EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 3, 2022, 1120-1143 ISSN 1307-5543 – ejpam.com Published by New York Business Global Bounds for the Convex Combination of Contra-harmonic and Harmonic Means by the Generalized Logarithmic Mean Annop Sonubon1, Somsak Orankitjaroen1,∗, Kamsing Nonlaopon2 1 Department of Mathematics, Faculty of Science, Mahidol University, Bangkok, 10400, Thailand 2 Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen, 40002, Thailand Abstract. We verify the optimal upper bound and optimal lower bound for the convex combina- tion of contra-harmonic and harmonic means by the generalized logarithmic mean Lp when p is of the linear form p = 2(1 − c)α + c and p is of the reciprocal of linear form p = 1/[2(1 − c)α + c] respectively. We prove that 1) L4α−1 = minc { L2(1−c)α+c | L2(1−c)α+c > αC + (1− α)H } for α ∈ (0, 1/2), 2) L 7 13−12α = maxc { L 1 2(1−c)α+c ∣∣∣L 1 2(1−c)α+c < αC + (1− α)H } for α ∈ (1/2, 1) where C(a, b) and H(a, b) are contra-harmonic and harmonic means. 2020 Mathematics Subject Classifications: 26D15, 26D20, 26E20 Key Words and Phrases: Inequality, Generalized logarithmic mean, Weighted arithmetic mean, Harmonic mean, Contra-harmonic mean 1. Introduction For any real number p, generalized logarithmic mean Lp(a, b) of two positive numbers a and b is defined by Lp(a, b) =  [ ap+1 − bp+1 (p+ 1)(a− b) ]1/p , a ̸= b, p ̸= 0, p ̸= −1; 1 e ( bb aa )1/(b−a) , a ̸= b, p = 0; b− a log b− log a , a ̸= b, p = −1; a, a = b. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4444 Email addresses: murfy.pm@gmail.com (A. Sonubon), ,,somsak.ora@mahidol.ac.th (S. Orankitjaroen), nkamsi@kku.ac.th (K. Nonlaopon) https://www.ejpam.com 1120 © 2022 EJPAM All rights reserved. A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1121 Mean Lp is continuous and strictly increasing with respect to p. It has many applications in physics involving a heat conductor problem and a mean temperature between two points at different temperature [6, 10]. Many classical bivariate means are special cases of generalized logarithmic means such as G(a, b) = L−2(a, b), L(a, b) = L−1(a, b), N(a, b) = L−1/2(a, b), I(a, b) = L0(a, b), and A(a, b) = L1(a, b), where G,L,N, I, and A are geometric, logarithmic, square-root, identric and arithmetic means, respectively. In view of generalized logarithmic means we have a well-known string of inequalities min{a, b} < H(a, b) < L−2(a, b) < L1(a, b) < S(a, b) < C(a, b) < max{a, b} for all distinct positive numbers a, b; here H(a, b) = 2ab a+ b , S(a, b) = √ a2 + b2 2 , C(a, b) = a2 + b2 a+ b are harmonic, root-square, and contra-harmonic means, respectively. Generalized logarithmic mean has been the subject of intensive research in particular those involving inequalities and monotonicity [1–4, 11–13, 18, 19]. Below we present some recent works concerning the optimal bound of certain means by generalized logarithmic means in one direction and weighted means by generalized logarithmic means in another. For a problem of finding sharp double inequalities between generalized logarithmic means and other means, recently it was found possible for Neuman-Sándor mean M and Yang mean U which are defined by M(a, b) =  a− b 2 sinh−1 ( a−b a+b ) , a ̸= b; a, a = b, U(a, b) =  a− b √ 2tan−1 ( a−b√ 2ab ) , a ̸= b; a, a = b. In case of Neuman-Sándor mean, Li, Long and Chu [8] in 2012 found the best largest value p = 1.8435 . . . and smallest value q = 2, where p is the unique solution of the equation (p+ 1)1/p = 2 log(1 + √ 2) such that the double inequalities Lp(a, b) < M(a, b) < Lq(a, b) hold for all distinct positive numbers a, b. In case of Yang mean, Qian and Chu [14] in 2016 found the best possible parameters p = 0.5451 . . . and q = 2, where p is the unique solution of the equation (p+1)1/p = √ 2π/2 such that the double inequalities Lp(a, b) < U(a, b) < Lq(a, b) A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1122 hold for all distinct positive numbers a, b. For a problem of finding optimal bound of weight either geometric or arithmetic means by generalized logarithmic means, there are many recent works in this direction. In case of weighted geometric mean, for α, β, γ ∈ (0, 1) and α + β + γ = 1, Chu and Long [4] in 2010 found the optimal bound for Aα(a, b)Gβ(a, b)Hγ(a, b). That is, they discovered that the largest value p = 6α+3β − 5 and the smallest value q = −2/(2α+ β) are the optimal values such that the double inequalities Lp(a, b) < Aα(a, b)Gβ(a, b)Hγ(a, b) < Lq(a, b) hold for all distinct positive numbers a, b. In 2011, Qian and Long [16] presented the sharp upper and lower bound for the weighted geometric mean of geometric and harmonic means by