# Sharp bounds for the Sándor–Yang means in terms of arithmetic and contra-harmonic means

Sharp bounds for the Sándor–Yang means in terms of arithmetic and contra-harmonic means Department of Mathematics, In the article, we provide several sharp upper and lower bounds for two Sándor–Yang Huzhou University, Huzhou, China Full list of author information is means in terms of combinations of arithmetic and contra-harmonic means. available at the end of the article MSC: Primary 26E60; secondary 26D07; 26D99 Keywords: Schwab–Borchardt mean; Sándor–Yang mean; Arithmetic mean; Contra-harmonic mean; Quadratic mean 1Preliminaries Let a, b >0with a = b. Then the arithmetic mean A(a, b)[1–4], the quadratic mean Q(a, b) , the contra-harmonic mean C(a, b)[6–9], the Neuman–Sándor mean NS(a, b)[10–12], the second Seiﬀert mean T(a, b)[13, 14], and the Schwab–Borchardt mean SB(a, b)[15, 16]of a and b are deﬁned by 2 2 2 2 a + b a + b a + b A(a, b)= , Q(a, b)= , C(a, b)= , (1.1) 2 2 a + b a – b NS(a, b)= , (1.2) –1 a–b 2 sinh ( ) a+b a – b T(a, b)= , (1.3) a–b 2 arctan( ) a+b 2 2 b –a , a < b, arccos(a/b) SB(a, b)= 2 2 a –b , a > b, –1 cosh (a/b) √ √ –1 –1 2 2 respectively, where sinh (x)= log(x + x +1) and cosh (x)= log(x + x – 1) are re- spectively the inverse hyperbolic sine and cosine functions. The Schwab–Borchardt mean SB(a, b) is strictly increasing, non-symmetric and homogeneous of degree one with re- spect to its variables. It can be expressed by the degenerated completely symmetric elliptic integral of the ﬁrst kind . Recently, the Schwab–Borchardt mean has attracted the at- tention of many researchers. In particular, many remarkable inequalities for the Schwab– Borchardt mean and its generated means can be found in the literature [18–38]. © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 2 of 13 Let X(a, b)and Y (a, b) denote symmetric bivariate means of a and b.ThenYang introduced the Sándor–Yang mean X(a,b) –1 SB[X(a,b),Y (a,b)] R (a, b)= Y (a, b)e XY and presented the explicit formulas for R (a, b)and R (a, b) as follows: QA AQ Q(a,b) –1 NS(a,b) R (a, b)= A(a, b)e , (1.4) QA A(a,b) –1 T(a,b) R (a, b)= Q(a, b)e . (1.5) AQ Very recently, the bounds involving the Sándor–Yang means have been the subject of in- tensive research. Numerous interesting results and inequalities for R (a, b)and R (a, b) QA AQ canbefound in theliterature[40–42]. Neuman  established the inequality R (a, b)< R (a, b) (1.6) AQ QA for a, b >0 with a = b. In , Xu proved that the double inequalities α C(a, b)+(1 – α )A(a, b)< R (a, b)< β C(a, b)+(1 – β )A(a, b), 1 1 QA 1 1 (1.7) α C(a, b)+(1 – α )A(a, b)< R (a, b)< β C(a, b)+(1 – β )A(a, b) 2 2 AQ 2 2 hold for all a, b >0 with a = b if and only if α ≤ (1 + 2) /e – 1 = 0.2794 . . . , β ≥ 1/3, 1 1 π/4–1 α ≤ 2e – 1 = 0.1410 . . . and β ≥ 1/6. 2 2 From (1.6)and (1.7), together the well-known inequalities 1 2 C(a, b)> Q(a, b)> A(a, b), Q(a, b)> C(a, b)+ A(a, b), 3 3 we clearly see that A(a, b)< R (a, b)< R (a, b)< Q(a, b)< C(a, b) (1.8) AQ QA for all a, b >0 with a = b. The main purpose of this paper is to ﬁnd the best possible parameters α , β ∈ (0, 1) i i (i = 1,2,3,4) such that the double inequalities α 1–α β 1–β 1 1 1 1 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b), QA α 1–α β 1–β 2 2 2 2 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b), AQ 1 2 1/3 2/3 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 3 3 3 3 1 2 1/3 2/3 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b), QA 3 3 3 3 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 3 of 13 1 5 1/6 5/6 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 4 4 6 6 1 5 1/6 5/6 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b) AQ 4 4 6 6 hold for all a, b >0 with a = b. 2 Lemmas In order to prove our main results, we need several lemmas, which we present in this section. Lemma 2.1 (see ) Let a, b ∈ R with a < b, f , g :[a, b] → R be continuous on [a, b] and diﬀerentiable on (a, b), and g (x) =0 on (a, b). If f (x)/g (x) is increasing (decreasing) on (a, b), then so are the functions f (x)– f (a) f (x)– f (b) , . g(x)– g(a) g(x)– g(b) If f (x)/g (x) is strictly monotone, then the monotonicity in the conclusion is also strict. ∞ ∞ k k Lemma 2.2 (see ) Let A(t)= a t and B(t)= b t be two real power series k k k=0 k=0 converging on (–r, r)(r >0) with b >0 for all k. If the non-constant sequence {a /b } k k k k=0 is increasing (decreasing) for all k, then the function t → A(t)/B(t) is strictly increasing (decreasing) on (0, r). Lemma 2.3 The function x coth(x)–1 φ(x)= 2 log[cosh(x)] √ √ √ is strictly increasing from (0, log(1 + 2) onto (1/3, [ 2 log(1 + 2) – 1]/ log 2). Proof Let φ (x)= x coth(x)–1, φ (x)=2 log[cosh(x)]. Then elaborate computations lead 1 2 to φ (x) φ (x)– φ (0 ) 1 1 1 φ(x)= = , (2.1) φ (x) φ (x)– φ (0) 2 2 2 φ (x) sinh(x)cosh (x)– x cosh(x) φ (x) 2sinh (x) 2n+1 ∞ 3 –8n–3 2n+1 sinh(3x)+ sinh(x)–4x cosh(x) n=0 (2n+1)! = = 2n ∞ 6(3 –1) 2n+1 2 sinh(3x)–6 sinh(x) n=0 (2n+1)! 