Bounds of the Neuman-Sándor Mean Using Power and Identric Means

and Applied Analysis 3 It is the aim of this paper to find the best possible lower power mean bound for the Neuman-Sándor mean M(a, b) and to present the sharp constants α and β such that the double inequality α < M(a, b) I (a, b) < β (17) holds for all a, b > 0 with a ̸ = b. 2. Main Results Theorem 1. p0 = (log 2)/ log [2 log(1 + √2)] = 1.224 . . . is the greatest value such that the inequality M(a, b) > Mp 0 (a, b) (18) holds for all a, b > 0 with a ̸ = b. Proof. From (1) and (2), we clearly see that both M(a, b) andMp(a, b) are symmetric and homogenous of degree one. Without loss of generality, we assume that b = 1 and a = x > 1. Let p0 = (log 2)/ log [2 log(1 + √2)], then from (1) and (2) one has logM(x, 1) − logMp 0 (x, 1) = log x − 1 2sinh ((x − 1) / (x + 1)) − 1 p0 log x p 0 + 1 2 . (19) Let f (x) = log x − 1 2sinh ((x − 1) / (x + 1)) − 1 p0 log x p 0 + 1 2 . (20) Then, simple computations lead to lim x→1 f (x) = 0, (21) lim x→+∞ f (x) = 1 p0 log 2 − log [2sinh (1)] = 0, (22) f 󸀠 (x) = (1 + x p 0 −1 ) f1 (x) (x − 1) (x0 + 1) sinh ((x − 1) / (x + 1)) , (23) where f1 (x) = − √2 (x − 1) (x p 0 + 1) (x + 1) (x0 + 1)√1 + x + sinh ( − 1 x + 1 ) , f1 (1) = 0, (24) lim x→+∞ f1 (x) = − √2 + sinh (1) = −0.5328 ⋅ ⋅ ⋅ < 0, (25) f 󸀠 1 (x) = √2 (x − 1) f2 (x) (x + 1) 2 (x0 + 1) 2 (1 + x) 3/2 , (26) where f2 (x) = 1 + x + 2x 2 + (p0 − 1) x p 0 −2 − x p 0 −1 + x p 0 +1 − (p0 − 1) x p 0 +2 − 2x 2p 0 −2 − x 2p 0 −1 − x 2p 0 , f2 (1) = 0, (27) lim x→+∞ f2 (x) = −∞, (28) f 󸀠 2 (x) = 1 + 4x + (p0 − 1) (p0 − 2) x p 0 −3 − (p0 − 1) x p 0 −2 + (p0 + 1) x p 0 − (p0 − 1) (p0 + 2) x p 0 +1 − 4 (p0−1) x 2p 0 −3 −(2p0−1) x 2p 0 −2 −2p0x 2p 0 −1 , f 󸀠 2 (1) = 4 (4 − 3p0) > 0, (29) lim x→+∞ f 󸀠 2 (x) = −∞, (30) f 󸀠󸀠 2 (x) = 4 + (p0 − 1) (p0 − 2) (p0 − 3) x p 0 −4 − (p0 − 1) (p0 − 2) x p 0 −3 + p0 (p0 + 1) x p 0 −1 − (p0 − 1) (p0 + 2) (p0 + 1) x p 0 − 4 (p0 − 1) (2p0 − 3) x 2p 0 −4 − 2 (2p0 − 1) (p0 − 1) x 2p 0 −3 − 2p0 (2p0 − 1) x 2p 0 −2 , f 󸀠󸀠 2 (1) = 4 (2p0 − 1) (4 − 3p0) > 0, (31) lim x→+∞ f 󸀠󸀠 2 (x) = −∞, (32) f 󸀠󸀠󸀠 2 (x) = (p0 − 1) x p 0 −5 f3 (x) , (33) where f3 (x) = − (2 − p0) (3 − p0) (4 − p0) − (2 − p0) (3 − p0) x + p0 (p0 + 1) x 3 − p0 (p0 + 1) (p0 + 2) x 4 − 8 (3−2p0) (2−p0) x p 0+2 (2p0−1) (3−2p0) x p 0 +1 − 4p0 (2p0 − 1) x p 0 +2 < − (2 − p0) (3 − p0) (4 − p0) − (2 − p0) (3 − p0) x + p0 (p0 + 1) x 4 − p0 (p0 + 1) (p0 + 2) x 4 − 8 (3 − 2p0) (2 − p0) x p 0 + 2 (2p0 − 1) (3 − 2p0) x p 0 +2 − 4p0 (2p0 − 1) x p 0 +2 = − (2 − p0) (3 − p0) (4 − p0) − (2 − p0) (3 − p0) x − p0(p0 + 1) 2 x 4 − 8 (3 − 2p0) (2 − p0) x p 0 − 2 (2p0 − 1) (4p0 − 3) x p 0 +2 < 0, (34) for x > 1. 