generalized logarithmic means: for all positive numbers a and b 1) L3α−5(a, b) = Gα(a, b)H1−α(a, b) = L−2/α(a, b) for α = 2/3, 2) L3α−5(a, b) ⩾ Gα(a, b)H1−α(a, b) ⩾ L−2/α(a, b) for α ∈ (0, 2/3), and L3α−5(a, b) ⩽ Gα(a, b)H1−α(a, b) ⩽ L−2/α for α ∈ (2/3, 1), with equality if and only if a = b, and the parameters 3α− 5 and −2/α cannot be improved in either case. Chunrong and Siqi [5] established the optimal bounds for Gα(a, b)N1−α(a, b) in term of Lp(a, b). They found that for any positive numbers a and b 1) L−(1+3α)/2(a, b) = Gα(a, b)N1−α(a, b) = L2/(α−2)(a, b) for α = 2/3, 2) L−(1+3α)/2(a, b) > Gα(a, b)N1−α(a, b) > L2/(α−2)(a, b) for α ∈ (0, 2/3), and L−(1+3α)/2(a, b) < Gα(a, b)N1−α(a, b) < L2/(α−2)(a, b) for α ∈ (2/3, 1), and the pa- rameters −(1 + 3α)/2 and 2/(α− 2) cannot be improved in either case. In the case of weighted arithmetic mean, Long and Chu [9] in 2010 proposed the inequalities involving generalized logarithmic means and weighted arithmetic means of arithmetic and geometric means: 1) L3α−2(a, b) = αA(a, b) + (1− α)G(a, b) for α = 1/2, 2) L3α−2(a, b) < αA(a, b) + (1− α)G(a, b) for α ∈ (0, 1/2), 3) L3α−2(a, b) > αA(a, b) + (1− α)G(a, b) for α ∈ (1/2, 1). Moreover, in each case, the bound L3α−2(a, b) for the sum of αA(a, b) + (1−α)G(a, b) is optimal. The harmonic and contra-harmonic means have recently been used to investigate the optimal bounds for means inequalities as mentioned in the following. In 2017, Qian, Zhang and Chu [17] discovered the greatest values α and λ, and the smallest values β and µ in [0,1/2] such that H[αa+ (1− α)b, αb+ (1− α)a] < TQ(a, b) < H[βa+ (1− β)b, βb+ (1− β)a], A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1123 G[λa+ (1− λ)b, λb+ (1− λ)a] < TQ(a, b) < H[µa+ (1− µ)b, µb+ (1− µ)a] hold for all a, b > 0 with a ̸= b, where TQ(a, b) = 2 π ∫ π/2 0 acos 2 θbsin 2 θdθ is the Toader-Qi mean of a and b. In 2018, Xu, Chu and Qian [20] found the optimal parameters αi, βi ∈ (0, 1) (i = 1, 2, 3, 4) to ensure that four double inequalities Cα1(a, b)A1−α1(a, b) < RSA(a, b) < Cβ1(a, b)A1−β1(a, b), Cα2(a, b)A1−α2(a, b) < RAS(a, b) < Cβ2(a, b)A1−β2(a, b), α3 [ 1 3 C(a, b) + 2 3 A(a, b) ] + (1− α3)C 1/3(a, b)A2/3(a, b) < RSA(a, b) < β3 [ 1 3 C(a, b) + 2 3 A(a, b) ] + (1− β3)C 1/3(a, b)A2/3(a, b), α4 [ 1 6 C(a, b) + 5 6 A(a, b) ] + (1− α4)C 1/6(a, b)A5/6(a, b) < RAS(a, b) < β4 [ 1 6 C(a, b) + 5 6 A(a, b) ] + (1− β4)C 1/6(a, b)A5/6(a, b) hold for all distinct positive numbers a, b and RSA(a, b) = 1 2 A(a, b) [√ 1 + u2 + sinh−1(u) u ] , RAS(a, b) = 1 2 A(a, b) [ 1 + (1 + u2) tan−1(u) u ] , where a > b > 0 and u = (a− b)/(a+ b). In 2019, Qian, He, Zhang and Chu [15] found the best values λ1 = λ1(ν), µ1 = µ1(ν), λ2 = λ2(ν) and µ2 = µ2(ν) on the interval [1/2, 1] such that the double inequalities Wλ1,ν(a, b) < RSA(a, b) < Wµ1,ν(a, b), Wλ2,ν(a, b) < RAS(a, b) < Wµ2,ν(a, b) hold for all distinct positive numbers a, b and ν ≥ 1/2 where Wλ,ν(a, b) = Cν [λa+ (1− λ)b, λb+ (1− λ)a]A1−ν(a, b). In 2022, Li, Miao and Guo [7] discovered the largest values αi and the smallest values βi (i = 1, 2, 3) such that the inequalities α1 C(a, b) + 1− α1 A(a, b) < 1 M(a, b) < β1 C(a, b) + 1− β1 A(a, b) , α2 C2(a, b) + 1− α2 A2(a, b) < 1 M2(a, b) < β2 C2(a, b) + 1− β2 A2(a, b) , A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1124 and α3C 2(a, b) + (1− α3)A 2(a, b) < M2(a, b) < β3C 2(a, b) + (1− β3)A 2(a, b) hold for all positive real numbers a and b with a ̸= b. The purpose of this paper is to present the inequalities with optimal upper bound and optimal lower bound of weighted arithmetic means of contra-harmonic and harmonic means by generalized logarithmic means Lp when p is of the linear form p = 2(1− c)α+ c and p is of the reciprocal of linear form p = 1/[2(1 − c)α + c] respectively and c is the value to be determined in both cases. Precisely, we prove that 1) L4α−1 = minc { L2(1−c)α+c | L2(1−c)α+c > αC + (1− α)H } for α ∈ (0, 1/2), 2) L 7 13−12α = maxc { L 1 2(1−c)α+c ∣∣∣L 1 2(1−c)α+c < αC + (1− α)H } for α ∈ (1/2, 1). Details of the results are Theorem 1 and Theorem 2 in section 3. Some complicated computations are carried out using MatlabR2021a software computer system. 