Since *l*_{
p
} is (*B*_{
p
} , *G*)-stabilizable, we then will be interested by bounds of *l*_{
p
} in terms of *B*_{
p
} and *G*.

It is worth noticing that, for given *p*, bounds of *lp* in the form {B}_{p}^{\alpha}{G}^{1-\alpha} (resp., *αB*_{
p
} + (1 − *α*)G) exist for some *α* ∈ [0, 1]. This follows from (1.2) with the relationships

L\left({a}^{2},\phantom{\rule{2.77695pt}{0ex}}{b}^{2}\right)=L\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)A\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right),{\left({l}_{p}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)\right)}^{p}=L\left({a}^{p},\phantom{\rule{2.77695pt}{0ex}}{b}^{p}\right),;{\left({B}_{p}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)\right)}^{p}=A\left({a}^{p},\phantom{\rule{2.77695pt}{0ex}}{b}^{p}\right)

(4.1)

valid for all *a, b* > 0 and *p* ≠ 0.

We begin by regarding bounds of *l*_{
p
} in a convex-geometric form {B}_{p}^{\alpha}{G}^{1-\alpha} as well:

**Theorem 4.1.** *Let α*, *β* ∈ [0, 1] *be such that*

{B}_{p}^{\alpha}{G}^{1-\alpha}<{l}_{p}<{B}_{p}^{\beta}{G}^{1-\beta}

(4.2)

*for some p. Then there holds*

{G}^{\frac{1-\alpha}{2}}{B}_{p/2}^{\frac{1+\alpha}{2}}={G}^{\frac{1-\alpha}{2}}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{\frac{1+\alpha}{2}}<{l}_{p}<{G}^{\frac{1-\beta}{2}}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{\frac{1+\beta}{2p}}={G}^{\frac{1-\beta}{2}}{B}_{p/2}^{\frac{1+\beta}{2}}.

(4.3)

*Proof*. Since *lp* is (*Bp*, *G*)-stabilizable then Theorem 3.2 gives

\mathcal{R}\left({B}_{p},\phantom{\rule{2.77695pt}{0ex}}{B}_{p}^{\alpha}{G}^{1-\alpha},\phantom{\rule{2.77695pt}{0ex}}G\right)<{l}_{p}<\mathcal{R}\left({B}_{p},\phantom{\rule{2.77695pt}{0ex}}{B}_{p}^{\beta}{G}^{1-\beta},\phantom{\rule{2.77695pt}{0ex}}G\right).

(4.4)

According to Lemma 2.3 we have, for all *a*, *b* > 0,

\begin{array}{ll}\hfill \mathcal{R}\left({B}_{p},\phantom{\rule{2.77695pt}{0ex}}{B}_{p}^{\alpha}{G}^{1-\alpha},\phantom{\rule{2.77695pt}{0ex}}G\right)\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)\phantom{\rule{2.77695pt}{0ex}}& =\phantom{\rule{2.77695pt}{0ex}}{B}_{p}\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right)\left({B}_{p}^{\alpha}{G}^{1-\alpha}\right)\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right)\phantom{\rule{2em}{0ex}}\\ ={B}_{p}^{1+\alpha}\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right){G}^{1-\alpha}\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right).\phantom{\rule{2em}{0ex}}\end{array}

(4.5)

For all *a*, *b* > 0, we can write G\left(\sqrt{a},\sqrt{b}\right)={G}^{1/2}\left(a,b\right) and it is easy to verify that,

{B}_{p}\left(\sqrt{a},\sqrt{b}\right)={\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/2p}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right)={B}_{p/2}^{1/2}\left(a,\phantom{\rule{2.77695pt}{0ex}}b\right),

(4.6)

from which the desired double inequality (4.3) follows. □

**Corollary 4.2.** *Let α*, *β* ∈ [0, 1] *be two real numbers such that*

{A}^{\alpha}{G}^{1-\alpha}<L<{A}^{\beta}{G}^{1-\beta}.

(4.7)

*Then there holds*

{G}^{\frac{1-\alpha}{2}}{\left(\frac{A+G}{2}\right)}^{\frac{1+\alpha}{2}}<L<{G}^{{}^{\frac{1-\beta}{2}}}{\left(\frac{A+G}{2}\right)}^{\frac{1+\beta}{2}}.

