# On the 1-parameter trigonometric and hyperbolic inequalities chains

Abd Raouf Chouikha *

## Abstract

In this paper, we introduce other classes of inequalities chains depending on a parameter, extending the classical ones. We show in particular how the Huygens and Wilker inequalities are related in chains. We thus improve recent results of Chen and Mortici [4].

*Key Words and phrases:* Trigonometric functions; Sinc function; Trigonometric Inequalities.<fnref n="1" />

## 1 Introduction and classical inequalities

Wilker [8] presented the following inequality which will appear of great importance since it inspired many researchers (see [2], [3], [4], [5], [10], [13] and references therein), for $x \in (0, \frac{\pi}{2})$,

$$ \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} > 2. \qquad (1) $$

Furthermore, we have an equivalence between inequalities

$$ \cos x < \left(\frac{\sin x}{x}\right)^3 \iff 1 < \left(\frac{\sin x}{x}\right)^2 \frac{\tan x}{x}, \quad 0 < x < \frac{\pi}{2}. $$

By using the arithmetic-geometric mean inequality, Baricz and Sandor [1] have pointed out that this inequality implies

$$ \frac{2 \sin x}{x} + \frac{\tan x}{x} > 3 \quad \text{and} \quad \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} > 2 $$

<footnote>*juju153@yahoo.com. 4, Cour des Quesblais 35430 Saint-Pere, FRANCE</footnote>

<footnote n="1">¹2010 Mathematics Subject Classification: 26D07, 33B10, 33B20, 26D15</footnote>
The following inequality is due to Huygens for $0 < x < \frac{\pi}{2}$

$$ \frac{\sin x}{x} + \frac{\tan x}{2x} > \frac{3}{2}. \qquad (2) $$

The following inequality valid for $x \in (0, \frac{\pi}{2})$ is known as Cusa inequality

$$ \frac{\sin x}{x} < \frac{2}{3} + \frac{\cos x}{3}. $$

Inspired by Wilker inequality, Wu and Srivastava [9] have introduced a similar inequality of Wilker type:

$$ \left( \frac{x}{\sin x} \right)^2 + \frac{x}{\tan x} > 2. \qquad (3) $$

These inequalities were widely extended to cases related to Bessel functions [2], Lemniscate functions [3], mixed trigonometric-polynomial problems [6], [7], [11], asymptotic expansions [8]-[10], generalized hyperbolic functions.

In [2] the authors compared these two inequalities as well as the similar inequalities and provided for $x \in (0, \frac{\pi}{2})$,

$$ \frac{\sin x}{x} + \frac{\tan x}{2x} + \frac{1}{2} < \left( \frac{\sin x}{x} \right)^2 + \frac{\tan x}{x}. \qquad (4) $$

$$ \frac{x}{\sin x} + \frac{x}{2\tan x} + \frac{1}{2} < \left( \frac{x}{\sin x} \right)^2 + \frac{x}{\tan x}. \qquad (5) $$

Analog results occur for the hyperbolic functions.

More precisely, the authors proved the following chain ([15], Theorem 2.1 or Theorem 4.1) for $x \in (0, \frac{\pi}{2})$,

$$ \frac{1}{2} \left[ \left( \frac{\sin x}{x} \right)^2 + \frac{\tan x}{x} \right] > \frac{1}{3} \left[ \frac{2 \sin x}{x} + \frac{\tan x}{x} \right] > \\ \frac{1}{2} \left[ \left( \frac{x}{\sin x} \right)^2 + \frac{x}{\tan x} \right] > \frac{1}{3} \left[ \frac{2x}{\sin x} + \frac{x}{\tan x} \right] > 1. \qquad (6) $$

