Two sharp double inequalities for Seiffert mean

Abstract

In this paper, we establish two new inequalities between the root-square, arithmetic, and Seiffert means.

The achieved results are inspired by the paper of Seiffert (Die Wurzel, 29, 221-222, 1995), and the methods from Chu et al. (J. Math. Inequal., 4, 581-586, 2010). The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

Mathematics Subject Classification (2010): 26E60.

1 Introduction

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

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

(1.1)

and

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

(1.2)

respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for S and T can be found in the literature [1–11].

Let $A\left(a,b\right)=\left(a+b\right)∕2,G\left(a,b\right)=\sqrt{ab}$, and H(a, b) = 2ab/(a + b) be the classical arithmetic, geometric, and harmonic means of two positive numbers a and b, respectively. In [1], Seiffert proved that

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

for all a, b > 0 with a ≠ b.

Taneja [5] presented that

$$\begin{array}{lll}\hfill G\left(a,b\right)& <\frac{2}{3}H\left(a,b\right)+\frac{1}{3}S\left(a,b\right)<\frac{1}{2}A\left(a,b\right)+\frac{1}{2}H\left(a,b\right)<\frac{1}{2}S\left(a,b\right)+\frac{1}{2}G\left(a,b\right)& \hfill \text{(1) }\\ <\frac{1}{3}H\left(a,b\right)+\frac{2}{3}S\left(a,b\right)<A\left(a,b\right)<S\left(a,b\right)-G\left(a,b\right)+H\left(a,b\right)& \hfill \text{(2) }\\ \hfill \text{(3) }\end{array}$$

for all a, b > 0 with a ≠ b.

In [2], the authors find the greatest value p and the least value q such that the double inequality H _p (a, b) < T(a, b) < H _q (a, b) for all a, b > 0 with a ≠ b. Here, $H_p\left(a,b\right)={\left[\left(a^p+{\left(ab\right)}^{p∕2}+b^p\right)∕3\right]}^{1∕p}$ is the power-type Heron mean of a and b.

Wang, Qiu, and Chu [3] established that

$$T\left(a,b\right)<L_{1∕3}\left(a,b\right)$$

for all a, b > 0 with a ≠ b, where L _p (a, b) = (a^p+1+ b^p+1) = (a^p + b^p ) is the Lehmer mean of a and b.

The purpose of the paper is to find the greatest values α₁ and α₂, and the least values β₁ and β₂, such that the double inequalities α₁S(a, b) + (1 - α₁)A(a, b) < T (a, b) < β₁S(a, b) + (1 - β₁)A(a, b) and $S^{{\alpha }_2}\left(a,b\right)A^{1-{\alpha }_2}\left(a,b\right)<T\left(a,b\right)<S^{{\beta }_2}\left(a,b\right)A^{1-{\beta }_2}\left(a,b\right)$ hold for all a, b > 0 with a ≠ b.

2 Main results

Theorem 2.1. The double inequality α₁S(a, b)+(1 - α₁)A(a, b) < T (a, b) < β₁S(a, b) + (1 - β₁)A(a, b) holds for all a, b > 0 with a ≠ b if and only if ${\alpha }_1\le \left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]=0.659\cdots$ and β₁ ≥ 2/3.

Proof. Firstly, we prove that

$$T\left(a,b\right)<\frac{2}{3}S\left(a,b\right)+\frac{1}{3}A\left(a,b\right),$$

(2.1)

$$T\left(a,b\right)>\frac{4-\pi }{\pi \left(\sqrt{2}-1\right)}S\left(a,b\right)+\frac{\sqrt{2}\pi -4}{\pi \left(\sqrt{2}-1\right)}A\left(a,b\right)$$

(2.2)

for all a, b > 0 with a ≠ b.

Without loss of generality, we assume that a > b. Let $t=\sqrt{a∕b}>1$ and $p\in \left\{2∕3,\left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]\right\}$, then from (1.1) and (1.2) we have

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

(2.3)

Let

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

(2.4)

then simple computations lead to

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

(2.5)

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

(2.6)

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

(2.7)

where

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

(2.8)

We divide the proof into two cases.

Case 1. p = 2/3. Then, we clearly see that

$$2p-1=3p^2-1=\frac{1}{3}>0,$$

(2.9)

and

$$\begin{array}{c}{\left[\sqrt{2}p\left(2p-1\right)\left(t+1\right)\sqrt{1+t^2}\right]}^2-{\left[\left(3p^2-1\right)t^2+2{\left(p-1\right)}^{2}t+3p^2-1\right]}^2\\ =-\frac{{\left(t-1\right)}^4}{81}<0\end{array}$$

(2.10)

for t > 1.

