A Best Possible Double Inequality for Power Mean
Abstract
We answer the question: for any p, q ∈ ℝ with p ≠ q and p ≠ −q, what are the greatest value λ = λ(p, q) and the least value μ = μ(p, q), such that the double inequality holds for all a, b > 0 with a ≠ b? Where Mp(a, b) is the pth power mean of two positive numbers a and b.
1. Introduction
It is well known that Mp(a, b) is continuous and strictly increasing with respect to p ∈ ℝ for fixed a, b > 0 with a ≠ b. Many classical means are special case of the power mean, for example, M−1(a, b) = H(a, b) = 2ab/(a + b), , and M1(a, b) = A(a, b) = (a + b)/2 are the harmonic, geometric, and arithmetic means of a and b, respectively. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities and properties for the power mean can be found in literature [1–15].
2. Main Result
In order to establish our main result, we need a lemma which we present in this section.
Lemma 2.1. Let p, q ≠ 0, p ≠ q and x > 1. Then
Proof. From (1.1), we have
Let
Equation (2.6) implies that
Therefore, inequality (2.1) follows from (2.3)–(2.5) and inequality (2.7), and inequality (2.2) follows from (2.3)–(2.5) and inequality (2.8).
Let
Then we clearly see that , and it is not difficult to verify that the identity holds for all a, b > 0 if . Let
Then we have Theorem 2.2 as follows.
Theorem 2.2. The double inequality
Proof. From (1.1), we clearly see that Mp(a, b) is symmetric and homogenous of degree 1. Without loss of generality, we assume that b = 1, a = x > 1 and p > q. We divide the proof of inequality (2.11) into three cases.
Case 1. (p, q) ∈ E1⋃ E2. Then from Lemma 2.1, we clearly see that
From (1.1), we get
Let
If (p, q) ∈ E1, then (2.15), (2.18), (2.21), (2.22), (2.24), (2.25), (2.27), and (2.28) lead to
We divide the discussion into two subcases.
Subcase 1.1. (p, q) ∈ E1. Then (2.26) and (2.29) together with inequalities (2.36) and (2.37) imply that I(x) is strictly decreasing in [1, +∞). In fact, if (q2 − p2 + 2pq)/(p + q) ≥ 0, then (2.29) and inequality (2.37) imply that J(x) < 0 for x ∈ [1, +∞). If (q2 − p2 + 2pq)/(p + q) < 0, then (2.29) and inequality (2.36) lead to the conclusion that J(x) < 0 for x ∈ [1, +∞).
From inequalities (2.34) and (2.35) together with the monotonicity of I(x), we know that there exists λ1 > 1 such that I(x) > 0 for x ∈ [1, λ1) and I(x) < 0 for x ∈ (λ1, +∞). Then (2.23) leads to the conclusion that H(x) is strictly increasing in [1, λ1] and strictly decreasing in [λ1, +∞).
It follows from (2.32) and (2.33) together with the piecewise monotonicity of H(x) that there exists λ2 > λ1 > 1 such that H(x) > 0 for [1, λ2] and H(x) < 0 for (λ2, +∞). Then (2.20) leads to the conclusion that G(x) is strictly increasing in [1, λ2] and strictly decreasing in [λ2, +∞).
From (2.17), (2.19) and (2.31) together with the piecewise monotonicity of G(x), we clearly see that there exists λ3 > λ2 > 1 such that F(x) is strictly increasing in [1, λ3] and strictly decreasing in [λ3, +∞).
Therefore, follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of F(x).
Subcase 1.2. (p, q) ∈ E2. Then (2.30) and (2.35) again hold, and (2.18), (2.21), (2.22), and (2.28) lead to
It follows from (2.29) and inequalities (q2 − p2 + 2pq)/(p + q) < 0 and (2.41) that J(x) > 0 for x ∈ [1, +∞). Then (2.26) and inequality (2.35) lead to the conclusion that I(x) > 0 for x ∈ [1, +∞). Therefore, H(x) is strictly increasing in [1, +∞) follows from (2.23).
It follows from (2.20) and (2.39) together with inequality (2.40) and the monotonicity of H(x) that there exists μ1 > 1 such that G(x) is strictly decreasing in [1, μ1] and strictly increasing in [μ1, +∞).