For the hyperbolic counterpart they proved the analog chain ([15], Theorem 2.2 or Theorem 4.2) for $x \in (0, \infty)$,

$$ \frac{1}{2} \left[ \left( \frac{\sinh x}{x} \right)^2 + \frac{\tanh x}{x} \right] > \frac{1}{3} \left[ \frac{2 \sinh x}{x} + \frac{\tanh x}{x} \right] > $$
$$ \frac{1}{2} \left[ \left( \frac{x}{\sinh x} \right)^2 + \frac{x}{\tanh x} \right] > \frac{1}{3} \left[ \frac{2x}{\sinh x} + \frac{x}{\tanh x} \right] > 1. \quad (7) $$

The following inequality which is the hyperbolic counterpart is due to Lazarevic, for $x \neq 0$

$$ \left( \frac{\sinh x}{x} \right)^3 > \cosh x. $$

Moreover, in [4] C-P. Chen and C. Mortici showed how the Huygens and Wilker inequalities are related. In this sense, they established a class of inequalities depending on a parameter n, where Huygens and Wilker inequalities are obtained when n = 1 and n = 2, respectively. More precisely they proved the following results

**Theorems 1,2,3** For every integer $n \ge 1$ and $0 < x < (0, \frac{\pi}{2})$, we have

$$ \left(\frac{\sin x}{x}\right)^n + \frac{n \tan x}{2x} - \frac{n+2}{2} > 0, \quad (8) $$

$$ \left(\frac{x}{\sin x}\right)^n + \frac{nx}{2\tan x} - \frac{n+2}{2} > 0. \quad (9) $$

For every integer $n \ge 2$ and $0 < x < (0, \frac{\pi}{2})$, we have

$$ \left(\frac{\sin x}{x}\right)^n - \frac{n}{3}\cos x + \frac{n-3}{3} > 0. \quad (10) $$

In fact, we provide the following chains depending in n which improves Theorem 1 of [4] and inequality (4) of [2]

**Theorem A** For every integer $n \ge 1$ and $x \in (0, \frac{\pi}{2})$, we have

$$ \begin{aligned} & 0 < \frac{\sin x}{x} + \frac{\tan x}{2x} - \frac{3}{2} < \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} - 2 < \dots \\ & < f(n, x) = \left(\frac{\sin x}{x}\right)^n + \frac{n \tan x}{2x} - \frac{n+2}{2}. \end{aligned} \quad (11) $$

This inequality becomes Huygens and Wilker inequality for n = 1 and n = 2, respectively.
**Theorem B** For every integer $n \ge 1$ and $x \in (0, \frac{\pi}{2})$, we have

$$ 
\begin{aligned}
0 < & \frac{x}{\sin x} + \frac{x}{2 \tan x} - \frac{3}{2} < \left( \frac{x}{\sin x} \right)^2 + \frac{x}{\tan x} - 2 < \dots \\
& < g(n, x) = \left( \frac{x}{\sin x} \right)^n + \frac{nx}{2 \tan x} - \frac{n+2}{2}.
\end{aligned} 
\quad (12) $$

Theorem B improves Theorem 2 of [4] and inequality (5) of [2]

**Theorem C** For every integer $n \ge 2$ and $x \in (0, \frac{\pi}{2})$, we have

$$ 
\begin{aligned}
0 < & \left( \frac{\sin x}{x} \right)^2 - \frac{2 \cos x}{3} - \frac{1}{3} < \left( \frac{\sin x}{x} \right)^3 - \cos x < \dots \\
& < \left( \frac{\sin x}{x} \right)^n - \frac{n \cos x}{3} + \frac{n-3}{3}.
\end{aligned} 
\quad (13) $$

Note that the case $n = 3$ corresponds to the Cusa inequality. Obviously, Theorem C is not true for $n = 1$. Theorem C asserts that inequalities of Cusa type hold true, for every integer $n \ge 2$. Theorem C improves Theorem 3 of [4]

