Sharp bounds for Seiffert mean in terms of root mean square

Abstract

We find the greatest value α and least value β in (1/ 2,1) such that the double inequality

$$S\left(\alpha a+\left(1-\alpha \right)b,\alpha b+\left(1-\alpha \right)a\right)<T\left(a,b\right)<S\left(\beta a+\left(1-\beta \right)b,\beta b+\left(1-\beta \right)a\right)$$

holds for all a,b > 0 with a ≠ b. Here, T(a, b) = (a-b)/[2 arctan((a-b)/(a + b))] and S(a, b) = [(a² + b²)/2]^{1/ 2}are the Seiffert mean and root mean square of a and b, respectively.

2010 Mathematics Subject Classification: 26E60.

1 Introduction

For a,b > 0 with a ≠ b the Seiffert mean T(a, b) and root mean square S(a, b) are defined by

$$T\left(a,b\right)=\frac{a-b}{2\text{arctan}\left(\frac{a-b}{a+b}\right)}$$

(1.1)

and

$$S\left(a,b\right)=\sqrt{\frac{a^2+b^2}{2}},$$

(1.2)

respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities and properties for T and S can be found in the literature [1–14].

Let $A\left(a,b\right)=\left(a+b\right)/2,G\left(a,b\right)=\sqrt{ab}$, 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 arithmetic, geometric, and p th power means of two positive numbers a and b, respectively. Then it is well known that

$$G\left(a,b\right)=M_0\left(a,b\right)<A\left(a,b\right)=M_1\left(a,b\right)<T\left(a,b\right)<S\left(a,b\right)=M_2\left(a,b\right)$$

for all a, b > 0 with a ≠ b.

Seiffert [1] proved that inequalities

$$A\left(a,b\right)<T\left(a,b\right)<S\left(a,b\right)$$

hold for all a, b > 0 with a ≠ b.

Chu et al. [5] found the greatest value p₁ and least value p₂ such that the double inequality H_{p 1}(a,b) < T(a,b) < H_{p 2}(a,b) holds for all a,b > 0 with a ≠ b, where H _p (a, b) = ((a^p+ (ab)^p/2+b^p)/3)^1/p(p ≠ 0) and $H_0\left(a,b\right)=\sqrt{ab}$ is the p th power-type Heron mean of a and b.

In [6], Wang et al. answered the question: What are the best possible parameters λ and μ such that the double inequality L _λ (a,b) < T(a,b) < L _μ (a,b) holds for all a,b > 0 with a ≠ b? where L _r (a,b) = (a^r+1+ b^r+1)/(a^r+ b^r) is the r th Lehmer mean of a and b.

Chu et al. [7] proved that inequalities

$$pT\left(a,b\right)+\left(1-p\right)G\left(a,b\right)<A\left(a,b\right)<qT\left(a,b\right)+\left(1-q\right)G\left(a,b\right)$$

hold for all a,b > 0 with a ≠ b if and only if p ≤ 3/5 and q ≥ π/4.

Hou and Chu [9] gave the best possible parameters α and β such that the double inequality

$$\alpha S\left(a,b\right)+\left(1-\alpha \right)\overline{H}\left(a,b\right)<T\left(a,b\right)<\beta S\left(a,b\right)+\left(1-\beta \right)\overline{H}\left(a,b\right)$$

holds for all a, b > 0 with a ≠ b.

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

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

Then it is not difficult to verify that f(x) is continuous and strictly increasing in [1/2,1]. Note that f(1/2) = A(a,b) < T(a,b) and f(1) = S(a, b) > T(a, b). Therefore, it is natural to ask what are the greatest value α and least value β in (1/2,1) such that the double inequality

$$S\left(\alpha a+\left(1-\alpha \right)b,\alpha b+\left(1-\alpha \right)a\right)<T\left(a,b\right)<S\left(\beta a+\left(1-\beta \right)b,\beta b+\left(1-\beta \right)a\right)$$

holds for all a, b > 0 with a ≠ b. The main purpose of this article is to answer these questions. Our main result is the following Theorem 1.1.

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

$$S\left(\alpha a+\left(1-\alpha \right)b,\alpha b+\left(1-\alpha \right)a\right)<T\left(a,b\right)<S\left(\beta a+\left(1-\beta \right)b,\beta b+\left(1-\beta \right)a\right)$$

(1.3)

holds for all a,b > 0 with a ≠ b if and only if $\alpha \le \left(1+\sqrt{16/{\pi }^2-1}\right)/2$ and $\beta \ge \left(3+\sqrt{6}\right)/6$.

2 Proof of Theorem 1.1

Proof of Theorem 1.1. Let $\lambda =\left(1+\sqrt{16/{\pi }^2-1}\right)/2$ and $\mu =\left(3+\sqrt{6}\right)/6$.

We first proof that inequalities

$$T\left(a,b\right)>S\left(\lambda a+\left(1-\lambda \right)b,\lambda b+\left(1-\lambda \right)a\right)$$

(2.1)

and

$$T\left(a,b\right)<S\left(\mu a+\left(1-\mu \right)b,\mu b+\left(1-\mu \right)a\right)$$

(2.2)

hold for all a, b > 0 with a ≠ b.

