A best-possible double inequality between Seiffert and harmonic means

Abstract

In this paper, we establish a new double inequality between the Seiffert and harmonic means.

The achieved results is inspired by the papers of Sándor (Arch. Math., 76, 34-40, 2001) and Hästö (Math. Inequal. Appl., 7, 47-53, 2004), and the methods from Wang et al. (J. Math. Inequal., 4, 581-586, 2010). The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

2010 Mathematics Subject Classification: 26E60.

1 Introduction

For a, b > 0 with a ≠ b, the Seiffert mean P(a, b) was introduced by Seiffert [1] as follows:

$$P\left(a,b\right)=\frac{a-b}{4arctan\sqrt{a∕b}-\pi }.$$

(1.1)

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature [1–9].

Let H(a, b) = 2ab/(a+b), $G\left(a,b\right)=\sqrt{ab}$, L(a, b) = (a - b)/(log a - log b), I(a, b) = 1/e(b^b /a^a )^1/(b-a), A(a, b) = (a+b)/2, C(a, b) = (a²+b²)/(a+b), and M _p (a, b) = ((a^p + b^p )/2)^1/p(p ≠ 0) and $M_0\left(a,b\right)=\sqrt{ab}$ be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and p-th power means of two different positive numbers a and b, respectively. Then, it is well known that

$$\begin{array}{cc}\hfill min\left\{a,b\right\}<H\left(a,b\right)& =M_{-1}\left(a,b\right)<G\left(a,b\right)=M_0\left(a,b\right)<L\left(a,b\right)\hfill \\ <I\left(a,b\right)<A\left(a,b\right)=M_1\left(a,b\right)<C\left(a,b\right)<max\left\{a,b\right\}.\hfill \end{array}$$

For all a, b > 0 with a ≠ b, Seiffert [1] established that L(a, b) < P(a, b) < I(a, b); Jagers [4] proved that M_1/2(a, b) < P (a, b) < M_2/3(a, b) and M_2/3(a, b) is the best-possible upper power mean bound for the Seiffert mean P(a, b); Seiffert [7] established that P(a, b) > A(a, b)G(a, b)/L(a, b) and P(a, b) > 2 A(a, b)/π; Sándor [6] presented that $\left(A\left(a,b\right)+G\left(a,b\right)\right)∕2<P\left(a,b\right)<\sqrt{A\left(a,b\right)\left(A\left(a,b\right)+G\left(a,b\right)\right)∕2}$ and $\sqrt[3]{A^2\left(a,b\right)G\left(a,b\right)}<P\left(a,b\right)<\left(G\left(a,b\right)+2A\left(a,b\right)\right)∕3$; Hästö [3] proved that P(a, b) > M_{log 2/ log π}(a, b) and M_{log 2/ log π}(a, b) is the best-possible lower power mean bound for the Seiffert mean P(a, b).

Very recently, Wang and Chu [8] found the greatest value α and the least value β such that the double inequality A^α (a, b)H^1-α(a, b) < P(a, b) < A^β (a, b)H^1-β(a, b) holds for a, b > 0 with a ≠ b; For any α ∈ (0, 1), Chu et al. [10] presented the best-possible bounds for P^α (a, b)G^1-α(a, b) in terms of the power mean; In [2], the authors proved that the double inequality αA(a, b) + (1 - α)H(a, b) < P(a, b) < βA(a, b) + (1 - β)H(a, b) holds for all a, b > 0 with a ≠ b if and only if α ≤ 2/π and β ≥ 5/6; Liu and Meng [5] proved that the inequalities

$${\alpha }_{1}C\left(a,b\right)+\left(1-{\alpha }_1\right)G\left(a,b\right)<P\left(a,b\right)<{\beta }_{1}C\left(a,b\right)+\left(1-{\beta }_1\right)G\left(a,b\right)$$

and

$${\alpha }_{2}C\left(a,b\right)+\left(1-{\alpha }_2\right)H\left(a,b\right)<P\left(a,b\right)<{\beta }_{2}C\left(a,b\right)+\left(1-{\beta }_2\right)H\left(a,b\right)$$

hold for all a, b > 0 with a ≠ b if and only if α₁ ≤ 2/9, β₁ ≥ 1/π, α₂ ≤ 1/π and β₂ ≥ 5/12.

