Abstract
Let \((M1,M2,M3)\) be three means in two variables chosen from \(H,G,L,I,A,Q,S,C\) so that \(M_{1}(a,b)<M_{2}(a,b)<M_{3}(a,b),0<a<b\). We consider the problem of finding \(\U{3b1})\, \(\U{3b2}\) \in R)\ for which \U{3b1} \(M_{1} (a,b)+(1-\U{3b1} )M_{3}(a,b)<M_{2}(a,b)<\U{3b2} M_{1}(a,b)+(1-\U{3b2})M_{3}(a,b)\). We solve the problem for the triplets \((G,L,A),(G,A,Q),(G,A,C),(G,Q,C),(A,Q,C),(A,S,C),(A,Q,S)\) and \((L,A,C)\). The Symbolic Algebra ProgramMapleis used to determine the range where some parameters can vary, or tofind the minimal polynomial for an algebraic number.
Authors
Mira-Cristiana Anisiu
Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania
Valeriu Anisiu
Babes-Bolyai University, Faculty of Mathematics and Computer Science, Kogalniceanu, Cluj-Napoca, Romania
Keywords
Two-variable means; weighted arithmetic mean; inequalities; sym-bolic computer algebra.
Paper coordinates
M.-C. Anisiu, V. Anisiu, Bilateral inequalities for means, Rev. Anal. Numér. Théor. Approx., 42 (2) (2013), 94-102.
About this paper
Journal
Revue d’analyse Numerique et de Theorie de l’Approximation
Publisher Name
Romanian Academy
Print ISSN
Online ISSN
1222-9024
google scholar link
[1] H. Alzer and S. L. Qiu, Inequalities for means in two variables, Arch. Math. (Basel),80(2003), pp. 201–215.
[2] H. Alzer and S. Ruscheweyh, On the intersection of two-parameter mean value families, Proc. A. M. S.,129(9) (2001), pp. 2655–2662.
[3] M. C. Anisiu and V. Anisiu, Refinement of some inequalities for means, Rev. Anal.Numer. Theor. Approx.,35(2006) no. 1, pp. 5–10.
[4] M. C. Anisiu and V. Anisiu, Logarithmic mean and weighted sum of geometric and anti-harmonic means, Rev. Anal. Numer. Theor. Approx.,41(2012) no. 2, pp. 95-98.
[5] P. S. Bullen,Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.
[6] C. Gini, Di una formula comprensiva delle medie, Metron,13(1938), pp. 3–22.
[7] M. Ivan and I. Rasa, Some inequalities for means, Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29, 2000, pp. 99–102.
[8] K. B. Stolarsky,Generalizations of the logarithmic mean, Math. Mag.,48(1975),pp. 87–92.[9]W. F. Xia and Y. 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.
[2] H. Alzer and S. Ruscheweyh, On the intersection of two-parameter mean value families, Proc. A. M. S.,129(9) (2001), pp. 2655–2662.
[3] M. C. Anisiu and V. Anisiu, Refinement of some inequalities for means, Rev. Anal.Numer. Theor. Approx.,35(2006) no. 1, pp. 5–10.
[4] M. C. Anisiu and V. Anisiu, Logarithmic mean and weighted sum of geometric and anti-harmonic means, Rev. Anal. Numer. Theor. Approx.,41(2012) no. 2, pp. 95-98.
[5] P. S. Bullen,Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.
[6] C. Gini, Di una formula comprensiva delle medie, Metron,13(1938), pp. 3–22.
[7] M. Ivan and I. Rasa, Some inequalities for means, Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29, 2000, pp. 99–102.
[8] K. B. Stolarsky,Generalizations of the logarithmic mean, Math. Mag.,48(1975),pp. 87–92.[9]W. F. Xia and Y. 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.
BILATERAL INEQUALITIES FOR MEANS
Abstract
Let ( M_(1),M_(2),M_(3)M_{1}, M_{2}, M_{3} ) be three means in two variables chosen from HH, G,L,I,A,Q,S,CG, L, I, A, Q, S, C so that M_(1)(a,b) < M_(2)(a,b) < M_(3)(a,b),quad0 < a < bM_{1}(a, b)<M_{2}(a, b)<M_{3}(a, b), \quad 0<a<b
We consider the problem of finding alpha,beta inR\alpha, \beta \in \mathbb{R} for which
alphaM_(1)(a,b)+(1-alpha)M_(3)(a,b) < M_(2)(a,b) < betaM_(1)(a,b)+(1-beta)M_(3)(a,b).\alpha M_{1}(a, b)+(1-\alpha) M_{3}(a, b)<M_{2}(a, b)<\beta M_{1}(a, b)+(1-\beta) M_{3}(a, b) .
We solve the problem for the triplets (G,L,A),(G,A,Q),(G,A,C),(G,Q,C)(G, L, A),(G, A, Q),(G, A, C),(G, Q, C), (A,Q,C),(A,S,C),(A,Q,S)(A, Q, C),(A, S, C),(A, Q, S) and (L,A,C)(L, A, C). The Symbolic Algebra Program Maple is used to determine the range where some parameters can vary, or to find the minimal polynomial for an algebraic number.
MSC 2000. 26D15, 26E60; 26-04
Keywords. Two-variable means, weighted arithmetic mean, inequalities, symbolic computer algebra.
MSC 2000. 26D15, 26E60; 26-04
Keywords. Two-variable means, weighted arithmetic mean, inequalities, symbolic computer algebra.