From (1.1) and (1.2), we clearly see that both T(a, b) and S(a, b) are symmetric and homogenous of degree 1. Without loss of generality we assume that a > b. Let t = a/b > 1 and p ∈ (1/2,1), then from (1.1) and (1.2) one has

$$\begin{array}{c}S\left(pa+\left(1-p\right)b,pb+\left(1-p\right)a\right)-T\left(a,b\right)\\ =b\frac{\sqrt{{\left[pt+\left(1-p\right)\right]}^2+{\left[\left(1-p\right)t+p\right]}^2}}{2\text{arctan}\left(\frac{t-1}{t+1}\right)}\\ \times \left\{\sqrt{2}\text{arctan}\left(\frac{t-1}{t+1}\right)-\frac{t-1}{\sqrt{{\left[pt+\left(1-p\right)\right]}^2+{\left[\left(1-p\right)t+p\right]}^2}}\right\}.\end{array}$$

(2.3)

Let

$$f\left(t\right)=\sqrt{2}\text{arctan}\left(\frac{t-1}{t+1}\right)-\frac{t-1}{\sqrt{{\left[pt+\left(1-p\right)\right]}^2+{\left[\left(1-p\right)t+p\right]}^2}},$$

(2.4)

then simple computations lead to

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

(2.5)

$$\underset{t\to +\infty }{\text{lim}}f\left(t\right)=\frac{\sqrt{2\pi }}{4}-\frac{1}{\sqrt{p^2+{\left(1-p\right)}^2}},$$

(2.6)

$$f^{\prime }\left(t\right)=\frac{f_1\left(t\right)}{{\left\{{\left[pt+\left(1-p\right)\right]}^2+{\left[\left(1-p\right)t+p\right]}^2\right\}}^{\frac{3}{2}}\left(t^2+1\right)},$$

(2.7)

where

$$f_1\left(t\right)=\sqrt{2}{\left\{{\left[pt+\left(1-p\right)\right]}^2+{\left[\left(1-p\right)t+p\right]}^2\right\}}^{\frac{3}{2}}-\left(t+1\right)\left(t^2+1\right).$$

(2.8)

Note that

$$\begin{array}{c}{\left\{\sqrt{2}{\left\{{\left[pt+\left(1-p\right)\right]}^2+{\left[\left(1-p\right)t+p\right]}^2\right\}}^{\frac{3}{2}}\right\}}^2-{\left[\left(t+1\right)\left(t^2+1\right)\right]}^2\\ ={\left(t-1\right)}^2{g}_1\left(t\right),\end{array}$$

(2.9)

where

$$\begin{array}{c}{g}_1\left(t\right)=\left(16p^6-48p^5+72p^4-64p^3+36p^2-12p+1\right)t^4-16p^2\\ \left(4p^2-4p+3\right){\left(p-1\right)}^2{t}^3+2\left(48p^6-144p^5+168p^4-96p^3+36p^2-12p+1\right)\\ \times t^2-16p^2\left(4p^2-4p+3\right){\left(p-1\right)}^{2}t+16p^6-48p^5+72p^4-64p^3+36p^2\\ -12p+1,\end{array}$$

(2.10)

$$g_1\left(1\right)=4\left(12p^2-12p+1\right).$$

(2.11)

Let $g_2\left(t\right)=g_1^{\prime }\left(t\right)/4,g_3\left(t\right)=g_2^{\prime }\left(t\right),g_4\left(t\right)=g_3^{\prime }\left(t\right)/6$. Then simple computations lead to

$$\begin{array}{c}{g}_2\left(t\right)=\left(16p^6-48p^5+72p^4-64p^3+36p^2-12p+1\right)t^3-12p^2\\ \left(4p^2-4p+3\right){\left(p-1\right)}^2{t}^2+\left(48p^6-144p^5+168p^4-96p^3+36p^2-12p+1\right)\\ t-4p^2\left(4p^2-4p+3\right){\left(p-1\right)}^2,\end{array}$$

(2.12)

$$g_2\left(1\right)=4\left(6p^2-12p+1\right),$$

(2.13)

$$\begin{array}{c}{g}_3\left(t\right)=3\left(16p^6-48p^5+72p^4-64p^3+36p^2-12p+1\right)t^2-24p^2\left(4p^2-4p+3\right)\\ {\left(p-1\right)}^{2}t+48p^6-144p^5+168p^4-96p^3+36p^2-12p+1,\end{array}$$

(2.14)

$$g_3\left(1\right)=4\left(6p^4-12p^3+18p^2-12p+1\right),$$

(2.15)

$$\begin{array}{cc}\hfill g_4\left(t\right)& =\left(16p^6-48p^5+72p^4-64p^3+36p^2-12p+1\right)t\hfill \\ \hfill -4p^2\left(4p^2-4p+3\right){\left(p-1\right)}^2,\end{array}$$

(2.16)

$$g_4\left(1\right)=12p^4-24p^3+24p^2-12p+1.$$

(2.17)

We divide the proof into two cases.