For fixed a, b > 0 with a ≠ b and x ∈ [0, 1/2], let

$$h\left(x\right)=H\left(xa+\left(1-x\right)b,xb+\left(1-x\right)a\right).$$

Then, it is not difficult to verify that h(x) is continuous and strictly increasing in [0, 1/2]. Note that h(0) = H(a, b) < P(a, b) and h(1/2) = A(a, b) > P(a, b). Therefore, it is natural to ask what are the greatest value α and least value β in (0, 1/2) such that the double inequality H(αa + (1 - α)b, αb + (1 - α)a) < P(a, b) < H(βa + (1 - β)b, βb + (1 - β)a) holds for all a, b > 0 with a ≠ b. The main purpose of this paper is to answer these questions. Our main result is the following Theorem 1.1.

Theorem 1.1. If α, β ∈ (0, 1/2), then the double inequality

$$H\left(\alpha a+\left(1-\alpha \right)b,\alpha b+\left(1-\alpha \right)a\right)<P\left(a,b\right)<H\left(\beta a+\left(1-\beta \right)b,\beta b+\left(1-\beta \right)a\right)$$

holds for all a, b > 0 with a ≠ b if and only if $\alpha \le \left(1-\sqrt{1-2∕\pi }\right)∕2$ and $\beta \ge \left(6-\sqrt{6}\right)∕12$.

2 Proof of Theorem 1.1

Proof of Theorem 1.1. Let $\lambda =\left(1-\sqrt{1-2∕\pi }\right)∕2$ and $\mu =\left(6-\sqrt{6}\right)∕12$. We first prove that inequalities

$$P\left(a,b\right)>H\left(\lambda a+\left(1-\lambda \right)b,\lambda b+\left(1-\lambda \right)a\right)$$

(2.1)

and

$$P\left(a,b\right)<H\left(\mu a+\left(1-\mu \right)b,\mu b+\left(1-\mu \right)a\right)$$

(2.2)

hold for all a, b > 0 with a ≠ b.

Without loss of generality, we assume that a > b. Let $t=\sqrt{a∕b}>1$ and p ∈ (0, 1/2); then, from (1.1), one has

$$\begin{array}{c}H\left(pa+\left(1-p\right)b,pb+\left(1-p\right)a\right)-P\left(a,b\right)\\ =\frac{2\left[p{t}^2+\left(1-p\right)\right]\left[\left(1-p\right)t^2+p\right]}{t^2+1}-\frac{t^2-1}{4arctant-\pi }\\ =\frac{2\left[p{t}^2+\left(1-p\right)\right]\left[\left(1-p\right)t^2+p\right]}{\left(t^2+1\right)\left(4arctant-\pi \right)}\\ \times \left\{4arctant-\frac{t^4-1}{2\left[p{t}^2+\left(1-p\right)\right]\left[\left(1-p\right)t^2+p\right]}-\pi \right\}.\end{array}$$

(2.3)

Let

$$f\left(t\right)=4arctant-\frac{t^4-1}{2\left[p{t}^2+\left(1-p\right)\right]\left[\left(1-p\right)t^2+p\right]}-\pi ,$$

(2.4)

then, simple computations lead to

$$f\mathsf{\text{(1) = 0,}}$$

(2.5)

$$\underset{t\to +\infty }{lim}f\left(t\right)=\pi -\frac{1}{2p\left(1-p\right)}$$

(2.6)

and

$$f^{\prime }\left(t\right)=\frac{f_1\left(t\right)}{\left(t^2+1\right){\left[p\left(1-p\right)t^4+\left(2p^2-2p+1\right)t^2+p\left(1-p\right)\right]}^2},$$

