Let $ B(n, p) $ denote a binomial random variable with parameters $ n $ and $ p $. Janson in [1] introduced the following conjecture suggested by Va\v{s}k Chvátal.

Conjecture 1 (Chvátal). For any fixed $ n\geq 2 $, as $ m $ ranges over $ \{0, \ldots, n\} $, the probability $ q_m: =P(B(n, m/n)\leq m) $ is the smallest when $ m $ is closest to $ \frac{2n}{3} $.

Conjecture 1 has applications in machine learning, such as the analysis of generalized boundaries of relative deviation bounds and unbounded loss functions ([2] and [3]). As to the probability of a binomial random variable exceeding its expectation, we refer to Doerr [2], Greenberg and Mohri [3], Pelekis and Ramon [4]. Janson [1] proved that Conjecture 1 holds for large $ n $. Barabesi et al. [5] and Sun [6] gave an affirmative answer to Conjecture 1 for general $ n\geq 2 $. Hereafter, we call Conjecture 1 by Chvátal's theorem.

Motivated by Chvátal's theorem, Li et al. [7] considered the infimum value problem on the probability that a random variable is not more than its expectation, when its distribution is the Poisson distribution, the geometric distribution or the Pascal distribution. Sun et al. [8] investigated the corresponding infimum value problem for the Gamma distribution among other things. Hu et al. [9] studied the corresponding problem for some infinitely divisible distributions including the inverse Gaussian, log-normal, Gumbel and logistic distributions. In this note, we consider the infimum value problem for the Weibull distribution and the Pareto distribution in Sections 2 and 3, respectively. Before presenting the main results, we give two remarks.

Remark 1.1 (ⅰ) In virtue of Chvátal's theorem, there is a natural question as follows:

For any fixed integer $ n\geq 2 $, how about the minimum value of the probability $ P(B(n, p)\leq np) $ for $ p\in (0, 1] $?

For small fixed $ n $, we may find the solution. Up to now, we don't know the solution for general $ n\geq 2 $. In fact, it was posed as an open question in the first version of [7] (i.e., [10]).

(ⅱ) Motivated by Chvátal's theorem, Li et al. [7] initiated the study on the infimum value problem on the probability that a random variable is not more than its expectation. In this topic, [8] is the second paper, this note is the third one and [9] is the fourth one.

Remark 1.2 Let $ X $ be a random variable with expectation $ EX $. Assume that the distribution of $ X $ contains some parameters $ \alpha $ and $ \beta $. The motivation to study $ \inf_{\alpha, \beta}P(X\leq EX) $ is that from it we can get $ \sup_{\alpha, \beta}P(X > EX) $. Obviously, if we wish the probability $ P(X > EX) $ is as large as possible, we should find $ \sup_{\alpha, \beta}P(X > EX) $ or equivalently $ \inf_{\alpha, \beta}P(X\leq EX) $. We wish that our work on this topic may find some applications in machine learning, statistics, finance and economics etc.

Let $ X $ be a Weibull random variable with parameters $ \alpha $ and $ \theta\, (\alpha > 0, \theta > 0) $ and the density function

$ f(x)=\frac{\alpha }{\theta }x^{\alpha -1}e^{-\frac{x^{\alpha}}{\theta }}, \, x>0. $

We know that its expectation $ EX=\theta ^{\frac{1}{\alpha }}\Gamma \left (\frac{1}{\alpha } +1\right), $ where $ \Gamma \left (\frac{1}{\alpha } +1\right) $ is the Gamma function, i.e., $ \Gamma \left (\frac{1}{\alpha } +1\right)=\int_0^{\infty}u^{\frac{1}{\alpha}}e^{-u}du $. For any given real number $ \kappa > 0, $ we have

$ P\left ( X\leq \kappa EX \right )=\displaystyle{\int}_{0}^{\kappa \theta ^{\frac{1}{\alpha }}\Gamma \left ( \frac{1}{\alpha } +1\right )}\frac{\alpha }{\theta }t^{\alpha -1}e^{-\frac{t^{\alpha}}{\theta }}dt. $

By taking the change of variable $ t= \left (\theta x \right)^{\frac{1}{\alpha }} $, we get

