Optimal Bounds for Neuman-Sándor Mean in Terms of the Convex Combination of Geometric and Quadratic Means

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1 Mathematica Aeterna, Vol. 5, 015, no. 1, Optimal Bounds for Neuman-Sándor Mean in Terms of the Convex Combination of Geometric Quadratic Means Liu Chunrong1 College of Mathematics Information Science, Hebei University, Baoding 07100, P. R. China, Wang Jing College of Mathematics Information Science, Hebei University, Baoding 07100, P. R. China, Abstract In this paper, we present the least value α the greatest value β such that the double inequality αg(a,b)+(1 α)q(a,b) < M(a,b) < βg(a,b)+(1 β)q(a,b) holds for all a,b > 0 with a b, where G(a,b), M(a,b) Q(a,b) are respectively the geometric, Neuman-Sándor quadratic means of a b. Mathematics Subject Classification: 6D15 Keywords: Inequality, Neuman-Sándor mean, geometric mean, quadratic mean 1 Introduction For a,b > 0 with a b the Neuman-Sándor mean M(a,b)[1 was defined by M(a,b) = a b sinh 1 ( a b, a+b ) (1.1) where sinh 1 (x) = log(x+ 1+x ) is the inverse hyperbolic sine function.

2 84 Liu Chunrong1 Wang Jing Recently, the Neuman-Sándor mean has been the subject of intensive research. In particular, many remarkable inequalities for the Neuman-Sándor mean M(a,b) can be found in the literature [1,. Let H(a,b) = (ab)/(a + b), G(a,b) = ab, L(a,b) = (a b)/(loga logb), P(a,b) = (a b)/(4arctan a/b π), A(a,b) = (a + b)/, T(a,b) = (a b)/[arctan(a b)/(a+b), Q(a,b) = (a +b )/, C(a,b) = (a + b )/(a+b) be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, contra-harmonic mean of a b, respectively. Then min{a,b} < H(a,b) < G(a,b) < L(a,b) < P(a,b) < A(a,b) < M(a,b) < T(a,b) < Q(a,b) < C(a,b) < max{a,b} hold for all a,b > 0 with a b. In [3, Neuman proved that the double inequalities (1.) αq(a,b)+(1 α)a(a,b) < M(a,b) < βq(a,b)+(1 β)a(a,b) (1.3) λc(a,b)+(1 λ)a(a,b) < M(a,b) < µc(a,b)+(1 µ)a(a,b) (1.4) hold for all a,b > 0 with a b if only if α [1 log(1 + )/[( 1)log(1 + ) = 0.349,β 1/3,λ [1 log(1 + )/log(1 + ) = µ 1/6. In [4, Li etc showed that the double inequality L p0 (a,b) < M(a,b) < L (a,b) (1.5) holds for all a,b > 0 with a b, where L p (a,b) = [(a p+1 b p+1 )/((p+1)(a b)) 1/p (p 1,0), L 0 (a,b) = 1/e(a a /b b ) 1/(a b) L 1 (a,b) = (a b)/(loga logb) is the p-th generalized logarithmic mean of a b, p 0 = is the unique solution of the equation (p+1) 1/p = log(1+ ). In[5, Chu etc proved that the double inequalities α 1 L(a,b)+(1 α 1 )Q(a,b) < M(a,b) < β 1 L(a,b)+(1 β 1 )Q(a,b) (1.6) α L(a,b)+(1 α )C(a,b) < M(a,b) < β L(a,b)+(1 β )C(a,b) (1.7) hold for all a,b > 0 with a b if only if α 1 /5,β 1 1 1/[ log(1+ ) = ,α 5/8 β 1 1/[log(1+ ) = The main purpose of this paper is to find the least value α the greatest value β such that the double inequality αg(a,b)+(1 α)q(a,b) < M(a,b) < βg(a,b)+(1 β)q(a,b) holds for all a,b > 0 with a b.