1. INTRODUCTION
We remind the definitions of the classical means, namely, for 0 < a < b0<a<b
- the arithmetic, geometric and harmonic ones
A=(a+b)/(2),quad G=sqrt(ab),quad H=(2ab)/(a+b)A=\frac{a+b}{2}, \quad G=\sqrt{a b}, \quad H=\frac{2 a b}{a+b}
as well as
- the Hölder and the anti-harmonic mean Q=((a^(2)+b^(2))/(2))^(1//2),C=(a^(2)+b^(2))/(a+b)Q=\left(\frac{a^{2}+b^{2}}{2}\right)^{1 / 2}, C=\frac{a^{2}+b^{2}}{a+b};
- the Pólya EG Szegő logarithmic mean, the exponential (or identric), and the weighted geometric mean
L=(b-a)/(ln b-ln a),quad I=(1)/(e)((b^(b))/(a^(a)))^(1//(b-a)),quad S=(a^(a)b^(b))^(1//(a+b))L=\frac{b-a}{\ln b-\ln a}, \quad I=\frac{1}{e}\left(\frac{b^{b}}{a^{a}}\right)^{1 /(b-a)}, \quad S=\left(a^{a} b^{b}\right)^{1 /(a+b)}
References on means and inequalities between them can be found in [5]. At first, the following inequalities between means were established
{:(1)H < G < L < I < A < Q < S < C:}\begin{equation*}
H<G<L<I<A<Q<S<C \tag{1}
\end{equation*}
followed by relations between some means and the arithmetic means of two others ([7], 3])
{:(2)L < (G+A)/(2)","quad(G+Q)/(2) < A < (G+C)/(2) < Q < (A+C)/(2) < S.:}\begin{equation*} L<\frac{G+A}{2}, \quad \frac{G+Q}{2}<A<\frac{G+C}{2}<Q<\frac{A+C}{2}<S . \tag{2} \end{equation*}
{:(2)L < (G+A)/(2)","quad(G+Q)/(2) < A < (G+C)/(2) < Q < (A+C)/(2) < S.:}\begin{equation*} L<\frac{G+A}{2}, \quad \frac{G+Q}{2}<A<\frac{G+C}{2}<Q<\frac{A+C}{2}<S . \tag{2} \end{equation*}
A more difficult problem is to obtain results of the type (2) for weighted arithmetic means and to determine the maximal interval for the parameter for which the inequalities hold.
We mention here an inequality proved by Alzer and Qiu for the means G,IG, I and AA.
Theorem 1. (1) The double inequality
{:(3)alpha A(a","b)+(1-alpha)G(a","b) < I(a","b) < beta A(a","b)+(1-beta)G(a","b):}\begin{equation*}
\alpha A(a, b)+(1-\alpha) G(a, b)<I(a, b)<\beta A(a, b)+(1-\beta) G(a, b) \tag{3}
\end{equation*}
holds true for all positive real numbers a!=ba \neq b, if and only if alpha <= 2//3\alpha \leq 2 / 3 and beta >= 2//e\beta \geq 2 / \mathrm{e}.
Results of this type continued to appear, recent ones are given in [9] for ( H,L,AH, L, A ) and ( H,I,AH, I, A ), and in 4 for ( G,L,CG, L, C ).
Let M_(1),M_(2),M_(3)M_{1}, M_{2}, M_{3} be three means out of the eight listed in (1) so that
{:(4)M_(1)(a","b) < M_(2)(a","b) < M_(3)(a","b).:}\begin{equation*}
M_{1}(a, b)<M_{2}(a, b)<M_{3}(a, b) . \tag{4}
\end{equation*}
We consider the problem of finding alpha,beta inR\alpha, \beta \in \mathbb{R} for which
{:(5)alphaM_(1)(a","b)+(1-alpha)M_(3)(a","b) < M_(2)(a","b):}\begin{equation*}
\alpha M_{1}(a, b)+(1-\alpha) M_{3}(a, b)<M_{2}(a, b) \tag{5}
\end{equation*}
and
{:(6)M_(2)(a","b) < betaM_(1)(a","b)+(1-beta)M_(3)(a","b).:}\begin{equation*}
M_{2}(a, b)<\beta M_{1}(a, b)+(1-\beta) M_{3}(a, b) . \tag{6}
\end{equation*}
The inequalities (5) and (6) are equivalent to
{:(7)alpha > (M_(3)(a,b)-M_(2)(a,b))/(M_(3)(a,b)-M_(1)(a,b))",":}\begin{equation*}
\alpha>\frac{M_{3}(a, b)-M_{2}(a, b)}{M_{3}(a, b)-M_{1}(a, b)}, \tag{7}
\end{equation*}
respectively
{:(8)beta < (M_(3)(a,b)-M_(2)(a,b))/(M_(3)(a,b)-M_(1)(a,b)).:}\begin{equation*}
\beta<\frac{M_{3}(a, b)-M_{2}(a, b)}{M_{3}(a, b)-M_{1}(a, b)} . \tag{8}
\end{equation*}
Basically, denoting by t=b//a,t > 1t=b / a, t>1, the problem reduces to find i n f f\inf f and s u p f\sup f, where
{:(9)f(t)=(M_(3)(1,t)-M_(2)(1,t))/(M_(3)(1,t)-M_(1)(1,t)).:}\begin{equation*}
f(t)=\frac{M_{3}(1, t)-M_{2}(1, t)}{M_{3}(1, t)-M_{1}(1, t)} . \tag{9}
\end{equation*}
The function ff is obviously bounded, 0 <= f(t) <= 10 \leq f(t) \leq 1. If s u p f\sup f is attained at some t in(1,oo)t \in(1, \infty), then alpha in(s u p f,oo)\alpha \in(\sup f, \infty); otherwise alpha in[s u p f,oo)\alpha \in[\sup f, \infty). Similarly, beta in(-oo,i n f f)\beta \in(-\infty, \inf f) if i n f f\inf f is attained in (1,oo)(1, \infty), and beta in(-oo,i n f f]\beta \in(-\infty, \inf f] otherwise. Symbolic Algebra Programs can be of great help to determine the range where the parameters can vary. Maple was used in [3] to find the interval for alpha\alpha in Theorem 9 below. We also use it to simplify the polynomials in the proof of Theorem 5 and to obtain the optimal value beta_(0)\beta_{0} of beta\beta.
Starting from the means listed in (1), we can formulate ((8)/(3))=56\binom{8}{3}=56 bilateral inequalities of the type (3). We shall choose seven of them, for which one of
(5) and (6) was already proved in [3], and we shall find the possible values of the parameter for the remaining one. Then, for L < A < CL<A<C we find the optimal intervals for alpha\alpha and beta\beta in order that both inequalities (5) and (6) hold. To this aim Maple is again very useful.
(5) and (6) was already proved in [3], and we shall find the possible values of the parameter for the remaining one. Then, for L < A < CL<A<C we find the optimal intervals for alpha\alpha and beta\beta in order that both inequalities (5) and (6) hold. To this aim Maple is again very useful.
2. BILATERAL INEQUALITIES
We consider means in two variables, but we prefer to use a simpler (and shorter) notation.
Let us denote for 0 < a < b,t=b//a,t > 10<a<b, t=b / a, t>1. It it obvious, due to the homogeneity, that, if M(a,b)M(a, b) is any mean from (1), it suffices to prove the inequalities for M(1,t)M(1, t). We shall write from now on M(t)M(t) instead of M(1,t)M(1, t).
