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Sharp bounds for Seiffert mean in terms of root mean square

Journal of Inequalities and Applications20122012:11

https://doi.org/10.1186/1029-242X-2012-11

• Accepted: 17 January 2012
• Published:

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)

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

2010 Mathematics Subject Classification: 26E60.

Keywords

• Seiffert mean
• root mean square
• power mean
• inequality

1 Introduction

For a,b > 0 with ab 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 .

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 +bp)/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)

for all a, b > 0 with ab.

Seiffert  proved that inequalities
$A\left(a,b\right)

hold for all a, b > 0 with ab.

Chu et al.  found the greatest value p1 and least value p2 such that the double inequality Hp 1(a,b) < T(a,b) < Hp 2(a,b) holds for all a,b > 0 with ab, 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 , 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 ab? where L r (a,b) = (ar+1+ br+1)/(a r + b r ) is the r th Lehmer mean of a and b.

Chu et al.  proved that inequalities
$pT\left(a,b\right)+\left(1-p\right)G\left(a,b\right)

hold for all a,b > 0 with ab if and only if p ≤ 3/5 and qπ/4.

Hou and Chu  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)

holds for all a, b > 0 with ab.

For fixed a, b > 0 with ab, 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)

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)
(1.3)

holds for all a,b > 0 with ab 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)
(2.2)

hold for all a, b > 0 with ab.

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
$\begin{array}{c}S\left(pa+\left(1-p\right)b,pb+\left(1-p\right)a\right)-T\left(a,b\right)\\ =b\frac{\sqrt{{\left[pt+\left(1-p\right)\right]}^{2}+{\left[\left(1-p\right)t+p\right]}^{2}}}{2\text{arctan}\left(\frac{t-1}{t+1}\right)}\\ ×\left\{\sqrt{2}\text{arctan}\left(\frac{t-1}{t+1}\right)-\frac{t-1}{\sqrt{{\left[pt+\left(1-p\right)\right]}^{2}+{\left[\left(1-p\right)t+p\right]}^{2}}}\right\}.\end{array}$
(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)
$f\left(1\right)=0,$
(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
$\begin{array}{c}{g}_{1}\left(t\right)=\left(16{p}^{6}-48{p}^{5}+72{p}^{4}-64{p}^{3}+36{p}^{2}-12p+1\right){t}^{4}-16{p}^{2}\\ \left(4{p}^{2}-4p+3\right){\left(p-1\right)}^{2}{t}^{3}+2\left(48{p}^{6}-144{p}^{5}+168{p}^{4}-96{p}^{3}+36{p}^{2}-12p+1\right)\\ ×{t}^{2}-16{p}^{2}\left(4{p}^{2}-4p+3\right){\left(p-1\right)}^{2}t+16{p}^{6}-48{p}^{5}+72{p}^{4}-64{p}^{3}+36{p}^{2}\\ -12p+1,\end{array}$
(2.10)
${g}_{1}\left(1\right)=4\left(12{p}^{2}-12p+1\right).$
(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
$\begin{array}{c}{g}_{2}\left(t\right)=\left(16{p}^{6}-48{p}^{5}+72{p}^{4}-64{p}^{3}+36{p}^{2}-12p+1\right){t}^{3}-12{p}^{2}\\ \left(4{p}^{2}-4p+3\right){\left(p-1\right)}^{2}{t}^{2}+\left(48{p}^{6}-144{p}^{5}+168{p}^{4}-96{p}^{3}+36{p}^{2}-12p+1\right)\\ t-4{p}^{2}\left(4{p}^{2}-4p+3\right){\left(p-1\right)}^{2},\end{array}$
(2.12)
${g}_{2}\left(1\right)=4\left(6{p}^{2}-12p+1\right),$
(2.13)
$\begin{array}{c}{g}_{3}\left(t\right)=3\left(16{p}^{6}-48{p}^{5}+72{p}^{4}-64{p}^{3}+36{p}^{2}-12p+1\right){t}^{2}-24{p}^{2}\left(4{p}^{2}-4p+3\right)\\ {\left(p-1\right)}^{2}t+48{p}^{6}-144{p}^{5}+168{p}^{4}-96{p}^{3}+36{p}^{2}-12p+1,\end{array}$
(2.14)
${g}_{3}\left(1\right)=4\left(6{p}^{4}-12{p}^{3}+18{p}^{2}-12p+1\right),$
(2.15)
$\begin{array}{cc}\hfill {g}_{4}\left(t\right)& =\left(16{p}^{6}-48{p}^{5}+72{p}^{4}-64{p}^{3}+36{p}^{2}-12p+1\right)t\hfill \\ \hfill \phantom{\rule{2.77695pt}{0ex}}-4{p}^{2}\left(4{p}^{2}-4p+3\right){\left(p-1\right)}^{2},\end{array}$
(2.16)
${g}_{4}\left(1\right)=12{p}^{4}-24{p}^{3}+24{p}^{2}-12p+1.$
(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
$16{p}^{6}-48{p}^{5}+72{p}^{4}-64{p}^{3}+36{p}^{2}-12p+1=\frac{1024-{\pi }^{6}}{{\pi }^{6}}>0.$
(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
${g}_{1}\left(t\right)=\frac{\left(17{t}^{2}+2t+17\right)}{108}{\left(t-1\right)}^{2}>0$
(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 ab.

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 ab.

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 ab.

For any $1/2, from (2.11) one has
${g}_{1}\left(1\right)=4\left(12{p}^{2}-12p+1\right)<0.$
(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 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).

Declarations

Acknowledgements

This research was supported by the Natural Science Foundation of China under Grant 11071069, the Natural Science Foundation of Hunan Province under Grant 09JJ6003 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

Authors’ Affiliations

(1)
Department of Mathematics and Computing Science, Hunan City University, Yiyang, 413000, China
(2)
Department of Mathematics, Hangzhou Normal University, Hangzhou, 310012, China

References 