We get also the hyperbolic analog

**Theorem D** For every integer $n \ge 1$ and $x \in (0, \infty)$, we have

$$ 
\begin{aligned}
0 < & \frac{\sinh x}{x} + \frac{\tanh x}{2x} - \frac{3}{2} < \left( \frac{\sinh x}{x} \right)^2 + \frac{\tanh x}{x} - 2 < \dots \\
& < f_h(n, x) = \left( \frac{\sinh x}{x} \right)^n + \frac{n \tanh x}{2x} - \frac{n+2}{2}.
\end{aligned} 
\quad (14) $$

**Theorem E** For every integer $n \ge 1$ and $x \in (0, \infty)$, we have

$$ 
\begin{aligned}
0 < & \frac{x}{\sinh x} + \frac{x}{2 \tanh x} - \frac{3}{2} < \left( \frac{x}{\sinh x} \right)^2 + \frac{x}{\tanh x} - 2 < \dots \\
& < g_h(n, x) = \left( \frac{x}{\sinh x} \right)^n + \frac{nx}{2 \tanh x} - \frac{n+2}{2}.
\end{aligned} 
\quad (15) $$
**Theorem F** For every integer $n \ge 2$ and $x \in (0, \infty)$, we have

$$ 
\begin{aligned}
0 < \left(\frac{\sinh x}{x}\right)^2 & - \frac{2 \cosh x}{3} - \frac{1}{3} < \left(\frac{\sinh x}{x}\right)^3 - \cosh x < \dots \\
& < \left(\frac{\sinh x}{x}\right)^n - \frac{n \cosh x}{3} + \frac{n-3}{3}.
\end{aligned} 
\quad (16) $$

Note that the case $n = 3$ corresponds to the Lazarevic inequality. However, Theorem F is not true for $n = 1$. In fact we may prove $\left(\frac{\sinh x}{x}\right) - \frac{\cosh x}{3} - \frac{2}{3} < 0$ (to see that it suffices to calculate the second derivative of $\sinh x - \frac{x \cosh x}{3} - \frac{2x}{3}$ It equals $\frac{1}{3}(\sinh x - x \cosh x) < 0$). Theorem F asserts that inequalities of Lazarevic type hold true, for every integer $n \ge 2$.

# 2 Proofs

## 2.1 Proof of Theorem A

The first inequality of (11) has been proved by [2] for $x \in (0, \frac{\pi}{2})$, (see (4) above)

$$ 
\begin{aligned}
& \frac{\sin x}{x} + \frac{\tan x}{2x} + \frac{1}{2} < \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} \iff \\
& 0 < \frac{\sin x}{x} + \frac{\tan x}{2x} - \frac{3}{2} < \left(\frac{\sin x}{x}\right)^2 + \frac{\tan x}{x} - 2 \iff \\
& 0 < \frac{\sin x}{x} \left[\frac{\sin x}{x} - 1\right] + \frac{\tan x}{2x} - \frac{1}{2}.
\end{aligned} 
\quad (17) $$

We will proceed by recurrence. Suppose up to order $p \ge 1$ inequality (11) is true for $1 \le n \le p$. Then we get

$$ 0 < \left(\frac{\sin x}{x}\right)^{p-1} + \frac{(p-1) \tan x}{2x} - \frac{p+1}{2} < \left(\frac{\sin x}{x}\right)^p + \frac{p \tan x}{2x} - \frac{p+2}{2}. $$

Compute the difference

$$ \left[ \left( \frac{\sin x}{x} \right)^{p+1} + \frac{(p+1) \tan x}{2x} - \frac{p+3}{2} \right] - \left[ \left( \frac{\sin x}{x} \right)^p + \frac{p \tan x}{2x} - \frac{p+2}{2} \right] = $$
$$ \left(\frac{\sin x}{x}\right)^p \left[\frac{\sin x}{x} - 1\right] + \frac{\tan x}{2x} - \frac{1}{2} $$

Moreover, since for $x \in (0, \frac{\pi}{2})$, $(\frac{\sin x}{x}) < 1$ and $(\frac{\sin x}{x})^p < (\frac{\sin x}{x})$. Then (17) implies

$$ \left(\frac{\sin x}{x}\right)^p \left[\frac{\sin x}{x} - 1\right] + \frac{\tan x}{2x} - \frac{1}{2} > \left(\frac{\sin x}{x}\right) \left[\frac{\sin x}{x} - 1\right] + \frac{\tan x}{2x} - \frac{1}{2} > 0. $$

Thus, inequality (11) is verified for $p + 1$.