Theorem 2. The double inequality
alpha G(t)+(1-alpha)A(t) < L(t) < beta G(t)+(1-beta)A(t),quad AA t > 1,\alpha G(t)+(1-\alpha) A(t)<L(t)<\beta G(t)+(1-\beta) A(t), \quad \forall t>1,
holds if and only if alpha >= 1\alpha \geq 1 and beta <= 2//3\beta \leq 2 / 3.
Proof. We denote, for t > 1t>1,
Proof. We denote, for t > 1t>1,
{:(10)f_(1)(t)=(A(t)-L(t))/(A(t)-G(t))=((t+1)ln t-2(t-1))/((t+1-2sqrtt)ln t).:}\begin{equation*}
f_{1}(t)=\frac{A(t)-L(t)}{A(t)-G(t)}=\frac{(t+1) \ln t-2(t-1)}{(t+1-2 \sqrt{t}) \ln t} . \tag{10}
\end{equation*}
Let us suppose that the first inequality in the theorem holds. From lim_(t rarr oo)f_(1)(t)=1\lim _{t \rightarrow \infty} f_{1}(t)=1 it follows obviously that alpha >= 1\alpha \geq 1. Conversely, if alpha >= 1\alpha \geq 1 it suffices to have
(A(t)-L(t))/(A(t)-G(t)) < 1,\frac{A(t)-L(t)}{A(t)-G(t)}<1,
which is true because L(t) > G(t)L(t)>G(t). We evaluate f_(1)(t)-2//3f_{1}(t)-2 / 3, where 2//3=lim_(t rarr1)f_(1)(t)2 / 3= \lim _{t \rightarrow 1} f_{1}(t) and show that it is positive. The denominator is obviously positive; we substitute u=sqrttu=\sqrt{t} in the numerator and obtain
f(u)=(u^(2)+4u+1)ln u-3u^(2)+3.f(u)=\left(u^{2}+4 u+1\right) \ln u-3 u^{2}+3 .
We have f(1)=f^(')(1)=f^('')(1)=0f(1)=f^{\prime}(1)=f^{\prime \prime}(1)=0 and f^(''')(u)=2(u-1)^(2)//u^(3) > 0f^{\prime \prime \prime}(u)=2(u-1)^{2} / u^{3}>0 for u > 1u>1, hence f_(1)(t) > 2//3f_{1}(t)>2 / 3 for t > 1t>1. It follows that L(t) < beta G(t)+(1-beta)A(t),AA t > 1L(t)<\beta G(t)+(1-\beta) A(t), \forall t>1 if and only if beta <= 2//3\beta \leq 2 / 3.
Theorem 3. The double inequality
alpha G(t)+(1-alpha)Q(t) < A(t) < beta G(t)+(1-beta)Q(t),quad AA t > 1,\alpha G(t)+(1-\alpha) Q(t)<A(t)<\beta G(t)+(1-\beta) Q(t), \quad \forall t>1,
holds if and only if alpha >= 1//2\alpha \geq 1 / 2 and beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2.
Proof. Let us consider for t > 1t>1, the function
Proof. Let us consider for t > 1t>1, the function
{:(11)f_(2)(t)=(Q(t)-A(t))/(Q(t)-G(t))=1-(sqrt(t^(2)+1)+sqrt(2t))/(sqrt2(sqrtt+1)^(2)).:}\begin{equation*}
f_{2}(t)=\frac{Q(t)-A(t)}{Q(t)-G(t)}=1-\frac{\sqrt{t^{2}+1}+\sqrt{2 t}}{\sqrt{2}(\sqrt{t}+1)^{2}} . \tag{11}
\end{equation*}
We have f_(2)(t) < 1//2f_{2}(t)<1 / 2, since sqrt(t^(2)+1)+sqrt(2t) > sqrt2//2(sqrtt+1)^(2)<=>sqrt(t^(2)+1) > sqrt2//2(t+1)<=>(t-1)^(2) > 0\sqrt{t^{2}+1}+\sqrt{2 t}>\sqrt{2} / 2(\sqrt{t}+1)^{2} \Leftrightarrow \sqrt{t^{2}+1}> \sqrt{2} / 2(t+1) \Leftrightarrow(t-1)^{2}>0. Since lim_(t rarr1)f_(2)(t)=1//2\lim _{t \rightarrow 1} f_{2}(t)=1 / 2, it follows that alpha G(t)+(1-alpha)Q(t) < A(t),AA t > 1\alpha G(t)+ (1-\alpha) Q(t)<A(t), \forall t>1 if and only if alpha >= 1//2\alpha \geq 1 / 2.
Let us suppose that A(t) < beta G(t)+(1-beta)Q(t),AA t > 1A(t)<\beta G(t)+(1-\beta) Q(t), \forall t>1. Since lim_(t rarr oo)f_(2)(t)=1-sqrt2//2\lim _{t \rightarrow \infty} f_{2}(t)=1-\sqrt{2} / 2, it follows that beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2. Conversely, if
beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2, it suffices to prove that f_(2)(t) > 1-sqrt2//2f_{2}(t)>1-\sqrt{2} / 2. This is equivalent with
beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2, it suffices to prove that f_(2)(t) > 1-sqrt2//2f_{2}(t)>1-\sqrt{2} / 2. This is equivalent with
(sqrt(t^(2)+1)+sqrt(2t))/((sqrtt+1)^(2)) < 1\frac{\sqrt{t^{2}+1}+\sqrt{2 t}}{(\sqrt{t}+1)^{2}}<1
i. e. sqrt(t^(2)+1)+sqrt(2t) < (sqrtt+1)^(2)\sqrt{t^{2}+1}+\sqrt{2 t}<(\sqrt{t}+1)^{2} and this is true because sqrt(t^(2)+1)+sqrt(2t) < t+1+sqrt(2t) < (sqrtt+1)^(2)\sqrt{t^{2}+1}+\sqrt{2 t}< t+1+\sqrt{2 t}<(\sqrt{t}+1)^{2}.
Theorem 4. The double inequality
alpha G(t)+(1-alpha)C(t) < A(t) < beta G(t)+(1-beta)C(t),quad AA t > 1,\alpha G(t)+(1-\alpha) C(t)<A(t)<\beta G(t)+(1-\beta) C(t), \quad \forall t>1,
holds if and only if alpha >= 2//3\alpha \geq 2 / 3 and beta <= 1//2\beta \leq 1 / 2.
