Sharp Power Mean Bounds for the Combination of Seiffert and Geometric Means

and Applied Analysis 3 The following sharp lower power mean bounds for 1/3 G a, b 2/3 H a, b , 2/3 G a, b 1/3 H a, b , and P a, b can be found in 4, 6 : 1 3 G a, b 2 3 H a, b > M−2/3 a, b , 2 3 G a, b 1 3 H a, b > M−1/3 a, b , P a, b > Mlog 2/ logπ a, b 1.8 for all a, b > 0 with a/ b. The purpose of this paper is to answer the question: for α ∈ 0, 1 , what are the greatest value p and the least value q such that the double inequality Mp a, b < P a, b G1−α a, b < Mq a, b holds for all a, b > 0 with a/ b. 2. Lemmas In order to prove our main result, we need several lemmas which we present in this section. Lemma 2.1. Let λ ∈ 0, 1/3 , x ∈ 1,∞ and h x 1−3λ x2λ 1− 1 3λ x2λ− 1 3λ x 1−3λ . Then there exists x0 > 1 such that h x < 0 for x ∈ 1, x0 , h x > 0 for x ∈ x0,∞ and h x0 0. Proof. Simple computations lead to h 1 −12λ < 0, 2.1 lim x→ ∞ h x ∞, 2.2 h′ x 1 − 3λ 2λ 1 x2λ − 2λ 1 3λ x2λ−1 − 1 3λ , h′ 1 −6λ 1 2λ < 0, 2.3 lim x→ ∞ h′ x ∞, 2.4 h′′ x 2λx2λ−2 1 2λ 1 − 3λ x 1 − 2λ 1 3λ > 0. 2.5 Inequality 2.5 implies that h′ x is strictly increasing in 1,∞ . Then 2.3 and 2.4 lead to that there exists x1 > 1 such that h′ x < 0 for x ∈ 1, x1 and h′ x > 0 for x ∈ x1,∞ . Hence, h x is strictly decreasing in 1, x1 and strictly increasing in x1,∞ . Therefore, Lemma 2.1 follows from 2.1 and 2.2 together with the monotonicity of h x . Lemma 2.2. If λ ∈ 1/√10, 1/3 , then the following statements are true: 1 300λ4 324λ3 29λ2 − 45λ − 8 < 0; 2 −1176λ5 − 24λ4 114λ3 10λ2 − 3λ − 1 < 0; 3 −24λ5 72λ4 178λ3 6λ2 − 25λ − 3 < 0. 4 Abstract and Applied Analysis Proof. Simple computations lead to 1 300λ4 324λ3 29λ2 −45λ−8 < 300× 1/3 4 324× 1/3 3 29× 1/3 2 −45/√10−8 590 − 243√10 /54 < 0; 2 −1176λ5 − 24λ4 114λ3 10λ2 − 3λ − 1 < −1176 × 1/√10 5 − 24 × 1/√10 4 114 × 1/3 3 10 × 1/3 2 − 3/√10 − 1 3070 − 1107√10 /750 < 0; 3 −24λ5 72λ4 178λ3 6λ2 − 25λ − 3 < −24 × 1/√10 5 72 × 1/3 4 178 × 1/3 3 6 × 1/3 2 − 25/√10 − 3 34750 − 17037√10 /6750 < 0. Lemma 2.3. If α ∈ 0, 3/√10 , then I a, b G1−α a, b < M2α/3 a, b 2.6 holds for all a, b > 0 with a/ b. Proof. Without loss of generality, we assume that a > b. Let t a/b > 1 and β α/3 ∈ 0, 1/ √ 10 . Then log M2α/3 a, b − log [ I a, b G1−α a, b ] log [ M2β a, b ] − log [ I3β a, b G1−3β a, b ] 1 2β log 1 t2β 2 − 3βt t − 1 log t − 1 − 3β 2 log t 3β. 2.7