2. Preliminaries In this section, we present four lemmas necessary in the proof of our main results in section 3. More specifically Lemma 1 is used in all Theorems whereas Lemma 2 to Lemma 4 are used only in Theorem 1. Lemma 1. If p ∈ R, t > 1 and F (t) := 1 p [ ln ( tp+1 − 1 ) − ln(p+ 1)− ln(t− 1) ] − ln [ α(t2 + 1) + 2(1− α)t ] + ln(t+ 1), (1) then F ′(t) = G(t) p (tp+1 − 1) (t2 − 1) [α(t2 + 1) + 2(1− α)t] , (2) where G(t) = (−3αp+ 2p− α) ( tp+3 − 1 ) + (5αp− 2p+ α− 2) ( tp+2 − t ) + (−αp+ α− 2) ( tp+1 − t2 ) − α(p+ 1) ( tp − t3 ) . Furthermore, G(1) = G′(1) = G′′(1) = 0. Proof. Differentiating F (t) yields (2) and by taking derivative of G, we obtain G′(t) = (−3αp+ 2p− α)(p+ 3)tp+2 + (5αp− 2p+ α− 2) [ (p+ 2)tp+1 − 1 ] + (−αp+ α− 2) [(p+ 1)tp − 2t]− α(p+ 1) ( ptp−1 − 3t2 ) , G′′(t) = (−3αp+ 2p− α)(p+ 3)(p+ 2)tp+1 + (5αp− 2p+ α− 2)(p+ 2)(p+ 1)tp + (−αp+ α− 2) [ (p+ 1)ptp−1 − 2 ] − α(p+ 1) [ p(p− 1)tp−2 − 6t ] , respectively. It follows immediately that G(1) = G ′ (1) = G ′′ (1) = 0. A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1125 Lemma 2. Let α ∈ (0, 1/8) and t > 1 and f(t) = 1 t exp [ t2 − 1 α(t2 + 1) + 2(1− α)t ] . Function f is strictly increasing for t satisfying the inequality α2t2 − 2(α2 − 3α+ 1)t+ α2 < 0. Proof. Differentiating f(t) with respect to t yields f ′(t) = t−1e t2−1 α(t2+1)+2(1−α)t {[ α(t2 + 1) + 2(1− α)t ] (2t)− (t2 − 1) [ 2αt+ 2(1− α) ][ α(t2 + 1) + 2(1− α)t ]2 } − t−2e t2−1 α(t2+1)+2(1−α)t . Simplifying f ′(t) and setting f ′(t) > 0, we obtain 2[α(t2 + 1) + 2(1− α)t]t2 − t(t2 − 1) [ 2αt+ 2(1− α) ] − [ α(t2 + 1) + 2(1− α)t ]2 > 0 or α2t2 − (2α2 − 6α+ 2)t+ α2 < 0. Lemma 3. For a, b > 0 and α, β ∈ (0, 1) with α > β, we have αC(a, b) + (1− α)H(a, b) > βC(a, b) + (1− β)H(a, b). Proof. Because C(a, b) > H(a, b) and α > β, the result follows immediately from the inequality (α− β)C(a, b) > (α− β)H(a, b). Lemma 4. For t > 1, we have L−1/2(1, t) > 1 4 C(1, t) + 3 4 H(1, t). Proof. The proposed inequality is 1 + 2 √ t+ t 4 > 1 4 ( t2 + 1 t+ 1 ) + 3 4 ( 2t t+ 1 ) which is equivalent to ( √ t− 1)2 > 0. A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1126 3. Main Results We first establish the optimal upper bound for the weighted arithmetic mean of contra- harmonic and harmonic means by generalized logarithmic means Lp where p has the linear form p = 2(1− c)α+ c and α ∈ (0, 1/2). Precisely, we have Theorem 1. Let a, b > 0 with a ̸= b. Then 1) L4α−1(a, b) = αC(a, b) + (1− α)H(a, b) for α = 1/2; 2) L4α−1(a, b) > αC(a, b) + (1− α)H(a, b) for α ∈ (0, 1/2), and the parameter 4α− 1 cannot be improved in the sense that L4α−1 = min c { L2(1−c)α+c | L2(1−c)α+c > αC + (1− α)H } for α ∈ (0, 1/2) i.e. c = −1; 3) L4α−1(a, b) < αC(a, b) + (1− α)H(a, b) for α ∈ (1/2, 1). Proof. 1) For α = 1/2, on one hand we have L4( 1 2)−1(a, b) = L1(a, b) = a+ b 2 . On the other hand, we have C(a, b) +H(a, b) 2 = a2+b2 a+b + 2ab a+b 2 = a+ b 2 . 2) Without loss of generality, we assume that b > a > 0 and set t = b/a > 1. The proposed inequality becomes[ t4α − 1 4α(t− 1) ]1/(4α−1) > α ( t2 + 1 t+ 1 ) + (1− α) ( 2t t+ 1 ) α ∈ (0, 1/2). (3) Inequality (3) is equivalent to F (t) > 0 in (1) with p = 4α− 1. Using Lemma 1, we have a formula for F ′(t), G′(t), G′′(t) where G(t) is the numerator of F ′(t) appearing in (2). Taking derivative of G′′(t), we have G ′′′ (t) 8αt4α−4 = (4α+ 2)(4α+ 1)(3α− 1)(1− 2α)t3 − 2α(4α+ 1)(4α− 1)(3− 5α)t2 + 2(2α2 − α+ 1)(4α− 1)(1− 2α)t − α(4α− 1)(1− 2α)(3− 4α) + 3αt4−4α. (4) We divide our proof into four cases: α ∈ [1/3, 1/2), α ∈ [1/8, 1/4), α ∈ [1/4, 1/3) and α ∈ (0, 1/8). A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1127 2.1) Case α ∈ [1/3, 1/2): Observe that the coefficients of t3 and t in (4) are positive while that of t2 and constant term are negative for α ∈ [1/3, 1/2). Since t > 1, we have t2 < t4−4α < t3 and consequently G ′′′ (t) 8αt4α−4 > [ (4α+ 2)(4α+ 1)(3α− 1)(1− 2α)− 2α(4α+ 1)(4α− 1)(3− 5α) + 3α ] t4−4α + [ 2(2α2 − α+ 1)(4α− 1)(1− 2α)− α(4α− 1)(1− 2α)(3− 4α) ] t > (1− 2α)(4α− 1)(−8α2 + 5α+ 2) ( t4−4α − t ) > 0. Together with G(1) = G′(1) = G′′(1) from Lemma 1, we conclude that F (t) > 0 for all α ∈ [1/3, 1/2). 2.2) Case α ∈ [1/8, 1/4): Since monotonicity property of Lp implies that L4α−1(1, t) ≥ L−1/2(1, t) for α ≥ 1/8, it is sufficient to prove instead the inequality L−1/2(1, t) > αC(1, t) + (1− α)H(1, t) or (1− 4α) √ t 2 + 4(1− 2α) √ t+ (1− 4α) > 0. which is true for t > 1 and α ∈ [1/8, 1/4). 2.3) Case α ∈ [1/4, 1/3): Monotonicity of Lp and Lemma 3 imply that L4α−1(1, t) ≥ L0(1, t) > L−1/2(1, t) for α ≥ 1/4, L0(1, t) = (1/e)t1+ 1 t−1 > t/e, (1/3)C(1, t) + (2/3)H(1, t) > αC(1, t) + (1− α)H(1, t) for α ∈ [1/4, 1/3). To prove this case, it is thus sufficient to show that there exist t1 < t2 < t3 s.t. (i) L−1/2(1, t) > (1/3)C(1, t) + (2/3)H(1, t), 1 < t < t2; (ii) t/e > (1/3)C(1, t) + (2/3)H(1, t), t3 < t; (iii) L0(1, t) > (1/3)C(1, t) + (2/3)H(1, t), t1 < t ≤ t3. Since the inequality in (i) is just 1 + 2 √ t+ t 4 > t2 + 4t+ 1 3(t+ 1) or √ t 2 − 4 √ t+ 1 < 0, which is true for 1 < t < (2 + √ 3)2 and we choose t2 = (2 + √ 3)2 ≈ 13.92 so that (i) is valid. Now, consider the inequality in (ii) or t e > t2 + 4t+ 1 3(t+ 1) or (3− e)t2 + (3− 4e)t− e > 0, which is true for t > 4e− 3 + √ 12e2 − 12e+ 9 6− 2e ≈ 28.28. A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1128 Hence, we choose t3 to be the right-side number so that (ii) is valid. Now, we consider t ≤ t3. Since t1/(t−1) is a decreasing function for t > 1, we have (1/e)t1+ 1 t−1 ≥ t/c where c = e/t3 1/(t3−1). To find t1 for (iii), it is enough to find it from t c > t2 + 4t+ 1 3(t+ 1) or (3− c)t2 + (3− 4c)t− e > 0, which is true for t > 4c− 3 + √ 12c2 − 12c+ 9 6− 2c ≈ 11.47. We therefore choose t1 = 4c−3+ √ 12c2−12c+9 6−2c so that (iii) is valid. Now t1 < t2 < t3 and satisfy (i), (ii) and (iii). 