(4.8)

*Proof*. Taking *p* = 1 in the above theorem, with the fact that *l*_{1} = *L* and *B*_{1} = *A*, we immediately obtain the announced result. □

Let us now examine the following examples in the aim to illustrate the above theoretical results.

*Example* 4.1. It is not hard to verify that *G < l*_{
p
} *< B*_{
p
} for every *p* > 0, with reversed double inequality for *p <* 0. Theorem 4.1 immediately gives (with *α* = 0, *β* = 1 for *p* > 0; *α* = 1, *β* = 0 for *p* < 0)

{G}^{1/2}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/2p}<{l}_{p}<{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/p}

(4.9)

for each real number *p* ≠ 0. It is easy to verify that the double inequality (4.9) refines the initial one. In particular, we have

G<{\left(\frac{AG+{G}^{2}}{2}\right)}^{1/2}<L<\frac{A+G}{2}<A,

(4.10)

which refines the arithmetic-logarithmic-geometric mean inequality *G* < *L* < *A*.

**Theorem 4.3.** *Let α* ∈ [0, 1] *be such that*

{B}_{p}^{\alpha}{G}^{1-\alpha}<\left(>\right){l}_{p}

(4.11)

*for some p* > (<)0, *respectively. Then one has*

{B}_{p}^{\frac{1+\alpha}{4}}{G}^{\frac{3-\alpha}{4}}<\left(>\right){l}_{p}.

(4.12)

*If moreover α <* (>)1/3 *then (4.12) refines (4.11)*.

*Proof*. Assume that

{B}_{p}^{\alpha}{G}^{1-\alpha}<{l}_{p}

(4.13)

for some *p* > 0. According to Theorem 4.1, the first inequality of (4.3) holds and the arithmetic-geometric mean inequality gives

{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{\frac{1+\alpha}{2p}}>{B}_{p}^{\frac{1+\alpha}{4}}{G}^{\frac{1+\alpha}{4}}.

(4.14)

The desired inequality follows after a simple reduction. Further, the inequality

{B}_{p}^{\alpha}{G}^{1-\alpha}<{B}_{p}^{\frac{1+\alpha}{4}}{G}^{\frac{3-\alpha}{4}}

(4.15)

for *p* > 0 is reduced to

{G}^{\frac{1-3\alpha}{4}}<{B}_{p}^{\frac{1-3\alpha}{4}}

(4.16)

which holds when *α <* 1/3. For the reversed inequalities, the same arguments as previous study, so completes the proof. □

If we get *p* = 1 in the above theorem, we immediately obtain the following result.

**Corollary 4.4.** *Let α* **∈** [0, 1] *be a real number satisfying that*

{A}^{\alpha}{G}^{1-\alpha}<L.

(4.17)

*Then one has*

{A}^{\frac{1+\alpha}{4}}{G}^{\frac{3-\alpha}{4}}<L.

(4.18)

*If moreover α <* 1/3 *then (4.18) refines (4.17)*.

Theorem 4.3 tells us that every given bound of *l*_{
p
} in a convex-geometric form yields another bound of *l*_{
p
} in an analogs, but different, form. Illustrating this latter point, we will deduce a better bound of *l*_{
p
} than the above ones. Precisely, we may state the next result.

**Theorem 4.5.** *Let p be a real number. If p* > 0 *then one has*

{B}_{p}^{1/3}{G}^{2/3}<{l}_{p}.

(4.19)

*If p <* 0 *then the above inequality is reversed. In particular the following inequality holds true*

{A}^{1/3}{G}^{2/3}<L.

(4.20)

*Proof*. Assume that *p* > 0. Starting from *G* < *l*_{
p
}(see Example 4.1), we are in the situation of Theorem 4.3 with *α* = 0, and so we have {B}_{p}^{1/4}{G}^{3/4}<{l}_{p}. Let us iterate successively this procedure: if in the step *n*, we have

{B}_{p}^{{\alpha}_{n}}{G}^{1-{\alpha}_{n}}<{l}_{p}

(4.21)

then in the step *n* + 1, we obtain

{B}_{p}^{\frac{1+{\alpha}_{n}}{4}}{G}^{\frac{3-{\alpha}_{n}}{4}}<{l}_{p},

(4.22)

that is,

{B}_{p}^{{\alpha}_{n+1}}{G}^{1-{\alpha}_{n+1}}<{l}_{p}\phantom{\rule{0.3em}{0ex}}\mathsf{\text{with}}\phantom{\rule{0.3em}{0ex}}{\alpha}_{n+1}=\frac{1+{\alpha}_{n}}{4},{\alpha}_{0}=\alpha .