Therefore, inequality (2.1) follows from (2.3)-(2.5) and (2.7)-(2.10).

Case 2. $p=\left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]=0.659\cdots$. Then, simple computations yield that

$$2p-1>0,$$

(2.11)

$$2-3p>0,$$

(2.12)

$$3p^2-1>0,$$

(2.13)

$$-p^4-8p^3+8p^2-1=-0.00456\cdots <0$$

(2.14)

and

$$\begin{array}{l}{\left[\sqrt{2}p(2p-1)(t+1)\sqrt{1+t^2}\right]}^2-{[(3p^2-1)t^2+2{(p-1)}^{2}t+3p^2-1]}^2\\ ={(t-1)}^2[(-p^4-8p^3+8p^2-1)t^2\\ +2(p^4-4p^3+6p^2-4p+1)t-p^4-8p^3+8p^2-1].\end{array}$$

(2.15)

Let

$$\begin{array}{ll}\hfill g\left(t\right)=& \left(-p^4-8p^3+8p^2-1\right)t^2+2\left(p^4-4p^3+6p^2-4p+1\right)t\\ -p^4-8p^3+8p^2-1,\end{array}$$

(2.16)

then from (2.11) and (2.12) together with (2.14), we get

$$g\left(1\right)=4p\left(2p-1\right)\left(2-3p\right)>0,$$

(2.17)

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

(2.18)

$$\begin{array}{c}{g}^{\prime }\left(t\right)=2\left(-p^4-8p^3+8p^2-1\right)t+2\left(p^4-4p^3+6p^2-4p+1\right),\\ g^{\prime }\left(1\right)=4p\left(2p-1\right)\left(2-3p\right)>0,\end{array}$$

(2.19)

$$\underset{t\to +\infty }{lim}{g}^{\prime }\left(t\right)=-\infty$$

(2.20)

and

$$g^{″}\left(t\right)=2\left(-p^4-8p^3+8p^2-1\right)=-0.00912\cdots <0.$$

(2.21)

It follows from (2.19)-(2.21) that there exists t₀ > 1 such that g'(t) > 0 for t ∈ [1, t₀) and g'(t) < 0 for t ∈ (t₀, ∞). Hence, g(t) is strictly increasing in [1, t₀] and strictly decreasing in [t₀, ∞).

From (2.17) and (2.18) together with the piecewise monotonicity of g(t), we clearly see that there exists t₁> t₀> 1 such that g(t) > 0 for t ∈ [1, t₁) and g(t) < 0 for t ∈ (t₁, ∞). Then, from (2.8), (2.11), (2.13), (2.15), and (2.16), we know that f₁(t) > 0 for t ∈ [1, t₁) and f₁(t) < 0 for t ∈ (t₁, ∞). It follows from (2.7) that f(t) is strictly increasing in [1, t₁] and strictly decreasing in [t₁, ∞).

Note that (2.6) becomes

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

(2.22)

Therefore, inequality (2.2) follows from (2.3)-(2.5) and (2.22) together with the piecewise monotonicity of f(t).

Secondly, we prove that 2S(a, b)/3 + A(a, b)/3 is the best possible upper convex combination bound of root-square and arithmetic means for the Seiffert mean T(a, b).

Letting x > 0 (x → 0) and making use of the Taylor expansion, one has

$$\begin{array}{l}{\beta }_{1}S(1,1+x)+(1-{\beta }_1)A(1,1+x)-T(1,1+x)\\ ={\beta }_1\left[[1+\frac{1}{2}x+\frac{1}{8}{x}^2+o(x^2)\right]+(1-{\beta }_1)\left(1+\frac{x}{2}\right)\\ -\left[1+\frac{1}{2}x+\frac{1}{12}{x}^2+o(x^2)\right]\\ =\frac{1}{24}(3{\beta }_1-2)x^2+o(x^2).\end{array}$$

(2.23)

Equation (2.23) implies that for any β₁ < 2/3, there exists δ₁ = δ₁(β₁) > 0, such that β₁S(1, 1 + x) + (1 - β₁)A(1, 1 + x) < T (1, 1 + x) for x ∈ (0, δ₁).

Finally, we prove that $\left(4-\pi \right)S\left(a,b\right)∕\left[\left(\sqrt{2}-1\right)\pi \right]+\left(\sqrt{2}\pi -4\right)A\left(a,b\right)∕\left[\left(\sqrt{2}-1\right)\pi \right]$ is the best possible lower convex combination bound of root-square and arithmetic means for the Seiffert mean T (a, b).

