# 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 ab, 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

$T\left(a,b\right)=\frac{a-b}{2arctan\phantom{\rule{2.77695pt}{0ex}}\left(\frac{a-b}{a+b}\right),}$
(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 .

Let $A\left(a,b\right)=\left(a+b\right)∕2,\phantom{\rule{2.77695pt}{0ex}}G\left(a,b\right)=\sqrt{ab}$, and H(a, b) = 2ab/(a + b) be the classical arithmetic, geometric, and harmonic means of two positive numbers a and b, respectively. In , Seiffert proved that

$A\left(a,b\right)

for all a, b > 0 with ab.

Taneja  presented that

for all a, b > 0 with a ≠ b.

In , 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 ab. 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  established that

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

for all a, b > 0 with ab, where L p (a, b) = (ap+1+ bp+1) = (ap + bp ) 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) hold for all a, b > 0 with ab.

## 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 ab 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 $p\in \left\{2∕3,\phantom{\rule{2.77695pt}{0ex}}\left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]\right\}$, then from (1.1) and (1.2) we have

$\begin{array}{c}T\left(a,b\right)-\left[pS\left(a,b\right)+\left(1-p\right)A\left(a,b\right)\right]\\ \phantom{\rule{1em}{0ex}}=\frac{b\left[\sqrt{2}p\sqrt{1+{t}^{2}}+\left(1-p\right)\left(1+t\right)\right]}{2arctan\left(\frac{t-1}{t+1}\right)}\\ \phantom{\rule{1em}{0ex}}×\phantom{\rule{2.77695pt}{0ex}}\left[\frac{t-1}{\sqrt{2}p\sqrt{1+{t}^{2}}+\left(1-p\right)\left(1+t\right)}-arctan\left(\frac{t-1}{t+1}\right)\right].\end{array}$
(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)

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

${f}_{1}\left(t\right)=\sqrt{2}p\left(2p-1\right)\left(t+1\right)\sqrt{{t}^{2}+1}-\left[\left(3{p}^{2}-1\right){t}^{2}+2{\left(p-1\right)}^{2}t+3{p}^{2}-1\right]$
(2.8)

We divide the proof into two cases.

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

$2p-1=3{p}^{2}-1=\frac{1}{3}>0,$
(2.9)

and

$\begin{array}{c}{\left[\sqrt{2}p\left(2p-1\right)\left(t+1\right)\sqrt{1+{t}^{2}}\right]}^{2}-{\left[\left(3{p}^{2}-1\right){t}^{2}+2{\left(p-1\right)}^{2}t+3{p}^{2}-1\right]}^{2}\\ \phantom{\rule{1em}{0ex}}=-\frac{{\left(t-1\right)}^{4}}{81}<0\end{array}$
(2.10)

for t > 1.

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

Case 2. $p=\left(4-\pi \right)∕\left[\left(\sqrt{2}-1\right)\pi \right]=0.659\cdots \phantom{\rule{0.3em}{0ex}}$. Then, simple computations yield that

$2p-1>0,$
(2.11)
$2-3p>0,$
(2.12)
$3{p}^{2}-1>0,$
(2.13)
$-{p}^{4}-8{p}^{3}+8{p}^{2}-1=-0.00456\cdots <0$
(2.14)

and

$\begin{array}{l}{\left[\sqrt{2}p\left(2p-1\right)\left(t+1\right)\sqrt{1+{t}^{2}}\right]}^{2}-{\left[\left(3{p}^{2}-1\right){t}^{2}+2{\left(p-1\right)}^{2}t+3{p}^{2}-1\right]}^{2}\\ \phantom{\rule{0.25em}{0ex}}={\left(t-1\right)}^{2}\left[\left(-{p}^{4}-8{p}^{3}+8{p}^{2}-1\right){t}^{2}\\ \phantom{\rule{0.25em}{0ex}}+2\left({p}^{4}-4{p}^{3}+6{p}^{2}-4p+1\right)t-{p}^{4}-8{p}^{3}+8{p}^{2}-1\right].\end{array}$
(2.15)

Let

$\begin{array}{ll}\hfill g\left(t\right)\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}& \left(-{p}^{4}-8{p}^{3}+8{p}^{2}-1\right){t}^{2}+2\left({p}^{4}-4{p}^{3}+6{p}^{2}-4p+1\right)t\phantom{\rule{2em}{0ex}}\\ -{p}^{4}-8{p}^{3}+8{p}^{2}-1,\phantom{\rule{2em}{0ex}}\end{array}$
(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)
$\begin{array}{c}{g}^{\prime }\left(t\right)=2\left(-{p}^{4}-8{p}^{3}+8{p}^{2}-1\right)t+2\left({p}^{4}-4{p}^{3}+6{p}^{2}-4p+1\right),\\ {g}^{\prime }\left(1\right)=4p\left(2p-1\right)\left(2-3p\right)>0,\end{array}$
(2.19)
$\underset{t\to +\infty }{lim}{g}^{\prime }\left(t\right)=-\infty$
(2.20)