Proof. For t > 1t>1 we define
Proof. For t > 1t>1 we define
{:(12)f_(3)(t)=(C(t)-A(t))/(C(t)-G(t))=((t-1)^(2))/(2(t^(2)+1-sqrtt(t+1))):}\begin{equation*}
f_{3}(t)=\frac{C(t)-A(t)}{C(t)-G(t)}=\frac{(t-1)^{2}}{2\left(t^{2}+1-\sqrt{t}(t+1)\right)} \tag{12}
\end{equation*}
Since lim_(t rarr1)f_(3)(t)=2//3\lim _{t \rightarrow 1} f_{3}(t)=2 / 3, from alpha G(t)+(1-alpha)C(t) < A(t),AA t > 1\alpha G(t)+(1-\alpha) C(t)<A(t), \forall t>1 it follows that alpha >= 2//3\alpha \geq 2 / 3. If alpha >= 2//3\alpha \geq 2 / 3, it is true that f_(3)(t) < alphaf_{3}(t)<\alpha, because f_(3)(t) < 2//3f_{3}(t)<2 / 3 is equivalent with
(sqrtt)/(t+sqrtt+1) < (1)/(3),\frac{\sqrt{t}}{t+\sqrt{t}+1}<\frac{1}{3},
or (sqrtt-1)^(2) > 0(\sqrt{t}-1)^{2}>0.
Similarly, it follows that f_(3)(t) > 1//2f_{3}(t)>1 / 2, since
Similarly, it follows that f_(3)(t) > 1//2f_{3}(t)>1 / 2, since
f_(3)(t)-(1)/(2)=(sqrtt(sqrtt-1)^(2))/(2(t^(2)+1-sqrtt(t+1)))=(sqrtt)/(2(t+sqrtt+1)) > 0f_{3}(t)-\frac{1}{2}=\frac{\sqrt{t}(\sqrt{t}-1)^{2}}{2\left(t^{2}+1-\sqrt{t}(t+1)\right)}=\frac{\sqrt{t}}{2(t+\sqrt{t}+1)}>0
The infimum of f_(3)f_{3} on ( 1,oo1, \infty ) is precisely 1//21 / 2, because lim_(t rarr oo)f_(3)(t)=1//2\lim _{t \rightarrow \infty} f_{3}(t)=1 / 2
Theorem 5. The double inequality
Theorem 5. The double inequality
alpha G(t)+(1-alpha)C(t) < Q(t) < beta G(t)+(1-beta)C(t),quad AA t > 1,\alpha G(t)+(1-\alpha) C(t)<Q(t)<\beta G(t)+(1-\beta) C(t), \quad \forall t>1,
holds if and only if alpha >= 1-sqrt2//2\alpha \geq 1-\sqrt{2} / 2 and beta < beta_(0)\beta<\beta_{0}, where beta_(0)~=0.3471574308 dots\beta_{0} \cong 0.3471574308 \ldots is the unique positive root of the polynomial
9x^(4)-26x^(3)+22x^(2)-2x-19 x^{4}-26 x^{3}+22 x^{2}-2 x-1
Proof. We have to find, for t > 1t>1, the extreme values of
{:(13)f_(4)(t)=(C(t)-Q(t))/(C(t)-G(t))=(2t^(2)-sqrt2sqrt(t^(2)+1)-sqrt2tsqrt(t^(2)+1)+2)/(2(t^(2)-t^((3)/(2))-sqrtt+1)):}\begin{equation*}
f_{4}(t)=\frac{C(t)-Q(t)}{C(t)-G(t)}=\frac{2 t^{2}-\sqrt{2} \sqrt{t^{2}+1}-\sqrt{2} t \sqrt{t^{2}+1}+2}{2\left(t^{2}-t^{\frac{3}{2}}-\sqrt{t}+1\right)} \tag{13}
\end{equation*}
Denoting by u=sqrttu=\sqrt{t}, we compute the derivative of
h(u)=f_(4)(u^(2))h(u)=f_{4}\left(u^{2}\right)
and we obtain
h^(')(u)=((u+1)(sqrt(2(u^(4)+1))(-u^(4)-4u^(2)-1)+u^(6)+2u^(5)+3u^(4)+3u^(2)+2u+1))/(sqrt(2(u^(4)+1))(u-1)^(3)(u^(2)+u+1)^(2)).h^{\prime}(u)=\frac{(u+1)\left(\sqrt{2\left(u^{4}+1\right)}\left(-u^{4}-4 u^{2}-1\right)+u^{6}+2 u^{5}+3 u^{4}+3 u^{2}+2 u+1\right)}{\sqrt{2\left(u^{4}+1\right)}(u-1)^{3}\left(u^{2}+u+1\right)^{2}} .
So, the roots of the derivative satisfy the algebraic equation
2(u^(4)+1)(u^(4)+4u^(2)+1)^(2)=(u^(6)+2u^(5)+3u^(4)+3u^(2)+2u+1)^(2).2\left(u^{4}+1\right)\left(u^{4}+4 u^{2}+1\right)^{2}=\left(u^{6}+2 u^{5}+3 u^{4}+3 u^{2}+2 u+1\right)^{2} .
After the simplification of a quartic polynomial whose roots are not in the interval ( 1,oo1, \infty ), we obtain the equation
{:(14)u^(8)-8u^(5)-10u^(4)-8u^(3)+1=0",":}\begin{equation*}
u^{8}-8 u^{5}-10 u^{4}-8 u^{3}+1=0, \tag{14}
\end{equation*}
which has a unique root u_(0)u_{0} in the interval ( 1,oo1, \infty ). This can be easily proved by using the Sturm sequence. Then u_(0)u_{0} will be the unique root of h^(')h^{\prime} in ( 1,oo1, \infty ).
Now h^(')(2) > 0,h^(')(3) < 0h^{\prime}(2)>0, h^{\prime}(3)<0, so 2 < u_(0) < 32<u_{0}<3 and hh is strictly increasing in the interval ( 1,u_(0)1, u_{0} ) and strictly decreasing in the interval ( u_(0),oou_{0}, \infty ). We also have lim_(u rarr1)h(u)=1//3,lim_(u rarr oo)h(u)=1-sqrt2//2\lim _{u \rightarrow 1} h(u)=1 / 3, \lim _{u \rightarrow \infty} h(u)=1-\sqrt{2} / 2 and therefore i n ff_(4)=i n f h=1-sqrt2//2,s u pf_(4)=s u p h=h(u_(0))=beta_(0)\inf f_{4}=\inf h= 1-\sqrt{2} / 2, \sup f_{4}=\sup h=h\left(u_{0}\right)=\beta_{0}.