(2.7)

where

$$\begin{array}{c}{f}_1\left(t\right)=4p^2{\left(1-p\right)}^2{t}^8-\left(2p^2-2p+1\right)t^7+8p\left(1-p\right)\left(2p^2-2p+1\right)t^6\\ +\left(2p^2-2p-1\right)t^5+4\left(6p^4-12p^3+10p^2-4p+1\right)t^4\\ +\left(2p^2-2p-1\right)t^3+8p\left(1-p\right)\left(2p^2-2p+1\right)t^2\\ -\left(2p^2-2p+1\right)t+4p^2{\left(1-p\right)}^2.\end{array}$$

(2.8)

Note that

$$f_1\left(1\right)=0,$$

(2.9)

$$\underset{t\to +\infty }{lim}{f}_1\left(t\right)=+\infty ,$$

(2.10)

$$\begin{array}{c}{f^{\prime }}_1\left(t\right)=32p^2{\left(1-p\right)}^2{t}^7-7\left(2p^2-2p+1\right)t^6+48p\left(1-p\right)\left(2p^2-2p+1\right)t^5\\ +5\left(2p^2-2p-1\right)t^4+16\left(6p^4-12p^3+10p^2-4p+1\right)t^3\\ +3\left(2p^2-2p-1\right)t^2+16p\left(1-p\right)\left(2p^2-2p+1\right)t\\ -\left(2p^2-2p+1\right),\end{array}$$

$$f_1^{\prime }\left(1\right)=0,$$

(2.11)

$$\underset{t\to +\infty }{lim}{f}_1^{\prime }\left(t\right)=+\infty .$$

(2.12)

Let $f_2\left(t\right)=f_1^{″}\left(t\right)∕2$, $f_3\left(t\right)=f_2^{\prime }\left(t\right)∕3$, $f_4\left(t\right)=f_3^{\prime }\left(t\right)∕4$, $f_5\left(t\right)=f_4^{\prime }\left(t\right)∕5$, $f_6\left(t\right)=f_5^{\prime }\left(t\right)∕6$ and $f_7\left(t\right)=f_6^{\prime }\left(t\right)∕7$. Then, simple computations lead to

$$\begin{array}{c}{f}_2\left(t\right)=112p^2{\left(1-p\right)}^2{t}^6-21\left(2p^2-2p+1\right)t^5+120p\left(1-p\right)\left(2p^2-2p+1\right)t^4\\ +10\left(2p^2-2p-1\right)t^3+24\left(6p^4-12p^3+10p^2-4p+1\right)t^2\\ +3\left(2p^2-2p-1\right)t+8p\left(1-p\right)\left(2p^2-2p+1\right),\end{array}$$

$$f_2\left(1\right)=-2\left(24p^2-24p+5\right),$$

(2.13)

$$\underset{t\to +\infty }{lim}{f}_2\left(t\right)=+\infty ,$$

(2.14)

$$\begin{array}{c}{f}_3\left(t\right)=224p^2{\left(1-p\right)}^2{t}^5-35\left(2p^2-2p+1\right)t^4+160p\left(1-p\right)\left(2p^2-2p+1\right)t^3\\ +10\left(2p^2-2p-1\right)t^2+16\left(6p^4-12p^3+10p^2-4p+1\right)t\\ +\left(2p^2-2p-1\right),\end{array}$$

$$f_3\left(1\right)=-6\left(24p^2-24p+5\right),$$

(2.15)

$$\underset{t\to +\infty }{lim}{f}_3\left(t\right)=+\infty ,$$

(2.16)

$$\begin{array}{c}{f}_4\left(t\right)=280p^2{\left(1-p\right)}^2{t}^4-35\left(2p^2-2p+1\right)t^3+120p\left(1-p\right)\left(2p^2-2p+1\right)t^2\\ +5\left(2p^2-2p-1\right)t+4\left(6p^4-12p^3+10p^2-4p+1\right),\end{array}$$

