Bilateral inequalities for means

Mira Anisiu
(original), paper

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.

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https://ictp.acad.ro/jnaat/journal/issue/view/2013-vol42-no2

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Revue d’analyse Numerique et de Theorie de l’Approximation

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Romanian Academy

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[1] H. Alzer and S. L. Qiu, Inequalities for means in two variables, Arch. Math. (Basel),80(2003), pp. 201–215.
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[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.
jnaat,+Journal+manager,+2013-2-Anisiu-Anisiu-15-10-19

BILATERAL INEQUALITIES FOR MEANS

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

Abstract

Let ( M 1 , M 2 , M 3 M 1 , M 2 , M 3 M_(1),M_(2),M_(3)M_{1}, M_{2}, M_{3}M1,M2,M3 ) be three means in two variables chosen from H H HHH, G , L , I , A , Q , S , C G , L , I , A , Q , S , C G,L,I,A,Q,S,CG, 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 ) , 0 < a < b M 1 ( a , b ) < M 2 ( a , b ) < M 3 ( a , b ) , 0 < a < b 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<bM1(a,b)<M2(a,b)<M3(a,b),0<a<b

We consider the problem of finding α , β ∈ R α , β ∈ R alpha,beta inR\alpha, \beta \in \mathbb{R}α,β∈R for which
α M 1 ( a , b ) + ( 1 − α ) M 3 ( a , b ) < M 2 ( a , b ) < β M 1 ( a , b ) + ( 1 − β ) M 3 ( a , b ) . α M 1 ( a , b ) + ( 1 − α ) M 3 ( a , b ) < M 2 ( a , b ) < β M 1 ( a , b ) + ( 1 − β ) M 3 ( a , b ) . 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) .αM1(a,b)+(1−α)M3(a,b)<M2(a,b)<βM1(a,b)+(1−β)M3(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 ) (G,L,A),(G,A,Q),(G,A,C),(G,Q,C)(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 ) (A,Q,C),(A,S,C),(A,Q,S)(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 ) (L,A,C)(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.

1. INTRODUCTION

We remind the definitions of the classical means, namely, for 0 < a < b 0 < a < b 0 < a < b0<a<b0<a<b
  • the arithmetic, geometric and harmonic ones
A = a + b 2 , G = a b , H = 2 a b a + b A = a + b 2 , G = a b , H = 2 a b a + b 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}A=a+b2,G=ab,H=2aba+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 = a 2 + b 2 2 1 / 2 , C = a 2 + b 2 a + b 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}Q=(a2+b22)1/2,C=a2+b2a+b;
  • the Pólya EG Szegő logarithmic mean, the exponential (or identric), and the weighted geometric mean
L = b − a ln ⁡ b − ln ⁡ a , I = 1 e ( b b a a ) 1 / ( b − a ) , S = ( a a b b ) 1 / ( a + b ) L = b − a ln ⁡ b − ln ⁡ a , I = 1 e b b a a 1 / ( b − a ) , S = a a b b 1 / ( a + b ) 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)}L=b−aln⁡b−ln⁡a,I=1e(bbaa)1/(b−a),S=(aabb)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 (1) H < G < L < I < A < Q < S < C {:(1)H < G < L < I < A < Q < S < C:}\begin{equation*} H<G<L<I<A<Q<S<C \tag{1} \end{equation*}(1)H<G<L<I<A<Q<S<C
followed by relations between some means and the arithmetic means of two others ([7], 3])
(2) L < G + A 2 , G + Q 2 < A < G + C 2 < Q < A + C 2 < S . (2) L < G + A 2 , G + Q 2 < A < G + C 2 < Q < A + C 2 < S . {:(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+A2,G+Q2<A<G+C2<Q<A+C2<S.
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 , I G , I G,IG, IG,I and A A AAA.
Theorem 1. (1) The double inequality
(3) α A ( a , b ) + ( 1 − α ) G ( a , b ) < I ( a , b ) < β A ( a , b ) + ( 1 − β ) G ( a , b ) (3) α A ( a , b ) + ( 1 − α ) G ( a , b ) < I ( a , b ) < β A ( a , b ) + ( 1 − β ) G ( a , b ) {:(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*}(3)αA(a,b)+(1−α)G(a,b)<I(a,b)<βA(a,b)+(1−β)G(a,b)
holds true for all positive real numbers a ≠ b a ≠ b a!=ba \neq ba≠b, if and only if α ≤ 2 / 3 α ≤ 2 / 3 alpha <= 2//3\alpha \leq 2 / 3α≤2/3 and β ≥ 2 / e β ≥ 2 / e beta >= 2//e\beta \geq 2 / \mathrm{e}β≥2/e.
Results of this type continued to appear, recent ones are given in [9] for ( H , L , A H , L , A H,L,AH, L, AH,L,A ) and ( H , I , A H , I , A H,I,AH, I, AH,I,A ), and in 4 for ( G , L , C G , L , C G,L,CG, L, CG,L,C ).
Let M 1 , M 2 , M 3 M 1 , M 2 , M 3 M_(1),M_(2),M_(3)M_{1}, M_{2}, M_{3}M1,M2,M3 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 ) . (4) M 1 ( a , b ) < M 2 ( a , b ) < M 3 ( a , b ) . {:(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*}(4)M1(a,b)<M2(a,b)<M3(a,b).
We consider the problem of finding α , β ∈ R α , β ∈ R alpha,beta inR\alpha, \beta \in \mathbb{R}α,β∈R for which
(5) α M 1 ( a , b ) + ( 1 − α ) M 3 ( a , b ) < M 2 ( a , b ) (5) α M 1 ( a , b ) + ( 1 − α ) M 3 ( a , b ) < M 2 ( a , b ) {:(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*}(5)αM1(a,b)+(1−α)M3(a,b)<M2(a,b)
and
(6) M 2 ( a , b ) < β M 1 ( a , b ) + ( 1 − β ) M 3 ( a , b ) . (6) M 2 ( a , b ) < β M 1 ( a , b ) + ( 1 − β ) M 3 ( a , b ) . {:(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*}(6)M2(a,b)<βM1(a,b)+(1−β)M3(a,b).
The inequalities (5) and (6) are equivalent to
(7) α > M 3 ( a , b ) − M 2 ( a , b ) M 3 ( a , b ) − M 1 ( a , b ) , (7) α > M 3 ( a , b ) − M 2 ( a , b ) M 3 ( a , b ) − M 1 ( a , b ) , {:(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*}(7)α>M3(a,b)−M2(a,b)M3(a,b)−M1(a,b),
respectively
(8) β < M 3 ( a , b ) − M 2 ( a , b ) M 3 ( a , b ) − M 1 ( a , b ) . (8) β < M 3 ( a , b ) − M 2 ( a , b ) M 3 ( a , b ) − M 1 ( a , b ) . {:(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*}(8)β<M3(a,b)−M2(a,b)M3(a,b)−M1(a,b).