## 2.2 Proof of Theorem B

The first inequality of (12) has been proved by [2] for $x \in (0, \frac{\pi}{2})$, (see (5) above)

$$ \begin{align} & \frac{x}{\sin x} + \frac{x}{2 \tan x} + \frac{1}{2} < \left(\frac{x}{\sin x}\right)^2 + \frac{x}{\tan x} \iff \nonumber \\ & 0 < \frac{x}{\sin x} + \frac{x}{2 \tan x} - \frac{3}{2} < \left(\frac{x}{\sin x}\right)^2 + \frac{x}{\tan x} - 2 \iff \nonumber \\ & 0 < \frac{x}{\sin x} \left[\frac{x}{\sin x} - 1\right] + \frac{x}{2 \tan x} - \frac{1}{2}. \tag{18} \end{align} $$

We will proceed by recurrence. Suppose up to order $p \ge 1$ inequality (12) is true for $1 \le n \le p$. Then we get

$$ 0 < \left(\frac{x}{\sin x}\right)^{p-1} + \frac{(p-1)x}{2 \tan x} - \frac{p+1}{2} < \left(\frac{x}{\sin x}\right)^p + \frac{px}{2 \tan x} - \frac{p+2}{2}. $$

Compute the difference

$$ \begin{align*} \left(\frac{x}{\sin x}\right)^{p+1} + \frac{(p+1)x}{2 \tan x} - \frac{p+3}{2} - \left(\frac{x}{\sin x}\right)^p + \frac{px}{2 \tan x} - \frac{p+2}{2} &= \\ \left[\frac{x}{\sin x}\right]^p \left[\frac{x}{\sin x} - 1\right] + \frac{x}{2 \tan x} - \frac{1}{2}. \end{align*} $$

Moreover, since for $x \in (0, \frac{\pi}{2})$, $(\frac{x}{\sin x}) > 1$ and $(\frac{x}{\sin x})^p > (\frac{x}{\sin x})$. Then (18) implies

$$ \left[\frac{x}{\sin x}\right]^p \left[\frac{x}{\sin x} - 1\right] + \frac{x}{2 \tan x} - \frac{1}{2} > \left(\frac{x}{\sin x}\right) \left[\frac{x}{\sin x} - 1\right] + \frac{x}{2 \tan x} - \frac{1}{2} > 0. $$

Thus, inequality (12) is verified for $p + 1$.
## 2.3 Proof of Theorem C

For the first inequality we compute the difference

$$ \left(\frac{\sin x}{x}\right)^2 - \frac{2 \cos x}{3} - \frac{1}{3} - \left[ \left(\frac{\sin x}{x}\right)^3 - \cos x \right] = \\ \left(\frac{\sin x}{x}\right)^2 \left[ 1 - \left(\frac{\sin x}{x}\right) \right] + \frac{\cos x - 1}{3}. $$

Moreover, for $x \in (0, \frac{\pi}{2})$, the following inequalities hold

$$ 1 - \frac{x^2}{6} < \frac{\sin x}{x} < 1 - \frac{x^2}{6} + \frac{x^4}{120}, \quad -\frac{x^2}{2} < \cos x - 1 < -\frac{x^2}{2} + \frac{x^4}{24}. $$

We then deduce

$$ \left(\frac{\sin x}{x}\right)^2 \left[1 - \left(\frac{\sin x}{x}\right)\right] + \frac{\cos x - 1}{3} < \left[1 - \frac{x^2}{6} + \frac{x^4}{120}\right]^2 \left(\frac{x^2}{6}\right) - \frac{x^2}{6} + \frac{x^4}{72} = \\ -\frac{x^4}{24} + \frac{x^6}{135} - \frac{x^8}{2160} + \frac{x^{10}}{86400} = -\frac{x^4}{86400} \left(3600 - 640x^2 + 40x^4 - x^6\right) < 0, \quad (19) $$

since $(3600 - 640x^2 + 40x^4 - x^6)$ is positive and has no root for $x \in (0, \frac{\pi}{2})$. This proved the first inequality of (13).