$ \begin{eqnarray*} P\left (X\leq \kappa EX \right )&=& \int_{0}^{\left (\kappa \Gamma \left ( \frac{1}{\alpha } +1\right )\right)^{\alpha }}\frac{\alpha }{\theta }\left ( \theta x \right )^{\frac{\alpha-1}{\alpha }}e^{-x} \frac{\theta ^{\frac{1}{\alpha }}}{\alpha }x^{\frac{1}{\alpha }-1}dx \\ &=&\int_{0}^{\left (\kappa \Gamma \left ( \frac{1}{\alpha } +1\right )\right)^{\alpha }}e^{-x}dx\\ & =&1-e^{-\left (\kappa \Gamma \left ( \frac{1}{\alpha } +1\right )\right)^{\alpha }}, \end{eqnarray*} $

which shows that $ P\left (X\leq \kappa EX \right) $ is independent of $ \theta $.

Define a function

$ \begin{eqnarray} g_{\kappa}( \alpha ):=1-e^{-\left (\kappa \Gamma \left ( \frac{1}{\alpha } +1\right )\right)^{\alpha }}, \quad \alpha > 0. \end{eqnarray} $

(2.1)

The main result of this section is

Proposition 2.1 (ⅰ) If $ \kappa \leq 1 $, then

$ \begin{eqnarray*} \inf\limits_{\alpha \in \left ( 0, \infty \right ) }g_{\kappa }\left ( \alpha \right )=\lim\limits_{\alpha\to \infty}g_{\kappa}(\alpha)= \left\{ \begin{array}{cl} 0, & \kappa < 1, \\ 1-e^{-e^{-\gamma }}, & \kappa = 1, \end{array} \right. \end{eqnarray*} $

where $ \gamma $ is the Euler's constant, i.e., $ \gamma=\sum_{n=1}^{\infty }\left [\frac{1}{n} -\ln\left (1+\frac{1}{n} \right)\right] $.

(ⅱ) If $ \kappa > 1 $, then

$ \min\limits_{\alpha \in \left ( 0, \infty \right ) }g_{\kappa }\left ( \alpha \right )=g_{\kappa }\left ( \alpha _{0}\left ( \kappa \right ) \right ), $

where $ \alpha _{0}\left (\kappa \right)=\frac{1}{x_{0}\left (\kappa \right) -1} $, and $ x_{0}\left (\kappa \right) $ is the unique null point of function $ \varphi _{\kappa }\left (x \right): = \left (x-1 \right)\psi \left (x \right)-\ln\left (\kappa \Gamma \left (x \right) \right) $ on $ \left (1, \infty \right), $ where $ \psi(x) $ is the digamma function (see Definition 2.3 below).

Note that $ \left (\kappa \Gamma \left (\frac{1}{\alpha } +1\right)\right)^{\alpha }=e^{\alpha \ln\left (\kappa \Gamma \left (\frac{1}{\alpha }+1 \right)\right)}. $ Let $ x=\frac{1}{\alpha }+1, $ and define function

$ \begin{eqnarray} h_{\kappa}(x):=\frac{\ln (\kappa \Gamma(x))}{x-1}, \quad x > 1. \end{eqnarray} $

(2.2)

Then

$ \begin{eqnarray} g_{\kappa }\left ( \alpha \right )=1-e^{-e^{h_{\kappa}\left ( x \right )}}, \end{eqnarray} $

(2.3)

and in order to finish the proof of Proposition 2.1, it is enough to prove the following lemma.

Lemma 2.2 (ⅰ) If $ \kappa \leq 1, $ then

$ \begin{eqnarray*} \inf\limits_{x\in \left ( 1, \infty \right ) }h_{\kappa}\left ( x \right )=\lim\limits_{x\to 1^+}h_{\kappa}(x)= \left\{ \begin{array}{ll} -\infty, & \kappa < 1, \\ -\gamma, & \kappa = 1, \end{array} \right. \end{eqnarray*} $

where $ \gamma $ is the Euler's constant.

(ⅱ) If $ \kappa > 1 $, then

$ \min\limits_{x\in \left ( 1, \infty \right ) }h_{\kappa}\left ( x \right )=h_{\kappa}\left ( x_{0}\left ( \kappa \right ) \right ), $

where $ x_{0}\left (\kappa \right) $ is the unique null point of function $ \varphi _{\kappa }\left (x \right): = \left (x-1 \right)\psi \left (x \right)-\ln\left (\kappa \Gamma \left (x \right) \right) $ on $ \left (1, \infty \right), $ where $ \psi(x) $ is the digamma function.