Basically, denoting by t = b / a , t > 1 t = b / a , t > 1 t=b//a,t > 1t=b / a, t>1t=b/a,t>1, 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
(9) f ( t ) = M 3 ( 1 , t ) − M 2 ( 1 , t ) M 3 ( 1 , t ) − M 1 ( 1 , t ) . (9) f ( t ) = M 3 ( 1 , t ) − M 2 ( 1 , t ) M 3 ( 1 , t ) − M 1 ( 1 , t ) . {:(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*}(9)f(t)=M3(1,t)−M2(1,t)M3(1,t)−M1(1,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. If sup f sup f s u p f\sup fsupf is attained at some t ∈ ( 1 , ∞ ) t ∈ ( 1 , ∞ ) t in(1,oo)t \in(1, \infty)t∈(1,∞), then α ∈ ( sup f , ∞ ) α ∈ ( sup f , ∞ ) alpha in(s u p f,oo)\alpha \in(\sup f, \infty)α∈(supf,∞); otherwise α ∈ [ sup f , ∞ ) α ∈ [ sup f , ∞ ) alpha in[s u p f,oo)\alpha \in[\sup f, \infty)α∈[supf,∞). Similarly, β ∈ ( − ∞ , inf f ) β ∈ ( − ∞ , inf f ) beta in(-oo,i n f f)\beta \in(-\infty, \inf f)β∈(−∞,inff) if inf f inf f i n f f\inf finff is attained in ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞), and β ∈ ( − ∞ , inf f ] β ∈ ( − ∞ , inf f ] beta in(-oo,i n f f]\beta \in(-\infty, \inf f]β∈(−∞,inff] 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 α α alpha\alphaα in Theorem 9 below. We also use it to simplify the polynomials in the proof of Theorem 5 and to obtain the optimal value β 0 β 0 beta_(0)\beta_{0}β0 of β β beta\betaβ.
Starting from the means listed in (1), we can formulate ( 8 3 ) = 56 ( 8 3 ) = 56 ((8)/(3))=56\binom{8}{3}=56(83)=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 < C L < A < C L < A < CL<A<CL<A<C we find the optimal intervals for α α alpha\alphaα and β β beta\betaβ 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 > 1 0 < a < b , t = b / a , t > 1 0 < a < b,t=b//a,t > 10<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 ) M(a,b)M(a, b)M(a,b) is any mean from (1), it suffices to prove the inequalities for M ( 1 , t ) M ( 1 , t ) M(1,t)M(1, t)M(1,t). We shall write from now on M ( t ) M ( t ) M(t)M(t)M(t) instead of M ( 1 , t ) M ( 1 , t ) M(1,t)M(1, t)M(1,t).
Theorem 2. The double inequality
α G ( t ) + ( 1 − α ) A ( t ) < L ( t ) < β G ( t ) + ( 1 − β ) A ( t ) , ∀ t > 1 , α G ( t ) + ( 1 − α ) A ( t ) < L ( t ) < β G ( t ) + ( 1 − β ) A ( t ) , ∀ t > 1 , 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,αG(t)+(1−α)A(t)<L(t)<βG(t)+(1−β)A(t),∀t>1,
holds if and only if α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1 and β ≤ 2 / 3 β ≤ 2 / 3 beta <= 2//3\beta \leq 2 / 3β≤2/3.
Proof. We denote, for t > 1 t > 1 t > 1t>1t>1,
(10) f 1 ( t ) = A ( t ) − L ( t ) A ( t ) − G ( t ) = ( t + 1 ) ln ⁡ t − 2 ( t − 1 ) ( t + 1 − 2 t ) ln ⁡ t . (10) f 1 ( t ) = A ( t ) − L ( t ) A ( t ) − G ( t ) = ( t + 1 ) ln ⁡ t − 2 ( t − 1 ) ( t + 1 − 2 t ) ln ⁡ t . {:(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*}(10)f1(t)=A(t)−L(t)A(t)−G(t)=(t+1)ln⁡t−2(t−1)(t+1−2t)ln⁡t.
Let us suppose that the first inequality in the theorem holds. From lim t → ∞ f 1 ( t ) = 1 lim t → ∞   f 1 ( t ) = 1 lim_(t rarr oo)f_(1)(t)=1\lim _{t \rightarrow \infty} f_{1}(t)=1limt→∞f1(t)=1 it follows obviously that α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1. Conversely, if α ≥ 1 α ≥ 1 alpha >= 1\alpha \geq 1α≥1 it suffices to have
A ( t ) − L ( t ) A ( t ) − G ( t ) < 1 , A ( t ) − L ( t ) A ( t ) − G ( t ) < 1 , (A(t)-L(t))/(A(t)-G(t)) < 1,\frac{A(t)-L(t)}{A(t)-G(t)}<1,A(t)−L(t)A(t)−G(t)<1,
which is true because L ( t ) > G ( t ) L ( t ) > G ( t ) L(t) > G(t)L(t)>G(t)L(t)>G(t). We evaluate f 1 ( t ) − 2 / 3 f 1 ( t ) − 2 / 3 f_(1)(t)-2//3f_{1}(t)-2 / 3f1(t)−2/3, where 2 / 3 = lim t → 1 f 1 ( t ) 2 / 3 = lim t → 1   f 1 ( t ) 2//3=lim_(t rarr1)f_(1)(t)2 / 3= \lim _{t \rightarrow 1} f_{1}(t)2/3=limt→1f1(t) and show that it is positive. The denominator is obviously positive; we substitute u = t u = t u=sqrttu=\sqrt{t}u=t in the numerator and obtain
f ( u ) = ( u 2 + 4 u + 1 ) ln ⁡ u − 3 u 2 + 3 . f ( u ) = u 2 + 4 u + 1 ln ⁡ u − 3 u 2 + 3 . 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 .f(u)=(u2+4u+1)ln⁡u−3u2+3.
We have f ( 1 ) = f ′ ( 1 ) = f ′ ′ ( 1 ) = 0 f ( 1 ) = f ′ ( 1 ) = f ′ ′ ( 1 ) = 0 f(1)=f^(')(1)=f^('')(1)=0f(1)=f^{\prime}(1)=f^{\prime \prime}(1)=0f(1)=f′(1)=f′′(1)=0 and f ′ ′ ′ ( u ) = 2 ( u − 1 ) 2 / u 3 > 0 f ′ ′ ′ ( u ) = 2 ( u − 1 ) 2 / u 3 > 0 f^(''')(u)=2(u-1)^(2)//u^(3) > 0f^{\prime \prime \prime}(u)=2(u-1)^{2} / u^{3}>0f′′′(u)=2(u−1)2/u3>0 for u > 1 u > 1 u > 1u>1u>1, hence f 1 ( t ) > 2 / 3 f 1 ( t ) > 2 / 3 f_(1)(t) > 2//3f_{1}(t)>2 / 3f1(t)>2/3 for t > 1 t > 1 t > 1t>1t>1. It follows that L ( t ) < β G ( t ) + ( 1 − β ) A ( t ) , ∀ t > 1 L ( t ) < β G ( t ) + ( 1 − β ) A ( t ) , ∀ t > 1 L(t) < beta G(t)+(1-beta)A(t),AA t > 1L(t)<\beta G(t)+(1-\beta) A(t), \forall t>1L(t)<βG(t)+(1−β)A(t),∀t>1 if and only if β ≤ 2 / 3 β ≤ 2 / 3 beta <= 2//3\beta \leq 2 / 3β≤2/3.
Theorem 3. The double inequality
α G ( t ) + ( 1 − α ) Q ( t ) < A ( t ) < β G ( t ) + ( 1 − β ) Q ( t ) , ∀ t > 1 , α G ( t ) + ( 1 − α ) Q ( t ) < A ( t ) < β G ( t ) + ( 1 − β ) Q ( t ) , ∀ t > 1 , 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,αG(t)+(1−α)Q(t)<A(t)<βG(t)+(1−β)Q(t),∀t>1,
holds if and only if α ≥ 1 / 2 α ≥ 1 / 2 alpha >= 1//2\alpha \geq 1 / 2α≥1/2 and β ≤ 1 − 2 / 2 β ≤ 1 − 2 / 2 beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2β≤1−2/2.