3 Optimal Bounds for Neuman-Sándor mean 85 Lemmas In order to establish our main result we need several lemmas, which we present in this section. Lemma.1 Let f(x) = 1/ 1+x, g(x) = 1/ 1 x, h(x) = 1 x 4, k(x) = 1/ 1 x 4. Then the inequalities 1 x < f(x) < 1 x x4, (.1) g(x) > 1+ x, (.) hold for all x (0,1). h(x) < 1 x4, (.3) k(x) > 1+ x4 (.4) Proof. The first inequality in (.1) is known (see [5, lemma.1). The second inequality in (.1) the inequalities (.), (.3) follow in turn from the inequalities ( 1 x + 3 ) 8 x4 f (x) = x 6 64(1+x ) [9x (x +1) +4(10 6x ) > 0, (.5) g (x) ( ( 1+ x 1 x4 ) = x 4 4(1 x ) (x +3) > 0, (.6) ) h (x) = x8 4 > 0 (.7) for all x (0,1). Making use of (.) with x replaced by x the inequality (.4) is obtained. Lemma. (see [5, lemma.4) Let ϕ(x) = x/[ 1+x ( sinh 1 (x) ) 1/sinh 1 (x). Then the inequality holds for all x (0,1). ϕ(x) < x x3 (.8)

4 86 Liu Chunrong1 Wang Jing Lemma.3 Let ψ(x) = log(x+ 1+x ). Then the double inequality holds for all x (0,1). x x3 6 < ψ(x) < x (.9) Proof. Let ψ 1 (x) = ψ(x) (x x3 6 ) (.10) Then simple computations lead to ψ 1(x) = x 0 +ψ 1(x) = 0, (.11) 1 1+x x (1 ). (.1) Making use of the first inequality in (.1) for (.1) cause the conclusion that ψ 1 (x) > 1 1+x 1 1+x = 0. (.13) for x (0,1). Therefore, the first inequality in (.9) follows from (.10) (.11) together with (.13). Let ψ (x) = x ψ(x). Then from x 0 +ψ (x) = 0 ψ (x) = 1 1/ 1+x > 0 the second inequality in (.9) is obtained. Lemma.4 Let λ = 1 1/ [ log(1+ ) = F(x) = 4λx 16 +(37λ )x 14 +(41 36λ)x 1 +(19λ 65)x 10 +4(46 15λ)x 8 (170λ+9)x 6 (10λ+161)x 4 +16(1 37λ)x +16(5λ 4). (.14) Then the inequality holds for all x (0,1/. F(x) < 0 (.15) Proof. Making use of the transform x = 1/t (t [4,+ )) for F(x) we get F(x) = t 8 F 1 (t), (.16) where F 1 (t) = 16(5λ 4)t 8 +16(1 37λ)t 7 (10λ+161)t 6 (170λ+9)t 5 +4(46 15λ)t 4 +(19λ 65)t 3 +(41 36λ)t +(37λ )t+4λ. (.17)

5 Optimal Bounds for Neuman-Sándor mean 87 Simple calculations of derivative yield F 1 (t) = [64(5λ 4)t7 +56(1 37λ)t 6 3(10λ+161)t 5 5(170λ+9)t 4 +8(46 15λ)t 3 +3(19λ 65)t +(41 36λ)t+(37λ ), F 1 (t) = [448(5λ 4)t6 +336(1 37λ)t 5 15(10λ+161)t 4 0(170λ+9)t 3 +4(46 15λ)t +6(19λ 65)t +(41 36λ), F 1 (t) = 1[448(5λ 4)t 5 +80(1 37λ)t 4 10(10λ+161)t 3 10(170λ+9)t +8(46 15λ)t+(19λ 65), F (4) 1 (t) = 4[110(5λ 4)t (1 37λ)t 3 15(10λ +161)t 10(170λ+9)t+4(46 15λ), F (5) 1 (t) = 40[448(5λ 4)t (1 37λ)t 3(10λ+161)t (170λ+9), (.18) (.19) (.0) (.1) (.) F (6) 1 (t) = 70[448(5λ 4)t +11(1 37λ)t (10λ+161) (.3) F (7) 1 (t) = 80640[8(5λ 4)t+(1 37λ). (.4) Noticing that 0 < λ < 1/5, from (.17)-(.4) we have F 1 (4) = 4[( λ+43000) < 0, (.5) F 1 (4) = [( λ ) < 0, (.6) F 1(4) = [(478668λ ) < 0, (.7) F 1 (4) = 1[(46671λ ) < 0, (.8) F (4) 1 (4) = 96[17855λ < 0, (.9) F (5) 1 (4) = 70[14098λ 8131 < 0, (.30) F (6) 1 (4) = 70[19144λ 3457 < 0, (.31) F (7) 1 (t) < (t 1) < 0 (.3) for all t [4,+ ). From(.3) we clearly see that F (6) 1 (t) is strictly decreasing in [4, + ). Therefore, the conclusion of lemma.4 follows easily from (.5)-(.31) (.16) together with the monotonicity of F (6) 1 (t).