Before giving the proof of Lemma 2.2, we need some preliminaries on the ploygamma function.

Definition 2.3([11, 1.16]) Let $ m $ be any nonnegative integers. $ m $-order ploygamma function $ \psi^{(m)} $ is defined by

$ \begin{eqnarray*} \psi ^{(m)}( z) := \frac{\mathrm{d}^{m} }{\mathrm{d} z^{m}}\psi (z)=\frac{\mathrm{d}^{m+1} }{\mathrm{d} z^{m+1}}\ln\Gamma (z), \text{Re} z>0. \end{eqnarray*} $

When $ m=0 $, $ \psi(z): =\psi ^{(0)}(z)=\frac{\mathrm{d} }{\mathrm{d} z}\ln\Gamma(z)=\frac{{\Gamma}'(z)}{\Gamma(z)} $ is called digamma function.

By [11, 1.7(3)] and [11, 1.9(10)], we know that

$ \begin{eqnarray} \psi(z)&=& -\gamma -\frac{1}{z}+\sum\limits_{n=1}^{\infty }\frac{z}{n(z+n)} =-\gamma +( z-1)\sum\limits_{n=0}^{\infty }\frac{1}{[( n+1)( z+n )]}, \end{eqnarray} $

(2.4)

$ \begin{eqnarray} \psi ^{( 1 )}( z) &=&{\psi }'( z)=\sum\limits_{n=0}^{\infty }\frac{1}{( z+n)^{2}}. \end{eqnarray} $

(2.5)

Proof of Lemma 2.2 By (2.2) and Definition 2.3, we have

$ \begin{eqnarray} {h}'_{\kappa }\left ( x \right )=\frac{\left ( x-1 \right )\psi \left ( x \right )-\ln\left ( \kappa \Gamma \left ( x \right ) \right )}{\left ( x-1 \right )^{2}}=\frac{\varphi _{\kappa }\left ( x \right )}{\left ( x-1 \right )^{2}}, \quad x> 1. \end{eqnarray} $

(2.6)

By (2.5), we get

$ \begin{eqnarray*} {\varphi}'_{\kappa }\left ( x \right )=\left ( x-1 \right )\psi ^{\left ( 1 \right )}\left ( x \right )=\left (x-1 \right )\sum\limits_{n=0}^{\infty }\frac{1}{\left ( x+n \right )^{2}}>0, \quad \forall x>1. \end{eqnarray*} $

It follows that the function $ \varphi _{\kappa } \left (x \right) $ is strictly increasing on the interval $ \left (1, \infty \right) $.

Thus, if $ \kappa \leq 1 $, we have $ \varphi _{\kappa }\left (x \right) > \varphi _{\kappa }\left (1 \right)=-\ln\kappa \geq 0, \quad \forall x > 1. $ Then, by (2.6) we get $ {h}'_{\kappa }\left (x \right) > 0, \quad \forall x > 1, $ which implies that the function $ h_{\kappa}\left (x \right) $ is strictly increasing on $ \left (1, +\infty \right). $ Hence the function $ h_{\kappa}\left (x \right) $ has no minimum value on $ \left (1, \infty \right) $ and

$ \inf\limits_{x\in \left ( 1, \infty \right ) }h_\kappa\left ( x \right )=\lim\limits_{x\rightarrow 1^+}h_{\kappa}\left ( x \right )=\lim\limits_{x\rightarrow 1^+}\frac{\ln\Gamma \left ( x \right )}{x-1}+\lim\limits_{x\rightarrow 1^+}\frac{\ln\kappa }{x-1}. $

By the L'Hospital's rule and (2.4), we have

$ \begin{eqnarray*} \lim\limits_{x\rightarrow 1^+}\frac{\ln\Gamma \left ( x \right )}{x-1}& =\lim\limits_{x\rightarrow 1^+}\frac{\Gamma'(x)}{\Gamma(x)}={\Gamma }'\left ( 1 \right )=\psi \left ( 1 \right )=-\gamma. \end{eqnarray*} $