4 Abstract and Applied Analysis Equation (33) and inequality (34) imply that f 2 (x) is strictly decreasing on [1, +∞). Then, the inequality (31) and (32) lead to the conclusion that there exists x1 > 1, such that f 󸀠 2 (x) is strictly increasing on [1, x1] and strictly decreasing on [x1, +∞). From (29) and (30) together with the piecewise monotonicity of f 2 (x), we clearly see that there exists x2 > x1 > 1, such that f2(x) is strictly increasing on [1, x2] and strictly decreasing on [x2, +∞). It follows from (26)–(28) and the piecewise monotonicity of f2(x) that there exists x3 > x2 > 1, such that f1(x), is strictly increasing on [1, x3] and strictly decreasing on [x3, +∞). From (23)–(25) and the piecewise monotonicity of f1(x) we see that there exists x4 > x3 > 1, such that f(x) is strictly increasing on (1, x4] and strictly decreasing on [x4, +∞). Therefore, M(x, 1) > Mp 0 (x, 1) for x > 1 follows easily from (19)–(22) and the piecewise monotonicity of f(x). Next, we prove that p0 = (log 2)/ log [2 log(1 + √2)] = 1.224 . . . is the greatest value such that M(x, 1) > Mp 0 (x, 1) for all x > 1. For any ε > 0 and x > 1, from (1) and (2), one has lim x→+∞ Mp 0 +ε (x, 1) M (x, 1) = lim x→+∞ [( 1 + x p 0 +ε 2 ) 1/(p 0 +ε) 2sinh ((x − 1) / (x + 1)) x − 1 ] = 2 −1/(p 0 +ε) × 2sinh (1) = 2 ε/p 0 (p 0 +ε) > 1. (35) Inequality (35) implies that for any ε > 0, there existsX = X(ε) > 1, such that M(x, 1) < Mp 0 +ε(x, 1) for x ∈ (X, +∞). Remark 2. 4/3 is the least value such that inequality (16) holds for all a, b > 0 with a ̸ = b, namely, M4/3(a, b) is the best possible upper power mean bound for the Neuman-Sándor meanM(a, b). In fact, for any ε ∈ (0, 4/3) and x > 0, one has M4/3−ε (1 + x, 1) − M (1 + x, 1) = [ (1 + x) 4/3−ε + 1 2 ] 1/(4/3−ε) − x 2sinh (x/ (2 + x)) . (36) Letting x → 0 and making use of Taylor expansion, we get [ (1 + x) 4/3−ε + 1 2 ] 1/(4/3−ε) − x 2sinh (x/ (2 + x)) = [1 + 4 − 3ε 6 x + (4 − 3ε) (1 − 3ε) 36 x 2 + o (x 2 )] 1/(4/3−ε) − x x − (1/2) x + (5/24) x + o (x) = [1 + 1 2 x + 1 − 3ε 24 x 2 + o (x 2 )] − [1 + 1 2 x + 1 24 x 2 + o (x 2 )] = − ε 8 x 2 + o (x 2 ) .