2n+1 2n+3 ∞ ∞ 3 –8n–3 3 –8n–11 2n+1 2n+3 x x n=1 n=0 (2n+1)! (2n+3)! = = . (2.2) 2n 2n+2 ∞ ∞ 6(3 –1) 6(3 –1) 2n+1 2n+3 x x n=1 n=0 (2n+1)! (2n+3)! Let 2n+3 2n+2 3 –8n –11 6(3 –1) a = , b = . (2.3) n n (2n +3)! (2n +3)! Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 4 of 13 Then b >0 (2.4) and 2n a a 4[(72n + 63)3 +1] n+1 n – = >0 (2.5) 2n+4 2n+2 b b 3(3 – 1)(3 –1) n+1 n for all n ≥ 0. It follows from Lemma 2.2 and (2.2)–(2.5)that φ (x)/φ (x) is strictly increasing on 1 2 (0, log(1 + 2)). Note that √ √ a 1 2 log(1 + 2) – 1 φ 0 = = , φ log(1 + 2) = = 0.3555 . . . . (2.6) b 3 log 2 Therefore, Lemma 2.3 follows from Lemma 2.1,(2.1), and (2.6) together with the mono- tonicity of φ (x)/φ (x). 1 2 Lemma 2.4 The function log sec(x)+ x cot(x)–1 ϕ(x)= 2 log sec(x) is strictly increasing from (0, π/4) onto (1/6, 1/2 – (4 – π)(4 log 2)). Proof Let ϕ (x)= log sec(x)+ x cot(x)–1, ϕ (x)=2 log[sec(x)], ϕ (x)= sin(x)– x cos(x), and 1 2 3 ϕ (x)=2sin (x). Then elaborate computations lead to ϕ (x) ϕ (x)– ϕ (0 ) 1 1 1 ϕ(x)= = , (2.7) ϕ (x) ϕ (x)– ϕ (0) 2 2 2 ϕ (x) ϕ (x) ϕ (x)– ϕ (0) 3 3 3 = = (2.8) ϕ (x) ϕ (x) ϕ (x)– ϕ (0) 4 4 4 and ϕ (x) x 1 1 = = × . (2.9) ϕ (x) 3 sin(2x) 6 sin(2x)/(2x) It is well known that the function x → sin(x)/x is strictly decreasing on (0, π/2), hence equation (2.9) leads to the conclusion that the function ϕ (x)/ϕ (x) is strictly increasing 3 4 on (0, π/4). Note that ϕ (x) 1 + 3 ϕ 0 = lim = , x→0 ϕ (x) 6 (2.10) π 1 4– π ϕ = – = 0.1903 . . . . 4 2 4 log 2 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 5 of 13 Therefore, Lemma 2.4 follows from Lemma 2.1 and (2.7)–(2.9) together with the mono- tonicity of ϕ (x)/ϕ (x). 3 4 Lemma 2.5 Let p ∈ (0, 1) and 2 10 6 2 4 2 2 f (x)=3p x +14p(1 – p)x +18p x –9(1 – p) x –2p(1 – p). Then the following statements are true: (1) If p = 3/10, then f (x)>0 for all x ∈ (1, 2); √ √ √ 3 3 (2) If p = 3[(1 + 2) /e – 2]/(4 – 3 2) = 0.2663 . . . , then there exists λ (= 1.0808 . . . ) ∈ (1, 2) such that f (x)<0 for x ∈ (1, λ ) and f (x)>0 for 0 0 x ∈ (λ , 2). Proof Part (1) follows easily from 2 8 6 4 2 f (x)= x –1 9x +9x + 107x + 161x +14 >0 for all x ∈ (1, 2) if p =3/10. √ √ √ 3 3 For part (2), if p = 3[(1 + 2) /e – 2]/(4 – 3 2), then numerical computations lead to 20p – 3 = 2.3273 ··· > 0, (2.11) f (1) = 3(10p – 3) = –1.008 ··· < 0, (2.12) f 2 = 1.6809 ··· > 0, (2.13) 2 9 5 2 3 2 f (x)=30p x +84p(1 – p)x +72p x – 18(1 – p) x. (2.14) It follows from (2.11)and (2.14)that 2 2 f (x)> 30p +84p(1 – p)+72p – 18(1 – p) x =6(20p –3)x > 0 (2.15) for all x ∈ (1, 2). Therefore, part (2) follows easily from (2.12), (2.13), (2.15), and the numerical results f (1.0808) < 0 and f (1.0809) > 0. Lemma 2.6 Let p ∈ (0, 1) and 2 11 6 2 5 2 g(x)=3p x +56p(1 – p)x +75p x – 72(1 – p) x –50p(1 – p). Then the following statements are true: (1) If p = 12/25, then g(x)>0 for all x ∈ (1, 2); √ √ √ 6 6 π/4–1 (2) If p =6[ 2e – 2]/(7 – 6 2) = 0.4210 . . . , then there exists μ (= 1.0577 . . . ) ∈ (1, 2) such that g(x)<0 for x ∈ (1, μ ) and g(x)>0 for 0 0 x ∈ (μ , 2). 0 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 6 of 13 Proof Part (1) follows easily from 10 9 8 7 6 5 4 g(x)= (x –1) 18x +18x +18x +18x +18x + 382x + 832x 3 2 + 832x + 832x + 832x + 325 >0 for all x ∈ (1, 2) if p = 12/25. √ √ √ 6 6 π/4–1 For part (2), if p =6[ 2e – 2]/(7–6 2) = 0.4210 . . . , then numerical computations lead to 20p – 3 = 5.4217 ··· > 0, (2.16) g(1) = 6(25p – 12) = –8.8367 ··· < 0, (2.17) g 2 = 13.6200 ··· > 0, (2.18) 2 10 5 2 4 g (x)=3 11p x + 112p(1 – p)x + 125p x – 24(1 – p) . (2.19) It follows from (2.16)and (2.19)that 2 2 2 g (x)>11p + 112p(1 – p) + 125p – 24(1 – p) = 24(20p – 3) > 0 (2.20) for x ∈ (1, 2). Therefore, part (2) follows easily from (2.17), (2.18), and (2.20) together with the numer- ical results g(1.0577) < 0 and g(1.0578) > 0. 3 Main results We are now in a position to state and prove our main results. Theorem 3.1 Thedoubleinequality α 1–α β 1–β 1 1 1 1 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b) (3.1) QA √ √ holds for all a, b >0 with a = bif and only if α ≤ 1/3 and β ≥ [ 2 log(1 + 2) – 1]/ log 2. 