We will proceed now by recurrence. Suppose up to order $p \ge 2$ inequality (13) is true for $1 \le n \le p$. Then we get:

$$ 0 < \left(\frac{\sin x}{x}\right)^{p-1} - \frac{(p-1) \cos x}{3} + \frac{p-4}{3} < \left(\frac{\sin x}{x}\right)^p - \frac{p \cos x}{3} + \frac{p-3}{3}. $$

Compute the difference

$$ \left(\frac{\sin x}{x}\right)^p - \frac{p \cos x}{3} + \frac{p-3}{3} - \left[ \left(\frac{\sin x}{x}\right)^{p+1} - \frac{(p+1) \cos x}{3} + \frac{p-2}{3} \right] = \\ \left(\frac{\sin x}{x}\right)^p \left[ 1 - \left(\frac{\sin x}{x}\right) \right] + \frac{\cos x - 1}{3}. $$

Moreover, since for $x \in (0, \frac{\pi}{2})$, $\left(\frac{\sin x}{x}\right) < 1$ and $\left(\frac{\sin x}{x}\right)^p < \left(\frac{\sin x}{x}\right)$. Then (19) implies

$$ \left(\frac{\sin x}{x}\right)^p \left[1 - \left(\frac{\sin x}{x}\right)\right] + \frac{\cos x - 1}{3} < \left(\frac{\sin x}{x}\right)^2 \left[1 - \left(\frac{\sin x}{x}\right)\right] + \frac{\cos x - 1}{3} < 0. $$

Thus, inequality (13) is verified for $p+1$.
## 2.4 Proof of Theorem D

The first inequality of (14) has been proved by [2] for $x \in (0, \infty)$, (see (7) above)

$$ 
\begin{aligned}
& \frac{\sinh x}{x} + \frac{\tanh x}{2x} + \frac{1}{2} < \left(\frac{\sinh x}{x}\right)^2 + \frac{\tanh x}{x} \iff \\
& 0 < \frac{\sinh x}{x} + \frac{\tanh x}{2x} - \frac{3}{2} < \left(\frac{\sinh x}{x}\right)^2 + \frac{\tanh x}{x} - 2 \iff \\
& 0 < \frac{\sinh x}{x} \left[ \frac{\sinh x}{x} - 1 \right] + \frac{\tanh x}{2x} - \frac{1}{2}. \qquad (20)
\end{aligned}
$$

We will proceed by recurrence. Suppose up to order $p \ge 1$ inequality (14) is true for $1 \le n \le p$. Then we get

$$ 0 < \left(\frac{\sinh x}{x}\right)^{p-1} + \frac{(p-1)\tanh x}{2x} - \frac{p+1}{2} < \left(\frac{\sinh x}{x}\right)^p + \frac{p\tanh x}{2x} - \frac{p+2}{2}. $$

Compute the difference

$$ \left[ \left( \frac{\sinh x}{x} \right)^{p+1} + \frac{(p+1)\tanh x}{2x} - \frac{p+3}{2} \right] - \left[ \left( \frac{\sinh x}{x} \right)^p + \frac{p\tanh x}{2x} - \frac{p+2}{2} \right] = \left( \frac{\sinh x}{x} \right)^p \left[ \frac{\sinh x}{x} - 1 \right] + \frac{\tanh x}{2x} - \frac{1}{2}. $$

Moreover, since for $x \in (0, \infty)$, $(\frac{\sinh x}{x}) > 1$ and $(\frac{\sinh x}{x})^p > (\frac{\sinh x}{x})$. Then (20) implies

$$ \left(\frac{\sinh x}{x}\right)^p \left[\frac{\sinh x}{x} - 1\right] + \frac{\tanh x}{2x} - \frac{1}{2} > \left(\frac{\sinh x}{x}\right) \left[\frac{\sinh x}{x} - 1\right] + \frac{\tanh x}{2x} - \frac{1}{2} > 0. $$

Thus, inequality (14) is verified for $p+1$.