From (2.17), (2.19) and (2.38) together with the piecewise monotonicity of G(x), we clearly see that there exists μ2 > μ1 > 1 such that F(x) is strictly decreasing in [1, μ2] and strictly increasing in [μ2, +∞).
Therefore, follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of F(x).
Case 2. (p, q) ∈ E3⋃ E5. Clearly, we have for (p, q) ∈ E3 and for (p, q) ∈ E5. Therefore, we need only to prove that
From (1.1), one has
Let
If r > 0 (or r < 0, resp.), then (2.47) leads to the conclusion that f(x) is strictly decreasing (or increasing, resp.) in [1, +∞). Therefore, inequalities (2.42) and (2.43) follow from (2.44)–(2.46) and the monotonicity of f(x).
Case 3. (p, q) ∈ E4⋃ E6. Then from Lemma 2.1, we clearly see that for (p, q) ∈ E4 and for (p, q) ∈ E6. Therefore, we need only to prove that
From (1.1), we get
Let
If (p, q) ∈ E4 (or E6, resp.), then (2.53) implies that f(x) is strictly increasing (or decreasing, resp.) in [1, +∞). Therefore, inequalities (2.48) and (2.49) follow from (2.50)–(2.52) and the monotonicity of f(x).
Next, we prove that Mλ(a, b) and Mμ(a, b) are the best possible lower and upper power mean bounds for the geometric mean of Mp(a, b) and Mq(a, b). We divide the proof into six cases.
Case A. (p, q) ∈ E1. Then for any ϵ ∈ (0, (p + q)/2) and x > 0, from (1.1), one has
Letting x → 0 and making use of Taylor expansion, we get
Equations (2.54) and (2.56) together with inequality (2.55) imply that for any ϵ ∈ (0, p + q/2), there exist δ1 = δ1(ϵ) > 0 and X1 = X1(p, q, ϵ) > 1 such that for x ∈ (0, δ1) and for x ∈ (X1, +∞).
Case B. (p, q) ∈ E2. Then for ϵ ∈ (0, −(p + q)/2) and x > 0, making use of (1.1) and Taylor expansion, we have
Equation (2.57) and inequality (2.58) imply that for any ϵ ∈ (0, −(p + q)/2), there exist δ2 = δ2(ϵ) > 0 and X2 = X2(p, q, ϵ) > 1 such that for x ∈ (0, δ2) and for x ∈ (X2, +∞).
Case C. (p, q) ∈ E3. Then for ϵ ∈ (0, p/2) and x > 0, making use of (1.1) and Taylor expansion, we have
Equation (2.59) leads to the conclusion that for any ϵ ∈ (0, p/2), there exist δ3 = δ3(ϵ) > 0 and X3 = X3(p, ϵ) > 1 such that for x ∈ (0, δ3) and for x ∈ (X3, +∞).
Case D. (p, q) ∈ E4. Then for ϵ ∈ (0, (p + q)/2) and x > 0, making use of (1.1) and Taylor expansion, we have
Equation (2.60) implies that for any ϵ ∈ (0, (p + q)/2), there exist δ4 = δ4(ϵ) > 0 and X4 = X4(p, q, ϵ) > 1 such that for x ∈ (0, δ4) and for x ∈ (X4, +∞).
Case E. (p, q) ∈ E5. Then for any ϵ ∈ (0, −q/2) and x > 0, making use of (1.1) and Taylor expansion, one has
Equation (2.61) leads to the conclusion that for any ϵ ∈ (0, −q/2), there exist δ5 = δ5(ϵ) > 0 and X5 = X5(q, ϵ) > 1 such that for x ∈ (0, δ5) and for x ∈ (X5, +∞).
Case F. (p, q) ∈ E6. Then for any ϵ ∈ (0, −(p + q)/2) and x > 0, making use of (1.1) and Taylor expansion, one has
Equation (2.62) shows that for any ϵ ∈ (0, −(p + q)/2), there exist δ6 = δ6(ϵ) > 0 and X6 = X6(p, q, ϵ) > 1 such that for x ∈ (0, δ6) and for x ∈ (X6, +∞).
Acknowledgments
This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.