2.4) Case α ∈ (0, 1/8): Monotonicity of Lp implies that L4α−1(1, t) > L−1(1, t) for α > 0. Hence we will seek t which satisfies L−1(1, t) > αC(1, t) + (1− α)H(1, t) or t− 1 ln t > α ( t2 + 1 t+ 1 ) + (1− α) ( 2t t+ 1 ) or 1 t exp [ t2 − 1 α(t2 + 1) + 2(1− α)t ] > 1. Setting f(t) := t−1 exp [ t2−1 α(t2+1)+2(1−α)t ] , we can see that f(1) = 1 and use Lemma 2 to conclude that f ′(t) > 0 where α2t2 − 2(α2 − 3α + 1)t + α2 < 0. Function f(t) is an increasing function when A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1129 t ∈ ( (α2 − 3α+ 1)− √ −6α3 + 11α2 − 6α+ 1 α2 , (α2 − 3α+ 1) + √ −6α3 + 11α2 − 6α+ 1 α2 ) , which implies that for α ∈ (0, 1/8), f(t) > 1 when t ∈ ( 1, α2 − 3α+ 1 α2 ] . Hence L4α−1(1, t) > αC(1, t) + (1− α)H(1, t) for 1 < t ≤ (α2 − 3α+ 1)/α2. (5) Now, observe that the qualities[ t4α − 1 4α(t− 1) ] 1 4α−1 > t ( 1 4α ) 1 4α−1 and αt+ (2− α) > α(1 + t2) + (1− α)(2t) 1 + t hold for all α ∈ (0, 1/8). We will show that t ( 1 4α ) 1 4α−1 > αt+ (2− α) or t [( 1 4α ) 1 4α−1 − α ] > 2− α for all t > (α2 − 3α+ 1)/α2. It is sufficient to show that( α2 − 3α+ 1 α2 )[( 1 4α ) 1 4α−1 − α ] > 2− α or ( 1 4α ) 1 4α−1 > α− α2 α2 − 3α+ 1 . However, this is a direct consequence of a simple string of inequalities,( 1 4α ) 1 4α−1 > 2α > α− α2 α2 − 3α+ 1 for all α ∈ (0, 1/8). Thus L4α−1(1, t) > αC(1, t) + (1− α)H(1, t) for t > (α2 − 3α+ 1)/α2. (6) From inequalities (5) and (6), we conclude that for α ∈ (0, 1/8) L4α−1(1, t) > αC(1, t) + (1− α)H(1, t) for all t > 1. (7) Finally, we will prove that the parameter 4α− 1 cannot be improved in this case. Suppose, to the contrary, that inequality (7) is true with parameter 2[1− (−1− ϵ)]α+ (−1− ϵ) for a sufficiently small ϵ > 0. That is L2[1−(−1−ϵ)]α+(−1−ϵ)(1, t)− [ αC(1, t) + (1− α)H(1, t) ] > 0 for all t > 1. A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1130 Hence L2[1−(−1−ϵ)]α+(−1−ϵ)(1, t)− [ αC(1, t) + (1− α)H(1, t) ] t > 0 for all t > 1. Taking limits to both sides of the above inequality leads to lim α→ [ ϵ 2(2+ϵ) ]+ ( lim t→+∞ L2[1−(−1−ϵ)]α+(−1−ϵ)(1, t)− [ αC(1, t) + (1− α)H(1, t) ] t ) = − ϵ 2(2 + ϵ) < 0, which is a contradiction. 3) As in 2), the proposed inequality becomes[ t4α − 1 4α(t− 1) ]1/(4α−1) < α ( t2 + 1 t+ 1 ) + (1− α) ( 2t t+ 1 ) α ∈ (1/2, 1). (8) Inequality (8) is equivalent to F (t) < 0 and the term G′′′(t) is still that in (4). It is sufficient to show that G′′′(t) < 0 for all α ∈ (1/2, 1). To that end, we divide our proof into three cases: α ∈ (1/2, 3/5], α ∈ (3/5, 3/4] and α ∈ (3/4, 1). (a) Case α ∈ (1/2, 3/5]: In (4), observe that the constant term is positive while the coefficients of t3, t2 and t are negative for α ∈ (1/2, 3/5]. Since t > 1, it follows that t4−4α < t2 < t3 and consequently G ′′′ (t) 8αt4α−4 < [ − (4α+ 2)(4α+ 1)(3α− 1)(2α− 1) + 2α(5α− 3)(4α+ 1)(4α− 1) + 3α ] t4−4α + [ − 2(2α2 − α+ 1)(4α− 1)(2α− 1) + α(4α− 1)(2α− 1)(3− 4α) ] < (2α− 1)(4α− 1) [ − (−8α2 + 5α+ 2)t4−4α − (8α2 − 5α+ 2) ] < (2α− 1)(4α− 1) [ − (−8α2 + 5α+ 2)− (8α2 − 5α+ 2) ] = −4(2α− 1)(4α− 1) < 0. (b) Case α ∈ (3/5, 3/4]: For such α, the coefficient of t2 and the constant term in (4) are positive while the coefficients of t3 and t are negative. Since t > 1, we have t2 < t3 and consequently G ′′′ (t) 8αt4α−4 < [ − (4α+ 2)(4α+ 1)(3α− 1)(2α− 1) + 2α(5α− 3)(4α+ 1)(4α− 1) ] t2 A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1131 + [−2(2α2 − α+ 1)(4α− 1)(2α− 1) + α(4α− 1)(2α− 1)(3− 4α)] + 3αt4−4α = −2(1− α)(4α+ 1)(8α2 − 5α+ 1)t2 − (2α− 1)(4α− 1)(8α2 − 5α+ 2) + 3αt4−4α. Now, consider the term on the right side of the above inequality. Since the coefficient −2(1− α)(4α + 1)(8α2 − 5α + 1) of t2 is negative for α ∈ (3/5, 3/4] and t4−4α < t2, we get G ′′′ (t) 8αt4α−4 < [ − 2(1− α)(4α+ 1)(8α2 − 5α+ 1) + 3α ] t4−4α − (2α− 1)(4α− 1)(8α2 − 5α+ 2) = (2α− 1)(4α− 1) [ (8α2 − 5α− 2)t4−4α − (8α2 − 5α+ 2) ] < (2α− 1)(4α− 1) [ (8α2 − 5α− 2)− (8α2 − 5α+ 2) ] = −4(2α− 1)(4α− 1) < 0. (c) Case α ∈ (3/4, 1): For α in this interval the coefficients of t3 and t in (4) are negative while the coefficient of t2 is positive. Because t > 1, we have t4−4α < t2 < t3. Thus G ′′′ (t) 8αt4α−4 < [ − α(4α+ 2)(4α+ 