## 2.5 Proof of Theorem E

The first inequality of (15) has been proved by [2] for $x \in (0, \infty)$, (see (7) above)

$$ 
\begin{aligned}
& \frac{x}{\sinh x} + \frac{x}{2\tanh x} + \frac{1}{2} < \left(\frac{x}{\sinh x}\right)^2 + \frac{x}{\tanh x} \iff \\
& 0 < \frac{x}{\sinh x} + \frac{x}{2\tanh x} - \frac{3}{2} < \left(\frac{x}{\sinh x}\right)^2 + \frac{x}{\tanh x} - 2 \iff
\end{aligned}
$$
$$0 < \frac{x}{\sinh x} \left[ \frac{x}{\sinh x} - 1 \right] + \frac{x}{2 \tanh x} - \frac{1}{2}. \quad (21)$$

We will proceed by recurrence. Suppose up to order $p \ge 1$ inequality (15) is true for $1 \le n \le p$. Then we get

$$0 < \left(\frac{x}{\sinh x}\right)^{p-1} + \frac{(p-1)x}{2\tanh x} - \frac{p+1}{2} < \left(\frac{x}{\sinh x}\right)^p + \frac{px}{2\tanh x} - \frac{p+2}{2}.$$

Compute the difference

$$\left(\frac{x}{\sinh x}\right)^{p+1} + \frac{(p+1)x}{2\tanh x} - \frac{p+3}{2} - \left(\frac{x}{\sinh x}\right)^p + \frac{px}{2\tanh x} - \frac{p+2}{2} = \\ \left[\frac{x}{\sinh x}\right]^p \left[\frac{x}{\sinh x} - 1\right] + \frac{x}{2\tanh x} - \frac{1}{2}.$$

Moreover, since for $x \in (0, \infty)$, $(\frac{x}{\sinh x}) < 1$ and $(\frac{x}{\sinh x})^p < (\frac{x}{\sinh x})$. Then (21) implies

$$\left[\frac{x}{\sinh x}\right]^p \left[\frac{x}{\sinh x} - 1\right] + \frac{x}{2\tanh x} - \frac{1}{2} > \left(\frac{x}{\sinh x}\right) \left[\frac{x}{\sinh x} - 1\right] + \frac{x}{2\tanh x} - \frac{1}{2} > 0.$$

Thus, inequality (15) is verified for $p+1$.

## 2.6 Proof of Theorem F

For the first inequality we compute the difference

$$\left(\frac{\sinh x}{x}\right)^2 - \frac{2 \cosh x}{3} - \frac{1}{3} - \left[\left(\frac{\sinh x}{x}\right)^3 - \cosh x\right] = \\ \left(\frac{\sinh x}{x}\right)^2 \left[1 - \left(\frac{\sinh x}{x}\right)\right] + \frac{\cosh x - 1}{3}.$$

Using the Lazarevic inequality

$$\left(\frac{\sinh x}{x}\right)^3 > \cosh x.$$

we deduce

$$\left(\frac{\sinh x}{x}\right)^2 \left[1 - \left(\frac{\sinh x}{x}\right)\right] + \frac{\cosh x - 1}{3} < \left(\frac{\sinh x}{x}\right)^2 \left[1 - \left(\frac{\sinh x}{x}\right)\right] + \frac{\left(\frac{\sinh x}{x}\right)^3 - 1}{3}$$
$$ = \left(\frac{\sinh x}{x}\right)^2 - \frac{2\left(\frac{\sinh x}{x}\right)^3}{3} - \frac{1}{3} = -\frac{1}{3}\left(2\frac{\sinh x}{x} + 1\right)\left(\frac{\sinh x}{x} - 1\right)^2 < 0. $$

for $x \in (0, \infty)$. This proved the first inequality of (16).