Proof. Let us consider for t > 1 t > 1 t > 1t>1t>1, the function
(11) f 2 ( t ) = Q ( t ) − A ( t ) Q ( t ) − G ( t ) = 1 − t 2 + 1 + 2 t 2 ( t + 1 ) 2 . (11) f 2 ( t ) = Q ( t ) − A ( t ) Q ( t ) − G ( t ) = 1 − t 2 + 1 + 2 t 2 ( t + 1 ) 2 . {:(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*}(11)f2(t)=Q(t)−A(t)Q(t)−G(t)=1−t2+1+2t2(t+1)2.
We have f 2 ( t ) < 1 / 2 f 2 ( t ) < 1 / 2 f_(2)(t) < 1//2f_{2}(t)<1 / 2f2(t)<1/2, since t 2 + 1 + 2 t > 2 / 2 ( t + 1 ) 2 ⇔ t 2 + 1 > 2 / 2 ( t + 1 ) ⇔ ( t − 1 ) 2 > 0 t 2 + 1 + 2 t > 2 / 2 ( t + 1 ) 2 ⇔ t 2 + 1 > 2 / 2 ( t + 1 ) ⇔ ( t − 1 ) 2 > 0 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}>0t2+1+2t>2/2(t+1)2⇔t2+1>2/2(t+1)⇔(t−1)2>0. Since lim t → 1 f 2 ( t ) = 1 / 2 lim t → 1   f 2 ( t ) = 1 / 2 lim_(t rarr1)f_(2)(t)=1//2\lim _{t \rightarrow 1} f_{2}(t)=1 / 2limt→1f2(t)=1/2, it follows that α G ( t ) + ( 1 − α ) Q ( t ) < A ( t ) , ∀ t > 1 α G ( t ) + ( 1 − α ) Q ( t ) < A ( t ) , ∀ t > 1 alpha G(t)+(1-alpha)Q(t) < A(t),AA t > 1\alpha G(t)+ (1-\alpha) Q(t)<A(t), \forall t>1αG(t)+(1−α)Q(t)<A(t),∀t>1 if and only if α ≥ 1 / 2 α ≥ 1 / 2 alpha >= 1//2\alpha \geq 1 / 2α≥1/2.
Let us suppose that A ( t ) < β G ( t ) + ( 1 − β ) Q ( t ) , ∀ t > 1 A ( t ) < β G ( t ) + ( 1 − β ) Q ( t ) , ∀ t > 1 A(t) < beta G(t)+(1-beta)Q(t),AA t > 1A(t)<\beta G(t)+(1-\beta) Q(t), \forall t>1A(t)<βG(t)+(1−β)Q(t),∀t>1. Since lim t → ∞ f 2 ( t ) = 1 − 2 / 2 lim t → ∞   f 2 ( t ) = 1 − 2 / 2 lim_(t rarr oo)f_(2)(t)=1-sqrt2//2\lim _{t \rightarrow \infty} f_{2}(t)=1-\sqrt{2} / 2limt→∞f2(t)=1−2/2, it follows that β ≤ 1 − 2 / 2 β ≤ 1 − 2 / 2 beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2β≤1−2/2. Conversely, if
β ≤ 1 − 2 / 2 β ≤ 1 − 2 / 2 beta <= 1-sqrt2//2\beta \leq 1-\sqrt{2} / 2β≤1−2/2, it suffices to prove that f 2 ( t ) > 1 − 2 / 2 f 2 ( t ) > 1 − 2 / 2 f_(2)(t) > 1-sqrt2//2f_{2}(t)>1-\sqrt{2} / 2f2(t)>1−2/2. This is equivalent with
t 2 + 1 + 2 t ( t + 1 ) 2 < 1 t 2 + 1 + 2 t ( t + 1 ) 2 < 1 (sqrt(t^(2)+1)+sqrt(2t))/((sqrtt+1)^(2)) < 1\frac{\sqrt{t^{2}+1}+\sqrt{2 t}}{(\sqrt{t}+1)^{2}}<1t2+1+2t(t+1)2<1
i. e. t 2 + 1 + 2 t < ( t + 1 ) 2 t 2 + 1 + 2 t < ( t + 1 ) 2 sqrt(t^(2)+1)+sqrt(2t) < (sqrtt+1)^(2)\sqrt{t^{2}+1}+\sqrt{2 t}<(\sqrt{t}+1)^{2}t2+1+2t<(t+1)2 and this is true because t 2 + 1 + 2 t < t + 1 + 2 t < ( t + 1 ) 2 t 2 + 1 + 2 t < t + 1 + 2 t < ( t + 1 ) 2 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}t2+1+2t<t+1+2t<(t+1)2.
Theorem 4. The double inequality
α G ( t ) + ( 1 − α ) C ( t ) < A ( t ) < β G ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , α G ( t ) + ( 1 − α ) C ( t ) < A ( t ) < β G ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , 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,αG(t)+(1−α)C(t)<A(t)<βG(t)+(1−β)C(t),∀t>1,
holds if and only if α ≥ 2 / 3 α ≥ 2 / 3 alpha >= 2//3\alpha \geq 2 / 3α≥2/3 and β ≤ 1 / 2 β ≤ 1 / 2 beta <= 1//2\beta \leq 1 / 2β≤1/2.
Proof. For t > 1 t > 1 t > 1t>1t>1 we define
(12) f 3 ( t ) = C ( t ) − A ( t ) C ( t ) − G ( t ) = ( t − 1 ) 2 2 ( t 2 + 1 − t ( t + 1 ) ) (12) f 3 ( t ) = C ( t ) − A ( t ) C ( t ) − G ( t ) = ( t − 1 ) 2 2 t 2 + 1 − t ( t + 1 ) {:(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*}(12)f3(t)=C(t)−A(t)C(t)−G(t)=(t−1)22(t2+1−t(t+1))
Since lim t → 1 f 3 ( t ) = 2 / 3 lim t → 1   f 3 ( t ) = 2 / 3 lim_(t rarr1)f_(3)(t)=2//3\lim _{t \rightarrow 1} f_{3}(t)=2 / 3limt→1f3(t)=2/3, from α G ( t ) + ( 1 − α ) C ( t ) < A ( t ) , ∀ t > 1 α G ( t ) + ( 1 − α ) C ( t ) < A ( t ) , ∀ t > 1 alpha G(t)+(1-alpha)C(t) < A(t),AA t > 1\alpha G(t)+(1-\alpha) C(t)<A(t), \forall t>1αG(t)+(1−α)C(t)<A(t),∀t>1 it follows that α ≥ 2 / 3 α ≥ 2 / 3 alpha >= 2//3\alpha \geq 2 / 3α≥2/3. If α ≥ 2 / 3 α ≥ 2 / 3 alpha >= 2//3\alpha \geq 2 / 3α≥2/3, it is true that f 3 ( t ) < α f 3 ( t ) < α f_(3)(t) < alphaf_{3}(t)<\alphaf3(t)<α, because f 3 ( t ) < 2 / 3 f 3 ( t ) < 2 / 3 f_(3)(t) < 2//3f_{3}(t)<2 / 3f3(t)<2/3 is equivalent with
t t + t + 1 < 1 3 , t t + t + 1 < 1 3 , (sqrtt)/(t+sqrtt+1) < (1)/(3),\frac{\sqrt{t}}{t+\sqrt{t}+1}<\frac{1}{3},tt+t+1<13,
or ( t − 1 ) 2 > 0 ( t − 1 ) 2 > 0 (sqrtt-1)^(2) > 0(\sqrt{t}-1)^{2}>0(t−1)2>0.