1)(3α− 1)(2α− 1) + 2α(5α− 3)(4α+ 1)(4α− 1) ] t2 + [ − 2(2α2 − α+ 1)(4α− 1)(2α− 1) + 3α ] t4−4α − α(4α− 1)(2α− 1)(4α− 3) = −2(1− α)(4α+ 1)(8α2 − 5α+ 1)t2 + (−32α4 + 40α3 − 32α2 + 17α− 2)t4−4α − α(4α− 1)(2α− 1)(4α− 3). Now consider the term on the right side of the last equality sign above. The coefficient −2(1 − α)(4α + 1)(8α2 − 5α + 1) of t2 is negative and t4−4α < t2 as α ∈ (3/4, 1). Hence G ′′′ (t) 8αt4α−4 < [ − 2(1− α)(4α+ 1)(8α2 − 5α+ 1) + (−32α4 + 40α3 − 32α2 + 17α− 2) ] t4−4α − α(4α− 1)(2α− 1)(4α− 3) = (2α− 1)(4α− 1) [ − (−4α2 + 3α+ 4)t4−4α − α(4α− 1)(2α− 1)(4α− 3) ] < (2α− 1)(4α− 1) [ − (−4α2 + 3α+ 4) − α(4α− 1)(2α− 1)(4α− 3) ] = −4(2α− 1)(4α− 1) < 0. A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1132 The proof is complete. The next theorem is concerned with determining the optimal lower bound for the weighted arithmetic mean of contra-harmonic and harmonic means by generalized loga- rithmic means Lp, where p is of the reciprocal of linear form p = 1/[2(1− c)α+ c] where α ∈ (1/2, 1). Theorem 2. Let a, b > 0 with a ̸= b. Then 1) L7/(13−12α)(a, b) = αC(a, b) + (1− α)H(a, b) for α = 1/2; 2) L7/(13−12α)(a, b) > αC(a, b) + (1− α)H(a, b) for α ∈ (0, 1/2); 3) L7/(13−12α)(a, b) < αC(a, b) + (1 − α)H(a, b) for α ∈ (1/2, 1), and the parameter 7/(13− 12α) cannot be improved in the sense that L 7 13−12α = max c { L 1 2(1−c)α+c ∣∣∣L 1 2(1−c)α+c < αC + (1− α)H } for α ∈ (1/2, 1) i.e. c = 13/7. Proof. 1) For α = 1/2, we have L7/[13−12(1/2)](a, b) = L1(a, b) = a+ b 2 = C(a, b) +H(a, b) 2 . 2) Treated as in Theorem 1.2), the proposed inequality becomes[ t 20−12α 13−12α − 1( 20−12α 13−12α ) (t− 1) ](13−12α)/7 > α ( t2 + 1 t+ 1 ) + (1− α) ( 2t t+ 1 ) α ∈ (0, 1/2), (9) for t = b/a > 1. Inequality (9) is equivalent to F (t) > 0 in (1) with p = 7/(13 − 12α). Using Lemma 1, we have a formula for F ′(t), G′(t), G′′(t) and by taking derivative of G′′(t), we obtain G′′′(t) = 6AtA−4 (13− 12α)3 H(t), (10) where A = 20− 12α 13− 12α , (11) and H(t) = 2(1− 2α)(7− 3α)(23− 18α)(11− 8α)t3 − 14(3α2 − 18α+ 10)(11− 8α)t2 + 14(6α2 − 15α+ 13)(1− 2α)t A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1133 − 7α(1− 2α)(19− 24α) + α(13− 12α)3t4−A. (12) Note that the coefficients of t3 and t4−A in (12) are positive but that of t2 is negative for α ∈ (0, 1/2). Since t4−A > t2 and t2 > 2t− 1, we further have H(t) > 2(1− 2α)(7− 3α)(23− 18α)(11− 8α)t(2t− 1) + [α(13− 12α)3 − 14(3α2 − 18α+ 10)(11− 8α)]t2 + 14(6α2 − 15α+ 13)(1− 2α)t− 7α(1− 2α)(19− 24α) = (1− 2α) [ (−864α3 + 6, 072α2 − 10, 723α+ 5, 544)t2 − 32(5− 3α)(9α2 − 29α+ 21)t− 7α(19− 24α) ] . Now consider the term on the right side of the above equality sign. The coefficient (1 − 2α)(−864α3+6, 072α2−10, 723α+5, 544) of t2 becomes positive for α ∈ (0, 1/2). Together with t < t2, we finally have H(t) > (1− 2α) [ (−864α3 + 6, 072α2 − 10, 723α+ 5, 544) − 32(5− 3α)(9α2 − 29α+ 21)− 7α(19− 24α) ] t = 168(1− α)(13− 12α)(1− 2α)t > 0. Therefore G′′′(t) > 0 for all α ∈ (0, 1/2). 3) Here the proposed inequality becomes[ t 20−12α 13−12α − 1( 20−12α 13−12α ) (t− 1) ](13−12α)/7 < α ( t2 + 1 t+ 1 ) + (1− α) ( 2t t+ 1 ) α ∈ (1/2, 1), (13) for t = b/a > 1. Inequality (13) is equivalent to F (t) < 0 in (1). Terms G′′′(t), A and H(t) are still of the forms (10), (11) and (12), respectively. It is then sufficient to show that H(t) < 0 for all α ∈ (1/2, 1). To that end, we divide our proof into two cases α ∈ (8/9, 1) and α ∈ (1/2, 8/9]. 3.1) Case α ∈ (8/9, 1): For such α, we have 4−A ∈ (−4, 0) and t4−A < 1 < 1 + (4−A)(t− 1) + (4−A)(3−A) 2 (t− 1)2 = (432α2 − 726α+ 304)t2 + (−432α2 + 600α− 192)t+ (144α2 − 186α+ 57) (13− 12α)2 . A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1134 Hence H(t) < 2(2α− 1) [ − (7− 3α)(23− 18α)(11− 8α)t3 + 2(−648α3 + 1, 509α2 − 1, 191α+ 385)t2 − (−1, 296α3 + 2, 598α2 − 1, 353α+ 91)t − 2α(24α− 19)(9α− 8) ] . Now consider the term on the right side of the above inequality sign. As the coefficient of t3 is negative for α ∈ (1/2, 1) and t3 > t(2t− 1), we have H(t) < 2(2α− 1) [ − (7− 3α)(23− 18α)(11− 8α)t(2t− 1) + 2(−648α3 + 1, 509α2 − 1, 191α+ 385)t2 − (−1, 296α3 + 2, 598α2 − 1, 353α+ 91)t − 2α(24α− 19)(9α− 8) ] = 2(2α− 1) [ (−432α3 − 1, 290α2 + 4, 484α− 2, 772)t2 + (864α3 − 444α2 − 2, 080α+ 1, 680)t − 2α(24α− 19)(9α− 8) ] . Examine the term on the right side of the equality sign above. The coefficient −432α3 − 1, 290α2 + 4, 484α − 2, 772 is an increasing function of α ∈ (1/2, 1) with negative value (-10) at α = 1. Since the coefficient of t2 is negative for α ∈ (8/9, 1) and t2 > 2t − 1, we obtain H(t) < 2(2α− 1) [ (−432α3 − 1, 290α2 + 4, 484α− 2, 772)(2t− 1) + (864α3 − 444α2 − 2, 080α+ 1, 680)t − 2α(24α− 19)(9α− 8) ] = 2(2α− 1) [ − 168(1− α)(23− 18α)t+ 252(1− α)(11− 8α) ] . Look at the term on the right side of the above inequality sign. Since the coefficient −168(1− α)(23− 18α) of t is negative and t > 1, we finally have H(t) < 2(2α− 1) [ − 168(1− α)(23− 18α)t+ 252(1− α)(11− 8α) ] < −168(2α− 1)(1− α)(13− 12α) < 0. 