(4.23)

It is easy to see that the real sequences (*α*_{
n
} ) _{
n
} converges to 1/3 for every given initial data *α*_{0} ∈ [0, 1]. The desired inequality follows by letting *n →* +*∞* in the recursive inequality

{B}_{p}^{{\alpha}_{n}}{G}^{1-{\alpha}_{n}}<{l}_{p}.

(4.24)

The proof is similar for *p <* 0. Taking *p* = 1 in (4.19) we obtain (4.20), so completes the proof. □

To understand the interest of the above theorem, let us observe the following example.

*Example* 4.2. Let us apply Theorem 4.3 to the previous inequality {B}_{p}^{1/3}{G}^{2/3}<\left(>\right){l}_{p}. Then, the next inequality

{l}_{p}^{3p}>{G}^{p}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{2}

(4.25)

holds true for each real number *p* (*p* ≠ 0). In particular, taking *p* = 1 we obtain

G{\left(\frac{A+G}{2}\right)}^{2}<{L}^{3},

(4.26)

which refines *A*^{1/3}*G*^{2/3} < *L*.

*Remark* 4.1. The inequality (4.20) was proved by Leach and Sholander [6], while (4.26) has been shown by Sāndor [10]. These two inequalities were proved by different methods therein while together obtained here via the same approach. In the same sense, other examples will be seen later (see Remarks 4.4, 4.5, and 5.1).

*Remark* 4.2. As well known, inequality (4.20) is the best possible in the sense that the constant *α* = 1/3 cannot be improved in *A*^{α}*G*^{1−α}< *L*. This latter point rejoins the fact that if we apply Corollary 4.4 to (4.20) we obtain the same inequality.

*Remark* 4.3. By virtue of the relationships (4.1), it has been possible to begin by stating and proving the results of the above corollaries and then to deduce those of the corresponding theorems (with discussion on *p*). Details of this latter point are omitted for the reader.

Now, we will be interested by bounds of *l*_{
p
} in a convex-arithmetic expression as well:

**Theorem 4.6.** *Let α*, *β* ∈ [0, 1] *be two real numbers such that*

\alpha {B}_{p}+\left(1-\alpha \right)G<{l}_{p}<\beta {B}_{p}+\left(1-\beta \right)G,

(4.27)

*for some real number p. Then there holds*

\begin{array}{c}\alpha {\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/p}+\left(1-\alpha \right){G}^{1/2}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/2p}<{l}_{p}\\ \phantom{\rule{1em}{0ex}}<\beta {\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/p}+\left(1-\beta \right){G}^{1/2}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/2p}.\end{array}

(4.28)

*Proof*. By the same arguments as previous, we have

\mathcal{R}\left({B}_{p},\alpha {B}_{p}+\left(1-\alpha \right)G,G\right)<{l}_{p}<\mathcal{R}\left({B}_{p},\beta {B}_{p}+\left(1-\beta \right)G,G\right).

(4.29)

Again, thanks to Lemma 2.3, we obtain

\begin{array}{c}\alpha {\left({B}_{p}\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right)\right)}^{2}+\left(1-\alpha \right){G}^{1/2}{B}_{p}\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right)<{l}_{p}\\ \phantom{\rule{1em}{0ex}}<\beta {\left({B}_{p}\left(\sqrt{a},\phantom{\rule{2.77695pt}{0ex}}\sqrt{b}\right)\right)}^{2}+\left(1-\beta \right){G}^{1/2}{B}_{p}\left(\sqrt{a},\sqrt{b}\right).\end{array}

(4.30)

By virtue of the identity (4.6), we obtain the desired result after simple manipulations. □

As in the above, taking *p* = 1 in the latter theorem we immediately obtain the following result.

**Corollary 4.7.** *Let α*, *β* ∈ [0, 1] *be two real numbers such that*

\alpha A+\left(1-\alpha \right)G<L<\beta A+\left(1-\beta \right)G.

(4.31)

*Then there holds*

\begin{array}{c}\alpha \left(\frac{A+G}{2}\right)+\left(1-\alpha \right){\left(\frac{AG+{G}^{2}}{2}\right)}^{1/2}<L\\ \phantom{\rule{1em}{0ex}}<\beta \left(\frac{A+G}{2}\right)+\left(1-\beta \right){\left(\frac{AG+{G}^{2}}{2}\right)}^{1/2}.\end{array}

(4.32)

Theorem 4.6 has many interesting consequences. For instance, we give the two following corollaries.