Similarly, it follows that f 3 ( t ) > 1 / 2 f 3 ( t ) > 1 / 2 f_(3)(t) > 1//2f_{3}(t)>1 / 2f3(t)>1/2, since
f 3 ( t ) − 1 2 = t ( t − 1 ) 2 2 ( t 2 + 1 − t ( t + 1 ) ) = t 2 ( t + t + 1 ) > 0 f 3 ( t ) − 1 2 = t ( t − 1 ) 2 2 t 2 + 1 − t ( t + 1 ) = t 2 ( t + t + 1 ) > 0 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)}>0f3(t)−12=t(t−1)22(t2+1−t(t+1))=t2(t+t+1)>0
The infimum of f 3 f 3 f_(3)f_{3}f3 on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) is precisely 1 / 2 1 / 2 1//21 / 21/2, because lim t → ∞ f 3 ( t ) = 1 / 2 lim t → ∞   f 3 ( t ) = 1 / 2 lim_(t rarr oo)f_(3)(t)=1//2\lim _{t \rightarrow \infty} f_{3}(t)=1 / 2limt→∞f3(t)=1/2
Theorem 5. The double inequality
α G ( t ) + ( 1 − α ) C ( t ) < Q ( t ) < β G ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , α G ( t ) + ( 1 − α ) C ( t ) < Q ( t ) < β G ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , 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,αG(t)+(1−α)C(t)<Q(t)<βG(t)+(1−β)C(t),∀t>1,
holds if and only if α ≥ 1 − 2 / 2 α ≥ 1 − 2 / 2 alpha >= 1-sqrt2//2\alpha \geq 1-\sqrt{2} / 2α≥1−2/2 and β < β 0 β < β 0 beta < beta_(0)\beta<\beta_{0}β<β0, where β 0 ≅ 0.3471574308 … β 0 ≅ 0.3471574308 … beta_(0)~=0.3471574308 dots\beta_{0} \cong 0.3471574308 \ldotsβ0≅0.3471574308… is the unique positive root of the polynomial
9 x 4 − 26 x 3 + 22 x 2 − 2 x − 1 9 x 4 − 26 x 3 + 22 x 2 − 2 x − 1 9x^(4)-26x^(3)+22x^(2)-2x-19 x^{4}-26 x^{3}+22 x^{2}-2 x-19x4−26x3+22x2−2x−1
Proof. We have to find, for t > 1 t > 1 t > 1t>1t>1, the extreme values of
(13) f 4 ( t ) = C ( t ) − Q ( t ) C ( t ) − G ( t ) = 2 t 2 − 2 t 2 + 1 − 2 t t 2 + 1 + 2 2 ( t 2 − t 3 2 − t + 1 ) (13) f 4 ( t ) = C ( t ) − Q ( t ) C ( t ) − G ( t ) = 2 t 2 − 2 t 2 + 1 − 2 t t 2 + 1 + 2 2 t 2 − t 3 2 − t + 1 {:(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*}(13)f4(t)=C(t)−Q(t)C(t)−G(t)=2t2−2t2+1−2tt2+1+22(t2−t32−t+1)
Denoting by u = t u = t u=sqrttu=\sqrt{t}u=t, we compute the derivative of
h ( u ) = f 4 ( u 2 ) h ( u ) = f 4 u 2 h(u)=f_(4)(u^(2))h(u)=f_{4}\left(u^{2}\right)h(u)=f4(u2)
and we obtain
h ′ ( u ) = ( u + 1 ) ( 2 ( u 4 + 1 ) ( − u 4 − 4 u 2 − 1 ) + u 6 + 2 u 5 + 3 u 4 + 3 u 2 + 2 u + 1 ) 2 ( u 4 + 1 ) ( u − 1 ) 3 ( u 2 + u + 1 ) 2 . h ′ ( u ) = ( u + 1 ) 2 u 4 + 1 − u 4 − 4 u 2 − 1 + u 6 + 2 u 5 + 3 u 4 + 3 u 2 + 2 u + 1 2 u 4 + 1 ( u − 1 ) 3 u 2 + u + 1 2 . 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}} .h′(u)=(u+1)(2(u4+1)(−u4−4u2−1)+u6+2u5+3u4+3u2+2u+1)2(u4+1)(u−1)3(u2+u+1)2.
So, the roots of the derivative satisfy the algebraic equation
2 ( u 4 + 1 ) ( u 4 + 4 u 2 + 1 ) 2 = ( u 6 + 2 u 5 + 3 u 4 + 3 u 2 + 2 u + 1 ) 2 . 2 u 4 + 1 u 4 + 4 u 2 + 1 2 = u 6 + 2 u 5 + 3 u 4 + 3 u 2 + 2 u + 1 2 . 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} .2(u4+1)(u4+4u2+1)2=(u6+2u5+3u4+3u2+2u+1)2.
After the simplification of a quartic polynomial whose roots are not in the interval ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ), we obtain the equation
(14) u 8 − 8 u 5 − 10 u 4 − 8 u 3 + 1 = 0 , (14) u 8 − 8 u 5 − 10 u 4 − 8 u 3 + 1 = 0 , {:(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*}(14)u8−8u5−10u4−8u3+1=0,
which has a unique root u 0 u 0 u_(0)u_{0}u0 in the interval ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ). This can be easily proved by using the Sturm sequence. Then u 0 u 0 u_(0)u_{0}u0 will be the unique root of h ′ h ′ h^(')h^{\prime}h′ in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ).
Now h ′ ( 2 ) > 0 , h ′ ( 3 ) < 0 h ′ ( 2 ) > 0 , h ′ ( 3 ) < 0 h^(')(2) > 0,h^(')(3) < 0h^{\prime}(2)>0, h^{\prime}(3)<0h′(2)>0,h′(3)<0, so 2 < u 0 < 3 2 < u 0 < 3 2 < u_(0) < 32<u_{0}<32<u0<3 and h h hhh is strictly increasing in the interval ( 1 , u 0 1 , u 0 1,u_(0)1, u_{0}1,u0 ) and strictly decreasing in the interval ( u 0 , ∞ u 0 , ∞ u_(0),oou_{0}, \inftyu0,∞ ). We also have lim u → 1 h ( u ) = 1 / 3 , lim u → ∞ h ( u ) = 1 − 2 / 2 lim u → 1   h ( u ) = 1 / 3 , lim u → ∞   h ( u ) = 1 − 2 / 2 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} / 2limu→1h(u)=1/3,limu→∞h(u)=1−2/2 and therefore inf f 4 = inf h = 1 − 2 / 2 , sup f 4 = sup h = h ( u 0 ) = β 0 inf f 4 = inf h = 1 − 2 / 2 , sup f 4 = sup h = h u 0 = β 0 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}inff4=infh=1−2/2,supf4=suph=h(u0)=β0.
Since h ( u 0 ) h u 0 h(u_(0))h\left(u_{0}\right)h(u0) 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];
9 x 4 − 26 x 3 + 22 x 2 − 2 x − 1 9 x 4 − 26 x 3 + 22 x 2 − 2 x − 1 9x^(4)-26x^(3)+22x^(2)-2x-19 x^{4}-26 x^{3}+22 x^{2}-2 x-19x4−26x3+22x2−2x−1
Notice that Maple is of course able to express the maximum h ( u 0 ) h u 0 h(u_(0))h\left(u_{0}\right)h(u0) in terms of radicals by executing the command:
> select ( u − > > select u − > > select(u- > :}>\operatorname{select}\left(\mathrm{u}->\right.>select(u−> is ( u > 0 ) ( u > 0 ) (u > 0)(\mathrm{u}>0)(u>0),[solve ( 9 ∗ x ∧ 4 − 26 ∗ x ∧ 3 + 22 ∗ x ∧ 2 − 2 ∗ x − 1 9 ∗ x ∧ 4 − 26 ∗ x ∧ 3 + 22 ∗ x ∧ 2 − 2 ∗ x − 1 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-19∗x∧4−26∗x∧3+22∗x∧2−2∗x−1, Explicit)]); but the resulting expression is cumbersome and we will not print it here.