$$f_4\left(1\right)=4\left(16p^4-32p^3-25p^2+41p-9\right),$$

(2.17)

$$\underset{t\to +\infty }{lim}{f}_4\left(t\right)=+\infty ,$$

(2.18)

$$\begin{array}{c}{f}_5\left(t\right)=224p^2{\left(1-p\right)}^2{t}^3-21\left(2p^2-2p+1\right)t^2+48p\left(1-p\right)\left(2p^2-2p+1\right)t\\ +\left(2p^2-2p-1\right),\end{array}$$

$$f_5\left(1\right)=2\left(64p^4-128p^3+20p^2+44p-11\right),$$

(2.19)

$$\underset{t\to +\infty }{lim}{f}_5\left(t\right)=+\infty ,$$

(2.20)

$$\begin{array}{c}{f}_6\left(t\right)=112p^2{\left(1-p\right)}^2{t}^2-7\left(2p^2-2p+1\right)t\\ +8p\left(1-p\right)\left(2p^2-2p+1\right),\end{array}$$

(2.21)

$$f_6\left(1\right)=96p^4-192p^3+74p^2+22p-7,$$

(2.22)

$$\underset{t\to +\infty }{lim}{f}_6\left(t\right)=+\infty ,$$

(2.23)

$$f_7\left(t\right)=32p^2{\left(1-p\right)}^{2}t-\left(2p^2-2p+1\right)$$

(2.24)

and

$$f_7\left(1\right)=32p^4-64p^3+30p^2+2p-1.$$

(2.25)

We divide the proof into two cases.

Case 1. $p=\lambda =\left(1-\sqrt{1-2∕\pi }\right)∕2$. Then equations (2.6), (2.13), (2.15), (2.17), (2.19), (2.22) and (2.25) become

$$\underset{t\to +\infty }{lim}f\left(t\right)=0,$$

(2.26)

$$f_2\left(1\right)=-\frac{2\left(5\pi -12\right)}{\pi }<0,$$

(2.27)

$$f_3\left(1\right)=-\frac{6\left(5\pi -12\right)}{\pi }<0,$$

(2.28)

$$f_4\left(1\right)=-\frac{2\left(18{\pi }^2-41\pi -8\right)}{{\pi }^2}<0,$$

(2.29)

$$f_5\left(1\right)=-\frac{2\left(11{\pi }^2-22\pi -16\right)}{{\pi }^2}<0,$$

(2.30)

$$f_6\left(1\right)=-\frac{7{\pi }^2-11\pi -24}{{\pi }^2}<0$$

(2.31)

and

$$f_7\left(1\right)=\frac{\pi +8-{\pi }^2}{{\pi }^2}>0.$$

(2.32)

From (2.24), we clearly see that f₇(t) is strictly increasing in [1, +∞), and then (2.32) leads to the conclusion that f₇(t) > 0 for t ∈ [1, +∞). Thus, f₆(t) is strictly increasing in [1, +∞).

It follows from (2.23) and (2.31) together with the monotonicity of f₆(t) that there exists t₁> 1 such that f₆(t) < 0 for t ∈ (1, t₁) and f₆(t) > 0 for t ∈ (t₁, +∞). Thus, f₅(t) is strictly decreasing in [1, t₁] and strictly increasing in [t₁, +∞).

From (2.20) and (2.30), together with the piecewise monotonicity of f₅(t), we clearly see that there exists t₂> t₁> 1 such that f₄(t) is strictly decreasing in [1, t₂] and strictly increasing in [t₂, +∞). Then, equation (2.18) and inequality (2.29) lead to the conclusion that there exists t₃> t₂> 1 such that f₃(t) is strictly decreasing in [1, t₃] and strictly increasing in [t₃, +∞).

