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 alpha,beta inR\alpha, \beta \in \mathbb{R} for which the inequality 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) holds for all a,b > 0,a!=ba, b>0, a \neq b, where G(a,b),L(a,b)G(a, b), L(a, b) and C(a,b)C(a, b) are respectively the geometric, logarithmic and anti-harmonic means of aa and bb. MSC 2000. 26-04, 26D15, 26E60. Keywords. Two-variable means, weighted arithmetic mean, inequalities, symbolic computer algebra.
1. INTRODUCTION
Given a,b > 0,a!=ba, b>0, a \neq b, the geometric, logarithmic and anti-harmonic means are defined by
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}
It is well-known that
{:(1)G < L < C:}\begin{equation*}
G<L<C \tag{1}
\end{equation*}
In this paper we find the values of the parameters alpha,beta inR\alpha, \beta \in \mathbb{R} for which the inequality
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)
holds for all positive numbers a!=ba \neq 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:}\begin{equation*}
H<G<L<I<A<Q<S<C \tag{2}
\end{equation*}
are proved in [2]. For those means Symbolic Algebra Program Maple was used to find the interval where the parameters alpha\alpha and beta\beta 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,LG, L and CC is among the more difficult ones.
2. MAIN RESULT
Theorem 1. The inequality
{:(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*}
holds for all positive numbers a!=ba \neq b if and only if alpha >= 1\alpha \geq 1 and beta < beta_(0)\beta<\beta_{0}, where beta_(0)=g(x_(0))=0.87002762 dots\beta_{0}=g\left(x_{0}\right)=0.87002762 \ldots, with x_(0)x_{0} the unique root of (7) which is greater than 1 , and with gg defined in (6).
Proof. The double inequality (3) is equivalent to
{:(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*}
Without loss of generality, we can consider 0 < a < b0<a<b. Denoting by t=b//at=b / a, t > 1t>1, due to the homogeneity of the means, the problem reduces to find i n f f\inf f and s u p f\sup f, where
{:(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*}
The function ff is obviously bounded, 0 <= f(t) <= 10 \leq f(t) \leq 1. We shall find i n f f\inf f and s u p f\sup f for t > 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)","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*}
In order to find i n f f=i n f g\inf f=\inf g and s u p f=s u p g\sup f=\sup g we shall show first that
{:(*)g^(')" has a unique root in "(1","oo).:}\begin{equation*}
g^{\prime} \text { has a unique root in }(1, \infty) . \tag{*}
\end{equation*}
Suppose for a moment that this is true and denote by x_(0)x_{0} this root. We have 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 / 9. It follows that gg has a minimal point in ( 1,oo1, \infty ), so this point must be x_(0)x_{0}. Furthermore, gg must be monotonic in (1,x_(0))\left(1, x_{0}\right) and (x_(0),oo)\left(x_{0}, \infty\right) and so 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)=1.
So, it remains to prove (*).
The derivative of gg is given by
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}}
where
{:[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*}
The equation g^(')(x)=0g^{\prime}(x)=0 is equivalent to h(x)=0h(x)=0, hence to
{:(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*}
where
{:[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*}
We have considered in (8) the positive root of the quadratic in ln x\ln x equation h(x)=0h(x)=0. Let us denote the left hand side of (8) by k(x)k(x).
We have to show that kk has a unique root in ( 1,oo1, \infty ). To this aim we compute k^(')(x)k^{\prime}(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^{\prime}(x) expressed in terms of d=sqrtpd=\sqrt{p}, where pp 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;]:}\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}
The numerator of k^(')(x)k^{\prime}(x) is given by
{:[ > 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}
{:[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}
Therefore the numerator ndkn d k of k^(')(x)k^{\prime}(x) is of the form p_(1)d+p_(0)p_{1} d+p_{0} ( p_(0)p_{0} and p_(1)p_{1} being polynomials) and a root of k^(')(x)k^{\prime}(x) must be a root of the polynomial p_(1)^(2)p-p_(0)^(2)p_{1}^{2} p-p_{0}^{2}.
We can factorize this polynomial using Maple:
{:[ > 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}
It follows that any root of k^(')(x)k^{\prime}(x) in (1,oo)(1, \infty) must be a root of the 10 th degree polynomial
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 .
But the polynomial PP has a unique root in (1,oo)(1, \infty). This can be verified using the Sturm sequence.
Indeed, Maple gives:
> sturm(P,x,0>\operatorname{sturm}(\mathrm{P}, \mathrm{x}, 0, infinity);
1
We conclude that k^(')k^{\prime} has a unique root r in(1,oo)r \in(1, \infty); actually r in(4,5)r \in(4,5) because k^(')(4) > 0,k^(')(5) < 0k^{\prime}(4)>0, k^{\prime}(5)<0. So, k^(') > 0k^{\prime}>0 in ( 1,r1, r ) and k^(') < 0k^{\prime}<0 in ( r,oor, \infty ). Since k(1)=0k(1)=0 and lim_(x rarr oo)k(x)=-oo\lim _{x \rightarrow \infty} k(x)=-\infty it follows that kk has a unique root in ( 1,oo1, \infty ), actually in ( r,oor, \infty ). So, we have proved (*).
The unique solution x_(0)x_{0} of g^(')(x)=0g^{\prime}(x)=0 can be easily approximated by using the command
Digits:=30:
x0:=fsolve(h(x), x=4\mathrm{x}=4..infinity);
x0:=7.27177296398582281915348781959x 0:=7.27177296398582281915348781959
giving g(x_(0))=0.87002762 dotsg\left(x_{0}\right)=0.87002762 \ldots
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.