and

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

$\begin{array}{l}{\beta }_{1}S\left(1,1+x\right)+\left(1-{\beta }_{1}\right)A\left(1,1+x\right)-T\left(1,1+x\right)\\ \phantom{\rule{0.25em}{0ex}}={\beta }_{1}\left[\left[1+\frac{1}{2}x+\frac{1}{8}{x}^{2}+o\left({x}^{2}\right)\right]+\left(1-{\beta }_{1}\right)\left(1+\frac{x}{2}\right)\\ \phantom{\rule{0.25em}{0ex}}-\left[1+\frac{1}{2}x+\frac{1}{12}{x}^{2}+o\left({x}^{2}\right)\right]\\ \phantom{\rule{0.25em}{0ex}}=\frac{1}{24}\left(3{\beta }_{1}-2\right){x}^{2}+o\left({x}^{2}\right).\end{array}$
(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) holds for all a, b > 0 with ab 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 ab.

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

$\begin{array}{c}logT\left(a,b\right)-\left[qlogS\left(a,b\right)+\left(1-q\right)logA\left(a,b\right)\right]\\ \phantom{\rule{1em}{0ex}}=log\frac{t-1}{2arctan\left(\frac{t-1}{t+1}\right)}-\frac{q}{2}log\left(\frac{1+{t}^{2}}{2}\right)-\left(1-q\right)log\left(\frac{1+t}{2}\right)\end{array}$
(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)

$\underset{t\to 1}{lim}F\left(t\right)=0,$
(2.29)
$\underset{t\to +\infty }{lim}F\left(t\right)=log\frac{4}{\pi }-\frac{q}{2}log2,$
(2.30)
${F}^{\prime }\left(t\right)=\frac{\left(2-q\right){t}^{2}+2qt+2-q}{\left({t}^{4}-1\right)arctan\phantom{\rule{2.77695pt}{0ex}}\left(\frac{t-1}{t+1}\right)}{F}_{1}\left(t\right)$
(2.31)

where

${F}_{1}\left(t\right)=arctan\phantom{\rule{2.77695pt}{0ex}}\left(\frac{t-1}{t+1}\right)-\frac{{t}^{2}-1}{\left(2-q\right){t}^{2}+2qt+2-q},$
(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

${F}_{2}\left(t\right)=\left({q}^{2}-6q+4\right){t}^{2}-2{q}^{2}t+{q}^{2}-6q+4,$
(2.36)
${F}_{2}\left(1\right)=4\left(2-3q\right),$
(2.37)
${F}_{2}^{\prime }\left(t\right)=2\left({q}^{2}-6q+4\right)t-2{q}^{2},$
(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 S2/3(a, b)A1/3(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

$\begin{array}{c}{S}^{{\alpha }_{2}}\left(1,\phantom{\rule{2.77695pt}{0ex}}1+x\right){A}^{1-{\alpha }_{2}}\left(1,\phantom{\rule{2.77695pt}{0ex}}1+x\right)-T\left(1,\phantom{\rule{2.77695pt}{0ex}}1+x\right)\\ \phantom{\rule{1em}{0ex}}=\left[1+\frac{{\alpha }_{2}}{2}x+\frac{{\alpha }_{{2}^{2}}}{8}{x}^{2}+o\left({x}^{2}\right)\right]\left[1+\frac{1-{\alpha }_{2}}{2}x+\frac{{\alpha }_{2}\left({\alpha }_{2}-1\right)}{8}{x}^{2}+o\left({x}^{2}\right)\right]\\ \phantom{\rule{1em}{0ex}}-\left[1+\frac{1}{2}x+\frac{1}{12}{x}^{2}+o\left({x}^{2}\right)\right]\\ \phantom{\rule{1em}{0ex}}=\frac{1}{24}\left(3{\alpha }_{2}-2\right){x}^{2}+o\left({x}^{2}\right).\end{array}$
(2.48)

Equation (2.48) implies that for any α2 > 2/3, there exists δ2 = δ2(α2) > 0, such that ${S}^{{\alpha }_{2}}\left(1,\phantom{\rule{2.77695pt}{0ex}}1+x\right){A}^{1-{\alpha }_{2}}\left(1,\phantom{\rule{2.77695pt}{0ex}}1+x\right)>T\left(1,\phantom{\rule{2.77695pt}{0ex}}1+x\right)$ for x (0, δ2).

Finally, we prove that [S(a, b)]4-2 logπ /log2[A(a, b)]2 log π/log 2-3is 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

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

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## Acknowledgements

This study is partly supported by the Natural Science Foundation of China (Grant no. 11071069), the Natural Science Foundation of Hunan Province (Grant no. 09JJ6003), and the Innovation Team Foundation of the Department of Education of Zhejiang Province(Grant no. T200924).

## Author information

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### Corresponding author

Correspondence to Yu-Ming Chu.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

Y-MC carried out the proof of Theorme 2.1 in this paper. M-KW carried out the proof of Theorem 2.2 in this paper. W-MG provieded the main idea of this paper. All authors read and approved the final manuscript.

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Chu, YM., Wang, MK. & Gong, WM. Two sharp double inequalities for Seiffert mean. J Inequal Appl 2011, 44 (2011). https://doi.org/10.1186/1029-242X-2011-44 