(0)p_{0}p0 and p 1 p 1 p_(1)p_{1}p1 being polynomials) and a root of k ′ ( x ) k ′ ( x ) k^(')(x)k^{\prime}(x)k′(x) must be a root of the polynomial p 1 2 p − p 0 2 p 1 2 p − p 0 2 p_(1)^(2)p-p_(0)^(2)p_{1}^{2} p-p_{0}^{2}p12p−p02.
We can factorize this polynomial using Maple:
> p 0 := coeff ( ndk , d , 0 ) : > p 1 := coeff ( ndk , d , 1 ) : > factor ( p 1 ∧ 2 ∗ p − p 0 ∧ 2 ) ; − ( x − 1 ) 4 ( x + 1 ) 4 ( x 2 + 1 ) 2 ( x 2 + x + 1 ) ( x 4 + 4 x 2 + 1 ) 2 ( x 10 − x 9 − 3 x 8 − 44 x 7 − 94 x 6 − 150 x 5 − 94 x 4 − 44 x 3 − 3 x 2 − x + 1 ) > p 0 := coeff ( ndk , d , 0 ) : > p 1 := coeff ( ndk , d , 1 ) : >  factor  p 1 ∧ 2 ∗ p − p 0 ∧ 2 ; − ( x − 1 ) 4 ( x + 1 ) 4 x 2 + 1 2 x 2 + x + 1 x 4 + 4 x 2 + 1 2 x 10 − x 9 − 3 x 8 − 44 x 7 − 94 x 6 − 150 x 5 − 94 x 4 − 44 x 3 − 3 x 2 − x + 1 {:[ > p0:=coeff(ndk","d","0):],[ > p1:=coeff(ndk","d","1):],[ > " factor "(p1^2**p-p0^2);],[quad-(x-1)^(4)(x+1)^(4)(x^(2)+1)^(2)(x^(2)+x+1)(x^(4)+4x^(2)+1)^(2)],[(x^(10)-x^(9)-3x^(8)-44x^(7)-94x^(6)-150x^(5)-94x^(4)-44x^(3)-3x^(2)-x+1)]:}\begin{aligned} & >\mathrm{p} 0:=\operatorname{coeff}(\mathrm{ndk}, \mathrm{~d}, 0): \\ & >\mathrm{p} 1:=\operatorname{coeff}(\mathrm{ndk}, \mathrm{~d}, 1): \\ & >\text { factor }\left(\mathrm{p} 1^{\wedge} 2 * \mathrm{p}-\mathrm{p} 0^{\wedge} 2\right) ; \\ & \quad-(x-1)^{4}(x+1)^{4}\left(x^{2}+1\right)^{2}\left(x^{2}+x+1\right)\left(x^{4}+4 x^{2}+1\right)^{2} \\ & \left(x^{10}-x^{9}-3 x^{8}-44 x^{7}-94 x^{6}-150 x^{5}-94 x^{4}-44 x^{3}-3 x^{2}-x+1\right) \end{aligned}>p0:=coeff(ndk, d,0):>p1:=coeff(ndk, d,1):> factor (p1∧2∗p−p0∧2);−(x−1)4(x+1)4(x2+1)2(x2+x+1)(x4+4x2+1)2(x10−x9−3x8−44x7−94x6−150x5−94x4−44x3−3x2−x+1)
It follows that any root of k ′ ( x ) k ′ ( x ) k^(')(x)k^{\prime}(x)k′(x) in ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞) must be a root of the 10 th degree polynomial
P = x 10 − x 9 − 3 x 8 − 44 x 7 − 94 x 6 − 150 x 5 − 94 x 4 − 44 x 3 − 3 x 2 − x + 1 . P = x 10 − x 9 − 3 x 8 − 44 x 7 − 94 x 6 − 150 x 5 − 94 x 4 − 44 x 3 − 3 x 2 − x + 1 . P=x^(10)-x^(9)-3x^(8)-44x^(7)-94x^(6)-150x^(5)-94x^(4)-44x^(3)-3x^(2)-x+1.P=x^{10}-x^{9}-3 x^{8}-44 x^{7}-94 x^{6}-150 x^{5}-94 x^{4}-44 x^{3}-3 x^{2}-x+1 .P=x10−x9−3x8−44x7−94x6−150x5−94x4−44x3−3x2−x+1.
But the polynomial P P PPP has a unique root in ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞). This can be verified using the Sturm sequence.
Indeed, Maple gives:
> sturm ( P , x , 0 > sturm ( P , x , 0 > sturm(P,x,0>\operatorname{sturm}(\mathrm{P}, \mathrm{x}, 0>sturm(P,x,0, infinity);
1
We conclude that k ′ k ′ k^(')k^{\prime}k′ has a unique root r ∈ ( 1 , ∞ ) r ∈ ( 1 , ∞ ) r in(1,oo)r \in(1, \infty)r∈(1,∞); actually r ∈ ( 4 , 5 ) r ∈ ( 4 , 5 ) r in(4,5)r \in(4,5)r∈(4,5) because k ′ ( 4 ) > 0 , k ′ ( 5 ) < 0 k ′ ( 4 ) > 0 , k ′ ( 5 ) < 0 k^(')(4) > 0,k^(')(5) < 0k^{\prime}(4)>0, k^{\prime}(5)<0k′(4)>0,k′(5)<0. So, k ′ > 0 k ′ > 0 k^(') > 0k^{\prime}>0k′>0 in ( 1 , r 1 , r 1,r1, r1,r ) and k ′ < 0 k ′ < 0 k^(') < 0k^{\prime}<0k′<0 in ( r , ∞ r , ∞ r,oor, \inftyr,∞ ). Since k ( 1 ) = 0 k ( 1 ) = 0 k(1)=0k(1)=0k(1)=0 and lim x → ∞ k ( x ) = − ∞ lim x → ∞   k ( x ) = − ∞ lim_(x rarr oo)k(x)=-oo\lim _{x \rightarrow \infty} k(x)=-\inftylimx→∞k(x)=−∞ it follows that k k kkk has a unique root in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ), actually in ( r , ∞ r , ∞ r,oor, \inftyr,∞ ). So, we have proved (*).
The unique solution x 0 x 0 x_(0)x_{0}x0 of g ′ ( x ) = 0 g ′ ( x ) = 0 g^(')(x)=0g^{\prime}(x)=0g′(x)=0 can be easily approximated by using the command
Digits:=30:
x0:=fsolve(h(x), x = 4 x = 4 x=4\mathrm{x}=4x=4..infinity);
x 0 := 7.27177296398582281915348781959 x 0 := 7.27177296398582281915348781959 x0:=7.27177296398582281915348781959x 0:=7.27177296398582281915348781959x0:=7.27177296398582281915348781959
giving g ( x 0 ) = 0.87002762 … g x 0 = 0.87002762 … g(x_(0))=0.87002762 dotsg\left(x_{0}\right)=0.87002762 \ldotsg(x0)=0.87002762…