1 1 Proof Clearly, inequality (3.1)can be rewrittenas α β 1 1 C(a, b) R (a, b) C(a, b) QA < < . (3.2) A(a, b) A(a, b) A(a, b) Since A(a, b), R (a, b), and C(a, b) are symmetric and homogenous of degree one, we QA assume that a > b >0. Let v =(a – b)/(a + b) ∈ (0, 1). Then from (1.1), (1.2), and (1.4)we know that inequality (3.2)isequivalentto –1 [ 1+ v sinh (v)]/v –1 α < < β . (3.3) 1 1 log(1 + v ) Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 7 of 13 –1 Let x = sinh (v). Then x ∈ (0, log(1 + 2)) and –1 [ 1+ v sinh (v)]/v –1 x coth(x)–1 = := φ(x). (3.4) log(1 + v ) 2 log[cosh(x)] Therefore, inequality (3.1)holds forall a, b >0 with a = b if and only if α ≤ 1/3 and √ √ β ≥ [ 2 log(1 + 2) – 1]/ log 2 follows from (3.2)–(3.4) and Lemma 2.3. Theorem 3.2 Thedoubleinequality α 1–α β 1–β 2 2 2 2 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b) (3.5) AQ holds for all a, b >0 with a = bif and only if α ≤ 1/6 and β ≥ 1/2 – (4 – π)/(4 log 2) = 2 2 0.1903 . . . . Proof Clearly, inequality (3.5)can be rewrittenas α β 2 2 C(a, b) R (a, b) C(a, b) AQ < < . (3.6) A(a, b) A(a, b) A(a, b) Since A(a, b), R (a, b), and C(a, b) are symmetric and homogenous of degree one, we AQ assume that a > b >0. Let v =(a – b)/(a + b) ∈ (0, 1). Then from (1.1), (1.3), and (1.5)we see that inequality (3.6)isequivalentto log 1+ v +[arctan(v)]/v –1 α < < β . (3.7) 2 2 log(1 + v ) Let x = arctan(v). Then x ∈ (0, π/4) and log 1+ v +[arctan(v)]/v –1 log(1 + v ) log sec(x)+ x cot(x)–1 = := ϕ(x). (3.8) 2 log sec(x) Therefore, inequality (3.5)holds forall a, b >0 with a = b if and only if α ≤ 1/6 and β ≥ 1/2 – (4 – π)/(4 log 2) = 0.1903 . . . follows from (3.6)–(3.8) and Lemma 2.4. Theorem 3.3 Thedoubleinequality 1 2 1/3 2/3 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 3 3 3 3 1 2 1/3 2/3 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b) QA 3 3 3 3 √ √ √ 3 3 holds for all a, b >0 with a = bif and only if α ≤ 3[(1 + 2) /e – 2]/(4 – 3 2) = 0.2663 . . . and β ≥ 3/10. 3 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 8 of 13 Proof Since R (a, b), A(a, b), and C(a, b) are symmetric and homogenous of degree one, QA without loss generality, we assume that a > b >0. Let v =(a – b)/(a + b), x = 1+ v ,and p ∈ (0, 1). Then v ∈ (0, 1), x ∈ (1, 2), and (1.1), (1.2), and (1.4)leadto R (a, b) QA log 1 2 1/3 2/3 p[ C(a, b)+ A(a, b)] + (1 – p)C (a, b)A (a, b) 3 3 –1 2 √ 1+ v sinh (v) 1 = – log p v +1 +(1– p) 1+ v –1 v 3 3 –1 x sinh ( x –1) 1 2 6 2 = √ – log p x + +(1– p)x – 1. (3.9) 3 3 x –1 Let –1 3 6 x sinh ( x –1) 1 2 6 2 F(x)= √ – log p x + +(1– p)x – 1. (3.10) 3 3 x –1 Then simple computations lead to √ √ √ √ 6 3 F 1 =0, F 2 = 2 log(1 + 2) – log p + 2(1 – p) – 1, (3.11) 3x F (x)= F (x), (3.12) 3/2 (x –1) where 10 6 4 6 √ x –1[–px +(1– p)x +4px +2(1 – p)] –1 F (x)= – sinh x –1 , 6 2 x[p(x +2)+3(1– p)x ] √ √ √ 6 6 ( 2048 – 2 2)p – 2048 F (1) = 0, F 2 = √ √ – log(1 + 2), (3.13) 1 1 3 3 3 2p –3 2–4p 3/2 2(x –1) F (x)=– f (x), (3.14) 2 6 2 x [p(x +2)+3(1– p)x ] where f (x) is deﬁned as in Lemma 2.5. We divide the proof into four cases. Case 1 p = 3/10. Then it follows from (3.9)–(3.14) and Lemma 2.5(1) that 3 1 2 7 1/3 2/3 R (a, b)< C(a, b)+ A(a, b) + C (a, b)A (a, b). QA 10 3 3 10 Case 2 0< p < 3/10. Let v >0 and v → 0 .Thenpower series expansionleads to –1 1+ v sinh (v) 1 – log p v +1 +(1– p) 1+ v –1 v 3 1 1 4 6 = – p v + O v . (3.15) 30 9 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 9 of 13 Equations (3.9), (3.10), and (3.15) lead to the conclusion that there exists 0 < δ <1 such that 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) QA 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (0, δ ). √ √ √ 3 3 Case 3 p = 3[(1 + 2) /e – 2]/(4 – 3 2). Then (3.13)leads to F 2 = –0.0039 ··· < 0. (3.16) Let λ = 1.0808 . . . be the number given in Lemma 2.5(2). Then we divide the discussion into two subcases. Subcase 1 x ∈ (1, λ ]. Then F (x)>0 for x ∈ (1, λ ] follows easily from (3.13)and (3.14) 0 1 0 together with Lemma 2.5(2). Subcase 2 x ∈ (λ , 2). Then Lemma 2.5(2) and (3.14)leadtothe conclusion that F (x)is 0 1 strictly decreasing on the interval [λ , 2). Then, from (3.16)and Subcase1,weknowthat √ √ 6 6 there exists λ ∈ (λ , 2) such that F (x)>0 for x ∈ [λ , λ )and F (x)<0 for x ∈ (λ , 2). 1 0 1 0 1 1 1 It follows from Subcases 1 and 2 together with (3.12)that F(x) is strictly increasing on (1, λ ] and strictly decreasing on [λ , 2). Therefore, 1 1 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) QA 3 3 follows from (3.9)–(3.11)and (3.16) together with the piecewise monotonicity of F(x). √ √ √ 3 3 Case 4 3[(1 + 2) /e – 2]/(4 – 3 2) < p <1. Then (3.11)leads to √ √ √ √ 6 3 F 2 = 2 log(1 + 2) – log p + 2(1 – p) – 1 < 0. (3.17) Equations (3.9)and (3.10) together with inequality (3.17) imply that there exists 0 < δ < 1such that 1 2 1/3 2/3 R (a, b)< p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) QA 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (1 – δ ,1). Theorem 3.4 Thedoubleinequality 1 5 1/6 5/6 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 4 4 6 6 1 5 1/6 5/6 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b) AQ 4 4 6 6 √ √ √ 6 6 (π/4–1) holds for all a, b >0 with a = bif and only if α ≤ 6[ 2e – 2]/(7 – 6 2) = 0.4210 . . . and β ≥ 12/25. 