We will proceed now by recurrence. Suppose up to order $p \ge 2$ inequality (16) is true for $2 \le n \le p$. Then we get :

$$ 0 < \left(\frac{\sinh x}{x}\right)^{p-1} - \frac{(p-1)\cosh x}{3} + \frac{p-4}{3} < \left(\frac{\sinh x}{x}\right)^p - \frac{p\cosh x}{3} + \frac{p-3}{3}. $$

Compute the difference

$$ \left(\frac{\sinh x}{x}\right)^p - \frac{p \cosh x}{3} + \frac{p-3}{3} - \left[ \left(\frac{\sinh x}{x}\right)^{p+1} - \frac{(p+1)\cosh x}{3} + \frac{p-2}{3} \right] = \left(\frac{\sinh x}{x}\right)^p \left[1 - \left(\frac{\sinh x}{x}\right)\right] + \frac{\cosh x - 1}{3}. $$

Consider at first

**Lemma 2.6** Let $p \ge 2$ be an integer then the X-polynomial

$$ P(X) = X^p (1-X) + \left(\frac{X^3-1}{3}\right) $$

is negative for $X \ge 0$.

Indeed,

$$ P(X) = \frac{1}{3} \left[ X^p (1-X) + (X-1) (X^2 + X + 1) \right] = $$

$$ \frac{1}{3}(X-1)\left[X^2 + X + 1 - 3X^p\right] = -\frac{1}{3}(X-1)^2\left[1 + 2X + 3X^2 + 3X^3 + \dots + 3X^{p-1}\right]. $$

Then using again the Lazarevic inequality, we derive

$$ \left(\frac{\sinh x}{x}\right)^p \left[1 - \left(\frac{\sinh x}{x}\right)\right] + \frac{\cosh x - 1}{3} < \left(\frac{\sinh x}{x}\right)^p \left[1 - \left(\frac{\sinh x}{x}\right)\right] + \frac{\left(\frac{\sinh x}{x}\right)^3 - 1}{3} = \\ \frac{1}{3} \left(\left(\frac{\sinh x}{x}\right) - 1\right) \left[\left(\frac{\sinh x}{x}\right)^2 + \left(\frac{\sinh x}{x}\right) + 1 - 3\left(\frac{\sinh x}{x}\right)^p\right] = \\ -\frac{1}{3} \left(\left(\frac{\sinh x}{x}\right) - 1\right)^2 \left[1 + 2\left(\frac{\sinh x}{x}\right) + 3\left(\frac{\sinh x}{x}\right)^2 + 3\left(\frac{\sinh x}{x}\right)^3 + \dots + \left(\frac{\sinh x}{x}\right)^{p-1}\right] < 0, $$

by Lemma 2.6 where $X = \frac{\sinh x}{x}$. Thus, inequality (16) is verified for $p+1$.
### 3 Further remarks

The interesting question is whether it is possible to generalize the previous chains of inequalities or to find new ones dependent on n.