Theorem 6. The double inequality
α A ( t ) + ( 1 − α ) C ( t ) < Q ( t ) < β A ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , α A ( t ) + ( 1 − α ) C ( t ) < Q ( t ) < β A ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , 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,αA(t)+(1−α)C(t)<Q(t)<βA(t)+(1−β)C(t),∀t>1,
holds if and only if α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2 and β ≤ 1 / 2 β ≤ 1 / 2 beta <= 1//2\beta \leq 1 / 2β≤1/2.
Proof. Let us consider, for t > 1 t > 1 t > 1t>1t>1
(15) f 5 ( t ) = C ( t ) − Q ( t ) C ( t ) − A ( t ) = 2 ( t 2 + 1 ) − ( t + 1 ) 2 ( t 2 + 1 ) ( t − 1 ) 2 . (15) f 5 ( t ) = C ( t ) − Q ( t ) C ( t ) − A ( t ) = 2 t 2 + 1 − ( t + 1 ) 2 t 2 + 1 ( t − 1 ) 2 . {:(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*}(15)f5(t)=C(t)−Q(t)C(t)−A(t)=2(t2+1)−(t+1)2(t2+1)(t−1)2.
From lim t → ∞ f 5 ( t ) = 2 − 2 lim t → ∞   f 5 ( t ) = 2 − 2 lim_(t rarr oo)f_(5)(t)=2-sqrt2\lim _{t \rightarrow \infty} f_{5}(t)=2-\sqrt{2}limt→∞f5(t)=2−2 it follows that α A ( t ) + ( 1 − α ) C ( t ) ⟨ Q ( t ) , ∀ t ⟩ α A ( t ) + ( 1 − α ) C ( t ) ⟨ Q ( t ) , ∀ t ⟩ alpha A(t)+(1-alpha)C(t)(:Q(t),AA t:)\alpha A(t)+(1-\alpha) C(t)\langle Q(t), \forall t\rangleαA(t)+(1−α)C(t)⟨Q(t),∀t⟩ 1 implies α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2. Now if α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2 we have to prove that f 5 ( t ) < 2 − 2 f 5 ( t ) < 2 − 2 f_(5)(t) < 2-sqrt2f_{5}(t)<2-\sqrt{2}f5(t)<2−2, which can be written as 2 ( t 2 + 1 ) ( t + 1 ) > − 2 t 2 + ( 4 + 2 2 ) t − 2 2 t 2 + 1 ( t + 1 ) > − 2 t 2 + ( 4 + 2 2 ) t − 2 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}2(t2+1)(t+1)>−2t2+(4+22)t−2. If − 2 t 2 + ( 4 + 2 2 ) t − 2 ≤ 0 − 2 t 2 + ( 4 + 2 2 ) t − 2 ≤ 0 -sqrt2t^(2)+(4+2sqrt2)t-sqrt2 <= 0-\sqrt{2} t^{2}+(4+2 \sqrt{2}) t-\sqrt{2} \leq 0−2t2+(4+22)t−2≤0, the inequality holds. Otherwise, squaring both sides it reduces to 4 ( 3 + 2 2 ) t ( t − 1 ) 2 > 0 4 ( 3 + 2 2 ) t ( t − 1 ) 2 > 0 4(3+2sqrt2)t(t-1)^(2) > 04(3+2 \sqrt{2}) t(t-1)^{2}>04(3+22)t(t−1)2>0.
We obtain also
f 5 ( t ) − 1 2 = 3 ( t 2 + 1 ) + 2 t − 2 ( t + 1 ) 2 ( t 2 + 1 ) 2 ( t − 1 ) 2 > 0 , f 5 ( t ) − 1 2 = 3 t 2 + 1 + 2 t − 2 ( t + 1 ) 2 t 2 + 1 2 ( t − 1 ) 2 > 0 , 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,f5(t)−12=3(t2+1)+2t−2(t+1)2(t2+1)2(t−1)2>0,
because ( 3 ( t 2 + 1 ) + 2 t ) 2 − 8 ( t + 1 ) 2 ( t 2 + 1 ) > 0 ⇔ ( t − 1 ) 4 > 0 3 t 2 + 1 + 2 t 2 − 8 ( t + 1 ) 2 t 2 + 1 > 0 ⇔ ( t − 1 ) 4 > 0 (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(3(t2+1)+2t)2−8(t+1)2(t2+1)>0⇔(t−1)4>0. We have lim t → 1 f 5 ( t ) = 1 / 2 lim t → 1   f 5 ( t ) = 1 / 2 lim_(t rarr1)f_(5)(t)=1//2\lim _{t \rightarrow 1} f_{5}(t)=1 / 2limt→1f5(t)=1/2, hence this is the infimum of f 5 f 5 f_(5)f_{5}f5 on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) and the second part of the theorem is also true.
Lemma 7. 3] For t > 1 t > 1 t > 1t>1t>1, the following inequality holds
(16) t t t + 1 > t − ln ⁡ t . (16) t t t + 1 > t − ln ⁡ t . {:(16)t^((t)/(t+1)) > t-ln t.:}\begin{equation*} t^{\frac{t}{t+1}}>t-\ln t . \tag{16} \end{equation*}(16)ttt+1>t−ln⁡t.
Proof. The inequality (16) is equivalent to
t t + 1 ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) t t + 1 ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) (t)/(t+1)ln t > ln(t-ln t)\frac{t}{t+1} \ln t>\ln (t-\ln t)tt+1ln⁡t>ln⁡(t−ln⁡t)
We consider the function
k ( t ) = ln ⁡ ( t − ln ⁡ t ) − t − 1 t ln ⁡ t , t > 1 k ( t ) = ln ⁡ ( t − ln ⁡ t ) − t − 1 t ln ⁡ t , t > 1 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>1k(t)=ln⁡(t−ln⁡t)−t−1tln⁡t,t>1
with
k ′ ( t ) = ( ln ⁡ t − 1 ) ln ⁡ t t 2 ( t − ln ⁡ t ) k ′ ( t ) = ( ln ⁡ t − 1 ) ln ⁡ t t 2 ( t − ln ⁡ t ) 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)}k′(t)=(ln⁡t−1)ln⁡tt2(t−ln⁡t)
It has lim t → 1 k ( t ) = 0 , lim t → ∞ k ( t ) = 0 lim t → 1   k ( t ) = 0 , lim t → ∞   k ( t ) = 0 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)=0limt→1k(t)=0,limt→∞k(t)=0 and a minimum at t 0 = e t 0 = e t_(0)=et_{0}=et0=e. It follows that k ( t ) < 0 k ( t ) < 0 k(t) < 0k(t)<0k(t)<0 on ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞), hence ( ( t − 1 ) / t ) ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) ( ( t − 1 ) / t ) ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) ((t-1)//t)ln t > ln(t-ln t)((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 ) t t + 1 ln ⁡ t > t − 1 t ln ⁡ t > ln ⁡ ( t − ln ⁡ t ) (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)tt+1ln⁡t>t−1tln⁡t>ln⁡(t−ln⁡t)
Theorem 8. The double inequality
α A ( t ) + ( 1 − α ) C ( t ) < S ( t ) < β A ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , α A ( t ) + ( 1 − α ) C ( t ) < S ( t ) < β A ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 , 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,αA(t)+(1−α)C(t)<S(t)<βA(t)+(1−β)C(t),∀t>1,
holds if and only if α ≥ 1 / 2 α ≥ 1 / 2 alpha >= 1//2\alpha \geq 1 / 2α≥1/2 and β ≤ 0 β ≤ 0 beta <= 0\beta \leq 0β≤0.