Since h(u_(0))h\left(u_{0}\right) is an algebraic number, we can easily find its minimal polynomial by performing the following commands in Maple:
>theta:=RootOf(u^8-8*u^5-10*u^4-8*u^3+1,u):
> M:= g(theta):
> sqrfree(evala(Norm(convert(Z-M,RootOf))),Z)[2][1][1];
9x^(4)-26x^(3)+22x^(2)-2x-19 x^{4}-26 x^{3}+22 x^{2}-2 x-1
Notice that Maple is of course able to express the maximum h(u_(0))h\left(u_{0}\right) in terms of radicals by executing the command:
> select(u- > :}>\operatorname{select}\left(\mathrm{u}->\right. is (u > 0)(\mathrm{u}>0),[solve ( 9**x^4-26**x^3+22**x^2-2**x-19 * x^{\wedge} 4-26 * x^{\wedge} 3+22 * x^{\wedge} 2-2 * x-1, Explicit)]); but the resulting expression is cumbersome and we will not print it here.
> select(u- > :}>\operatorname{select}\left(\mathrm{u}->\right. is (u > 0)(\mathrm{u}>0),[solve ( 9**x^4-26**x^3+22**x^2-2**x-19 * x^{\wedge} 4-26 * x^{\wedge} 3+22 * x^{\wedge} 2-2 * x-1, Explicit)]); but the resulting expression is cumbersome and we will not print it here.
Theorem 6. The double inequality
alpha A(t)+(1-alpha)C(t) < Q(t) < beta A(t)+(1-beta)C(t),quad AA t > 1,\alpha A(t)+(1-\alpha) C(t)<Q(t)<\beta A(t)+(1-\beta) C(t), \quad \forall t>1,
holds if and only if alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} and beta <= 1//2\beta \leq 1 / 2.
Proof. Let us consider, for t > 1t>1
Proof. Let us consider, for t > 1t>1
{:(15)f_(5)(t)=(C(t)-Q(t))/(C(t)-A(t))=(2(t^(2)+1)-(t+1)sqrt(2(t^(2)+1)))/((t-1)^(2)).:}\begin{equation*}
f_{5}(t)=\frac{C(t)-Q(t)}{C(t)-A(t)}=\frac{2\left(t^{2}+1\right)-(t+1) \sqrt{2\left(t^{2}+1\right)}}{(t-1)^{2}} . \tag{15}
\end{equation*}
From lim_(t rarr oo)f_(5)(t)=2-sqrt2\lim _{t \rightarrow \infty} f_{5}(t)=2-\sqrt{2} it follows that alpha A(t)+(1-alpha)C(t)(:Q(t),AA t:)\alpha A(t)+(1-\alpha) C(t)\langle Q(t), \forall t\rangle 1 implies alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}. Now if alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} we have to prove that f_(5)(t) < 2-sqrt2f_{5}(t)<2-\sqrt{2}, which can be written as sqrt(2(t^(2)+1))(t+1) > -sqrt2t^(2)+(4+2sqrt2)t-sqrt2\sqrt{2\left(t^{2}+1\right)}(t+1)>-\sqrt{2} t^{2}+(4+2 \sqrt{2}) t-\sqrt{2}. If -sqrt2t^(2)+(4+2sqrt2)t-sqrt2 <= 0-\sqrt{2} t^{2}+(4+2 \sqrt{2}) t-\sqrt{2} \leq 0, the inequality holds. Otherwise, squaring both sides it reduces to 4(3+2sqrt2)t(t-1)^(2) > 04(3+2 \sqrt{2}) t(t-1)^{2}>0.
We obtain also
f_(5)(t)-(1)/(2)=(3(t^(2)+1)+2t-2(t+1)sqrt(2(t^(2)+1)))/(2(t-1)^(2)) > 0,f_{5}(t)-\frac{1}{2}=\frac{3\left(t^{2}+1\right)+2 t-2(t+1) \sqrt{2\left(t^{2}+1\right)}}{2(t-1)^{2}}>0,
because (3(t^(2)+1)+2t)^(2)-8(t+1)^(2)(t^(2)+1) > 0<=>(t-1)^(4) > 0\left(3\left(t^{2}+1\right)+2 t\right)^{2}-8(t+1)^{2}\left(t^{2}+1\right)>0 \Leftrightarrow(t-1)^{4}>0. We have lim_(t rarr1)f_(5)(t)=1//2\lim _{t \rightarrow 1} f_{5}(t)=1 / 2, hence this is the infimum of f_(5)f_{5} on ( 1,oo1, \infty ) and the second part of the theorem is also true.
Lemma 7. 3] For t > 1t>1, the following inequality holds
{:(16)t^((t)/(t+1)) > t-ln t.:}\begin{equation*}
t^{\frac{t}{t+1}}>t-\ln t . \tag{16}
\end{equation*}
Proof. The inequality (16) is equivalent to
(t)/(t+1)ln t > ln(t-ln t)\frac{t}{t+1} \ln t>\ln (t-\ln t)
We consider the function
k(t)=ln(t-ln t)-(t-1)/(t)ln t,quad t > 1k(t)=\ln (t-\ln t)-\frac{t-1}{t} \ln t, \quad t>1
with
k^(')(t)=((ln t-1)ln t)/(t^(2)(t-ln t))k^{\prime}(t)=\frac{(\ln t-1) \ln t}{t^{2}(t-\ln t)}
It has lim_(t rarr1)k(t)=0,lim_(t rarr oo)k(t)=0\lim _{t \rightarrow 1} k(t)=0, \lim _{t \rightarrow \infty} k(t)=0 and a minimum at t_(0)=et_{0}=e. It follows that k(t) < 0k(t)<0 on (1,oo)(1, \infty), hence ((t-1)//t)ln t > ln(t-ln t)((t-1) / t) \ln t>\ln (t-\ln t). It follows that
(t)/(t+1)ln t > (t-1)/(t)ln t > ln(t-ln t)\frac{t}{t+1} \ln t>\frac{t-1}{t} \ln t>\ln (t-\ln t)
Theorem 8. The double inequality
alpha A(t)+(1-alpha)C(t) < S(t) < beta A(t)+(1-beta)C(t),quad AA t > 1,\alpha A(t)+(1-\alpha) C(t)<S(t)<\beta A(t)+(1-\beta) C(t), \quad \forall t>1,
holds if and only if alpha >= 1//2\alpha \geq 1 / 2 and beta <= 0\beta \leq 0.