4 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 10 of 13 Proof Since R (a, b), A(a, b), and C(a, b) are symmetric and homogenous of degree one, AQ without loss generality, we assume that a > b >0. Let v =(a – b)/(a + b), x = 1+ v ,and p ∈ (0, 1). Then v ∈ (0, 1), x ∈ (1, 2) and (1.1), (1.3), and (1.5)leadto R (a, b) AQ log 1 5 1/6 5/6 p[ C(a, b)+ A(a, b)] + (1 – p)C (a, b)A (a, b) 6 6 √ √ arctan(v) 1 2 2 = log 1+ v + – log p v +1 +(1– p) 1+ v –1 v 6 arctan( x –1) 1 5 =3 log(x)+ √ – log p x + +(1– p)x – 1. (3.18) 6 6 x –1 Let arctan( x –1) 1 5 G(x)=3 log(x)+ √ – log p x + +(1– p)x – 1. (3.19) 6 6 x –1 Then simple computations lead to √ √ √ π 7 6 6 G 1 =0, G 2 = log( 2) + – log p + 2(1 – p) – 1, (3.20) 4 6 3x G (x)= G (x), (3.21) 3/2 (x –1) where 6 11 6 5 x –1[–px +4(1 – p)x +7px +2(1 – p)] G (x)= – arctan x –1 , 5 6 x [p(x +5)+6(1– p)x] 5[ 2(1 – p)+ p] π G (1) = 0, G 2 = √ – , (3.22) 1 1 6 2(1 – p)+7p 3/2 (x –1) G (x)=– g(x), (3.23) 1 2 6 6 x [p(x +5)+6(1– p)x] where g(x) is deﬁned as in Lemma 2.6. We divide the proof into four cases. Case 1 p = 12/25. Then it follows from (3.18)–(3.23) and Lemma 2.6(1) that 12 1 2 13 1/3 2/3 R (a, b)< C(a, b)+ A(a, b) + C (a, b)A (a, b). AQ 25 3 3 25 Case 2 0< p < 12/25. Let v >0 and v → 0 ,thenpower series expansionleads to √ √ arctan(v) 1 2 2 log 1+ v + – log p v +1 +(1– p) 1+ v –1 v 6 1 5 4 6 = – p v + O v . (3.24) 30 72 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 11 of 13 Equations (3.18), (3.19), and (3.24) lead to the conclusion that there exists 0 < δ <1 such that 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) AQ 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (0, δ ). √ √ √ 6 6 (π/4–1) Case 3 p =6[ 2e – 2]/(7 – 6 2). Then, from (3.20)and (3.22)togetherwithnu- merical computations, we get √ √ 6 6 G 2 =0, G 2 = –0.0033 ··· < 0. (3.25) Let μ = 1.0577 . . . be the number given in Lemma 2.6(2). Then we divide the discussion into two subcases. Subcase 1 x ∈ (1, μ ]. Then G (x)>0 for x ∈ (1, μ ] follows easily from (3.22)and (3.23) 0 1 0 together with Lemma 2.6(2). Subcase 2 x ∈ (μ , 2). Then Lemma 2.6(2) and (3.23)leadtothe conclusion that G (x) 0 1 is strictly decreasing on the interval [μ , 2). Then, from (3.25) and Subcase 1, we know that there exists μ ∈ (μ , 2) such that G (x)>0 for x ∈ [μ , μ )and G (x)<0 for x ∈ 1 0 1 0 1 1 (μ , 2). It follows from Subcases 1 and 2 together with (3.21)that G(x) is strictly increasing on (1, μ ] and strictly decreasing on [μ , 2). Therefore, 1 1 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) AQ 3 3 follows from (3.18)–(3.20)and (3.25) together with the piecewise monotonicity of G(x). √ √ √ 6 6 (π/4–1) Case 4 6[ 2e – 2]/(7 – 6 2) < p <1. Then (3.21)leads to √ √ √ π 7 6 6 G 2 = log( 2) + – log p + 2(1 – p) – 1 < 0. (3.26) 4 6 Equations (3.18)and (3.19) together with inequality (3.26) imply that there exists 0 < δ <1 such that 1 2 1/3 2/3 R (a, b)< p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) AQ 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (1 – δ ,1). 4 Results and discussion In this paper, we provide the optimal upper and lower bounds for the Sándor–Yang means R (a, b)and R (a, b) in terms of combinations of the arithmetic mean A(a, b)and the QA AQ contra-harmonic mean C(a, b). Our approach may have further applications in the theory of bivariate means. 5Conclusion In the article, we ﬁnd several best possible bounds for the Sándor–Yang means R (a, b) QA and R (a, b). These results are improvements and reﬁnements of the previous results. AQ Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 12 of 13 Funding The research was supported by the Natural Science Foundation of China (Grants Nos. 61673169, 61374086, 11371125, 11401191), the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11626101) and the Natural Science Foundation of the Department of Education of Zhejiang Province (Grant No. Y201635325). Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. All authors read and approved the ﬁnal manuscript. Author details 1 2 School of Economics and Management, Wenzhou Broadcast and TV University, Wenzhou, China. Department of Mathematics, Huzhou University, Huzhou, China. School of Distance Education, Huzhou Broadcast and TV University, Huzhou, China. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Received: 4 January 2018 Accepted: 24 May 2018 References 1. Chu, Y.-M., Wang, M.-K., Qiu, S.-L.: Optimal combinations bounds of root-square and arithmetic means for Toader mean. Proc. Indian Acad. Sci. Math. Sci. 122(1), 41–51 (2012) Available online at https://link.springer.com/article/10.1007%2Fs12044-012-0062-y 2. Hua, Y., Qi, F.: The best bounds for Toader mean in terms of the centroidal and arithmetic means. Filomat 28(4), 775–780 (2014) Available at http://www.pmf.ni.ac.rs/ﬁlomat 3. 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# Sharp bounds for the Sándor–Yang means in terms of arithmetic and contra-harmonic means