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 (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 {:(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*}(17)f6(t)=C(t)−S(t)C(t)−A(t)=2t2+1−(t+1)ttt+1(t−1)2
We have
f 6 ( t ) − 1 2 = 3 ( t 2 + 1 ) + 2 t − 4 ( t + 1 ) t t t + 1 2 ( t − 1 ) 2 = 2 ( t + 1 ) ( t − 1 ) 2 ⋅ g ( t ) f 6 ( t ) − 1 2 = 3 t 2 + 1 + 2 t − 4 ( t + 1 ) t t t + 1 2 ( t − 1 ) 2 = 2 ( t + 1 ) ( t − 1 ) 2 ⋅ g ( t ) 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)f6(t)−12=3(t2+1)+2t−4(t+1)ttt+12(t−1)2=2(t+1)(t−1)2⋅g(t)
where
g ( t ) = 3 ( t 2 + 1 ) + 2 t 4 ( t + 1 ) − t t t + 1 g ( t ) = 3 t 2 + 1 + 2 t 4 ( t + 1 ) − t t t + 1 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}}g(t)=3(t2+1)+2t4(t+1)−ttt+1
Then
g ′ ( t ) = − g 1 ( t ) 4 ( t + 1 ) 2 g ′ ( t ) = − g 1 ( t ) 4 ( t + 1 ) 2 g^(')(t)=-(g_(1)(t))/(4(t+1)^(2))g^{\prime}(t)=-\frac{g_{1}(t)}{4(t+1)^{2}}g′(t)=−g1(t)4(t+1)2
where
(18) g 1 ( t ) = 4 t t t + 1 ( t + 1 + ln ⁡ t ) + 1 − 3 t 2 − 6 t (18) g 1 ( t ) = 4 t t t + 1 ( t + 1 + ln ⁡ t ) + 1 − 3 t 2 − 6 t {:(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*}(18)g1(t)=4ttt+1(t+1+ln⁡t)+1−3t2−6t
Using the fact that S > Q S > Q S > QS>QS>Q, i. e. t t / ( t + 1 ) > ( t 2 + 1 ) / 2 t t / ( t + 1 ) > t 2 + 1 / 2 t^(t//(t+1)) > sqrt((t^(2)+1)//2)t^{t /(t+1)}>\sqrt{\left(t^{2}+1\right) / 2}tt/(t+1)>(t2+1)/2, we obtain that g 1 ( t ) > 2 ( t 2 + 1 ) g 2 ( t ) g 1 ( t ) > 2 t 2 + 1 g 2 ( t ) g_(1)(t) > sqrt(2(t^(2)+1))g_(2)(t)g_{1}(t)>\sqrt{2\left(t^{2}+1\right)} g_{2}(t)g1(t)>2(t2+1)g2(t), where
g 2 ( t ) = 2 ( t + 1 + ln ⁡ t ) − ( 3 t 2 + 6 t − 1 ) / 2 ( t 2 + 1 ) g 2 ( t ) = 2 ( t + 1 + ln ⁡ t ) − 3 t 2 + 6 t − 1 / 2 t 2 + 1 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)}g2(t)=2(t+1+ln⁡t)−(3t2+6t−1)/2(t2+1)
The derivative of g 2 g 2 g_(2)g_{2}g2 is
g 2 ′ ( t ) = ( t + 1 ) ( 2 ( t 2 + 1 ) ) 3 − ( 3 t 4 + 7 t 2 + 6 t ) t ( t 2 + 1 ) 2 ( t 2 + 1 ) g 2 ′ ( t ) = ( t + 1 ) 2 t 2 + 1 3 − 3 t 4 + 7 t 2 + 6 t t t 2 + 1 2 t 2 + 1 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)}}g2′(t)=(t+1)(2(t2+1))3−(3t4+7t2+6t)t(t2+1)2(t2+1)
In order to establish its sign we consider the polynomial
P ( t ) = ( t + 1 ) 2 ( 2 ( t 2 + 1 ) ) 6 − ( 3 t 4 + 7 t 2 + 6 t ) 2 = t 6 ( t − 1 ) 2 ( 8 t 6 + 32 t 5 + 52 t 4 + 36 t 3 + 19 t 2 + 14 t − 1 ) P ( t ) = ( t + 1 ) 2 2 t 2 + 1 6 − 3 t 4 + 7 t 2 + 6 t 2 = t 6 ( t − 1 ) 2 8 t 6 + 32 t 5 + 52 t 4 + 36 t 3 + 19 t 2 + 14 t − 1 {:[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}P(t)=(t+1)2(2(t2+1))6−(3t4+7t2+6t)2=t6(t−1)2(8t6+32t5+52t4+36t3+19t2+14t−1)
The expression from the last parenthesis is obviously decreasing for t ≥ 1 t ≥ 1 t >= 1t \geq 1t≥1 and it is positive for t = 10 t = 10 t=10t=10t=10. It follows that it is positive on ( 1 , 10 ) ( 1 , 10 ) (1,10)(1,10)(1,10), hence on this interval P P PPP is also positive. Therefore g 2 ′ ( t ) > 0 , g 2 ( t ) > g 2 ( 1 ) = 0 g 2 ′ ( t ) > 0 , g 2 ( t ) > g 2 ( 1 ) = 0 g_(2)^(')(t) > 0,g_(2)(t) > g_(2)(1)=0g_{2}^{\prime}(t)>0, g_{2}(t)>g_{2}(1)=0g2′(t)>0,g2(t)>g2(1)=0, so g 1 g 1 g_(1)g_{1}g1 is positive too for 1 < t < 10 1 < t < 10 1 < t < 101<t<101<t<10.
Let us consider now that t ≥ 10 t ≥ 10 t >= 10t \geq 10t≥10. Using (16) in (18) we obtain that g 1 ( t ) > g 3 ( t ) g 1 ( t ) > g 3 ( t ) g_(1)(t) > g_(3)(t)g_{1}(t)> g_{3}(t)g1(t)>g3(t), where
g 3 ( t ) = t 2 − 2 t + 2 − ( 2 ln ⁡ t + 1 ) 2 . g 3 ( t ) = t 2 − 2 t + 2 − ( 2 ln ⁡ t + 1 ) 2 . g_(3)(t)=t^(2)-2t+2-(2ln t+1)^(2).g_{3}(t)=t^{2}-2 t+2-(2 \ln t+1)^{2} .g3(t)=t2−2t+2−(2ln⁡t+1)2.