**Corollary 4.8.** *Let α* ∈ [0, 1] *be such that*

L<\alpha A+\left(1-\alpha \right)G.

(4.33)

*Then we have*

L<\frac{1+\alpha}{4}A+\frac{3-\alpha}{4}G.

(4.34)

*If α* > 1=3 *then (4.34) refines (4.33)*.

*Proof*. According to Theorem 4.6, we have

L<\alpha \left(\frac{A+G}{2}\right)+\left(1-\alpha \right){\left(\frac{AG+{G}^{2}}{2}\right)}^{1/2}.

(4.35)

If we write

{\left(\frac{AG+{G}^{2}}{2}\right)}^{1/2}={G}^{1/2}{\left(\frac{A+G}{2}\right)}^{1/2}

(4.36)

and we apply the arithmetic-geometric mean inequality, i.e.,

{G}^{1/2}{\left(\frac{A+G}{2}\right)}^{1/2}<\frac{1}{2}G+\frac{1}{2}\frac{A+G}{2},

(4.37)

we obtain the announced result after substituting this latter inequality in (4.35). If *α* > 1/3, it is easy to see by similar manner as previous that (4.34) refines (4.33) and the proof is completed. □

**Corollary 4.9.**
*The following inequality holds true*

L<\frac{1}{3}A+\frac{2}{3}G.

(4.38)

*Proof*. Similarly to the above, it is sufficient to see that the sequence (*α*_{
n
}) defined by

{\alpha}_{n+1}=\frac{1+{\alpha}_{n}}{4},\phantom{\rule{1em}{0ex}}\mathsf{\text{with}}{\alpha}_{0}\in \left[0,1\right],

(4.39)

converges to 1/3 and the desired result follows as previous. We omit the routine detail here. □

*Remark* 4.4. The inequality (4.38) was differently proved by Carlson [12] and here obtained by the same approach as (4.20) and (4.26).

Let us illustrate the above theoretical examples with the following examples.

*Example* 4.3. Consider the above mean-inequality *L <* (1/3)*A* + (2/3)*G* which corresponds to *α* = 1/3 in Corollary 4.8. With this, the obtained refinement is given by

L<\frac{1}{3}\left(\frac{A+G}{2}\right)+\frac{2}{3}{\left(\frac{AG+{G}^{2}}{2}\right)}^{1/2}<\frac{1}{3}A+\frac{2}{3}G.

(4.40)

Of course, we can combine some the above results to improve the lower and upper bounds of *L*. The following example explains this situation.

*Example* 4.4. Let us consider the following double inequality

{A}^{1/3}{G}^{2/3}<L<\frac{1}{3}A+\frac{2}{3}G.

(4.41)

Combining Theorems 4.1 and 4.6 we immediately obtain

{G}^{1/3}{\left(\frac{1}{2}A+\frac{1}{2}G\right)}^{2/3}<L<\frac{1}{3}\left(\frac{1}{2}A+\frac{1}{2}G\right)+\frac{2}{3}{\left(\frac{1}{2}AG+\frac{1}{2}{G}^{2}\right)}^{1/2}.

(4.42)

The reader can easily verify that this latter double inequality refines the initial one, so proving our desired aim.

**Theorem 4.10.** *Let α* ∈ [0, 1] *be such that*

{l}_{p}<\alpha {B}_{p}+\left(1-\alpha \right)G

(4.43)

*for some p ≤* 1. *Then there holds*

{l}_{p}<\frac{1+\alpha}{4}{B}_{p}+\frac{3-\alpha}{4}G.

(4.44)

*Proof*. If (4.43) holds then Theorem 4.6 gives

{l}_{p}<\alpha {\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/p}+\left(1-\alpha \right){G}^{1/2}{\left(\frac{{B}_{p}^{p}+{G}^{p}}{2}\right)}^{1/2p}.

(4.45)

This, with *p ≤* 1 and the monotonicity of power means, yields

{l}_{p}<\alpha \left(\frac{{B}_{p}+G}{2}\right)+\left(1-\alpha \right){G}^{1/2}{\left(\frac{{B}_{p}+G}{2}\right)}^{1/2}.