It follows from (2.16) and (2.28) together with the piecewise monotonicity of f₃(t) we conclude that there exists t₄> t₃> 1 such that f₂(t) is strictly decreasing in [1, t₄] and strictly increasing in [t₄, +∞). Then, equation (2.14) and inequality (2.27) lead to the conclusion that there exists t₅> t₄> 1 such that $f_1^{\prime }\left(t\right)$ is strictly decreasing in [1, t₅] and strictly increasing in [t₅, +∞).

From equations (2.11) and (2.12), together with the piecewise monotonicity of $f_1^{\prime }\left(t\right)$, we know that there exists t₆> t₅> 1 such that f₁(t) is strictly decreasing in [1, t₆] and strictly increasing in [t₆, +∞). Then, equations (2.7)-(2.10) lead to the conclusion that there exists t₇> t₆> 1 such that f(t) is strictly decreasing in [1, t₇] and strictly increasing in [t₇, +∞).

Therefore, inequality (2.1) follows from equations (2.3)-(2.5) and (2.26) together with the piecewise monotonicity of f(t).

Case 2. $p=\mu =\left(6-\sqrt{6}\right)∕12$. Then, equations (2.13), (2.15), (2.17), (2.19) and (2.21) become

$$f_{\mathsf{\text{2}}}\mathsf{\text{(1) = 0,}}$$

(2.33)

$$f_{\mathsf{\text{3}}}\mathsf{\text{(1) = 0,}}$$

(2.34)

$$f_4\left(1\right)=\frac{17}{18}>0,$$

(2.35)

$$f_5\left(1\right)=\frac{17}{9}>0$$

(2.36)

and

$$f_6\left(t\right)=\frac{1}{36}\left(175t^2-147t+35\right)>0$$

(2.37)

for t > 1.

From inequality (2.37), we know that f₅(t) is strictly increasing in [1, +∞), and then inequality (2.36) leads to the conclusion that f₅(t) > 0 for t ∈ [1, +∞). Thus, f₄(t) is strictly increasing in [1, +∞).

It follows from inequality (2.35) and the monotonicity of f₄(t) that f₃(t) is strictly increasing in [1, +∞).

Therefore, inequality (2.2) follows easily from equations (2.3)-(2.5), (2.7), (2.9), (2.11), (2.33), and (2.34) together with the monotonicity of f₃(t).

Next, we prove that $\lambda =\left(1-\sqrt{1-2∕\pi }\right)∕2$ is the best-possible parameter such that inequality (2.1) holds for all a, b > 0 with a ≠ b. In fact, if $\left(1-\sqrt{1-2∕\pi }\right)∕2=\lambda <p<1∕2$, then equation (2.6) leads to

$$\underset{t\to +\infty }{lim}f\left(t\right)=\pi -\frac{1}{2p\left(1-p\right)}>0.$$

(2.38)

Inequality (2.38) implies that there exists T = T(p) > 1 such that

$$f\left(t\right)>0$$

(2.39)

for t ∈ (T, +∞).

From equations (2.3) and (2.4), together with inequality (2.39), we clearly see that P(a, b) < H(pa + (1 - p)b, pb + (1 - p)a) for a/b ∈ (T², +∞).

Finally, we prove that $\mu =\left(6-\sqrt{6}\right)∕12$ is the best-possible parameter such that inequality (2.2) holds for all a, b > 0 with a ≠ b. In fact, if $0<p<\mu =\left(6-\sqrt{6}\right)∕12$, then equation (2.13) leads to

$$f_2\left(1\right)=-2\left(24p^2-24p+5\right)<0.$$

(2.40)

Inequality (2.40) implies that there exists δ = δ (p) > 0 such that

$$f_2\left(t\right)<0$$

(2.41)

for t ∈ (1, 1 + δ).

Therefore, P(a, b) > H(pa + (1 - p)b, pb + (1 - p)a) for a/b ∈ (1, (1 + δ)²) follows from equations (2.3)-(2.5), (2.7), (2.9), and (2.11) together with inequality (2.41).