For
g 4 ( t ) = t 2 − 2 t + 2 − 2 ln ⁡ t − 1 g 4 ( t ) = t 2 − 2 t + 2 − 2 ln ⁡ t − 1 g_(4)(t)=sqrt(t^(2)-2t+2)-2ln t-1g_{4}(t)=\sqrt{t^{2}-2 t+2}-2 \ln t-1g4(t)=t2−2t+2−2ln⁡t−1
the sign of g 4 ′ g 4 ′ g_(4)^(')g_{4}^{\prime}g4′ is given by t 2 − t − 2 t 2 − 2 t + 2 t 2 − t − 2 t 2 − 2 t + 2 t^(2)-t-2sqrt(t^(2)-2t+2)t^{2}-t-2 \sqrt{t^{2}-2 t+2}t2−t−2t2−2t+2; but ( t 2 − t ) 2 − 4 ( t 2 − 2 t + 2 ) = ( t − 10 ) 4 + 38 ( t − 10 ) 3 + 537 ( t − 10 ) 2 + 3348 ( t − 10 ) + 7772 > 0 t 2 − t 2 − 4 t 2 − 2 t + 2 = ( t − 10 ) 4 + 38 ( t − 10 ) 3 + 537 ( t − 10 ) 2 + 3348 ( t − 10 ) + 7772 > 0 (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(t2−t)2−4(t2−2t+2)=(t−10)4+38(t−10)3+537(t−10)2+3348(t−10)+7772>0 for t ≥ 10 t ≥ 10 t >= 10t \geq 10t≥10. It follows that g 3 ( t ) ≥ g 3 ( 10 ) = 3.45 … > 0 g 3 ( t ) ≥ g 3 ( 10 ) = 3.45 … > 0 g_(3)(t) >= g_(3)(10)=3.45 dots > 0g_{3}(t) \geq g_{3}(10)=3.45 \ldots>0g3(t)≥g3(10)=3.45…>0, hence g 1 g 1 g_(1)g_{1}g1 is positive for t ≥ 10 t ≥ 10 t >= 10t \geq 10t≥10 too.
In conclusion, g 1 ( t ) > 0 g 1 ( t ) > 0 g_(1)(t) > 0g_{1}(t)>0g1(t)>0 on ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞), therefore g ′ ( t ) < 0 g ′ ( t ) < 0 g^(')(t) < 0g^{\prime}(t)<0g′(t)<0 on ( 1 , ∞ ) ( 1 , ∞ ) (1,oo)(1, \infty)(1,∞). The function g g ggg being decreasing, g ( t ) < g ( 1 ) = 0 g ( t ) < g ( 1 ) = 0 g(t) < g(1)=0g(t)<g(1)=0g(t)<g(1)=0 for t > 1 t > 1 t > 1t>1t>1 and f 6 ( t ) < 1 / 2 f 6 ( t ) < 1 / 2 f_(6)(t) < 1//2f_{6}(t)<1 / 2f6(t)<1/2 for t > 1 t > 1 t > 1t>1t>1.
The second part of the theorem follows from lim t → ∞ f 6 ( t ) = 0 lim t → ∞   f 6 ( t ) = 0 lim_(t rarr oo)f_(6)(t)=0\lim _{t \rightarrow \infty} f_{6}(t)=0limt→∞f6(t)=0 and f 6 ( t ) > 0 f 6 ( t ) > 0 f_(6)(t) > 0f_{6}(t)>0f6(t)>0, ∀ t > 1 ∀ t > 1 AA t > 1\forall t>1∀t>1.
Theorem 9. The double inequality
α A ( t ) + ( 1 − α ) S ( t ) < Q ( t ) < β A ( t ) + ( 1 − β ) S ( t ) , ∀ t > 1 α A ( t ) + ( 1 − α ) S ( t ) < Q ( t ) < β A ( t ) + ( 1 − β ) S ( t ) , ∀ t > 1 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>1αA(t)+(1−α)S(t)<Q(t)<βA(t)+(1−β)S(t),∀t>1
holds if and only if α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2 and β ≤ 0 β ≤ 0 beta <= 0\beta \leq 0β≤0.
Proof. We shall prove that the first inequality holds for α = 2 − 2 α = 2 − 2 alpha=2-sqrt2\alpha=2-\sqrt{2}α=2−2 (hence a fortiori for α ≥ 2 − 2 α ≥ 2 − 2 alpha >= 2-sqrt2\alpha \geq 2-\sqrt{2}α≥2−2 ).
Let us denote
H ( t , α ) = Q ( t ) − α A ( t ) − ( 1 − α ) S ( t ) = 1 2 2 + 2 t 2 − 1 2 α ( 1 + t ) − ( 1 − α ) t t 1 + t H ( t , α ) = Q ( t ) − α A ( t ) − ( 1 − α ) S ( t ) = 1 2 2 + 2 t 2 − 1 2 α ( 1 + t ) − ( 1 − α ) t t 1 + t 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}}H(t,α)=Q(t)−αA(t)−(1−α)S(t)=122+2t2−12α(1+t)−(1−α)tt1+t.
and
h 1 ( t ) = ( 2 + 1 ) H ( t , 2 − 2 ) , h 1 ( t ) = ( 2 + 1 ) H ( t , 2 − 2 ) , h_(1)(t)=(sqrt2+1)H(t,2-sqrt2),h_{1}(t)=(\sqrt{2}+1) H(t, 2-\sqrt{2}),h1(t)=(2+1)H(t,2−2),
where H H HHH is given in (19). We have to prove that h 1 ( t ) > 0 h 1 ( t ) > 0 h_(1)(t) > 0h_{1}(t)>0h1(t)>0 for t > 1 t > 1 t > 1t>1t>1. It follows that
h 1 ( t ) = ( 2 + 1 ) 2 ( t 2 + 1 ) 2 − 2 ( t + 1 ) 2 − t t t + 1 . h 1 ( t ) = ( 2 + 1 ) 2 t 2 + 1 2 − 2 ( t + 1 ) 2 − t t t + 1 . 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}} .h1(t)=(2+1)2(t2+1)2−2(t+1)2−ttt+1.
We put in the inequality ( 1 + x ) q < 1 + q x ( 1 + x ) q < 1 + q x (1+x)^(q) < 1+qx(1+x)^{q}<1+q x(1+x)q<1+qx, which holds for x > 0 , 0 < q < 1 x > 0 , 0 < q < 1 x > 0,0 < q < 1x>0,0<q<1x>0,0<q<1, x = t − 1 x = t − 1 x=t-1x=t-1x=t−1 and q = t / ( t + 1 ) q = t / ( t + 1 ) q=t//(t+1)q=t /(t+1)q=t/(t+1). It follows that
t t t + 1 < t 2 + 1 t + 1 t t t + 1 < t 2 + 1 t + 1 t^((t)/(t+1)) < (t^(2)+1)/(t+1)t^{\frac{t}{t+1}}<\frac{t^{2}+1}{t+1}ttt+1<t2+1t+1
and
h 1 ( t ) > ( 1 + 2 ) 2 ( t + 1 ) ( ( t + 1 ) 2 + 2 t 2 − 2 ( t 2 + 2 ( 2 − 1 ) t + 1 ) ) h 1 ( t ) > ( 1 + 2 ) 2 ( t + 1 ) ( t + 1 ) 2 + 2 t 2 − 2 t 2 + 2 ( 2 − 1 ) t + 1 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)h1(t)>(1+2)2(t+1)((t+1)2+2t2−2(t2+2(2−1)t+1))
Let us denote the positive expressions
h 2 ( t ) = ( t + 1 ) 2 + 2 t 2 , h 3 ( t ) = 2 ( t 2 + 2 ( 2 − 1 ) t + 1 ) h 2 ( t ) = ( t + 1 ) 2 + 2 t 2 , h 3 ( t ) = 2 t 2 + 2 ( 2 − 1 ) t + 1 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)h2(t)=(t+1)2+2t2,h3(t)=2(t2+2(2−1)t+1)
it follows easily that h 2 2 ( t ) − h 3 2 ( t ) = 4 t ( t − 1 ) 2 h 2 2 ( t ) − h 3 2 ( t ) = 4 t ( t − 1 ) 2 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}h22(t)−h32(t)=4t(t−1)2, therefore h 1 ( t ) > 0 h 1 ( t ) > 0 h_(1)(t) > 0h_{1}(t)>0h1(t)>0.
