Sharp bounds by the power mean for the generalized Heronian mean

Abstract

In this article, we answer the question: For p, ω ∈ ℝ with ω > 0 and p(ω - 2) ≠ 0, what are the greatest value r₁ = r₁(p, ω) and the least value r₂ = r₂(p, ω) such that the double inequality $M_{r_1}\left(a,b\right)<H_{p,\omega }\left(a,b\right)<M_{r_2}\left(a,b\right)$ holds for all a, b > 0 with a ≠ b? Here H_p,ω(a, b) and M _r (a, b) denote the generalized Heronian mean and r th power mean of two positive numbers a and b, respectively.

2010 Mathematics Subject Classification: 26E60.

1 Introduction

In the recent past, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities can be found in the literature [1–26].

The power mean M _r (a, b) of order r of two positive numbers a and b is defined by

$$M_r\left(a,b\right)=\left\{\begin{array}{cc}{\left(\frac{a^r+b^r}{2}\right)}^{1/r},\hfill & r\ne 0,\hfill \\ \sqrt{ab},\hfill & r=0.\hfill \end{array}\right.$$

(1.1)

It is well-known that M _r (a, b) is continuous and strictly increasing with respect to r ∈ ℝ for fixed a, b > 0 with a ≠ b. Let A(a, b) = (a + b)/2, $G\left(a,b\right)=\sqrt{ab}$, H(a, b) = 2ab/(a + b), I(a, b) = 1/e(b^b/a^a )^1/(b-a)(b ≠ a), I(a, b) = a (b = a), and L(a, b) = (b-a)/(log b- log a) (b ≠ a), L(a, b) = a (b = a) be the arithmetic, geometric, harmonic, identric, and logarithmic means of two positive numbers a and b, respectively. Then

$$\begin{array}{c}\text{min}\left\{a,b\right\}\le H\left(a,b\right)=M_{-1}\left(a,b\right)\le G\left(a,b\right)=M_0\left(a,b\right)\hfill \\ \le L\left(a,b\right)\le I\left(a,b\right)\le A\left(a,b\right)=M_1\left(a,b\right)\le \text{max}\left\{a,b\right\}\hfill \end{array}$$

(1.2)

for all a, b > 0, and each inequality becomes equality if and only if a = b.

The classical Heronian mean He(a, b) of two positive numbers a and b is defined by ([27], see also [28])

$$He\left(a,b\right)=\frac{2}{3}A\left(a,b\right)+\frac{1}{3}G\left(a,b\right)=\frac{a+\sqrt{ab}+b}{3}.$$

(1.3)

In [27], Alzer and Janous established the following sharp double inequality (see also [[28], p. 350]):

$$M_{\text{log}2/\text{log}3}\left(a,b\right)<He\left(a,b\right)<M_{2/3}\left(a,b\right)$$

for all a, b > 0 with a ≠ b.

Mao [29] proved that

$$M_{1/3}\left(a,b\right)<\frac{1}{3}A\left(a,b\right)+\frac{2}{3}G\left(a,b\right)<M_{1/2}\left(a,b\right)$$

for all a, b > 0 with a ≠ b, and M_{1/ 3}(a, b) is the best possible lower power mean bound for the sum $\frac{1}{3}A\left(a,b\right)+\frac{2}{3}G\left(a,b\right)$.

For any α ∈ (0, 1), Janous [30] found the greatest value p and the least value q such that M _p (a, b) < αA(a, b) + (1 - α)G(a, b) < M _q (a, b) for all a, b > 0 with a ≠ b.