6 88 Liu Chunrong1 Wang Jing Lemma.5 Let λ = 1 1/ [ log(1+ ) = H(x) = 8λx 14 +1(14λ 3)x 1 +15(11 1λ)x 10 +(809λ 403)x 8 +( λ)x 6 +3(361λ 71)x 4 +4(17 335λ)x +3(5λ 4). (.33) Then the inequality holds for all x (1/,1). H(x) < 0 (.34) Proof. Let x = t (t (1/4,1)). Then H(x) = 8λt 7 +1(14λ 3)t 6 +15(11 1λ)t 5 +(809λ 403)t 4 +( λ)t 3 +3(361λ 71)t +4(17 335λ)t +3(5λ 4) = H 1 (t). (.35) Simple calculations of derivative yield H 1(t) = 56λt 6 +7(14λ 3)t 5 +75(11 1λ)t 4 +4(809λ 403)t 3 +3( λ)t +6(361λ 71)t+4(17 335λ), (.36) H 1 (t) = 6[56λt5 +60(14λ 3)t 4 +50(11 1λ)t 3 +(809λ 403)t +( λ)t+(361λ 71), H 1 (t) = 6[80λt4 +40(14λ 3)t (11 1λ)t +4(809λ 403)t+( λ). (.37) (.38) Whereafter, making use of the transform t = 1/u (u (1,4)) for H 1 (t) one has H 1 (t) = 6u 4 H (u), (.39) where H (u) = ( λ)u 4 +4(809λ 403)u (11 1λ)u +40(14λ 3)u+80λ. Again calculations of derivative result in (.40) H (u) = 4[( λ)u3 +3(809λ 403)u +75(11 1λ)u+60(14λ 3), (.41) H (u) = 1[( λ)u +(809λ 403)u+5(11 1λ), (.4) H (u) = 4[( λ)u+(809λ 403). (.43)

7 Optimal Bounds for Neuman-Sándor mean 89 Noticing that 49/50 < λ < 1/5, from (.35) (.37) (.40) (.43) one has H 1(t) = 45 (6013λ+190) < 0, t t 1 H 1(t) = 100λ < 0, (.44) 4 H t (t) = 1 ( λ) < 0, 51 4 (.45) t 1 H 1 (t) = 1768λ < 0, H t (t) = 3 (743λ 1750) < 0, 64 t 1 H 1 (t) = 31λ > 0, (.46) 4 u 1 +H (u) = 1953λ+5 > 0, (.47) u 1 +H (u) = 4(143 81λ) > 0, (.48) u 1 +H (u) = 96( 85λ) > 0, (.49) H (u) > 4[351u 47 > 0 (.50) for all u (1,4). From (.50) we clearly see that H (u) is strictly increasing in (1,4). Thus H (u) > 0 for u (1,4) follows from (.47)-(.49) the monotonicity of H (u). From (.39) H (u) > 0 we know that H 1 (t) > 0 for t (1/4,1), hence H 1 (t) is strictly increasing in (1/4,1). It follows from (.46) together withthemonotonicityofh 1 (t)thatthereexistst 0 (1/4,1)suchthatH 1 (t) < 0 for t (1/4,t 0 ) H 1(t) > 0 for t (t 0,1), thus H 1(t) is strictly decreasing in (1/4,t 0 ) strictly increasing in [t 0,1). From (.45) the monotonicity of H 1 (t) we affirm H 1 (t) < 0 for t (1/4,1), so that H 1(t) is strictly decreasing in (1/4,1). Therefore, the inequality H(x) < 0 follows from (.44) (.35) together with the monotonicity of H 1 (t). 3 Main Results Theorem 3.1 The double inequality αg(a,b)+(1 α)q(a,b) < M(a,b) < βg(a,b)+(1 β)q(a,b) (1) holds for all a,b > 0 with a b if only if α 1/3 β 1 1/ [ log(1+ ) =