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Department of Mathematics, In the article, we provide several sharp upper and lower bounds for two Sándor–Yang Huzhou University, Huzhou, China Full list of author information is means in terms of combinations of arithmetic and contra-harmonic means. available at the end of the article MSC: Primary 26E60; secondary 26D07; 26D99 Keywords: Schwab–Borchardt mean; Sándor–Yang mean; Arithmetic mean; Contra-harmonic mean; Quadratic mean 1Preliminaries Let a, b >0with a = b. Then the arithmetic mean A(a, b)[1–4], the quadratic mean Q(a, b) , the contra-harmonic mean C(a, b)[6–9], the Neuman–Sándor mean NS(a, b)[10–12], the second Seiﬀert mean T(a, b)[13, 14], and the Schwab–Borchardt mean SB(a, b)[15, 16]of a and b are deﬁned by 2 2 2 2 a + b a + b a + b A(a, b)= , Q(a, b)= , C(a, b)= , (1.1) 2 2 a + b a – b NS(a, b)= , (1.2) –1 a–b 2 sinh ( ) a+b a – b T(a, b)= , (1.3) a–b 2 arctan( ) a+b 2 2 b –a , a < b, arccos(a/b) SB(a, b)= 2 2 a –b , a > b, –1 cosh (a/b) √ √ –1 –1 2 2 respectively, where sinh (x)= log(x + x +1) and cosh (x)= log(x + x – 1) are re- spectively the inverse hyperbolic sine and cosine functions. The Schwab–Borchardt mean SB(a, b) is strictly increasing, non-symmetric and homogeneous of degree one with re- spect to its variables. It can be expressed by the degenerated completely symmetric elliptic integral of the ﬁrst kind . Recently, the Schwab–Borchardt mean has attracted the at- tention of many researchers. In particular, many remarkable inequalities for the Schwab– Borchardt mean and its generated means can be found in the literature [18–38]. © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 2 of 13 Let X(a, b)and Y (a, b) denote symmetric bivariate means of a and b.ThenYang introduced the Sándor–Yang mean X(a,b) –1 SB[X(a,b),Y (a,b)] R (a, b)= Y (a, b)e XY and presented the explicit formulas for R (a, b)and R (a, b) as follows: QA AQ Q(a,b) –1 NS(a,b) R (a, b)= A(a, b)e , (1.4) QA A(a,b) –1 T(a,b) R (a, b)= Q(a, b)e . (1.5) AQ Very recently, the bounds involving the Sándor–Yang means have been the subject of in- tensive research. Numerous interesting results and inequalities for R (a, b)and R (a, b) QA AQ canbefound in theliterature[40–42]. Neuman  established the inequality R (a, b)< R (a, b) (1.6) AQ QA for a, b >0 with a = b. In , Xu proved that the double inequalities α C(a, b)+(1 – α )A(a, b)< R (a, b)< β C(a, b)+(1 – β )A(a, b), 1 1 QA 1 1 (1.7) α C(a, b)+(1 – α )A(a, b)< R (a, b)< β C(a, b)+(1 – β )A(a, b) 2 2 AQ 2 2 hold for all a, b >0 with a = b if and only if α ≤ (1 + 2) /e – 1 = 0.2794 . . . , β ≥ 1/3, 1 1 π/4–1 α ≤ 2e – 1 = 0.1410 . . . and β ≥ 1/6. 2 2 From (1.6)and (1.7), together the well-known inequalities 1 2 C(a, b)> Q(a, b)> A(a, b), Q(a, b)> C(a, b)+ A(a, b), 3 3 we clearly see that A(a, b)< R (a, b)< R (a, b)< Q(a, b)< C(a, b) (1.8) AQ QA for all a, b >0 with a = b. The main purpose of this paper is to ﬁnd the best possible parameters α , β ∈ (0, 1) i i (i = 1,2,3,4) such that the double inequalities α 1–α β 1–β 1 1 1 1 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b), QA α 1–α β 1–β 2 2 2 2 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b), AQ 1 2 1/3 2/3 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 3 3 3 3 1 2 1/3 2/3 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b), QA 3 3 3 3 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 3 of 13 1 5 1/6 5/6 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 4 4 6 6 1 5 1/6 5/6 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b) AQ 4 4 6 6 hold for all a, b >0 with a = b. 2 Lemmas In order to prove our main results, we need several lemmas, which we present in this section. Lemma 2.1 (see ) Let a, b ∈ R with a < b, f , g :[a, b] → R be continuous on [a, b] and diﬀerentiable on (a, b), and g (x) =0 on (a, b). If f (x)/g (x) is increasing (decreasing) on (a, b), then so are the functions f (x)– f (a) f (x)– f (b) , . g(x)– g(a) g(x)– g(b) If f (x)/g (x) is strictly monotone, then the monotonicity in the conclusion is also strict. ∞ ∞ k k Lemma 2.2 (see ) Let A(t)= a t and B(t)= b t be two real power series k k k=0 k=0 converging on (–r, r)(r >0) with b >0 for all k. If the non-constant sequence {a /b } k k k k=0 is increasing (decreasing) for all k, then the function t → A(t)/B(t) is strictly increasing (decreasing) on (0, r). Lemma 2.3 The function x coth(x)–1 φ(x)= 2 log[cosh(x)] √ √ √ is strictly increasing from (0, log(1 + 2) onto (1/3, [ 2 log(1 + 2) – 1]/ log 2). Proof Let φ (x)= x coth(x)–1, φ (x)=2 log[cosh(x)]. Then elaborate computations lead 1 2 to φ (x) φ (x)– φ (0 ) 1 1 1 φ(x)= = , (2.1) φ (x) φ (x)– φ (0) 2 2 2 φ (x) sinh(x)cosh (x)– x cosh(x) φ (x) 2sinh (x) 2n+1 ∞ 3 –8n–3 2n+1 sinh(3x)+ sinh(x)–4x cosh(x) n=0 (2n+1)! = = 2n ∞ 6(3 –1) 2n+1 2 sinh(3x)–6 sinh(x) n=0 (2n+1)! 