The following sharp bounds for L, I, (LI)^{1/ 2}and (L + I)/ 2 in terms of power mean are given in [10, 21–25, 31, 32]:

$$\begin{array}{c}{M}_0\left(a,b\right)<L\left(a,b\right)<M_{1/3}\left(a,b\right),\\ M_{2/3}\left(a,b\right)<I\left(a,b\right)<M_{\text{log}2}\left(a,b\right),\\ M_0\left(a,b\right)<\sqrt{L\left(a,b\right)I\left(a,b\right)}<M_{1/2}\left(a,b\right),\\ M_{\text{log}2/\left(1+\text{log}2\right)}\left(a,b\right)<\frac{1}{2}\left[L\left(a,b\right)+I\left(a,b\right)\right]<M_{1/2}\left(a,b\right)\end{array}$$

for all a, b > 0 with a ≠ b.

In [6, 7] the authors established the following sharp inequalities:

$$\begin{array}{c}{M}_{-1/3}\left(a,b\right)<\frac{2}{3}G\left(a,b\right)+\frac{1}{3}H\left(a,b\right)<M_0\left(a,b\right),\hfill \\ M_{-2/3}\left(a,b\right)<\frac{1}{3}G\left(a,b\right)+\frac{2}{3}H\left(a,b\right)<M_0\left(a,b\right),\hfill \\ M_0\left(a,b\right)<A^{\alpha }\left(a,b\right)L^{1-\alpha }\left(a,b\right)<M_{\left(1+2\alpha \right)/3}\left(a,b\right),\hfill \\ M_0\left(a,b\right)<G^{\alpha }\left(a,b\right)L^{1-\alpha }\left(a,b\right)<M_{\left(1-\alpha \right)/3}\left(a,b\right)\hfill \end{array}$$

for all for all a, b > 0 with a ≠ b and α ∈ (0, 1).

For ω ≥ 0 and p ∈ ℝ the generalized Heronian mean H_p,ω(a, b) of two positive numbers a and b was introduced in [33] as follows:

$$H_{p,\omega }\left(a,b\right)=\left\{\begin{array}{cc}{\left[\frac{a^p+\omega {\left(ab\right)}^{p/2}+b^p}{\omega +2}\right]}^{1/p},\hfill & p\ne 0,\hfill \\ \sqrt{ab},\hfill & p=0.\hfill \end{array}\right.$$

(1.4)

It is not difficult to verify that H_p,ω(a, b) is continuous with respect to p ∈ ℝ for fixed a, b > 0 and ω ≥ 0, strictly increasing with respect to p ∈ ℝ for fixed a, b > 0 with a ≠ b and ω ≥ 0, strictly decreasing with respect to ω ≥ 0 for fixed a, b > 0 with a ≠ b and p > 0 and strictly increasing with respect to ω ≥ 0 for fixed a, b > 0 with a ≠ b and p < 0.

From (1.1) and (1.3) together with (1.4) we clearly see that H_p,0(a, b) = M _p (a, b), $H_{p,2}\left(a,b\right)=M_{\frac{p}{2}}\left(a,b\right)$, H_0,ω(a, b) = M₀(a, b) and H_1,1(a, b) = H _e (a, b) for all a, b > 0 and ω ≥ 0.

The purpose of this article is to answer the question: For p, ω ∈ ℝ with ω > 0 and p(ω - 2) ≠ 0, what are the greatest value r₁ = r₁(p, ω) and the least value r₂ = r₂(p, ω) such that the double inequality $M_{r_1}\left(a,b\right)<H_{p,\omega }\left(a,b\right)<M_{r_2}\left(a,b\right)$ holds for all a, b > 0 with a ≠ b?

2 Main result

In order to establish our main results we need the following Lemma 2.1.

Lemma 2.1. (see [30]). (ω + 2)²> 2^ω+2for ω ∈ (0, 2), and (ω + 2)²< 2^ω+2for ω ∈ (2, +∞).