8 90 Liu Chunrong1 Wang Jing Proof. Without loss of generality, we assume that a > b > 0. Let x = (a b)/(a+b) (0,1) λ = 1 1/ [ log(1+ ) = Then G(a, b) A(a,b) = 1 x, M(a,b) A(a,b) = Firstly, we prove that Equations (3.1) lead to x sinh 1 (x), Q(a,b) A(a,b) = 1+x. (3.1) 1 3 G(a,b)+ Q(a,b) < M(a,b). (3.) 3 G(a, b) 3A(a,b) + Q(a,b) 3A(a,b) M(a,b) A(a, b) = 1 1 x + x 1+x 3 3 sinh 1 (x) = d(x) (3.3) Simple computations yield x 0 +d(x) = 0, (3.4) d x (x) = 3 1 x + x 3 1+x 1 sinh 1 (x) x + [ 1+x sinh 1 (x) = 3 xf(x) 1 3 xg(x)+ϕ(x), (3.5) where f(x),g(x) ϕ(x) are defined as in Lemma.1 Lemma., respectively. From (3.5), lemma.1 lemma. one has d (x) < x x( x4 ) 1 x x(1+ 3 )+( x x3 ) = x3 x (1 60 ) < 0 (3.6) for all x (0, 1). Therefore, inequality (3.) follows from (3.3) (3.4) together with (3.6). Secondly, we prove that λg(a,b)+(1 λ)q(a,b) > M(a,b). (3.7)

9 Optimal Bounds for Neuman-Sándor mean 91 Equations (3.1) lead to λg(a, b) A(a,b) + (1 λ)q(a,b) M(a,b) A(a, b) A(a, b) = λ 1 x +(1 λ) x 1+x log(x+ 1+x ) D(x) = log(x+ 1+x ), (3.8) where D(x) = [ λ 1 x +(1 λ) 1+x log(x+ 1+x ) x. (3.9) Some tedious, but not difficult, calculations lead to D (x) = x x 0 +D(x) = 0, D(x) = 0, (3.10) ( 1 λ 1+x x 1 ) λ log(x+ 1+x ) 1 x + λ(1 x ) 1 x 4 λ, (3.11) D (x) = [ (x) = 0, (x) =, (3.1) x 0 +D x 1 D 1 λ (1+x ) 3/ λ log(x+ 1+x (1 x ) ) 3/ λx(3+x ) (1+x ) 1 x 4 + (1 λ)x 1+x, (3.13) D ( 1 3 ) = 7 > 7 (x) = 0, (x) =, (3.14) x 0 +D x 1 D ( 1 λ 5 10 λ ) log( ) λ 10 1λ log( ) (3.15) 5 = > 0, [ D 1 λ (x) = 3x (1+x ) λ log(x+ 1+x 5/ (1 x ) ) 5/ λ(x6 +8x 4 x +4) + (1 λ)( x ), (1+x )(1 x 4 ) 3/ (1+x ) (3.16)

10 9 Liu Chunrong1 Wang Jing [ (1 λ)(4x D (4) 1) (x) = 3 λ(4x +1) log(x+ 1+x (1+x ) 7/ (1 x ) ) 7/ λx(x8 +33x 6 7x 4 +75x 7) (1+x )(1 x 4 ) 5/ + (1 λ)x(x 13). (1+x ) 3 (3.17) In order to discuss D (4) (x) is positive or negative, we divide the range of variable x into two intervals (0,1/ (1/,1). For x (0,1/, (3.17) is rewritten into [ (1 λ)(4x D (4) 1) (x) = 3 f(x) λ(4x +1) (1+x ) 3 (1 x ) g(x) ψ(x) 3 + 7λx(1+x4 ) λx3 (x 6 +33x 4 +75) (1+x )(1 x 4 ) 3h(x) k(x) (1+x )(1 x 4 ) + (1 λ)x(x 13), (1+x ) 3 (3.18) where f(x),g(x),ψ(x),h(x) k(x) are defined as in Lemma.1.3, respectively. From (3.18), lemma.1 lemma.3 one has D (4) (x) < 3 = [ (1 λ)(4x 1) (1+x ) 3 (1 x ) λ(4x +1) (x x3 (1 x ) 3 6 ) λx3 (x 6 +33x 4 +75) (1+ x4 (1+x )(1 x 4 ) )+ 7λx(1+x4 ) (1+x )(1 x 4 ) 3 (1 x4 )+ (1 λ)x(x 13) (1+x ) x 3 4(1+x ) 4 (1 x ) 3F(x), (3.19) where F(x) is defined as in lemma.4. It fllows from (3.19) lemma.4 that D (4) (x) < 0. (3.0) For x (1/,1), (3.17) is rewritten into D (4) (x) = 3 [ (1 λ)(4x 1) f(x)ψ(x) λ(4x +1) (1+x ) 3 (1 x ) g(x)ψ(x) 3 λx(x8 +33x 6 7x 4 +75x 7) k(x) (1+x )(1 x 4 ) + (1 λ)x(x 13), (1+x ) 3 (3.1)