For any ${\alpha }_1>\left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]$, it follows from (1.1) and (1.2) that

$$\underset{x\to +\infty }{lim}\frac{{\alpha }_{1}S\left(1,x\right)+\left(1-{\alpha }_1\right)A\left(1,x\right)}{T\left(1,x\right)}=\frac{\left(\sqrt{2}-1\right){\alpha }_1+1}{4}\pi >1.$$

(2.24)

Inequality (2.24) implies that for any ${\alpha }_1>\left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]$, there exists X₁ = X₁(α₁) > 1 such that α₁S(1, x) + (1 α₁)A(1, x) > T (1, x) for x ∈ (X₁, ∞).

Theorem 2.2. The double inequality $S^{{\alpha }_2}\left(a,b\right)A^{1-{\alpha }_2}\left(a,b\right)<T\left(a,b\right)<S^{{\beta }_2}\left(a,b\right)A^{1-{\beta }_2}\left(a,b\right)$ holds for all a, b > 0 with a ≠ b if and only if α₂ ≤ 2/3 and β₂ ≥ 4 - 2 log = log 2 = 0.697⋯.

Proof. Firstly, we prove that

$$T\left(a,b\right)>S^{2∕3}\left(a,b\right)A^{1∕3}\left(a,b\right),$$

(2.25)

$$T\left(a,b\right)<{\left[S\left(a,b\right)\right]}^{4-2log\pi ∕log2}{\left[A\left(a,b\right)\right]}^{2log\pi ∕log2-3}$$

(2.26)

for all a, b > 0 with a ≠ b.

Without loss of generality, we assume that a > b. Let $t=\sqrt{a∕b}>1$ and q ∈ {2/3, 4 - 2 log π /log 2}, then from (1.1) and (1.2), we have

$$\begin{array}{c}logT\left(a,b\right)-\left[qlogS\left(a,b\right)+\left(1-q\right)logA\left(a,b\right)\right]\\ =log\frac{t-1}{2arctan\left(\frac{t-1}{t+1}\right)}-\frac{q}{2}log\left(\frac{1+t^2}{2}\right)-\left(1-q\right)log\left(\frac{1+t}{2}\right)\end{array}$$

(2.27)

Let

$$F\left(t\right)=log\frac{t-1}{2arctan\left(\frac{t-1}{t+1}\right)}-\frac{q}{2}log\left(\frac{1+t^2}{2}\right)-\left(1-q\right)log\left(\frac{1+t}{2}\right),$$

(2.28)

then simple computations lead to

$$\underset{t\to 1}{lim}F\left(t\right)=0,$$

(2.29)

$$\underset{t\to +\infty }{lim}F\left(t\right)=log\frac{4}{\pi }-\frac{q}{2}log2,$$

(2.30)

$$F^{\prime }\left(t\right)=\frac{\left(2-q\right)t^2+2qt+2-q}{\left(t^4-1\right)arctan\left(\frac{t-1}{t+1}\right)}{F}_1\left(t\right)$$

(2.31)

where

$$F_1\left(t\right)=arctan\left(\frac{t-1}{t+1}\right)-\frac{t^2-1}{\left(2-q\right)t^2+2qt+2-q},$$

(2.32)

$$F_1\left(1\right)=0,$$

(2.33)

$$\underset{t\to +\infty }{lim}{F}_1\left(t\right)=\frac{\pi }{4}-\frac{1}{2-q},$$

(2.34)

$$F_1^{\prime }\left(t\right)=\frac{{\left(t-1\right)}^2}{\left(1+t^2\right){\left[\left(2-q\right)t^2+2qt+2-q\right]}^2}{F}_2\left(t\right),$$

(2.35)

where

$$F_2\left(t\right)=\left(q^2-6q+4\right)t^2-2q^{2}t+q^2-6q+4,$$

(2.36)

$$F_2\left(1\right)=4\left(2-3q\right),$$

(2.37)

$$F_2^{\prime }\left(t\right)=2\left(q^2-6q+4\right)t-2q^2,$$

(2.38)

$$F_2^{\prime }\left(1\right)=4\left(2-3q\right).$$

(2.39)

We divide the proof into two cases.

Case 1. q = 2/3. Then, we clearly see that

$$2-3q=0,$$

(2.40)

$$q^2-6q+4=\frac{4}{9}>0.$$

(2.41)

From (2.38)-(2.41), we know that $F_2^{\prime }\left(t\right)>0$ for t ∈ (1, ∞). Hence, F₂(t) is strictly increasing in [1, ∞). It follows from (2.35), (2.37), (2.40), and the monotonicity of F₂(t) that F₁(t) is strictly increasing in [1, ∞).