3.2) Case α ∈ (1/2, 8/9]: For α in this interval, the coefficient of t3 in (12) is negative and 4−A ∈ [0, 2). Hence t4−A < t2. These consequences and t3 > t(2t− 1) imply that A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1135 H(t) < −2(2α− 1)(7− 3α)(23− 18α)(11− 8α)t(2t− 1) + [ α(13− 12α)3 − 14(3α2 − 18α+ 10)(11− 8α) ] t2 − 14(6α2 − 15α+ 13)(2α− 1)t− 7α(2α− 1)(24α− 19) = (864α3 − 6, 072α2 + 10, 723α− 5, 544)t2 + 32(5− 3α)(9α2 − 29α+ 21)t− 7α(24α− 19). Examine the term on the right side of the inequality sign shown above. Notice that the coefficient 864α3 − 6, 072α2 + 10, 723α − 5, 544 is an increasing function for α ∈ (1/2, 1) with negative value (-29) at α = 1. The coefficient of t2 is negative for α ∈ (1/2, 8/9]. We have t2 > t and then H(t) < [ (864α3 − 6, 072α2 + 10, 723α− 5, 544) + 32(5− 3α)(9α2 − 29α+ 21) ] t− 7α(24α− 19) = (−1, 848α2 + 4, 067α− 2, 184)t− 7α(24α− 19). Consider the term on the right side of the equality sign above. Coefficient −1, 848α2 + 4, 067α− 2, 184 is increasing for α ∈ (1/2, 8/9] with negative value (-29.03) at α = 8/9, it is therefore negative on the whole interval (1/2, 8/9]. Since t > 1, we get H(t) < (−1, 848α2 + 4, 067α− 2, 184)− 7α(24α− 19) < −168(1− α)(13− 12α) < 0. Finally, we will prove that the parameter 7/(13− 12α) cannot be improved in this case. Suppose, to the contrary, that inequality (13) is true for the parameter 1 2[1− (1 + k + ϵ)]α+ (1 + k + ϵ) for a sufficiently small ϵ > 0. That is L 1 2[1−(1+k+ϵ)]α+(1+k+ϵ) (1, t) < αC(1, t) + (1− α)H(1, t) for all t > 1. Taking logarithm of both sides of the above inequality, we get ln [ L 1 2[1−(13/7+ϵ)]α+(13/7+ϵ) (1, t) ] − ln [ αC(1, t) + (1− α)H(1, t) ] < 0. With the notation in Lemma 1, this is just F (t) < 0 for all t > 1 where p = 1 2 [1− (13/7 + ϵ)]α+ (13/7 + ϵ) . (14) From Lemma 1, G(1) = G′(1) = G′′(1) = 0. Taking derivative of G′′(t), we have A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1136 G′′′(t) = (−3αp+ 2p− α)(p+ 3)(p+ 2)(p+ 1)tp + (5αp− 2p+ α− 2)(p+ 2)(p+ 1)ptp−1 + (−αp+ α− 2)(p+ 1)p(p− 1)tp−2 − α(p+ 1) [ p(p− 1)(p− 2)tp−3 − 6 ] . Hence G′′′(1) = 2(p+ 1)p(p− 12α+ 5) for α ∈ (1/2, 1). However, lim α→1− 2(p+ 1)p(p− 12α+ 5) = 686(7ϵ− 8)ϵ (7ϵ− 1)3 > 0. Therefore G(t) < 0 or F (t) < 0 in a small neighborhood of 1 if ϵ < 1/7. This contradict to statement (14). The proof is complete. Remark 1. In statement 3) of Theorem 1, L4α−1 is not the optimal lower bound of the considered weighted arithmetic mean for α ∈ (1/2, 1). Neither is L7/(13−12α) the optimal upper bound of the one for α ∈ (0, 1/2) in statement 2) of Theorem 2. Due to monotonicity property of generalized logarithmic means, we expect a sharper result. Partial results are shown in the following theorem. Theorem 3. Let a, b > 0 with a ̸= b and k = 2/(2 ln 2− 1). Then 1) L1/[−2kα+(k+1)](a, b) = αC(a, b) + (1− α)H(a, b) = L2kα+(1−k)(a, b) for α = 1/2; 2) L1/[−2kα+(k+1)](a, b) > αC(a, b) + (1− α)H(a, b) for α ∈ (0, 2/k) ≈ (0, 0.38); 3) L2kα+(1−k)(a, b) < αC(a, b) + (1− α)H(a, b) for α ∈ ((k + 2)/2k, 1) ≈ (0.7, 1). Proof. 1) This is obvious after inserting α = 1/2 into the left and right sides of the statement. 2) Treated as in Theorem 1(2), the proposed inequality becomes t1+ { 1/[−2kα+(k+1)] } − 1( 1 + { 1/ [−2kα+ (k + 1)] }) (t− 1) −2kα+(k+1) > (α) t2 + 1 t+ 1 + (1− α) 2t t+ 1 (15) for all α ∈ (0, 2/k) and t = b/a > 1. Inequality (15) is equivalent to F (t) > 0 in (1) with p = 1/ [−2kα+ (k + 1)]. Using Lemma 1, we have formulae for F ′(t), G′(t), G′′(t). Taking derivative of G′′(t), we have G′′′(t) = 2 + k − 2kα (2kα− k − 1)4 t −2+(6α−3)k −1+(2α−1)k J(t), A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1137 where J(t) = (1− 2α)(2− kα)(6kα− 3k − 4)(4kα− 2k − 3)t3 − (2α2k − 5kα− 6α+ 2k + 4)(−4kα+ 2k + 3)t2 + (1− 2α)k(2α2k − 5kα+ 2k + 2)t− α(1− 2α)k(−4kα+ 2k + 1) + 6α(1 + k − 2kα)3t −2+(6α−3)k −1+(2α−1)k . For α ∈ (0, 2/k), exponent [−1 + (4α− 2)k]/[−1 + (2α− 1)k] > 1 and hence t −2+(6α−3)k −1+(2α−1)k = t [ t −1+(4α−2)k −1+(2α−1)k ] > t { 1 + [−1 + (4α− 2)k −1 + (2α− 1)k ] (t− 1) } = [−1 + (4α− 2)k −1 + (2α− 1)k ] t2 + [ (1− 2α)k −1 + (2α− 1)k ] t. Furthermore coefficient 6α(1 + k − 2kα)3 > 0 for such α. Therefore J(t) > (1− 2α) [ (2− kα)(6kα− 3k − 4)(4kα− 2k − 3)t3 + (48α3k3 − 48α2k3 − 64α2k2 + 12αk3 + 40αk2 + 39kα− 4k2 − 14k − 12)t2 − k(24α3k2 − 24α2k2 − 26α2k + 6αk2 + 17kα+ 6α− 2k − 2)t − αk(−4kα+ 2k + 1) ] . Consider the term on the right side of the inequality sign above. Since