Proof. We define
Proof. We define
{:(17)f_(6)(t)=(C(t)-S(t))/(C(t)-A(t))=2(t^(2)+1-(t+1)t^((t)/(t+1)))/((t-1)^(2)):}\begin{equation*}
f_{6}(t)=\frac{C(t)-S(t)}{C(t)-A(t)}=2 \frac{t^{2}+1-(t+1) t^{\frac{t}{t+1}}}{(t-1)^{2}} \tag{17}
\end{equation*}
We have
f_(6)(t)-(1)/(2)=(3(t^(2)+1)+2t-4(t+1)t^((t)/(t+1)))/(2(t-1)^(2))=(2(t+1))/((t-1)^(2))*g(t)f_{6}(t)-\frac{1}{2}=\frac{3\left(t^{2}+1\right)+2 t-4(t+1) t^{\frac{t}{t+1}}}{2(t-1)^{2}}=\frac{2(t+1)}{(t-1)^{2}} \cdot g(t)
where
g(t)=(3(t^(2)+1)+2t)/(4(t+1))-t^((t)/(t+1))g(t)=\frac{3\left(t^{2}+1\right)+2 t}{4(t+1)}-t^{\frac{t}{t+1}}
Then
g^(')(t)=-(g_(1)(t))/(4(t+1)^(2))g^{\prime}(t)=-\frac{g_{1}(t)}{4(t+1)^{2}}
where
{:(18)g_(1)(t)=4t^((t)/(t+1))(t+1+ln t)+1-3t^(2)-6t:}\begin{equation*}
g_{1}(t)=4 t^{\frac{t}{t+1}}(t+1+\ln t)+1-3 t^{2}-6 t \tag{18}
\end{equation*}
Using the fact that S > QS>Q, i. e. t^(t//(t+1)) > sqrt((t^(2)+1)//2)t^{t /(t+1)}>\sqrt{\left(t^{2}+1\right) / 2}, we obtain that g_(1)(t) > sqrt(2(t^(2)+1))g_(2)(t)g_{1}(t)>\sqrt{2\left(t^{2}+1\right)} g_{2}(t), where
g_(2)(t)=2(t+1+ln t)-(3t^(2)+6t-1)//sqrt(2(t^(2)+1))g_{2}(t)=2(t+1+\ln t)-\left(3 t^{2}+6 t-1\right) / \sqrt{2\left(t^{2}+1\right)}
The derivative of g_(2)g_{2} is
g_(2)^(')(t)=((t+1)(sqrt(2(t^(2)+1)))^(3)-(3t^(4)+7t^(2)+6t))/(t(t^(2)+1)sqrt(2(t^(2)+1)))g_{2}^{\prime}(t)=\frac{(t+1)\left(\sqrt{2\left(t^{2}+1\right)}\right)^{3}-\left(3 t^{4}+7 t^{2}+6 t\right)}{t\left(t^{2}+1\right) \sqrt{2\left(t^{2}+1\right)}}
In order to establish its sign we consider the polynomial
{:[P(t)=(t+1)^(2)(sqrt(2(t^(2)+1)))^(6)-(3t^(4)+7t^(2)+6t)^(2)],[=t^(6)(t-1)^(2)((8)/(t^(6))+(32)/(t^(5))+(52)/(t^(4))+(36)/(t^(3))+(19)/(t^(2))+(14 )/(t)-1)]:}\begin{aligned}
P(t) & =(t+1)^{2}\left(\sqrt{2\left(t^{2}+1\right)}\right)^{6}-\left(3 t^{4}+7 t^{2}+6 t\right)^{2} \\
& =t^{6}(t-1)^{2}\left(\frac{8}{t^{6}}+\frac{32}{t^{5}}+\frac{52}{t^{4}}+\frac{36}{t^{3}}+\frac{19}{t^{2}}+\frac{14}{t}-1\right)
\end{aligned}
The expression from the last parenthesis is obviously decreasing for t >= 1t \geq 1 and it is positive for t=10t=10. It follows that it is positive on (1,10)(1,10), hence on this interval PP is also positive. Therefore g_(2)^(')(t) > 0,g_(2)(t) > g_(2)(1)=0g_{2}^{\prime}(t)>0, g_{2}(t)>g_{2}(1)=0, so g_(1)g_{1} is positive too for 1 < t < 101<t<10.
Let us consider now that t >= 10t \geq 10. Using (16) in (18) we obtain that g_(1)(t) > g_(3)(t)g_{1}(t)> g_{3}(t), where
g_(3)(t)=t^(2)-2t+2-(2ln t+1)^(2).g_{3}(t)=t^{2}-2 t+2-(2 \ln t+1)^{2} .
For
g_(4)(t)=sqrt(t^(2)-2t+2)-2ln t-1g_{4}(t)=\sqrt{t^{2}-2 t+2}-2 \ln t-1
the sign of g_(4)^(')g_{4}^{\prime} is given by t^(2)-t-2sqrt(t^(2)-2t+2)t^{2}-t-2 \sqrt{t^{2}-2 t+2}; but (t^(2)-t)^(2)-4(t^(2)-2t+2)=(t-10)^(4)+38(t-10)^(3)+537(t-10)^(2)+3348(t-10)+7772 > 0\left(t^{2}-t\right)^{2}-4\left(t^{2}-2 t+2\right)= (t-10)^{4}+38(t-10)^{3}+537(t-10)^{2}+3348(t-10)+7772>0 for t >= 10t \geq 10. It follows that g_(3)(t) >= g_(3)(10)=3.45 dots > 0g_{3}(t) \geq g_{3}(10)=3.45 \ldots>0, hence g_(1)g_{1} is positive for t >= 10t \geq 10 too.
In conclusion, g_(1)(t) > 0g_{1}(t)>0 on (1,oo)(1, \infty), therefore g^(')(t) < 0g^{\prime}(t)<0 on (1,oo)(1, \infty). The function gg being decreasing, g(t) < g(1)=0g(t)<g(1)=0 for t > 1t>1 and f_(6)(t) < 1//2f_{6}(t)<1 / 2 for t > 1t>1.