11 Optimal Bounds for Neuman-Sándor mean 93 where f(x),g(x),h(x) ψ(x) are defined as in lemma.1.3, respectively. From (3.1), lemma.1 lemma.3 together with the fact that x 8 +33x 6 7x 4 +75x 7 > (0) 8 +33(0) (1/) 7 = 19/4 > 0, one has D (4) (x) < 3 [ (1 λ)(4x 1) (1 x (1+x ) 3 + 3x4 8 )x λ(4x +1) (1 x ) (1+ 3 x x3 )(x 6 ) λx(x8 +33x 6 7x 4 +75x 7) (1+x )(1 x 4 ) (1+ x4 )+ (1 λ)x(x 13) (1+x ) x 3 = 8(1 x 4 ) 3H(x), (3.) where H(x) is defined as in Lemma.5. It follows from (3.) Lemma.5 that D (4) (x) < 0. (3.3) Synthesizing the above two cases we affirm that D (4) (x) < 0 for all x (0,1), hence the function D (x) is concave in (0,1). It follows from (3.14) (3.15) together with the concavity of D (x) that there exists x 0 (0,1) such that D (x) > 0 for x (0,x 0 ) D (x) < 0 for x (x 0,1), hence D (x) is strictly increasing in (0,x 0 ) strictly decreasing in [x 0,1). From (3.1) together with the monotonicity of D (x) we know that there exists x 1 (x 0,1) such that D (x) > 0 for x (0,x 1 ) D (x) < 0 for x (x 1,1), so that D(x) is strictly increasing in (0,x 1 ) strictly decreasing in [x 1,1). It follows from (3.10) together with the monotonicity of D(x) that D(x) > 0 (3.4) for all x (0,1). Therefore, the inequality (3.7) follows from (3.8) (3.4). At least, we prove that 1/3G(a,b) + /3Q(a,b) is the best possible lower convex combination bound λg(a,b) +(1 λ)q(a,b) is the best possible upper convex combination bound of the geometric quadratic means for the Neuman-Sándor mean. From equations (3.1) one has Q(a,b) M(a,b) Q(a,b) G(a,b) 1+x log(x+ 1+x = ) x ( 1+x 1 x )log(x+ 1+x ) = B(x). (3.5) It is easy to calculate that x 0 +B(x) = 1 3, (3.6)

12 94 Liu Chunrong1 Wang Jing x 1 B(x) = λ. (3.7) If α < 1/3, then equations (3.5) (3.6) lead to conclusion that there exists δ 1 = δ 1 (α) (0,1) such that M(a,b) < αg(a,b) + (1 α)q(a,b) for (a b)/(a+b) (0,δ 1 ). If β > λ, then equations (3.5) (3.7) imply the conclusion that there exists δ = δ (β) (0,1) such that M(a,b) > βg(a,b) + (1 β)q(a,b) for (a b)/(a+b) (1 δ,1). References [1 E. Neuman J. Sándor, On the Schwab-Borchardt mean, Math. Pannon. 14, (003), [ E. Neuman J. Sándor, On the Schwab-Borchardt mean II, Math. Pannon. 17, 1 (006), [3 E. Neuman, A note on a certain bivariate mean, J. Math. Inequal. 6, 4 (01), [4 Y.-M. Li, B.-Y. Long Y.-M. Chu, Sharp bounds for the Neuman- Sándor mean in terms of generalized logarithmic mean, J. Math. Inequal. 6, 4 (01), [5 Y.-M. Chu, T.-H. Zhao B.-Y. Liu, Optimal bound for Neuman-Sándor mean in terms of the convex combination of logarithmic quadratic or contra-harmonic means, J. Math. Inequal. 8, (014), Received: January, 015