Consider again inequalities chains proved by [15] (Theorem 2.1 or Theorem 4.1) for $x \in (0, \frac{\pi}{2})$,

$$ 
\begin{aligned}
& \frac{1}{2} \left[ \left( \frac{\sin x}{x} \right)^2 + \frac{\tan x}{x} \right] > \frac{1}{3} \left[ \frac{2 \sin x}{x} + \frac{\tan x}{x} \right] > \\
& \frac{1}{2} \left[ \left( \frac{x}{\sin x} \right)^2 + \frac{x}{\tan x} \right] > \frac{1}{3} \left[ \frac{2x}{\sin x} + \frac{x}{\tan x} \right] > 1
\end{aligned} 
\quad (6) $$

as well as [15] (Theorem 2.2 or Theorem 4.2) for $x \in (0, \infty)$,

$$ 
\begin{aligned}
& \frac{1}{2} \left[ \left( \frac{\sinh x}{x} \right)^2 + \frac{\tanh x}{x} \right] > \frac{1}{3} \left[ \frac{2 \sinh x}{x} + \frac{\tanh x}{x} \right] > \\
& \frac{1}{2} \left[ \left( \frac{x}{\sinh x} \right)^2 + \frac{x}{\tanh x} \right] > \frac{1}{3} \left[ \frac{2x}{\sinh x} + \frac{x}{\tanh x} \right] > 1.
\end{aligned} 
\quad (7) $$

We will examine the two cases separately

#### 3.1 Trigonometric case

Recall at first (11) and (12)

$$ f(n, x) = \left(\frac{\sin x}{x}\right)^n + \frac{n \tan x}{2x} - \frac{n+2}{2}, \quad g(n, x) = \left(\frac{x}{\sin x}\right)^n + \frac{nx}{2 \tan x} - \frac{n+2}{2}. $$

It is interesting to ask if we have analog chains for other values of $n \neq 1, 2$ like as those provided Theorems A,B. Unfortunately, the answer is no. We cannot provide more generalized chains. Indeed, one proved $f(1, x) - g(n, x) > 0$, for $n = 1, 2$. Otherwise, thanks to Maple we may also prove :

$$ 
\begin{aligned}
f(2, x) - g(n, x) &> 0, \quad \text{for } n \le 3, \\
f(3, x) - g(n, x) &> 0, \quad \text{for } n \le 4, \\
f(4, x) - g(n, x) &> 0, \quad \text{for } n \le 5, \\
f(5, x) - g(n, x) &> 0, \quad \text{for } n \le 7, \dots
\end{aligned} 
$$

We may expect that

$$ 
\begin{aligned}
f(p, x) - g(n, x) &> 0, \quad \text{for } 1 \le n \le p+1, \\
f(p, x) - g(n, x) &> 0, \quad \text{for } 5 \le n \le p+2.
\end{aligned} 
$$
## 3.2 Hyperbolic case

Recall (14) and (15)

$$f_h(n, x) = \left(\frac{\sinh x}{x}\right)^n + \frac{n \tanh x}{2x} - \frac{n+2}{2}, \quad g_h(n, x) = \left(\frac{x}{\sinh x}\right)^n + \frac{nx}{2 \tanh x} - \frac{n+2}{2}.$$

In this case too unfortunately, we have not analog chains for other values of $n \neq 1, 2$ like as those provided Theorems D,E. We have only $f_h(1, x) - g_h(n, x) > 0$, for $n = 1, 2$. Otherwise, thanks to Maple we may prove : - the difference $f_h(p, x) - g_h(n, x)$ has always zero for $p = 1, 2, 3$.

- However we have

$$f_h(4, x) - g_h(n, x) > 0, \quad \text{for } n \leq 6,$$

$$f_h(5, x) - g_h(n, x) > 0, \quad \text{for } n \leq 7,$$

$$f_h(6, x) - g_h(n, x) > 0, \quad \text{for } n \leq 8, \dots$$

We may expect that

$$f_h(p, x) - g_h(n, x) > 0, \quad \text{for } 4 \leq p, \ n \leq p+2.$$

# 4 Concluding remarks and perspectives

Following ideas of [4] C-P. Chen and C. Mortici who proposed to study the above families of inequalities depending on a (positive) real parameter $n$ in order to extend their Theorems 1, 2, 3. They proved Theorems 4, 5 as extensions of Theorems 1,2. We may hope to replace $n$ by real parameters for which the chains introduced in the above Theorems A-F hold true.

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