2n+1 2n+3 ∞ ∞ 3 –8n–3 3 –8n–11 2n+1 2n+3 x x n=1 n=0 (2n+1)! (2n+3)! = = . (2.2) 2n 2n+2 ∞ ∞ 6(3 –1) 6(3 –1) 2n+1 2n+3 x x n=1 n=0 (2n+1)! (2n+3)! Let 2n+3 2n+2 3 –8n –11 6(3 –1) a = , b = . (2.3) n n (2n +3)! (2n +3)! Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 4 of 13 Then b >0 (2.4) and 2n a a 4[(72n + 63)3 +1] n+1 n – = >0 (2.5) 2n+4 2n+2 b b 3(3 – 1)(3 –1) n+1 n for all n ≥ 0. It follows from Lemma 2.2 and (2.2)–(2.5)that φ (x)/φ (x) is strictly increasing on 1 2 (0, log(1 + 2)). Note that √ √ a 1 2 log(1 + 2) – 1 φ 0 = = , φ log(1 + 2) = = 0.3555 . . . . (2.6) b 3 log 2 Therefore, Lemma 2.3 follows from Lemma 2.1,(2.1), and (2.6) together with the mono- tonicity of φ (x)/φ (x). 1 2 Lemma 2.4 The function log sec(x)+ x cot(x)–1 ϕ(x)= 2 log sec(x) is strictly increasing from (0, π/4) onto (1/6, 1/2 – (4 – π)(4 log 2)). Proof Let ϕ (x)= log sec(x)+ x cot(x)–1, ϕ (x)=2 log[sec(x)], ϕ (x)= sin(x)– x cos(x), and 1 2 3 ϕ (x)=2sin (x). Then elaborate computations lead to ϕ (x) ϕ (x)– ϕ (0 ) 1 1 1 ϕ(x)= = , (2.7) ϕ (x) ϕ (x)– ϕ (0) 2 2 2 ϕ (x) ϕ (x) ϕ (x)– ϕ (0) 3 3 3 = = (2.8) ϕ (x) ϕ (x) ϕ (x)– ϕ (0) 4 4 4 and ϕ (x) x 1 1 = = × . (2.9) ϕ (x) 3 sin(2x) 6 sin(2x)/(2x) It is well known that the function x → sin(x)/x is strictly decreasing on (0, π/2), hence equation (2.9) leads to the conclusion that the function ϕ (x)/ϕ (x) is strictly increasing 3 4 on (0, π/4). Note that ϕ (x) 1 + 3 ϕ 0 = lim = , x→0 ϕ (x) 6 (2.10) π 1 4– π ϕ = – = 0.1903 . . . . 4 2 4 log 2 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 5 of 13 Therefore, Lemma 2.4 follows from Lemma 2.1 and (2.7)–(2.9) together with the mono- tonicity of ϕ (x)/ϕ (x). 3 4 Lemma 2.5 Let p ∈ (0, 1) and 2 10 6 2 4 2 2 f (x)=3p x +14p(1 – p)x +18p x –9(1 – p) x –2p(1 – p). Then the following statements are true: (1) If p = 3/10, then f (x)>0 for all x ∈ (1, 2); √ √ √ 3 3 (2) If p = 3[(1 + 2) /e – 2]/(4 – 3 2) = 0.2663 . . . , then there exists λ (= 1.0808 . . . ) ∈ (1, 2) such that f (x)<0 for x ∈ (1, λ ) and f (x)>0 for 0 0 x ∈ (λ , 2). Proof Part (1) follows easily from 2 8 6 4 2 f (x)= x –1 9x +9x + 107x + 161x +14 >0 for all x ∈ (1, 2) if p =3/10. √ √ √ 3 3 For part (2), if p = 3[(1 + 2) /e – 2]/(4 – 3 2), then numerical computations lead to 20p – 3 = 2.3273 ··· > 0, (2.11) f (1) = 3(10p – 3) = –1.008 ··· < 0, (2.12) f 2 = 1.6809 ··· > 0, (2.13) 2 9 5 2 3 2 f (x)=30p x +84p(1 – p)x +72p x – 18(1 – p) x. (2.14) It follows from (2.11)and (2.14)that 2 2 f (x)> 30p +84p(1 – p)+72p – 18(1 – p) x =6(20p –3)x > 0 (2.15) for all x ∈ (1, 2). Therefore, part (2) follows easily from (2.12), (2.13), (2.15), and the numerical results f (1.0808) < 0 and f (1.0809) > 0. Lemma 2.6 Let p ∈ (0, 1) and 2 11 6 2 5 2 g(x)=3p x +56p(1 – p)x +75p x – 72(1 – p) x –50p(1 – p). Then the following statements are true: (1) If p = 12/25, then g(x)>0 for all x ∈ (1, 2); √ √ √ 6 6 π/4–1 (2) If p =6[ 2e – 2]/(7 – 6 2) = 0.4210 . . . , then there exists μ (= 1.0577 . . . ) ∈ (1, 2) such that g(x)<0 for x ∈ (1, μ ) and g(x)>0 for 0 0 x ∈ (μ , 2). 0 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 6 of 13 Proof Part (1) follows easily from 10 9 8 7 6 5 4 g(x)= (x –1) 18x +18x +18x +18x +18x + 382x + 832x 3 2 + 832x + 832x + 832x + 325 >0 for all x ∈ (1, 2) if p = 12/25. √ √ √ 6 6 π/4–1 For part (2), if p =6[ 2e – 2]/(7–6 2) = 0.4210 . . . , then numerical computations lead to 20p – 3 = 5.4217 ··· > 0, (2.16) g(1) = 6(25p – 12) = –8.8367 ··· < 0, (2.17) g 2 = 13.6200 ··· > 0, (2.18) 2 10 5 2 4 g (x)=3 11p x + 112p(1 – p)x + 125p x – 24(1 – p) . (2.19) It follows from (2.16)and (2.19)that 2 2 2 g (x)>11p + 112p(1 – p) + 125p – 24(1 – p) = 24(20p – 3) > 0 (2.20) for x ∈ (1, 2). Therefore, part (2) follows easily from (2.17), (2.18), and (2.20) together with the numer- ical results g(1.0577) < 0 and g(1.0578) > 0. 3 Main results We are now in a position to state and prove our main results. Theorem 3.1 Thedoubleinequality α 1–α β 1–β 1 1 1 1 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b) (3.1) QA √ √ holds for all a, b >0 with a = bif and only if α ≤ 1/3 and β ≥ [ 2 log(1 + 2) – 1]/ log 2. 