Theorem 2.1. For all a, b > 0 with a ≠ b we have

$$M_{\frac{2}{\omega +2}p}\left(a,b\right)<H_{p,\omega }\left(a,b\right)<M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}\left(a,b\right)$$

for (p, ω) ∈ {(p, ω): p > 0, ω > 2} ∪ {(p, ω): p < 0, 0 < ω < 2} and

$$M_{\frac{2}{\omega +2}p}\left(a,b\right)>H_{p,\omega }\left(a,b\right)>M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}\left(a,b\right)$$

for (p, ω) ∈ {(p, ω): p > 0, 0 < ω < 2} ∪ {(p, ω): p < 0, ω > 2}, and the parameters $\frac{2}{\omega +2}p$ and $\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p$ are the best possible in either case.

Proof. Without loss of generality, we can assume that a > b and put $t=\frac{a}{b}>1$.

Firstly, we compare the value of $M_{\frac{2}{\omega +2}p}\left(a,b\right)$ with that of H_p,ω(a, b). From (1.1) and (1.4) we have

$$\begin{array}{c}\text{log}\left[M_{\frac{2}{\omega +2}p}\left(a,b\right)\right]-\text{log}\left[H_{p,\omega }\left(a,b\right)\right]\\ =\frac{\omega +2}{2p}\text{log}\frac{1+t^{\frac{2}{\omega +2}p}}{2}-\frac{1}{p}\text{log}\frac{1+\omega t^{\frac{p}{2}}+t^p}{\omega +2}.\end{array}$$

(2.1)

Let

$$f\left(t\right)=\frac{\omega +2}{2p}\text{log}\frac{1+t^{\frac{2p}{2+\omega }}}{2}-\frac{1}{p}\text{log}\frac{1+\omega t^{\frac{p}{2}}+t^p}{\omega +2}.$$

(2.2)

Then simple computations lead to

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

(2.3)

$$\begin{array}{c}{f}^{\prime }\left(t\right)=\frac{t^{\frac{2p}{\omega +2}}g\left(t\right)}{2t\left(1+t^{\frac{2p}{\omega +2}}\right)\left(1+\omega t^{\frac{p}{2}}+t^p\right)},\hfill \\ g\left(t\right)=-2t^{\frac{\omega p}{\omega +2}}+\omega t^{\frac{p}{2}}-\omega t^{\frac{\omega -2}{2\left(\omega +2\right)}p}+2,\hfill \end{array}$$

(2.4)

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

(2.5)

$$g^{\prime }\left(t\right)=\omega p{t}^{\frac{\left(\omega -2\right)p}{2\left(\omega +2\right)}-1}h\left(t\right),$$

(2.6)

$$\begin{array}{c}h\left(t\right)=-\frac{2}{\omega +2}{t}^{\frac{p}{2}}+\frac{1}{2}{t}^{\frac{2p}{\omega +2}}-\frac{\omega -2}{2\left(\omega +2\right)},\hfill \\ h\left(1\right)=0,\hfill \end{array}$$

(2.7)

$$h^{\prime }\left(t\right)=\frac{p}{\omega +2}{t}^{\frac{2p}{\omega +2}-1}\left[1-t^{\frac{\left(\omega -2\right)p}{2\left(\omega +2\right)}}\right].$$

(2.8)

We divide the comparison into two cases.

Case 1. If (p, ω) ∈ {(p, ω): p > 0, ω > 2} ∪ {(p, ω): p < 0, 0 < ω < 2}, then from (2.8) we clearly see that

$$h^{\prime }\left(t\right)<0$$

(2.9)

for t > 1.

Therefore, $M_{\frac{2}{\omega +2}p}\left(a,b\right)<H_{p,\omega }\left(a,b\right)$ follows from (2.1)-(2.7) and (2.9).

Case 2. If (p, ω) ∈ {(p, ω): p > 0, 0 < ω < 2} ∪ {(p, ω): p < 0, ω > 2}, then (2.8) leads to

$$h^{\prime }\left(t\right)>0$$

(2.10)

for t > 1.

Therefore, $M_{\frac{2}{\omega +2}p}\left(a,b\right)>H_{p,\omega }\left(a,b\right)$ follows from (2.1)-(2.7) and (2.10).