t2 > 2t − 1 and coefficient (2− kα)(6kα− 3k − 4)(4kα− 2k − 3) of t3 is positive, we have J(t) > (1− 2α) [ (100α2k2 − 90αk2 − 121kα+ 20k2 + 54k + 36)t2 + (−56α2k2 + 48αk2 + 74kα− 10k2 − 32k − 24)t− αk(−4kα+ 2k + 1) ] . Because t2 > t and coefficient 100α2k2 − 90αk2 − 121kα + 20k2 + 54k + 36 of t2 on the right side of the above inequality is positive for α ∈ (0, 2/k), we obtain J(t) > (1− 2α) [ (44α2k2 − 42αk2 − 47kα+ 10k2 + 22k + 12)t − αk(−4kα+ 2k + 1) ] . Observe that coefficient 44α2k2 − 42αk2 − 47kα+10k2 +22k+12 of t is positive for such α. Since t > 1, we finally have J(t) > 2(1− 2α)(12kα− 5k − 6)(2kα− k − 1) > 0. Therefore G′′′(t) > 0 for all α ∈ (0, 2/k). A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1138 3) Treated as in Theorem 1(2), the proposed inequality becomes{ t2kα+(2−k) − 1[ 2kα+ (2− k) ] (t− 1) }1/[2kα+(1−k)] < α ( t2 + 1 t+ 1 ) + (1− α) ( 2t t+ 1 ) (16) for all α ∈ ((k + 2)/2k, 1) and t = b/a > 1. Inequality (16) is equivalent to F (t) > 0 in (1) with p = 2kα+(1−k). Using Lemma 1, we have formulae for F ′(t), G′(t), G′′(t). Taking derivative of G′′(t), we have G′′′(t) = (2kα− k + 2)t2kα−k−2K(t), where K(t) = −(2α− 1)(3kα− 2k + 2)(2kα− k + 4)(2kα− k + 3)t3 + (10α2k − 9kα+ 6α+ 2k − 4)(2kα− k + 3)(2kα− k + 1)t2 − (2α2k − kα+ 2)(2kα− k + 1)(2α− 1)kt − α(2kα− k + 1)(2kα− k − 1)(2α− 1)k + 6αtk−2kα+2. (17) Since t3 > t(2t− 1) and the coefficient of t3 in (17) is negative for α ∈ ((k + 2)/2k, 1), we have K(t) < −(2kα− k + 3) (4α3k2 − 12α2k2 + 42α2k + 9αk2 − 49kα− 2k2 + 26α+ 14k − 2)t2 + 2(2α− 1)(2kα− k + 2)(2α2k2 − 3αk2 + 10kα+ k2 − 7k + 6)t − α(2kα− k + 1)(2kα− k − 1)(2α− 1)k + 6αtk−2kα+2. (18) Since t2 > 2t − 1 and the coefficient of t2 in (18) is negative for α ∈ ((k + 2)/2k, 1), it follows that K(t) < (16α3k3 − 96α3k2 − 24α2k3 + 184α2k2 + 12αk3 − 228α2k − 112αk2 − 2k3 + 250kα+ 22k2 − 108α− 68k + 48)t + (−16α3k3 + 96α3k2 + 24α2k3 − 176α2k2 − 12αk3 + 180α2k + 104αk2 + 2k3 − 198kα− 20k2 + 78α+ 54k − 36) + 6αtk−2kα+2. (19) The coefficient of t in (19) is negative and tk−2kα+2 < 1 for α ∈ ((k + 2)/2k, 1). Therefore, K(t) < (8α2k2 − 48α2k − 8αk2 + 52kα+ 2k2 − 30α− 14k + 12) + 6αtk−2kα+2 < (8α2k2 − 48α2k − 8αk2 + 52kα+ 2k2 − 30α− 14k + 12) + 6α = 2(k − 6)(2α− 1)(2kα− k + 1) < 0. As a result, G′′′(t) < 0 for all α ∈ ((k + 2)/2k, 1). A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1139 Conjecture 1. Let a, b > 0 with a ̸= b and k = 2/(2 ln 2− 1). 1) L1/[−2kα+(k+1)](a, b) > αC(a, b) + (1− α)H(a, b) for α ∈ (0, 1/2); 2) L2kα+(1−k)(a, b) < αC(a, b) + (1− α)H(a, b) for α ∈ (1/2, 1). If conjecture 1 is correct, L1/[−2kα+(k+1)] will be the optimal upper bound for the considered weighted arithmetic mean forα ∈ (0, 1/2) and L2kα+(1−k) will be the optimal lower bound for the considered weighted arithmetic mean for α ∈ (1/2, 1), as shown in the following theorem. Theorem 4. Let a, b > 0 with a ̸= b and k = 2/(2 ln 2− 1). 1) If L1/[−2kα+(k+1)](a, b) > αC(a, b) + (1−α)H(a, b) for α ∈ (0, 1/2), then the param- eter 1/ [−2kα+ (k + 1)] cannot be improved in the sense that L 1 −2kα+(k+1) = min c { L 1 2(1−c)α+c ∣∣∣L 1 2(1−c)α+c > αC + (1− α)H } for α ∈ (0, 1/2) i.e. c = 1 + k; 2) If L2kα+(1−k)(a, b) < αC(a, b) + (1 − α)H(a, b) for α ∈ (1/2, 1), then the parameter 2kα+ (1− k) cannot be improved in the sense that L2kα+(1−k) = max c { L2(1−c)α+c | L2(1−c)α+c < αC + (1− α)H } for α ∈ (1/2, 1) i.e. c = 1− k. Proof. 1) Suppose, to the contrary, that inequality (15) is true for the parameter 1 2[1− (1 + k + ϵ)]α+ (1 + k + ϵ) for a sufficiently small ϵ > 0. That is L 1 2[1−(1+k+ϵ)]α+(1+k+ϵ) (1, t) > αC(1, t) + (1− α)H(1, t) for all t > 1. Hence 1 t { L 1 2[1−(1+k+ϵ)]α+(1+k+ϵ) (1, t) } ≥ 1 t { αC(1, t) + (1− α)H(1, t) } for all t > 1. Taking limits on both sides of the above inequality lead to lim t→+∞ 1 t { L 1 2[1−(1+k+ϵ)]α+(1+k+ϵ) (1, t) } ≥ lim t→+∞ 1 t [ αC(1, t) + (1− α)H(1, t) ] , A. Sonubon, S. Orankitjaroen, K. Nonlaopon / Eur. J. Pure Appl. Math, 15 (3) (2022), 1120-1143 1140 which is equivalent to [ (1− 2α)(k + ϵ) + 1 (1− 2α)(k + ϵ) + 2 ](1−2α)(k+ϵ)+1 ≥ α. Taking logarithm of both sides of the above inequality, we get Q(α) := [(1− 2α)(k + ϵ) + 1] ln [ (1− 2α)(k + ϵ) + 1 (1− 2α)(k + ϵ) + 2 ] − lnα ≥ 0. Sine Q(α) ≥ 0 for all α ∈ (0, 1/2) and Q((1/2)−) = 0, it immediately follows that Q′((1/2)−) ≤ 0. However, Q′(α) = −2(k + ϵ) (1− 2α)(k + ϵ) + 2 − 2(k + ϵ) ln [ (1− 2α)(k + ϵ) + 1 (1− 2α)(k + ϵ) + 2 ] − 1 α leading to lim α→(1/2)− Q′(α) = 2ϵ k > 0, which is a contradiction. 