The second part of the theorem is obvious, since
f 7 ( t ) = S ( t ) − Q ( t ) S ( t ) − A ( t ) f 7 ( t ) = S ( t ) − Q ( t ) S ( t ) − A ( t ) f_(7)(t)=(S(t)-Q(t))/(S(t)-A(t))f_{7}(t)=\frac{S(t)-Q(t)}{S(t)-A(t)}f7(t)=S(t)−Q(t)S(t)−A(t)
satisfies f 7 ( t ) > 0 , ∀ t > 1 f 7 ( t ) > 0 , ∀ t > 1 f_(7)(t) > 0,AA t > 1f_{7}(t)>0, \forall t>1f7(t)>0,∀t>1 and lim t → 1 f 7 ( t ) = 0 lim t → 1   f 7 ( t ) = 0 lim_(t rarr1)f_(7)(t)=0\lim _{t \rightarrow 1} f_{7}(t)=0limt→1f7(t)=0.
Theorem 10. The double inequality
α L ( t ) + ( 1 − α ) C ( t ) < A ( t ) < β L ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 α L ( t ) + ( 1 − α ) C ( t ) < A ( t ) < β L ( t ) + ( 1 − β ) C ( t ) , ∀ t > 1 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>1αL(t)+(1−α)C(t)<A(t)<βL(t)+(1−β)C(t),∀t>1
holds if and only if α ≥ 3 / 4 α ≥ 3 / 4 alpha >= 3//4\alpha \geq 3 / 4α≥3/4 and β ≤ 1 / 2 β ≤ 1 / 2 beta <= 1//2\beta \leq 1 / 2β≤1/2.
Proof. We have to find the extreme values of
f 8 ( t ) = C ( t ) − A ( t ) C ( t ) − L ( t ) f 8 ( t ) = C ( t ) − A ( t ) C ( t ) − L ( t ) f_(8)(t)=(C(t)-A(t))/(C(t)-L(t))f_{8}(t)=\frac{C(t)-A(t)}{C(t)-L(t)}f8(t)=C(t)−A(t)C(t)−L(t)
for t > 1 t > 1 t > 1t>1t>1, where f 8 f 8 f_(8)f_{8}f8 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 ) = 1 2 ⋅ ( t − 1 ) 2 ln ⁡ t ln ⁡ t + t 2 ln ⁡ t − t 2 + 1 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}f8(t)=12⋅(t−1)2ln⁡tln⁡t+t2ln⁡t−t2+1
We obtain
f 8 ′ ( t ) = − ( t − 1 ) h 3 ( t ) 2 t ( ln ⁡ t + t 2 ln ⁡ t − t 2 + 1 ) 2 f 8 ′ ( t ) = − ( t − 1 ) h 3 ( t ) 2 t ln ⁡ t + t 2 ln ⁡ t − t 2 + 1 2 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}}f8′(t)=−(t−1)h3(t)2t(ln⁡t+t2ln⁡t−t2+1)2
where
h 4 ( t ) = t 3 − 2 ( t 2 ln ⁡ t ) 2 + 2 t 2 ln ⁡ t − t 2 − 2 t ln ⁡ t − 2 t ( ln ⁡ t ) 2 − t + 1 h 4 ( t ) = t 3 − 2 t 2 ln ⁡ t 2 + 2 t 2 ln ⁡ t − t 2 − 2 t ln ⁡ t − 2 t ( ln ⁡ t ) 2 − t + 1 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+1h4(t)=t3−2(t2ln⁡t)2+2t2ln⁡t−t2−2tln⁡t−2t(ln⁡t)2−t+1
Now, h 4 ( 1 ) = h 4 ′ ( 1 ) = h 4 ′ ′ ( 1 ) = h 4 ′ ′ ′ ( 1 ) = 0 h 4 ( 1 ) = h 4 ′ ( 1 ) = h 4 ′ ′ ( 1 ) = h 4 ′ ′ ′ ( 1 ) = 0 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)=0h4(1)=h4′(1)=h4′′(1)=h4′′′(1)=0 and
h 4 ( 4 ) ( t ) = 8 ( t − 1 ) ln ⁡ t t 3 > 0 for t > 1 . h 4 ( 4 ) ( t ) = 8 ( t − 1 ) ln ⁡ t t 3 > 0  for  t > 1 . 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 .h4(4)(t)=8(t−1)ln⁡tt3>0 for t>1.
Therefore h 4 > 0 h 4 > 0 h_(4) > 0h_{4}>0h4>0 in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ) and f 8 ′ < 0 f 8 ′ < 0 f_(8)^(') < 0f_{8}^{\prime}<0f8′<0 in ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ). The function f 8 f 8 f_(8)f_{8}f8 being strictly decreasing on ( 1 , ∞ 1 , ∞ 1,oo1, \infty1,∞ ), inf f 8 = lim t → ∞ f 8 ( t ) = 1 / 2 inf f 8 = lim t → ∞   f 8 ( t ) = 1 / 2 i n ff_(8)=lim_(t rarr oo)f_(8)(t)=1//2\inf f_{8}=\lim _{t \rightarrow \infty} f_{8}(t)=1 / 2inff8=limt→∞f8(t)=1/2 and sup f 8 = lim t → 1 f 8 ( t ) = 3 / 4 f 8 = lim t → 1   f 8 ( t ) = 3 / 4 f_(8)=lim_(t rarr1)f_(8)(t)=3//4f_{8}=\lim _{t \rightarrow 1} f_{8}(t)= 3 / 4f8=limt→1f8(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 ∈ R a , b > 0 , u , v ∈ R a,b > 0,u,v inRa, b>0, u, v \in \mathbb{R}a,b>0,u,v∈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 ) = 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)={[((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}Gu,v(a,b)={(au+buav+bv)1u−v,u≠vexp⁡(aulog⁡a+bulog⁡bau+bu),u=v
namely S = G 1 , 1 , C = G 2 , 1 S = G 1 , 1 , C = G 2 , 1 S=G_(1,1),C=G_(2,1)S=G_{1,1}, C=G_{2,1}S=G1,1,C=G2,1; two belong to the class of Stolarsky means [8] defined for a , b > 0 , a ≠ b , u , v ∈ R a , b > 0 , a ≠ b , u , v ∈ R a,b > 0,a!=b,u,v inRa, b>0, a \neq b, u, v \in \mathbb{R}a,b>0,a≠b,u,v∈R,
E r , s ( a , b ) = { ( s r a r − b r a s − b s ) 1 r − s , r s ( 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 a b , r = s ≠ 0 E r , s ( a , b ) = s r a r − b r a s − b s 1 r − s ,      r s ( 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 a b ,      r = s ≠ 0 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}Er,s(a,b)={(srar−bras−bs)1r−s,rs(r−s)≠0exp⁡(−1r+arlog⁡a−brlog⁡bar−br),r=s≠0(1sas−bslog⁡a−log⁡b)1s,r=0,s≠0ab,r=s≠0
namely L = E 1 , 0 , I = E 1 , 1 L = E 1 , 0 , I = E 1 , 1 L=E_(1,0),I=E_(1,1)L=E_{1,0}, I=E_{1,1}L=E1,0,I=E1,1. 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 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}H=G−1,0=E−2,−1,G=G0,0=E0,0,A=G1,0=E2,1 and Q = G 2 , 0 = E 4 , 2 Q = G 2 , 0 = E 4 , 2 Q=G_(2,0)=E_(4,2)Q=G_{2,0}=E_{4,2}Q=G2,0=E4,2. As it was shown in [2], the families of Gini means G u , v G u , v G_(u,v)G_{u, v}Gu,v and Stolarsky means E r , s E r , s E_(r,s)E_{r, s}Er,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 L L LLL is a Stolarsky mean, while C C CCC 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.
[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.
2013

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