On Huygens' Inequalities and the Theory of Means

Abstract

By using the theory of means, various refinements of Huygens' trigonometric and hyperbolic inequalities will be proved. New Huygens' type inequalities will be provided, too.

1. Introduction

The famous Huygens' trigonometric inequality (see e.g., [1–3]) states that for all 𝑥∈(0,𝜋/2) one has 2sin𝑥+tan𝑥>3𝑥. (1.1)

The hyperbolic version of inequality (1.1) has been established recently by Neuman and Sándor [3]: 2sinh𝑥+tanh𝑥>3𝑥,for𝑥>0. (1.2)

Let 𝑎,𝑏>0 be two positive real numbers. The logarithmic and identric means of 𝑎 and 𝑏 are defined by 𝐿=𝐿(𝑎,𝑏)∶=𝑏−𝑎()()()1ln𝑏−ln𝑎for𝑎≠𝑏;𝐿𝑎,𝑎=𝑎,𝐼=𝐼𝑎,𝑏∶=𝑒𝑏^𝑏𝑎^{𝑎1/(𝑏−𝑎)}(for𝑎≠𝑏);𝐼(𝑎,𝑎)=𝑎, (1.3) respectively. Seiffert's mean 𝑃 is defined by 𝑃=𝑃(𝑎,𝑏)∶=𝑎−𝑏)/(()()2arcsin((𝑎−𝑏𝑎+𝑏))for𝑎≠𝑏,𝑃𝑎,𝑎=𝑎. (1.4)

Let 𝐴=𝐴(𝑎,𝑏)∶=𝑎+𝑏2,𝐺=𝐺(𝑎,𝑏)=√()1𝑎𝑏,𝐻=𝐻𝑎,𝑏=2𝑎+1𝑏 (1.5) denote the arithmetic, geometric, and harmonic means of 𝑎 and 𝑏, respectively. These means have been also in the focus of many research papers in the last decades. For a survey of results, see, for example, [4–6]. In what follows, we will assume 𝑎≠𝑏.

Now, by remarking that letting 𝑎=1+sin𝑥, 𝑏=1−sin𝑥, where 𝑥∈(0,𝜋/2), in 𝑃,𝐺,and𝐴, we find that 𝑃=sin𝑥𝑥,𝐺=cos𝑥,𝐴=1, (1.6) so Huygens' inequality (1.1) may be written also as 𝑃>3𝐴𝐺22𝐺+𝐴=3𝐴+1𝐺=𝐻(𝐴,𝐴,𝐺). (1.7) Here 𝐻(𝑎,𝑏,𝑐) denotes the harmonic mean of the numbers 𝑎,𝑏,𝑐: 𝐻()1𝑎,𝑏,𝑐=3𝑎+1𝑏+1𝑐. (1.8)

On the other hand, by letting 𝑎=𝑒^𝑥, 𝑏=𝑒^−𝑥 in 𝐿,𝐺,and𝐴, we find that 𝐿=sinh𝑥𝑥,𝐺=1,𝐴=cosh𝑥, (1.9) so Huygens' hyperbolic inequality (1.2) may be written also as 𝐿>3𝐴𝐺22𝐴+𝐺=3𝐺+1𝐴=𝐻(𝐺,𝐺,𝐴). (1.10)

2. First Improvements

Suppose 𝑎,𝑏>0, 𝑎≠𝑏.

Theorem 2.1. One has 𝑃>𝐻(𝐿,𝐴)>3𝐴𝐺(),2𝐺+𝐴=𝐻𝐴,𝐴,𝐺𝐿>𝐻(𝑃,𝐺)>3𝐴𝐺().2𝐴+𝐺=𝐻𝐺,𝐺,𝐴 (2.1)(2.2)

Proof. The inequalities 𝑃>𝐻(𝐿,𝐴) and 𝐿>𝐻(𝑃,𝐺) have been proved in paper [7] (see Corollary 3.2). In fact, stronger relations are valid, as we will see in what follows.
Now, the interesting fact is that the second inequality of (2.1), that is, 2𝐿𝐴/(𝐿+𝐴)>3𝐴𝐺/(2𝐺+𝐴) becomes, after elementary transformations, exactly inequality (1.10), while the second inequality of (2.2), that is, 2𝑃𝐺/(𝑃+𝐺)>3𝐴𝐺/(2𝐴+𝐺) becomes inequality (1.7).