Secondly, we compare the value of $M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}\left(a,b\right)$ with that of H_p,ω(a, b). From (1.1) and (1.4) we have

$$\begin{array}{c}\text{log}\left[M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}\left(a,b\right)\right]-\text{log}\left[H_{p,\omega }\left(a,b\right)\right]\\ =\frac{\text{log}\left(\omega +2\right)}{p\text{log}2}\text{log}\frac{1+t^{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}}{2}-\frac{1}{p}\text{log}\frac{1+\omega t^{\frac{p}{2}}+t^p}{\omega +2}.\end{array}$$

(2.11)

Let

$$F\left(t\right)=\frac{\text{log}\left(\omega +2\right)}{p\text{log}2}\text{log}\frac{1+t^{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}}{2}-\frac{1}{p}\text{log}\frac{1+\omega t^{\frac{p}{2}}+t^p}{\omega +2}.$$

(2.12)

Then simple computations lead to

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

(2.13)

$$F^{\prime }\left(t\right)=\frac{t^{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}G\left(t\right)}{t\left(1+t^{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}\right)\left(1+\omega t^{\frac{p}{2}}+t^p\right)},$$

(2.14)

$$G\left(t\right)=-t^{\left(1-\frac{\text{log}2}{\text{log}\left(\omega +2\right)}\right)p}+\frac{\omega }{2}{t}^{\frac{p}{2}}-\frac{\omega }{2}{t^{\frac{1}{2}}}^{\left(1-\frac{2\text{log}2}{\text{log}\left(\omega +2\right)}\right)p}+1,$$

(2.15)

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

(2.16)

$$G^{\prime }\left(t\right)=p{t}^{\frac{1}{2}\left(1-\frac{2\text{log}2}{\text{log}\left(\omega +2\right)}\right)p-1}H\left(t\right),$$

(2.17)

$$H\left(t\right)=\left(\frac{\text{log}2}{\text{log}\left(\omega +2\right)}-1\right)t^{\frac{p}{2}}+\frac{\omega }{4}{t}^{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p}-\frac{\omega }{4}\left(1-\frac{2\text{log}2}{\text{log}\left(\omega +2\right)}\right),$$

(2.18)

$$H\left(1\right)=\frac{\left(\omega +2\right)\text{log}2}{2\text{log}\left(\omega +2\right)}-1,$$

(2.19)

$$\begin{array}{cc}\hfill H^{\prime }\left(t\right)& =\frac{\text{log}2-\text{log}\left(\omega +2\right)}{2\text{log}\left(\omega +2\right)}p\left[t^{\frac{1}{2}\left(1-\frac{2\text{log}2}{\text{log}\left(\omega +2\right)}\right)p}-\frac{\omega \text{log}2}{2\left(\text{log}\left(\omega +2\right)-\text{log}2\right)}\right]\hfill \\ \times t^{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p-1}.\hfill \end{array}$$

(2.20)

We divide the comparison into four cases.

Case A. If p > 0 and ω > 2, then from (2.15) and (2.18)-(2.20) together with Lemma 2.1 we clearly see that

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

(2.21)

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

(2.22)

$$H\left(1\right)>0,$$

(2.23)

and there exists a₁> 1 such that

$$H^{\prime }\left(t\right)>0$$

(2.24)

for t ∈ [1, a₁) and

$$H^{\prime }\left(t\right)<0$$

(2.25)

for t ∈ (a₁, +∞).

From (2.24) and (2.25) we know that H(t) is strictly increasing in [1, a₁] and strictly decreasing in [a₁, +∞). Then (2.22) and (2.23) together with the monotonicity of H(t) imply that there exists a₂> 1 such that H(t) > 0 for t ∈ [1, a₂) and H(t) < 0 for t ∈ (a₂, +∞). It follows from (2.17) that G(t) is strictly increasing in [1, a₂] and strictly decreasing in [a₂, +∞).