The second part of the theorem follows from lim_(t rarr oo)f_(6)(t)=0\lim _{t \rightarrow \infty} f_{6}(t)=0 and f_(6)(t) > 0f_{6}(t)>0, AA t > 1\forall t>1.
Theorem 9. The double inequality
alpha A(t)+(1-alpha)S(t) < Q(t) < beta A(t)+(1-beta)S(t),quad AA t > 1\alpha A(t)+(1-\alpha) S(t)<Q(t)<\beta A(t)+(1-\beta) S(t), \quad \forall t>1
holds if and only if alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} and beta <= 0\beta \leq 0.
Proof. We shall prove that the first inequality holds for alpha=2-sqrt2\alpha=2-\sqrt{2} (hence a fortiori for alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} ).
Proof. We shall prove that the first inequality holds for alpha=2-sqrt2\alpha=2-\sqrt{2} (hence a fortiori for alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2} ).
Let us denote
H(t,alpha)=Q(t)-alpha A(t)-(1-alpha)S(t)=(1)/(2)sqrt(2+2t^(2))-(1)/(2)alpha(1+t)-(1-alpha)t^((t)/(1+t))H(t, \alpha)=Q(t)-\alpha A(t)-(1-\alpha) S(t)=\frac{1}{2} \sqrt{2+2 t^{2}}-\frac{1}{2} \alpha(1+t)-(1-\alpha) t^{\frac{t}{1+t}}.
and
H(t,alpha)=Q(t)-alpha A(t)-(1-alpha)S(t)=(1)/(2)sqrt(2+2t^(2))-(1)/(2)alpha(1+t)-(1-alpha)t^((t)/(1+t))H(t, \alpha)=Q(t)-\alpha A(t)-(1-\alpha) S(t)=\frac{1}{2} \sqrt{2+2 t^{2}}-\frac{1}{2} \alpha(1+t)-(1-\alpha) t^{\frac{t}{1+t}}.
and
h_(1)(t)=(sqrt2+1)H(t,2-sqrt2),h_{1}(t)=(\sqrt{2}+1) H(t, 2-\sqrt{2}),
where HH is given in (19). We have to prove that h_(1)(t) > 0h_{1}(t)>0 for t > 1t>1. It follows that
h_(1)(t)=((sqrt2+1)sqrt(2(t^(2)+1)))/(2)-(sqrt2(t+1))/(2)-t^((t)/(t+1)).h_{1}(t)=\frac{(\sqrt{2}+1) \sqrt{2\left(t^{2}+1\right)}}{2}-\frac{\sqrt{2}(t+1)}{2}-t^{\frac{t}{t+1}} .
We put in the inequality (1+x)^(q) < 1+qx(1+x)^{q}<1+q x, which holds for x > 0,0 < q < 1x>0,0<q<1, x=t-1x=t-1 and q=t//(t+1)q=t /(t+1). It follows that
t^((t)/(t+1)) < (t^(2)+1)/(t+1)t^{\frac{t}{t+1}}<\frac{t^{2}+1}{t+1}
and
h_(1)(t) > ((1+sqrt2))/(2(t+1))((t+1)sqrt(2+2t^(2))-sqrt2(t^(2)+2(sqrt2-1)t+1))h_{1}(t)>\frac{(1+\sqrt{2})}{2(t+1)}\left((t+1) \sqrt{2+2 t^{2}}-\sqrt{2}\left(t^{2}+2(\sqrt{2}-1) t+1\right)\right)
Let us denote the positive expressions
h_(2)(t)=(t+1)sqrt(2+2t^(2)),quadh_(3)(t)=sqrt2(t^(2)+2(sqrt2-1)t+1)h_{2}(t)=(t+1) \sqrt{2+2 t^{2}}, \quad h_{3}(t)=\sqrt{2}\left(t^{2}+2(\sqrt{2}-1) t+1\right)
it follows easily that h_(2)^(2)(t)-h_(3)^(2)(t)=4t(t-1)^(2)h_{2}^{2}(t)-h_{3}^{2}(t)=4 t(t-1)^{2}, therefore h_(1)(t) > 0h_{1}(t)>0.
The second part of the theorem is obvious, since
The second part of the theorem is obvious, since
f_(7)(t)=(S(t)-Q(t))/(S(t)-A(t))f_{7}(t)=\frac{S(t)-Q(t)}{S(t)-A(t)}
satisfies f_(7)(t) > 0,AA t > 1f_{7}(t)>0, \forall t>1 and lim_(t rarr1)f_(7)(t)=0\lim _{t \rightarrow 1} f_{7}(t)=0.
Theorem 10. The double inequality
Theorem 10. The double inequality
alpha L(t)+(1-alpha)C(t) < A(t) < beta L(t)+(1-beta)C(t),quad AA t > 1\alpha L(t)+(1-\alpha) C(t)<A(t)<\beta L(t)+(1-\beta) C(t), \quad \forall t>1
holds if and only if alpha >= 3//4\alpha \geq 3 / 4 and beta <= 1//2\beta \leq 1 / 2.
Proof. We have to find the extreme values of
Proof. We have to find the extreme values of
f_(8)(t)=(C(t)-A(t))/(C(t)-L(t))f_{8}(t)=\frac{C(t)-A(t)}{C(t)-L(t)}
for t > 1t>1, where f_(8)f_{8} is given by
f_(8)(t)=(1)/(2)*((t-1)^(2)ln t)/(ln t+t^(2)ln t-t^(2)+1)f_{8}(t)=\frac{1}{2} \cdot \frac{(t-1)^{2} \ln t}{\ln t+t^{2} \ln t-t^{2}+1}
We obtain
f_(8)^(')(t)=-((t-1)h_(3)(t))/(2t(ln t+t^(2)ln t-t^(2)+1)^(2))f_{8}^{\prime}(t)=-\frac{(t-1) h_{3}(t)}{2 t\left(\ln t+t^{2} \ln t-t^{2}+1\right)^{2}}
where
h_(4)(t)=t^(3)-2(t^(2)ln t)^(2)+2t^(2)ln t-t^(2)-2t ln t-2t(ln t)^(2)-t+1h_{4}(t)=t^{3}-2\left(t^{2} \ln t\right)^{2}+2 t^{2} \ln t-t^{2}-2 t \ln t-2 t(\ln t)^{2}-t+1
Now, h_(4)(1)=h_(4)^(')(1)=h_(4)^('')(1)=h_(4)^(''')(1)=0h_{4}(1)=h_{4}^{\prime}(1)=h_{4}^{\prime \prime}(1)=h_{4}^{\prime \prime \prime}(1)=0 and
h_(4)^((4))(t)=(8(t-1)ln t)/(t^(3)) > 0" for "t > 1.h_{4}^{(4)}(t)=\frac{8(t-1) \ln t}{t^{3}}>0 \text { for } t>1 .