Another improvements of (1.7), respectively, (1.10) are provided by

Theorem 2.2. One has the inequalities: 𝑃>³𝐴²𝐺>3𝐴𝐺,2𝐺+𝐴𝐿>³𝐺²𝐴>3𝐴𝐺.2𝐴+𝐺 (2.3)(2.4)

Proof. The first inequality of (2.3) is proved in [6], while the first inequality of (2.8) is a well-known inequality due to Leach and Sholander [8] (see [4] for many related references). The second inequalities of (2.3) and (2.4) are immediate consequences of the arithmetic-geometric inequality applied for 𝐴, 𝐴, 𝐺 and 𝐴, 𝐺, 𝐺, respectively.

Remark 2.3. By (2.3) and (1.6), we can deduce the following improvement of the Huygens' inequality (1.1): sin𝑥𝑥>√³cos𝑥>3cos𝑥𝜋2cos𝑥+1,𝑥∈0,2. (2.5)
From (2.1) and (1.6), we get sin𝑥𝑥>2𝐿^∗𝐿^∗>+13cos𝑥𝜋2cos𝑥+1,𝑥∈0,2. 2.5^′
Similarly, by (2.4) and (1.9), we get sinh𝑥𝑥>√³cosh𝑥>3cosh𝑥2cosh𝑥+1,𝑥>0. (2.6)
From (2.2) and (1.9), we get sinh𝑥𝑥>2𝑃^∗𝑃^∗>+13cosh𝑥2cosh𝑥+1,𝑥>0. 2.6^′
Here, 𝐿^∗=𝐿(1+sin𝑥,1−sin𝑥), 𝑃^∗=𝑃(𝑒^𝑥,𝑒^−𝑥).
We note that the first inequality of (2.5) has been discovered by Adamović and Mitrinović (see [3]), while the first inequality of (2.6) by Lazarević (see [3]).
Now, we will prove that inequalities (2.2) of Theorem 2.1 and (2.4) of Theorem 2.2 may be compared in the following way.

Theorem 2.4. One has 𝐿>³𝐺²𝐴>𝐻(𝑃,𝐺)>3𝐴𝐺.2𝐴+𝐺 (2.7)

Proof. We must prove the second inequality of (2.7). For this purpose, we will use the inequality (see [6]): 𝑃<2𝐴+𝐺3. (2.8) This implies 𝐺/𝑃>3𝐺/(𝐺+2𝐴), so (1/2)(1+𝐺/𝑃)>(2𝐺+𝐴)/(𝐺+2𝐴).
Now, we will prove that 2𝐺+𝐴>𝐺+2𝐴³𝐺𝐴. (2.9)
By letting 𝑥=𝐺/𝐴∈(0,1), inequality (2.9) becomes 2𝑥+1>𝑥+2³√𝑥. (2.10)
Put 𝑥=𝑎³, where 𝑎∈(0,1). After elementary transformations, inequality (2.10) becomes (𝑎+1)(𝑎−1)³<0, which is true.

Note. The Referee suggested the following alternative proof: since 𝑃<(2𝐴+𝐺)/3 and the harmonic mean increases in both variables, it suffices to prove stronger inequality ³𝐴²𝐺>𝐻((2𝐴+𝐺)/3,𝐺) which can be written as (2.9).

Remark 2.5. The following refinement of inequalities 2.6^′ is true: sinh𝑥𝑥>√³cosh𝑥>2𝑃^∗𝑃^∗>+13cosh𝑥2cosh𝑥+1,𝑥>0. (2.11)
Unfortunately, a similar refinement to (2.7) for the mean 𝑃 is not possible, as by numerical examples one can deduce that generally 𝐻(𝐿,𝐴) and ³𝐴²𝐺 are not comparable. However, in a particular case, the following result holds true.

Theorem 2.6. Assume that 𝐴/𝐺≥4. Then one has 𝑃>𝐻(𝐿,𝐴)>³𝐴²𝐺>3𝐴𝐺.2𝐺+𝐴 (2.12)

First, prove one the following auxiliary results.

Lemma 2.7. For any 𝑥≥4, one has ³(𝑥+1)²2√³𝑥−1>𝑥√³4. (2.13)

Proof. A computer computation shows that (2.13) is true for 𝑥=4. Now put 𝑥=𝑎³ in (2.13). By taking logarithms, the inequality becomes 𝑓(𝑎)𝑎=2ln³+12−9ln𝑎+3ln(2𝑎−1)>0. (2.14)
An easy computation implies 𝑎(2𝑎−1)𝑎³+1𝑓^′(𝑎)=3(𝑎−1)𝑎²+𝑎−3. (2.15)
As ³4²+√³4−3=2√³2+(√³2)²−3=(√³2−1)(√³2+3)>0, we get that 𝑓′(𝑎)>0 for 𝑎≥√³4. This means that 𝑓(𝑎)>𝑓(√³4)>0, as the inequality is true for 𝑎=√³4.