1 1 Proof Clearly, inequality (3.1)can be rewrittenas α β 1 1 C(a, b) R (a, b) C(a, b) QA < < . (3.2) A(a, b) A(a, b) A(a, b) Since A(a, b), R (a, b), and C(a, b) are symmetric and homogenous of degree one, we QA assume that a > b >0. Let v =(a – b)/(a + b) ∈ (0, 1). Then from (1.1), (1.2), and (1.4)we know that inequality (3.2)isequivalentto –1 [ 1+ v sinh (v)]/v –1 α < < β . (3.3) 1 1 log(1 + v ) Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 7 of 13 –1 Let x = sinh (v). Then x ∈ (0, log(1 + 2)) and –1 [ 1+ v sinh (v)]/v –1 x coth(x)–1 = := φ(x). (3.4) log(1 + v ) 2 log[cosh(x)] Therefore, inequality (3.1)holds forall a, b >0 with a = b if and only if α ≤ 1/3 and √ √ β ≥ [ 2 log(1 + 2) – 1]/ log 2 follows from (3.2)–(3.4) and Lemma 2.3. Theorem 3.2 Thedoubleinequality α 1–α β 1–β 2 2 2 2 C (a, b)A (a, b)< R (a, b)< C (a, b)A (a, b) (3.5) AQ holds for all a, b >0 with a = bif and only if α ≤ 1/6 and β ≥ 1/2 – (4 – π)/(4 log 2) = 2 2 0.1903 . . . . Proof Clearly, inequality (3.5)can be rewrittenas α β 2 2 C(a, b) R (a, b) C(a, b) AQ < < . (3.6) A(a, b) A(a, b) A(a, b) Since A(a, b), R (a, b), and C(a, b) are symmetric and homogenous of degree one, we AQ assume that a > b >0. Let v =(a – b)/(a + b) ∈ (0, 1). Then from (1.1), (1.3), and (1.5)we see that inequality (3.6)isequivalentto log 1+ v +[arctan(v)]/v –1 α < < β . (3.7) 2 2 log(1 + v ) Let x = arctan(v). Then x ∈ (0, π/4) and log 1+ v +[arctan(v)]/v –1 log(1 + v ) log sec(x)+ x cot(x)–1 = := ϕ(x). (3.8) 2 log sec(x) Therefore, inequality (3.5)holds forall a, b >0 with a = b if and only if α ≤ 1/6 and β ≥ 1/2 – (4 – π)/(4 log 2) = 0.1903 . . . follows from (3.6)–(3.8) and Lemma 2.4. Theorem 3.3 Thedoubleinequality 1 2 1/3 2/3 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 3 3 3 3 1 2 1/3 2/3 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b) QA 3 3 3 3 √ √ √ 3 3 holds for all a, b >0 with a = bif and only if α ≤ 3[(1 + 2) /e – 2]/(4 – 3 2) = 0.2663 . . . and β ≥ 3/10. 3 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 8 of 13 Proof Since R (a, b), A(a, b), and C(a, b) are symmetric and homogenous of degree one, QA without loss generality, we assume that a > b >0. Let v =(a – b)/(a + b), x = 1+ v ,and p ∈ (0, 1). Then v ∈ (0, 1), x ∈ (1, 2), and (1.1), (1.2), and (1.4)leadto R (a, b) QA log 1 2 1/3 2/3 p[ C(a, b)+ A(a, b)] + (1 – p)C (a, b)A (a, b) 3 3 –1 2 √ 1+ v sinh (v) 1 = – log p v +1 +(1– p) 1+ v –1 v 3 3 –1 x sinh ( x –1) 1 2 6 2 = √ – log p x + +(1– p)x – 1. (3.9) 3 3 x –1 Let –1 3 6 x sinh ( x –1) 1 2 6 2 F(x)= √ – log p x + +(1– p)x – 1. (3.10) 3 3 x –1 Then simple computations lead to √ √ √ √ 6 3 F 1 =0, F 2 = 2 log(1 + 2) – log p + 2(1 – p) – 1, (3.11) 3x F (x)= F (x), (3.12) 3/2 (x –1) where 10 6 4 6 √ x –1[–px +(1– p)x +4px +2(1 – p)] –1 F (x)= – sinh x –1 , 6 2 x[p(x +2)+3(1– p)x ] √ √ √ 6 6 ( 2048 – 2 2)p – 2048 F (1) = 0, F 2 = √ √ – log(1 + 2), (3.13) 1 1 3 3 3 2p –3 2–4p 3/2 2(x –1) F (x)=– f (x), (3.14) 2 6 2 x [p(x +2)+3(1– p)x ] where f (x) is deﬁned as in Lemma 2.5. We divide the proof into four cases. Case 1 p = 3/10. Then it follows from (3.9)–(3.14) and Lemma 2.5(1) that 3 1 2 7 1/3 2/3 R (a, b)< C(a, b)+ A(a, b) + C (a, b)A (a, b). QA 10 3 3 10 Case 2 0< p < 3/10. Let v >0 and v → 0 .Thenpower series expansionleads to –1 1+ v sinh (v) 1 – log p v +1 +(1– p) 1+ v –1 v 3 1 1 4 6 = – p v + O v . (3.15) 30 9 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 9 of 13 Equations (3.9), (3.10), and (3.15) lead to the conclusion that there exists 0 < δ <1 such that 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) QA 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (0, δ ). √ √ √ 3 3 Case 3 p = 3[(1 + 2) /e – 2]/(4 – 3 2). Then (3.13)leads to F 2 = –0.0039 ··· < 0. (3.16) Let λ = 1.0808 . . . be the number given in Lemma 2.5(2). Then we divide the discussion into two subcases. Subcase 1 x ∈ (1, λ ]. Then F (x)>0 for x ∈ (1, λ ] follows easily from (3.13)and (3.14) 0 1 0 together with Lemma 2.5(2). Subcase 2 x ∈ (λ , 2). Then Lemma 2.5(2) and (3.14)leadtothe conclusion that F (x)is 0 1 strictly decreasing on the interval [λ , 2). Then, from (3.16)and Subcase1,weknowthat √ √ 6 6 there exists λ ∈ (λ , 2) such that F (x)>0 for x ∈ [λ , λ )and F (x)<0 for x ∈ (λ , 2). 