REFERENCES

[1] H. Alzer and S.-L. Qiu, Inequalities for means in two variables, Arch. Math. (Basel), 80 (2003), pp. 201-215.
[2] M.-C. Anisiu and V. Anisiu, Bilateral inequalities for means, communication, 9th Joint Conference on Mathematics and Computer Science (MACS 2012), Siófok, Hungary, February 9-12, Abstracts, p. 15, 2012.
[3] Y.-M.Chu, Y.-F. Qiu, M.-K. Wang and G.-D. Wang, The optimal convex combination bounds of arithmetic and harmonic means for the Seiffert's mean, J. Inequal. Appl., 2010, Article ID 436457, 2010, 7 pages, doi:10.1155/2010/436457.
[4] W.-F. XIA and Y.-M. Chu, Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means, Rev. Anal. Numér. Théor. Approx., 39 (2010) no. 2, pp. 176-183. 중
Received by the editors: July 12, 2012.
  1. *"T. Popoviciu" Institute of Numerical Analysis, P.O. Box 68, 400110 Cluj-Napoca, Romania, e-mail: mira@math.ubbcluj.ro.
    † † †\dagger† "Babeş-Bolyai" University, Faculty of Mathematics and Computer Science, 1 Kogălniceanu St., 400084 Cluj-Napoca, Romania, e-mail: anisiu@math.ubbcluj.ro.
2012

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