Proof of the theorem. We will apply the inequality: 𝐿>³𝐺𝐴+𝐺2², (2.16) due to the author [9]. This implies 12𝐴1+𝐿<121+³4𝐴³𝐺(𝐴+𝐺)²=𝑁. (2.17)
By letting 𝑥=𝐴/𝐺 in (2.13), we can deduce 𝑁<³𝐴𝐺. (2.18)
So 12𝐴1+𝐿<³𝐴𝐺. (2.19)
This immediately gives 𝐻(𝐿,𝐴)>³𝐴²𝐺.

Remark 2.8. If cos𝑥≤1/4, 𝑥∈(0,𝜋/2), then sin𝑥𝑥>2𝐿^∗𝐿^∗>+1³√cos𝑥>3cos𝑥,2cos𝑥+1 (2.20) which is a refinement, in this case, of inequality 2.5^′.

3. Further Improvements

Theorem 3.1. One has 𝑃>√𝐿𝐴>³𝐴²𝐺>𝐴𝐺𝐿>3𝐴𝐺,2𝐺+𝐴𝐿>√𝐺𝑃>³𝐺²𝐴>𝐴𝐺𝑃>3𝐴𝐺.2𝐴+𝐺 (3.1)(3.2)

Proof. The inequalities 𝑃>√𝐿𝐴 and 𝐿>√𝐺𝑃 are proved in [10]. We will see, that further refinements of these inequalities are true. Now, the second inequality of (3.1) follows by the first inequality of (2.3), while the second inequality of (3.2) follows by the first inequality of (2.4). The last inequality is in fact an inequality by Carlson [11]. For the inequalities on 𝐴𝐺/𝑃, we use (2.3) and (2.8).

Remark 3.2. One has sin𝑥𝑥>√𝐿^∗>√³cos𝑥>cos𝑥𝐿^∗>3cos𝑥𝜋2cos𝑥+1,𝑥∈0,2,sinh𝑥𝑥>√𝑃^∗>√³cosh𝑥>cosh𝑥𝑃^∗>3cosh𝑥2cosh𝑥+1,𝑥>0, (3.3)(3.4) where 𝐿^∗ and 𝑃^∗ are the same as in 2.6^′ and 2.5^′.

Theorem 3.3. One has 𝑃>√𝐿𝐴>𝐻(𝐴,𝐿)>𝐴𝐿𝐼>𝐴𝐺𝐿>3𝐴𝐺,2𝐺+𝐴𝐿>𝐿⋅𝐼−𝐺>√𝐴−𝐿𝐼𝐺>√𝑃𝐺>³𝐺²𝐴>3𝐴𝐺.2𝐴+𝐺 (3.5)(3.6)

Proof. The first two inequalities of (3.5) one followed by the first inequality of (3.1) and the fact that 𝐺(𝑥,𝑦)>𝐻(𝑥,𝑦) with 𝑥=𝐿, 𝑦=𝐴.
Now, the inequality 𝐻(𝐴,𝐿)>𝐴𝐿/𝐼 may be written also as 𝐼>𝐴+𝐿2, (3.7) which has been proved in [4] (see also [12]).
Further, by Alzer's inequality 𝐿²>𝐺𝐼 (see [13]) one has 𝐿𝐼>𝐺𝐿 (3.8) and by Carlson’s inequality 𝐿<(2𝐺+𝐴)/3 (see [11]), we get 𝐴𝐿𝐼>𝐴𝐺𝐿>3𝐴𝐺,2𝐺+𝐴 (3.9) so (3.5) is proved.
The first two inequalities of (3.6) have been proved by the author in [5]. Since 𝐼>𝑃 (see [14]) and by (3.2), inequalities (3.6) are completely proved.

Remark 3.4. One has the following inequalities: sin𝑥𝑥>√𝐿^∗>2𝐿^∗𝐿^∗>𝐿+1𝐼^∗∗>cos𝑥𝐿^∗>3cos𝑥𝜋2cos𝑥+1,𝑥∈0,2, (3.10) where 𝐼^∗=𝐼(1+sin𝑥,1−sin𝑥); sinh𝑥𝑥>sinh𝑥𝑥𝑒^{𝑥coth𝑥−1}−1cosh𝑥−sinh𝑥/𝑥>𝑒^{(𝑥coth𝑥−1)/2}>√𝑃^∗>√³cosh𝑥>3cosh𝑥.2cosh𝑥+1 (3.11)

