Optimal Lower Power Mean Bound for the Convex Combination of Harmonic and Logarithmic Means

and Applied Analysis 3 Alzer and Qiu 27 found the sharp bound of 1/2 L a, b I a, b in terms of the power mean as follows: Mc a, b < 1 2 L a, b I a, b 1.8 for all a, b > 0 with a/ b, with the best possible parameter c log 2/ 1 log 2 . The main purpose of this paper is to find the least value λ ∈ 0, 1 and the greatest value p p α such that αH a, b 1 − α L a, b > Mp a, b for α ∈ λ, 1 and all a, b > 0 with a/ b. 2. Lemmas In order to establish our main result we need three lemmas, which we present in this section. Lemma 2.1. Let α ∈ 1/4, 1 , p 1−4α /3 ∈ −1, 0 , and f t −4αp p 1 2 p 2 tp−1 2 1− α p2 1 − p2 tp−2 2 1 − α p 1 − p 2 2 − p tp−3 12 1 − α 1 − p . Then f t > 0 for t ∈ 1, ∞ . Proof. Simple computations lead to f 1 64 81 1 − α 2 ( 56α2 23α 11 ) > 0, 2.1 lim t→ ∞ f t 12 1 − α 1 − p 8 1 − α 1 2α > 0, 2.2 f ′ t −2p1 − p)tp−4f1 t , 2.3 where f1 t −2α ( p 1 )2( p 2 ) t2 1 − α pp 12 − pt 1 − α 1 − p2 − p3 − p, f1 1 4 27 1 − α ( 148α2 − 11α 25 ) > 0, 2.4 lim t→ ∞ f1 t −∞, 2.5 f 1 t −4α ( p 1 )2( p 2 ) t 1 − α pp 12 − p − 4 27 1 − α 2 16α 7 − 4α t 4α − 1 4α 5 < 0 2.6 for t ∈ 1, ∞ . Inequality 2.6 implies that f1 t is strictly decreasing in 1, ∞ , then from 2.4 and 2.5 we know that λ1 > 1 exists such that f1 t > 0 for t ∈ 1, λ1 and f1 t < 0 for t ∈ λ1, ∞ . Hence, equation 2.3 leads to the conclusion that f t is strictly increasing in 1, λ1 and strictly decreasing in λ1, ∞ . Therefore, Lemma 2.1 follows from 2.1 and 2.2 together with the piecewise monotonicity of f t . 4 Abstract and Applied Analysis Lemma 2.2. Let α ∈ 1/4, 1 , p 1 − 4α /3 ∈ −1, 0 , and g t − 1 − α p 1 p 2 2 p 3 t p 1 p3 − αp3 − 19αp2 3p2 − 34αp 2p − 8α tp−1 1 − α p p3 − 8p2 − p 4 tp−2 1 − α 1 − p p3 5p2 − 14p 4 tp−3 4 1 − α 7 − 4p − 4p 1 − α t−1 4α 1 p t−2, then g t > 0 for t ∈ 1, ∞ . Proof. Let g1 t − 1 − α p 1 p 2 2 p 3 t3 p 1 p3 − αp3 − 19αp2 3p2 − 34αp 2p − 8α t2 1 − α p p3 − 8p2 − p 4 t 1 − α 1 − p p3 5p2 − 14p 4 4 1 − α 7 − 4p t3−p − 4p 1 − α t2−p 4α 1 p t1−p. Then simple computations lead to g t tp−3g1 t , 2.7 g1 1 16 27 1 − α ( 80α2 110α − 1 ) > 0, 2.8 g ′ 1 t −3 1 − α p 1 p 2 2 p 3 t2 2 p 1 p3 − αp3 − 19αp2 3p2 − 34αp 2p − 8α t 1 − α p p3 − 8p2 − p 4 4 1 − α 7 − 4p 3 − p t2−p − 4p 1 − α 2 − p t1−p 4α 1 − p2 t−p, g1 1 32 27 1 − α ( −16α3 38α2 176α − 9 ) > 0, 2.9 g ′′ 1 t −6 1−α p 1 p 2 2 p 3 t 2 p 1 p3 −αp3 − 19αp2 3p2 − 34αp 2p− 8α 4 1− α 7 − 4p 3 − p 2 − p t1−p − 4p 1 − α 2 − p 1 − p t−p − 4αp 1 − p2 t−p−1, g′′ 1 1 8 81 1 − α ( −128α4 896α3 288α2 5294α − 437 )