Case 1. $p=\lambda =\left(1+\sqrt{16/{\pi }^2-1}\right)/2$. Then equations (2.6), (2.11), (2.13), (2.15), and (2.17) lead to

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

(2.18)

$$g_1\left(1\right)=-\frac{4\left(5{\pi }^2-48\right)}{{\pi }^2}<0,$$

(2.19)

$$g_2\left(1\right)=-\frac{2\left(5{\pi }^2-48\right)}{{\pi }^2}<0,$$

(2.20)

$$g_3\left(1\right)=-\frac{2\left(7{\pi }^4-48{\pi }^2-192\right)}{{\pi }^4}<0,$$

(2.21)

$$g_4\left(1\right)=-\frac{2\left({\pi }^4-96\right)}{{\pi }^4}<0.$$

(2.22)

Note that

$$16p^6-48p^5+72p^4-64p^3+36p^2-12p+1=\frac{1024-{\pi }^6}{{\pi }^6}>0.$$

(2.23)

From (2.10), (2.12), (2.14), (2.16), and (2.23) we clearly see that

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

(2.24)

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

(2.25)

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

(2.26)

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

(2.27)

From equation (2.16) and inequality (2.23) we clearly see that g₄(t) is strictly increasing in [1, + ∞), then inequality (2.22) and equation (2.27) lead to the conclusion that there exists t₀ > 1 such that g₄(t) < 0 for t ∈ (1,t₀) and g₄(t) > 0 for t ∈ (t₀,+∞). Hence, g₃(t) is strictly decreasing in [1, t₀] and strictly increasing in [t₀, +∞).

It follows from (2.21) and (2.26) together with the piecewise monotonicity of g₃(t) that there exists t₁ > t₀ > 1 such that g₂(t) is strictly decreasing in [1,t₁] and strictly increasing in [t₁,+∞).

From (2.20) and (2.25) together with the piecewise monotonicity of g₂(t) we conclude that there exists t₂ > t₁ > 1 such that g₁(t) is strictly decreasing in [1,t₂] and strictly increasing in [t₂,+∞).

Equations (2.7)-(2.9), (2.19), and (2.24) together with the piecewise monotonicity of g₁(t) imply that there exists t₃ > t₂ > 1 such that f(t) is strictly decreasing in [1,t₃] and strictly increasing in [t₃, +∞).

Therefore, inequality (2.1) follows from equations (2.3)-(2.5) and (2.18) together with the piecewise monotonicity of f(t).

Case 2. $p=\mu =\left(3+\sqrt{6}\right)/6$. Then equation (2.10) becomes

$$g_1\left(t\right)=\frac{\left(17t^2+2t+17\right)}{108}{\left(t-1\right)}^2>0$$

(2.28)

for t > 1.

Equations (2.7)-(2.10) and inequality (2.28) lead to the conclusion that f(t) is strictly increasing in [1, +∞).

Therefore, inequality (2.2) follows from equations (2.3)-(2.5) and the monotonicity of f(t).

From the monotonicity of f(x) = S(xa + (1 - x)b, xb + (1- x)a) in [1/2,1] and inequalities (2.1) and (2.2) we know that inequality (1.3) holds for all $\alpha \le \left(1+\sqrt{16/{\pi }^2-1}\right)/2,\beta \ge \left(3+\sqrt{6}\right)/6$ and a, b > 0 with a ≠ b.

Next, we prove that $\lambda =\left(1+\sqrt{16/{\pi }^2-1}\right)/2$ is the best possible parameter in (1/2,1) such that inequality (2.1) holds for all a, b > 0 with a ≠ b.

For any $1>p>\lambda =\left(1+\sqrt{16/{\pi }^2-1}\right)/2$, from (2.6) one has

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

(2.29)

Equations (2.3) and (2.4) together with inequality (2.29) imply that for any $1>p>\lambda =\left(1+\sqrt{16/{\pi }^2-1}\right)/2$ there exists T₀ = T₀(p) > 1 such that

$$S\left(pa+\left(1-p\right)b,pb+\left(1-p\right)a\right)>T\left(a,b\right)$$

for a/b ∈ (T₀, + ∞).

Finally, we prove that $\mu =\left(3+\sqrt{6}\right)/6$ is the best possible parameter in (1/2,1) such that inequality (2.2) holds for all a, b > 0 with a ≠ b.

For any $1/2<p<\mu =\left(3+\sqrt{6}\right)/6$, from (2.11) one has

$$g_1\left(1\right)=4\left(12p^2-12p+1\right)<0.$$

(2.30)

From inequality (2.30) and the continuity of g₁(t) we know that there exists δ = δ(p) > 0 such that

$$g_1\left(t\right)<0$$

(2.31)

fort ∈ (1,1 + δ).

Equations (2.3)-(2.5) and (2.7)-(2.10) together with inequality (2.31) imply that for any $1/2<p<\mu =\left(3+\sqrt{6}\right)/6$ there exists δ = δ(p) > 0 such that

$$T\left(a,b\right)>S\left(pa+\left(1-p\right)b,pb+\left(1-p\right)a\right)$$

for a/b ∈ (1,1 + δ).