Theorem 3.5. One has 𝑃>³𝐴𝐴+𝐺2²>𝐴𝐴+2𝐺3>√𝐴𝐿>𝐻(𝐴,𝐿)>𝐴𝐿𝐼>3𝐴𝐺,2𝐺+𝐴𝐿>³𝐺𝐴+𝐺2²>√𝐼𝐺>𝐺2𝐴+𝐺3>√𝑃𝐺>³𝐺²𝐴>3𝐴𝐺.2𝐴+𝐺 (3.12)(3.13)

Proof. In (3.12), we have to prove the first three inequalities, the rest are contained in (3.5).
The first inequality of (3.12) is proved in [6]. For the second inequality, put 𝐴/𝐺=𝑡>1 By taking logarithms, we have to prove that 𝑔(𝑡)=4ln𝑡+12−3ln𝑡+23−ln𝑡>0. (3.14)
As 𝑔′(𝑡)𝑡(𝑡+1)(𝑡+2)=2(𝑡−1)>0, 𝑔(𝑡) is strictly increasing, so 𝑔(𝑡)>𝑔(1)=0. (3.15)
The third inequality of (3.12) follows by Carlson's relation 𝐿<(2𝐺+𝐴)/3 (see [11]).
The first inequality of (3.13) is proved in [9], while the second one in [15]. The third inequality follows by 𝐼>(2𝐴+𝐺)/3 (see [12]), while the fourth one by relation (2.9). The fifth one is followed by (2.3).

Remark 3.6. The first three inequalities of (3.12) offer a strong improvement of the first inequality of (3.1); the same is true for (3.13) and (3.2).

4. New Huygens Type Inequalities

The main result of this section is contained in the following:

Theorem 4.1. One has 𝑃>³𝐴𝐴+𝐺2²>3𝐴(𝐴+𝐺)>𝐴()5𝐴+𝐺2𝐺+𝐴>2𝐴+𝐺3𝐴𝐺,2𝐺+𝐴𝐿>³𝐺𝐴+𝐺2²>3𝐺(𝐴+𝐺)>𝐺()5𝐺+𝐴2𝐴+𝐺>2𝐺+𝐴3𝐴𝐺.2𝐴+𝐺 (4.1)(4.2)

Proof. The first inequalities of (4.1), respectively, (4.2) are the first ones in relations (3.12), respectively, (3.13).
Now, apply the geometric mean-harmonic mean inequality: ³𝑥𝑦²=√³1𝑥⋅𝑦⋅𝑦>3𝑥+1𝑦+1𝑦1=3𝑥+2𝑦, (4.3) for 𝑥=𝐴, 𝑦=(𝐴+𝐺)/2 in order to deduce the second inequality of (4.1). The last two inequalities become, after certain transformation, (𝐴−𝐺)²>0. (4.4)

The proof of (4.2) follows on the same lines, and we omit the details.

Theorem 4.2. For all 𝑥∈(0,𝜋/2), one has 𝑥sin𝑥+4tan2>3𝑥. (4.5) For all 𝑥>0, one has 𝑥sinh𝑥+4tanh2>3𝑥. (4.6)

Proof. Apply (1.6) for 𝑃>(3𝐴(𝐴+𝐺))/(5𝐴+𝐺) of (4.1).
As cos𝑥+1=2cos²(𝑥/2) and sin𝑥=2sin(𝑥/2)cos(𝑥/2), we get inequality (4.5). A similar argument applied to (4.6), by an application of (4.2) and the formulae cosh𝑥+1=2cosh²(𝑥/2) and sinh𝑥=2sinh(𝑥/2)cosh(𝑥/2).

Remarks 4.3. By (4.1), inequality (4.5) is a refinement of the classical Huygens inequality (1.1): 𝑥2sin𝑥+tan𝑥>sin𝑥+4tan2>3𝑥. 4.3^′
Similarly, (4.6) is a refinement of the hyperbolic Huygens inequality (1.2): 𝑥2sinh𝑥+tanh𝑥>sinh𝑥+4tanh2>3𝑥. 4.4^′
We will call (4.5) as the second Huygens inequality, while (4.6) as the second hyperbolic Huygens inequality.
In fact, by (4.1) and (4.2) refinements of these inequalities may be stated, too.
The inequality 𝑃>𝐴(2𝐺+𝐴)/(2𝐴+𝐺) gives sin𝑥𝑥>2cos𝑥+1,cos𝑥+2 (4.7)
or written equivalently: sin𝑥𝑥+3𝜋cos𝑥+2>2,𝑥∈0,2. (4.8)

Acknowledgments

The author is indebted to Professor Edward Neuman for his support and discussions on this topic. He also thanks the Referee for a careful reading of the paper and a new proof of Theorem 2.4.