Therefore h_(4) > 0h_{4}>0 in ( 1,oo1, \infty ) and f_(8)^(') < 0f_{8}^{\prime}<0 in ( 1,oo1, \infty ). The function f_(8)f_{8} being strictly decreasing on ( 1,oo1, \infty ), i n ff_(8)=lim_(t rarr oo)f_(8)(t)=1//2\inf f_{8}=\lim _{t \rightarrow \infty} f_{8}(t)=1 / 2 and sup f_(8)=lim_(t rarr1)f_(8)(t)=3//4f_{8}=\lim _{t \rightarrow 1} f_{8}(t)= 3 / 4.
3. FINAL REMARKS
From the eight means considered in this paper, two enter the class of Gini means [6] defined for a,b > 0,u,v inRa, b>0, u, v \in \mathbb{R},
G_(u,v)(a,b)={[((a^(u)+b^(u))/(a^(v)+b^(v)))^((1)/(u-v))",",u!=v],[exp((a^(u)log a+b^(u)log b)/(a^(u)+b^(u)))",",u=v]:}G_{u, v}(a, b)= \begin{cases}\left(\frac{a^{u}+b^{u}}{a^{v}+b^{v}}\right)^{\frac{1}{u-v}}, & u \neq v \\ \exp \left(\frac{a^{u} \log a+b^{u} \log b}{a^{u}+b^{u}}\right), & u=v\end{cases}
namely S=G_(1,1),C=G_(2,1)S=G_{1,1}, C=G_{2,1}; two belong to the class of Stolarsky means [8] defined for a,b > 0,a!=b,u,v inRa, b>0, a \neq b, u, v \in \mathbb{R},
E_(r,s)(a,b)={[((s)/(r)(a^(r)-b^(r))/(a^(s)-b^(s)))^((1)/(r-s))",",rs(r-s)!=0],[exp(-(1)/(r)+(a^(r)log a-b^(r)log b)/(a^(r)-b^(r)))",",r=s!=0],[((1)/(s)(a^(s)-b^(s))/(log a-log b))^((1)/(s))",",r=0","s!=0],[sqrt(ab)",",r=s!=0]:}E_{r, s}(a, b)= \begin{cases}\left(\frac{s}{r} \frac{a^{r}-b^{r}}{a^{s}-b^{s}}\right)^{\frac{1}{r-s}}, & r s(r-s) \neq 0 \\ \exp \left(-\frac{1}{r}+\frac{a^{r} \log a-b^{r} \log b}{a^{r}-b^{r}}\right), & r=s \neq 0 \\ \left(\frac{1}{s} \frac{a^{s}-b^{s}}{\log a-\log b}\right)^{\frac{1}{s}}, & r=0, s \neq 0 \\ \sqrt{a b}, & r=s \neq 0\end{cases}
namely L=E_(1,0),I=E_(1,1)L=E_{1,0}, I=E_{1,1}. The other four are in both classes, namely H=G_(-1,0)=E_(-2,-1),G=G_(0,0)=E_(0,0),A=G_(1,0)=E_(2,1)H=G_{-1,0}=E_{-2,-1}, G=G_{0,0}=E_{0,0}, A=G_{1,0}=E_{2,1} and Q=G_(2,0)=E_(4,2)Q=G_{2,0}=E_{4,2}. As it was shown in [2], the families of Gini means G_(u,v)G_{u, v} and Stolarsky means E_(r,s)E_{r, s} have in common only the power means. So even if general results will be proved for these two classes of means, not all the inequalities from this paper will be consequences; for example, in the last theorem LL is a Stolarsky mean, while CC is a Gini one.
REFERENCES
[1] H. Alzer and S. L. Qiu, Inequalities for means in two variables, Arch. Math. (Basel), 80 (2003), pp. 201-215.
[2] H. Alzer and S. Ruscheweyh, On the intersection of two-parameter mean value families, Proc. A. M. S., 129(9) (2001), pp. 2655-2662.
[3] M. C. Anisiu and V. Anisiu, Refinement of some inequalities for means, Rev. Anal. Numér. Théor. Approx., 35 (2006) no. 1, pp. 5-10. 띠
[4] M. C. Anisiu and 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. 중
[5] P. S. Bullen, Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.
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[8] K. B. Stolarsky, Generalizations of the logarithmic mean, Math. Mag., 48 (1975), pp. 87-92.
[9] 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. 즈
[2] H. Alzer and S. Ruscheweyh, On the intersection of two-parameter mean value families, Proc. A. M. S., 129(9) (2001), pp. 2655-2662.
[3] M. C. Anisiu and V. Anisiu, Refinement of some inequalities for means, Rev. Anal. Numér. Théor. Approx., 35 (2006) no. 1, pp. 5-10. 띠
[4] M. C. Anisiu and 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. 중
[5] P. S. Bullen, Handbook of Means and Their Inequalities, Series: Mathematics and Its Applications, vol. 560, 2nd ed., Kluwer Academic Publishers Group, Dordrecht, 2003.
[6] C. Gini, Di una formula comprensiva delle medie, Metron, 13 (1938), pp. 3-22.
[7] M. Ivan and I. Raşa, Some inequalities for means, Tiberiu Popoviciu Itinerant Seminar of Functional Equations, Approximation and Convexity, Cluj-Napoca, May 23-29, 2000, pp. 99-102.
[8] K. B. Stolarsky, Generalizations of the logarithmic mean, Math. Mag., 48 (1975), pp. 87-92.
[9] 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. 즈
Received by the editors: October 24, 2013.
- "T. Popoviciu" Institute of Numerical Analysis, Romanian Academy, 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.
- "T. Popoviciu" Institute of Numerical Analysis, Romanian Academy, P.O. Box 68, 400110 Cluj-Napoca, Romania, e-mail: mira@math.ubbcluj.ro.