2) Suppose, to the contrary, that inequality (16) is true for the parameter 2[1− (1− k − ϵ)]α+ (1− k − ϵ) for a sufficiently small ϵ > 0. That is L2[1−(1−k−ϵ)]α+(1−k−ϵ)(1, t) < αC(1, t) + (1− α)H(1, t) for all t > 1. Hence 1 t { L2[1−(1−k−ϵ)]α+(1−k−ϵ)(1, t) } < 1 t [ αC(1, t) + (1− α)H(1, t) ] for all t > 1. Taking limits on both sides of the above inequality lead to lim t→+∞ 1 t { L2[1−(−1−ϵ)]α+(−1−ϵ)(1, t) } ≤ lim t→+∞ 1 t [ αC(1, t) + (1− α)H(1, t) ] , which is equivalent to[ 1 (2α− 1)(k + ϵ)α+ 2 ]1/[(2α−1)(k+ϵ)α+1] ≤ α. Taking logarithm of both sides of the above inequality, we get R(α) := − ln [(2α− 1)(k + ϵ) + 2]− [(2α− 1)(k + ϵ) + 1] ln(α) ≤ 0. Sine R(α) ≤ 0 for all α ∈ (1/2, 1) and R((1/2)+) = 0, it immediately follows that R′((1/2)+) ≤ 0. However, REFERENCES 1141 R′(α) = − (k + ϵ) { 4α [ 1 + (α− 1/2)(k + ϵ) ] ln(α) + 4(α− 1/2)2(k + ϵ) + (8α− 3) } + 2 α [(2α− 1)(k + ϵ) + 2] leading to lim α→(1/2)+ R′(α) = 2ϵ k > 0, which is a contradiction. 4. Conclusions In this paper, we seek the optimal upper and lower bounds of weighted arithmetic means of contra-harmonic and harmonic means by generalized logarithmic means Lp when p is of the linear form p = 2(1 − c)α + c and p is of the reciprocal of linear form p = 1/[2(1− c)α+ c] respectively. When p has a linear form, we found that L4α−1 = min c { L2(1−c)α+c | L2(1−c)α+c > αC + (1− α)H } for α ∈ (0, 1/2). When p has a reciprocal of linear form, we found that L 7 13−12α = max c { L 1 2(1−c)α+c ∣∣∣L 1 2(1−c)α+c < αC + (1− α)H } for α ∈ (1/2, 1). We also show that, if conjecture 1 is correct, then L1/[−2kα+(k+1)] will be the optimal upper bound for the considered weighted arithmetic mean for α ∈ (0, 1/2) and L2kα+(1−k) will be the optimal lower bound for the considered weighted arithmetic mean for α ∈ (1/2, 1). Acknowledgements The first author would like to thank the Development and Promotion for Science and Technology talents project (DPST) in Thailand for the support in this work. References [1] A A K Abuhany, S Salem, and I M Salman. On Steffensen’s Integral Inequality with Applications, Estimation and Testing. J. Rajasthan Acad. Phys. Sci., 5:1–12, 2006. [2] C P Chen. The monotonicity of the ratio between generalized logarithmic means. J. Math. Anal. Appl., 345:86–89, 2008. [3] C P Chen and F Qi. Monotonicity properties for generalized logarithmic means. Aust. J. Math. Anal., 1:1–4, 2004. REFERENCES 1142 [4] Y M Chu and B Y Long. Best possible inequalities between generalized logarithmic mean and classical means. Abstr. Appl. Anal., Article ID 303286:13 pages, 2010. [5] L Chunrong and L Siqi. Best possible inequalities between generalized logarithmic mean and weighted geometric mean of geometric, square-root, and root-square means. J. Math. Inequal., 8:899–914, 2014. [6] P Kahlig and J Matkowski. Functional equations involving the logarithmic mean. Zeithchür Angewandte Mathematik und Mechanik, 76:385–390, 1996. [7] W H Li, P Miao, and B N Guo. Bounds for the Neuman–Sándor mean in terms of the arithmetic and contraharmonic mean. Axioms., 11:12 pages, 2022. [8] Y M Li, B Y Long, and Y M Chu. Sharp bounds for the Neuman-Sándor mean in terms of generalized logarithmic mean. J. Math. Inequal., 6:567–577, 2012. [9] B Y Long and Y M Chu. Optimal inequalities for generalized logarithmic, arithmetic, and geometric means. J. Inequal. Appl., Article ID 806825:10 pages, 2010. [10] W H McAdam. Heat Transmission. McGraw-Hill, New York, 1954. [11] B Mond, C E M Pearce, and J Pec̆arić. The logarithmic mean is a mean. J. Math. Commun., 345:86–89, 2008. [12] F Qi. Refinement, extensions and generalizations of the second Kershaw’s double inequality. Math. Inequalities Appl., 11:457–465, 2008. [13] F Qi, S X Chen, and C P Chen. Monotonicity of the ratio between the generalized logarithmic means. Math. Inequalities Appl., 10:559–564, 2007. [14] W M Qian and Y M Chu. Best possible bounds for Yang mean using generalized logarithmic mean. Math. Probl. Eng., Article ID 8901258:7 pages, 2016. [15] W M Qian, Z Y He, H W Zhang, and Y M Chu. Sharp bounds for Neuman means in terms of two-parameter contraharmonic and arithmetic mean. J. Inequal. Appl., 168:13 pages, 2019. [16] W M Qian and B Y Long. Sharp bounds by the generalized logarithmic mean for the geometric weighted mean of the geometric and harmonic means. J. Appl. Math., Article ID 480689:8 pages, 2012. [17] W M Qian, X H Zhang, and Y M Chu. Sharp Bounds for the Toader-qi mean in terms of harmonic and geometric means. J. Math. Inequal., 11:121–127, 2017. [18] H N Shi. Schur-convex functions related to Hadamard-type inequalities. J. Math. Inequal., 1:127–136, 2007. [19] K B Stolarsky. The power and generalized logarithmic means. Am. Math. Mon., 87:545–548, 1980. REFERENCES 1143 [20] H Z Xu, Y M Chu, and W M Qian. Sharp bounds for the Sándor–Yang means in terms of arithmetic and contra-harmonic means. J. Inequal. Appl., 127:13 pages, 2018.