1 0 1 0 1 1 1 It follows from Subcases 1 and 2 together with (3.12)that F(x) is strictly increasing on (1, λ ] and strictly decreasing on [λ , 2). Therefore, 1 1 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) QA 3 3 follows from (3.9)–(3.11)and (3.16) together with the piecewise monotonicity of F(x). √ √ √ 3 3 Case 4 3[(1 + 2) /e – 2]/(4 – 3 2) < p <1. Then (3.11)leads to √ √ √ √ 6 3 F 2 = 2 log(1 + 2) – log p + 2(1 – p) – 1 < 0. (3.17) Equations (3.9)and (3.10) together with inequality (3.17) imply that there exists 0 < δ < 1such that 1 2 1/3 2/3 R (a, b)< p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) QA 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (1 – δ ,1). Theorem 3.4 Thedoubleinequality 1 5 1/6 5/6 α C(a, b)+ A(a, b) +(1– α )C (a, b)A (a, b) 4 4 6 6 1 5 1/6 5/6 < R (a, b)< β C(a, b)+ A(a, b) +(1– β )C (a, b)A (a, b) AQ 4 4 6 6 √ √ √ 6 6 (π/4–1) holds for all a, b >0 with a = bif and only if α ≤ 6[ 2e – 2]/(7 – 6 2) = 0.4210 . . . and β ≥ 12/25. 4 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 10 of 13 Proof Since R (a, b), A(a, b), and C(a, b) are symmetric and homogenous of degree one, AQ without loss generality, we assume that a > b >0. Let v =(a – b)/(a + b), x = 1+ v ,and p ∈ (0, 1). Then v ∈ (0, 1), x ∈ (1, 2) and (1.1), (1.3), and (1.5)leadto R (a, b) AQ log 1 5 1/6 5/6 p[ C(a, b)+ A(a, b)] + (1 – p)C (a, b)A (a, b) 6 6 √ √ arctan(v) 1 2 2 = log 1+ v + – log p v +1 +(1– p) 1+ v –1 v 6 arctan( x –1) 1 5 =3 log(x)+ √ – log p x + +(1– p)x – 1. (3.18) 6 6 x –1 Let arctan( x –1) 1 5 G(x)=3 log(x)+ √ – log p x + +(1– p)x – 1. (3.19) 6 6 x –1 Then simple computations lead to √ √ √ π 7 6 6 G 1 =0, G 2 = log( 2) + – log p + 2(1 – p) – 1, (3.20) 4 6 3x G (x)= G (x), (3.21) 3/2 (x –1) where 6 11 6 5 x –1[–px +4(1 – p)x +7px +2(1 – p)] G (x)= – arctan x –1 , 5 6 x [p(x +5)+6(1– p)x] 5[ 2(1 – p)+ p] π G (1) = 0, G 2 = √ – , (3.22) 1 1 6 2(1 – p)+7p 3/2 (x –1) G (x)=– g(x), (3.23) 1 2 6 6 x [p(x +5)+6(1– p)x] where g(x) is deﬁned as in Lemma 2.6. We divide the proof into four cases. Case 1 p = 12/25. Then it follows from (3.18)–(3.23) and Lemma 2.6(1) that 12 1 2 13 1/3 2/3 R (a, b)< C(a, b)+ A(a, b) + C (a, b)A (a, b). AQ 25 3 3 25 Case 2 0< p < 12/25. Let v >0 and v → 0 ,thenpower series expansionleads to √ √ arctan(v) 1 2 2 log 1+ v + – log p v +1 +(1– p) 1+ v –1 v 6 1 5 4 6 = – p v + O v . (3.24) 30 72 Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 11 of 13 Equations (3.18), (3.19), and (3.24) lead to the conclusion that there exists 0 < δ <1 such that 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) AQ 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (0, δ ). √ √ √ 6 6 (π/4–1) Case 3 p =6[ 2e – 2]/(7 – 6 2). Then, from (3.20)and (3.22)togetherwithnu- merical computations, we get √ √ 6 6 G 2 =0, G 2 = –0.0033 ··· < 0. (3.25) Let μ = 1.0577 . . . be the number given in Lemma 2.6(2). Then we divide the discussion into two subcases. Subcase 1 x ∈ (1, μ ]. Then G (x)>0 for x ∈ (1, μ ] follows easily from (3.22)and (3.23) 0 1 0 together with Lemma 2.6(2). Subcase 2 x ∈ (μ , 2). Then Lemma 2.6(2) and (3.23)leadtothe conclusion that G (x) 0 1 is strictly decreasing on the interval [μ , 2). Then, from (3.25) and Subcase 1, we know that there exists μ ∈ (μ , 2) such that G (x)>0 for x ∈ [μ , μ )and G (x)<0 for x ∈ 1 0 1 0 1 1 (μ , 2). It follows from Subcases 1 and 2 together with (3.21)that G(x) is strictly increasing on (1, μ ] and strictly decreasing on [μ , 2). Therefore, 1 1 1 2 1/3 2/3 R (a, b)> p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) AQ 3 3 follows from (3.18)–(3.20)and (3.25) together with the piecewise monotonicity of G(x). √ √ √ 6 6 (π/4–1) Case 4 6[ 2e – 2]/(7 – 6 2) < p <1. Then (3.21)leads to √ √ √ π 7 6 6 G 2 = log( 2) + – log p + 2(1 – p) – 1 < 0. (3.26) 4 6 Equations (3.18)and (3.19) together with inequality (3.26) imply that there exists 0 < δ <1 such that 1 2 1/3 2/3 R (a, b)< p C(a, b)+ A(a, b) +(1– p)C (a, b)A (a, b) AQ 3 3 for all a > b >0 with (a – b)/(a + b) ∈ (1 – δ ,1). 4 Results and discussion In this paper, we provide the optimal upper and lower bounds for the Sándor–Yang means R (a, b)and R (a, b) in terms of combinations of the arithmetic mean A(a, b)and the QA AQ contra-harmonic mean C(a, b). Our approach may have further applications in the theory of bivariate means. 5Conclusion In the article, we ﬁnd several best possible bounds for the Sándor–Yang means R (a, b) QA and R (a, b). These results are improvements and reﬁnements of the previous results. AQ Xu et al. Journal of Inequalities and Applications (2018) 2018:127 Page 12 of 13 Funding The research was supported by the Natural Science Foundation of China (Grants Nos. 61673169, 61374086, 11371125, 11401191), the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11626101) and the Natural Science Foundation of the Department of Education of Zhejiang Province (Grant No. Y201635325). Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. All authors read and approved the ﬁnal manuscript. 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Published: May 30, 2018