From (2.16) and (2.21) together with the monotonicity of G(t) we know that there exists a₃> 1 such that G(t) > 0 for t ∈ (1, a₃) and G(t) < 0 for t ∈ (a₃, +∞). Then (2.14) leads to that F (t) is strictly increasing in [1, a₃] and strictly decreasing in [a₃, +∞).

Therefore, $M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}}\left(a,b\right)>H_{p,\omega }\left(a,b\right)$ follows from (2.11)-(2.13) and the monotonicity of F (t).

Case B. If p > 0 and 0 < ω < 2, then (2.15) and (2.18)-(2.20) together with Lemma 2.1 lead to

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

(2.26)

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

(2.27)

$$H\left(1\right)<0,$$

(2.28)

and there exists b₁> 1 such that

$$H^{\prime }\left(t\right)<0$$

(2.29)

for t ∈ [1, b₁) and

$$H^{\prime }\left(t\right)>0$$

(2.30)

for t ∈ (b₁, +∞).

From (2.27)-(2.30) we clearly see that there exists b₂> 1 such that H(t) < 0 for t ∈ [1, b₂) and H(t) > 0 for t ∈ (b₂, +∞). Then (2.17) implies that G(t) is strictly decreasing in [1, b₂] and strictly increasing in [b₂, +∞). It follows from (2.16) and (2.26) together with the monotonicity of G(t) that there exists b₃> 1 such that G(t) < 0 for t ∈ (1, b₃) and G(t) > 0 for t ∈ (b₃, +∞). Then (2.14) leads to that F(t) is strictly decreasing in [1, b₃] and strictly increasing in [b₃, +∞).

Therefore, $M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}}\left(a,b\right)<H_{p,\omega }\left(a,b\right)$ follows from (2.11)-(2.13) and the monotonicity of F (t).

Case C. If p < 0 and ω > 2, then it follows from (2.15) and (2.18)-(2.20) together with Lemma 2.1 that

$$\underset{t\to +\infty }{\text{lim}}G\left(t\right)=1,$$

(2.31)

$$\underset{t\to +\infty }{\text{lim}}H\left(t\right)=\frac{\omega }{4}\left(\frac{2\text{log}2}{\text{log}\left(\omega +2\right)}-1\right)<0,$$

(2.32)

$$H\left(1\right)>0,$$

(2.33)

$$H^{\prime }\left(t\right)<0$$

(2.34)

for t ∈ [1, +∞).

From (2.32)-(2.34) we clearly see that there exists c₁> 1 such that H(t) > 0 for t ∈ [1, c₁) and H(t) < 0 for t ∈ (c₁, +∞). Then (2.17) implies that G(t) is strictly decreasing in [1, c₁] and strictly increasing in [c₁, +∞).

It follows from (2.16) and (2.31) together with the monotonicity of G(t) that there exists c₂> 1 such that G(t) < 0 for t ∈ (1, c₂) and G(t) > 0 for t ∈ (c₂, +∞). Then (2.14) leads to that F (t) is strictly decreasing in [1, c₂] and strictly increasing in [c₂, +∞).

Therefore, $M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}}\left(a,b\right)<H_{p,\omega }\left(a,b\right)$ follows from (2.11)-(2.13) and the monotonicity of F (t).

Case D. If p < 0 and 0 < ω < 2, then (2.15) and (2.18)-(2.20) together with Lemma 2.1 lead to

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

(2.35)

$$\underset{t\to +\infty }{\text{lim}}H\left(t\right)=\frac{\omega }{4}\left(\frac{2\text{log}2}{\text{log}\left(\omega +2\right)}-1\right)>0,$$

(2.36)

$$H\left(1\right)<0,$$

(2.37)

$$H^{\prime }\left(t\right)>0$$

(2.38)

for t > 1.