Therefore, inequality (2.25) follows from (2.27)-(2.29), (2.31), (2.33), and the monotonicity of F₁(t).

Case 2. q = 4 - 2 log π /log 2 = 0:697⋯. Then, simple computations lead to

$$log\frac{4}{\pi }-\frac{q}{2}log2=0,$$

(2.42)

$$\frac{\pi }{4}-\frac{1}{2-q}=0.0179\cdots >0,$$

(2.43)

$$q^2-6q+4=0.303\cdots >0,$$

(2.44)

$$2-3q=-0.09102\cdots <0.$$

(2.45)

It follows from (2.36) and (2.38) together with (2.44) that

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

(2.46)

$$\underset{t\to +\infty }{lim}{F}_2^{\prime }\left(t\right)=+\infty .$$

(2.47)

From (2.38) and (2.44), we clearly see that $F_2^{\prime }\left(t\right)$ is strictly increasing in [1, ∞). Then, (2.39) and (2.45) together with (2.47) imply that there exists λ₀ > 1 such that $F_2^{\prime }\left(t\right)<0$ for t ∈ [1, λ₀) and $F_2^{\prime }\left(t\right)>0$ for t ∈ (λ₀, ∞). Hence, F₂(t) is strictly decreasing in [1, λ₀] and strictly increasing in [λ₀, ∞).

From (2.37), (2.45), (2.46), and the piecewise monotonicity of F₂(t), we know that there exists λ₁ > λ₀ > 1 such that F₂(t) < 0 for t ∈ [1, λ₁) and F₂(t) > 0 for t ∈ (λ₁, ∞). Then (2.35) implies that F₁(t) is strictly decreasing in [1, λ₁] and strictly increasing in [λ₁, ∞).

From (2.33), (2.34), (2.43), and the piecewise monotonicity of F₁(t), we conclude that there exists λ₂ > λ₁ > 1 such that F₁(t) < 0 for t ∈ (1, λ₂) and F₁(t) > 0 for t ∈ (λ₂, ∞). Then, (2.31) implies that F(t) is strictly decreasing in (1, λ₂] and strictly increasing in [λ₂, ∞).

Therefore, inequality (2.26) follows from (2.27)-(2.30) and (2.42) together with the piecewise monotonicity of F(t).

Secondly, we prove that S^2/3(a, b)A^1/3(a, b) is the best possible lower geo-metric combination bound of root-square and arithmetic means for the Seiffert mean T(a, b).

Letting x > 0 (x → 0) and making use of the Taylor expansion, one has

$$\begin{array}{c}{S}^{{\alpha }_2}\left(1,1+x\right)A^{1-{\alpha }_2}\left(1,1+x\right)-T\left(1,1+x\right)\\ =\left[1+\frac{{\alpha }_2}{2}x+\frac{{\alpha }_{2^2}}{8}{x}^2+o\left(x^2\right)\right]\left[1+\frac{1-{\alpha }_2}{2}x+\frac{{\alpha }_2\left({\alpha }_2-1\right)}{8}{x}^2+o\left(x^2\right)\right]\\ -\left[1+\frac{1}{2}x+\frac{1}{12}{x}^2+o\left(x^2\right)\right]\\ =\frac{1}{24}\left(3{\alpha }_2-2\right)x^2+o\left(x^2\right).\end{array}$$

(2.48)

Equation (2.48) implies that for any α₂ > 2/3, there exists δ₂ = δ₂(α₂) > 0, such that $S^{{\alpha }_2}\left(1,1+x\right)A^{1-{\alpha }_2}\left(1,1+x\right)>T\left(1,1+x\right)$ for x ∈ (0, δ₂).

Finally, we prove that [S(a, b)]^{4-2 logπ /log2}[A(a, b)]^{2 log π/log 2-3}is the best possible upper geometric combination bound of root-square and arithmetic means for the Seiffert mean T (a, b).

For any β₂ < 4 - 2 log π /log 2 and x > 0, from (1.1) and (1.2), one has

$$\underset{x\to +\infty }{lim}\frac{S^{{\beta }_2}\left(1,x\right)A^{1-{\beta }_2}\left(1,x\right)}{T\left(1,x\right)}=2^{{\beta }_2∕2}\times \frac{\pi }{4}<1.$$

(2.49)

Inequality (2.49) implies that for any β₂ < 4 - 2 log π /log 2, there exists X₂ = X₂(β₂) > 1 such that $T\left(1,x\right)>S^{{\beta }_2}\left(1,x\right)A^{\left(1-{\beta }_2\right)}\left(1,x\right)$ for x ∈ (X₂, +∞).