(4.46)

The arithmetic-geometric mean inequality gives

{G}^{1/2}{\left(\frac{{B}_{p}+G}{2}\right)}^{1/2}<\frac{1}{2}G+\frac{1}{2}\frac{{B}_{p}+G}{2},

(4.47)

and the desired inequality follows by combining (4.46) and (4.47) with a simple reduction. □

Taking *p* = −1 in the above theorem, with the fact that *l*_{−1} = *L** = *G*^{2}/*L* and *B*_{−1} = *H* = *G*^{2}/*A*, we immediately obtain the next result.

**Corollary 4.11.** *Let α* ∈ [0, 1] *be such that*

\frac{1}{L}<\frac{\alpha}{A}+\frac{1-\alpha}{G}.

(4.48)

*Then one has*

\frac{1}{L}<\frac{1+\alpha}{4}\frac{1}{A}+\frac{3-\alpha}{4}\frac{1}{G}.

(4.49)

*If moreover α* > 1/3 *then (4.49) refines (4.48)*.

**Theorem 4.12.** *For all real number p ≤* 1 *with p* ≠ 0, *we have*

{l}_{p}<\frac{1}{3}{B}_{p}+\frac{2}{3}G.

(4.50)

*In particular, the following inequality holds*

\frac{1}{L}<\frac{1}{3}\frac{1}{A}+\frac{2}{3}\frac{1}{G}.

(4.51)

*Proof*. We left it to the reader as an interesting exercise. □

We end this section by stating another result showing how to obtain a lower bound of the logarithmic mean *L* when we start from an upper bound of its dual *L**. In fact, since *L** is (*A*, *H*)-stabilizable then we search bounds of *L** in terms of *A* and *H*. Precisely, we have the following.

**Theorem 4.13.**
*Let α be a real number satisfying that*

{L}^{*}<\alpha A+\left(1-\alpha \right)H.

(4.52)

*Then we have*

{L}^{*}<\left(\frac{1+\alpha}{4}\right)A+\left(\frac{3-\alpha}{4}\right)H.

(4.53)

*If moreover α* > 1/3 *then (4.53) refines (4.52)*.

*Proof*. Since *L** is (*A*, *H*)-stabilizable then we obtain, with Proposition 2.2,

{L}^{*}<\mathcal{R}\left(A,\alpha A+\left(1-\alpha \right)H,H\right)=\alpha \mathcal{R}\left(A,\phantom{\rule{2.77695pt}{0ex}}A,\phantom{\rule{2.77695pt}{0ex}}H\right)+\left(1-\alpha \right)\mathcal{R}\left(A,\phantom{\rule{2.77695pt}{0ex}}H,\phantom{\rule{2.77695pt}{0ex}}H\right).

(4.54)

Thanks to relationships (2.5) for obtaining

{L}^{*}<\alpha \left(\frac{A+H}{2}\right)+\left(1-\alpha \right){\left(\frac{3}{4}A+\frac{1}{4}H\right)}^{*}.

(4.55)

Due to the point-wise convexity of the mean-map *m* ↦ *m**, with *A** = *H* and *H** = *A*, we obtain

{L}^{*}<\alpha \left(\frac{A+H}{2}\right)+\left(1-\alpha \right)\left(\frac{3}{4}H+\frac{1}{4}A\right),

(4.56)

which after reduction yields the desired result.

**Corollary 4.14.**
*The following inequality holds true*

\frac{1}{L}<\frac{2}{3}\frac{1}{A}+\frac{1}{3}\frac{1}{H}.

(4.57)

*Proof*. Similarly to the same idea as in the above we have

{L}^{*}<\left(\frac{1+{\alpha}_{n}}{4}\right)A+\left(\frac{3-{\alpha}_{n}}{4}\right)H,

(4.58)

where (*α*_{
n
} ) _{
n
} is the sequence defined by the same recursive relation as in the proof of Corollary 4.9. Letting *n →* +*∞* we obtain

{L}^{*}<\frac{1}{3}A+\frac{2}{3}H.

(4.59)

The general relation *m** = *G*^{2}/*m* valid for all mean *m*, gives in particular, *L** = *G*^{2}/*L, H* = *G*^{2}/*A* and *A* = *G*^{2}/*H*. Substituting this in the latter inequality, we obtain the desired result. □

*Remark* 4.5. The inequality (4.57) was differently proved by Chen [5] and shown here by the same approach as (4.20), (4.26), and (4.38), so proving the interest of this study. Further, we notice that it is easy to verify that (4.57) is stronger than (4.51).