## 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 [111].

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 [1], Seiffert proved that

$A\left(a,b\right)

for all a, b > 0 with ab.

Taneja [5] presented that

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 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 [3] 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, +∞).