From (2.17) and (2.36)-(2.38) we clearly see that there exists d₁> 1 such that G(t) is strictly increasing in [1, d₁] and strictly decreasing in [d₁, +∞). It follows from (2.14), (2.16), (2.35) and the monotonicity of G(t) that there exists d₂> 1 such that F (t) is strictly increasing in [1, d₂] and strictly decreasing in [d₂, +∞).

Therefore, $M_{\frac{\text{log}2}{\text{log}\left(\omega +2\right)}}\left(a,b\right)>H_{p,\omega }\left(a,b\right)$ follows from (2.11)-(2.13) and the monotonicity of F(t).

Thirdly, we prove that the parameter $\frac{2}{\omega +2}p$ is the best possible in either case.

For any p, r ∈ ℝ with pr ≠ 0, ω ≥ 0 and x > 0, one has

$$\begin{array}{c}\text{log}\left[M_r\left(1,1+x\right)\right]-\text{log}\left[H_{p,\omega }\left(1,1+x\right)\right]\hfill \\ =\frac{1}{r}\text{log}\frac{1+{\left(1+x\right)}^r}{2}-\frac{1}{p}\text{log}\frac{1+\omega {\left(1+x\right)}^{\frac{p}{2}}+{\left(1+x\right)}^p}{\omega +2}.\hfill \end{array}$$

(2.39)

Let x → 0, then the Taylor expansion leads to

$$\begin{array}{c}\frac{1}{r}\text{log}\frac{1+{\left(1+x\right)}^r}{2}-\frac{1}{p}\text{log}\frac{1+\omega {\left(1+x\right)}^{\frac{p}{2}}+{\left(1+x\right)}^p}{\omega +2}\hfill \\ =\frac{\left(\omega +2\right)r-2p}{4\left(\omega +2\right)}{x}^2+o\left(x^2\right).\hfill \end{array}$$

(2.40)

If (p, ω) ∈{(p, ω): p > 0, ω > 2} ∪ {(p, ω): p < 0, 0 < ω < 2}, then equations (2.39) and (2.40) imply that for any $r>\frac{2}{\omega +2}p$ there exists δ₁ = δ₁(r, p, ω) > 0 such that M _r (1, 1 + x) > H_p,ω(1, 1 + x) for x ∈ (0, δ₁).

If (p, ω) {(p, ω): p > 0, 0 < ω < 2} ∪ {(p, ω): p < 0, ω > 2}, then from (2.39) and (2.40) we know that for any $r<\frac{2}{\omega +2}p$ there exists δ₂ = δ₂(r, p, ω) > 0 such that M _r (1, 1 + x) < H_{p, ω}(1, 1 + x) for x ∈ (0, δ₂).

Finally, we prove that the parameter $\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p$ is the optimal parameter in either case.

For any p, r ∈ ℝ with pr > 0, ω ≥ 0 and x > 0 we have

$$\begin{array}{c}\underset{x\to +\infty }{\text{lim}}\left[\text{log}{M}_r\left(1,x\right)-\text{log}{H}_{p,\omega }\left(1,x\right)\right]\hfill \\ =\frac{1}{p}\text{log}\left(\omega +2\right)-\frac{1}{r}\text{log}2.\hfill \end{array}$$

(2.41)

If (p, ω) ∈ {(p, ω): p > 0, ω > 2} ∪ {(p, ω): p < 0, 0 < ω < 2}, then equation (2.41) implies that for any $r<\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p$ there exists X₁ = X₁(r, p, ω) > 1 such that M _r (1, x) < H_{p, ω}(1, x) for x ∈ (X₁, +∞).

If (p, ω) ω ∈ {(p, ω): p > 0, 0 < ω < 2} ∪ {(p, ω): p < 0, ω > 2}, then equation (2.41) leads to that for any $r>\frac{\text{log}2}{\text{log}\left(\omega +2\right)}p$ there exists X₂ = X₂(r, p, ω) > 1 such that M _r (1, x) > H_{p, ω}(1, x) for x ∈ (X₂, +∞).