Thus,

$ \begin{eqnarray*} \lim\limits_{x\in (1, \infty) }h_{\kappa}(x)= \left\{ \begin{array}{ll} -\infty, & \kappa < 1, \\ -\gamma, & \kappa = 1. \end{array} \right. \end{eqnarray*} $

If $ \kappa > 1 $, then $ \varphi _{\kappa }\left (1 \right)=-\ln\kappa < 0. $ By [11, 1.18(1)] (Stirling formula) and [11, 1.18(7)], when $ z\rightarrow \infty $ we have

$ \begin{eqnarray*} \ln\Gamma \left ( z \right )&= \left ( z-\frac{1}{2} \right )\ln z-z+\frac{\ln\left ( 2\pi \right )}{2}+o\left ( 1 \right ), \\ \psi \left ( z \right )&= \ln z-\frac{1}{2z}+o\left ( \frac{1}{z} \right ), \quad \left | \arg z \right |< \pi . \end{eqnarray*} $

Then

$ \begin{eqnarray*} \lim\limits_{x\rightarrow \infty}\varphi_{\kappa}(x)&=&\lim\limits_{x\rightarrow \infty}[(x-1)\psi(x)-\ln \Gamma(x)-\ln \kappa]\\ &=&\lim\limits_{x\rightarrow \infty}\left[ (x-1)\left(\ln x-\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)\right.\\ &&\quad\quad\quad\left.-\left(\left(x-\frac{1}{2}\right)\ln x -x +\frac{\ln (2\pi)}{2}+o(1)\right)-\ln \kappa\right]\\ &=&\lim\limits_{x\rightarrow \infty}\left[ x-\frac{1}{2}\ln x +\frac{1}{2x}-\frac{1}{2}-\frac{\ln\left ( 2\pi \right )}{2}-\ln\kappa +o\left ( 1 \right ) \right]\\ &=&\infty. \end{eqnarray*} $

Since the function $ \varphi _{\kappa } \left (x \right) $ is continuous, by the zero point theorem, there exists $ x_{0}\left (\kappa \right) \in (1, \infty) $ which depends on $ \kappa $ satisfying that $ \varphi _{\kappa } \left (x_{0}\left (\kappa \right) \right)=0. $ Moreover, combining with the monotonicity of the function $ \varphi _{\kappa } \left (x \right) $ on the interval $ \left (1, \infty \right) $, we know that $ x_{0}\left (\kappa \right) $ is the unique null point of the function $ \varphi _{\kappa } \left (x \right) $ and

$ \begin{eqnarray*} \varphi _{\kappa } \left ( x \right )&<&0, \quad \forall x\in \left ( 1, x_{0}\left ( \kappa \right ) \right );\\ \varphi _{\kappa } \left ( x \right )&>&0, \quad \forall x\in \left ( x_{0}\left ( \kappa \right ), +\infty \right ). \end{eqnarray*} $

Then, by (2.6) we get

$ \begin{eqnarray*} {h}'_{\kappa }\left ( x \right )&<&0, \quad \forall x\in \left ( 1, x_{0}\left ( \kappa \right ) \right );\\ {h}'_{\kappa }\left ( x \right )&>&0, \quad \forall x\in \left ( x_{0}\left ( \kappa \right ), +\infty \right ). \end{eqnarray*} $

Thus, the function $ h_{\kappa }\left (x \right) $ is strictly decreasing on $ \left (1, x_{0}\left (\kappa \right) \right) $ and strictly increasing on $ \left (x_{0}\left (\kappa \right), +\infty \right) $, which implies that

$ h_{\kappa }\left ( x \right )\geq h_{\kappa }\left ( x_{0}\left ( \kappa \right ) \right ), \quad \forall x> 1. $

Therefore,

$ \min\limits_{x\in \left ( 1, \infty \right ) }h_{\kappa}\left ( x \right )=h_{\kappa}\left ( x_{0}\left ( \kappa \right ) \right ). $

The proof is complete.

Let $ X $ be a Pareto random variable with parameters $ a $ and $ \theta\, (a > 0, \theta > 0) $ and the density function

$ f(x)=\theta a^{\theta }x^{-(\theta +1)}I_{(a, \infty)}(x). $

When $ \theta > 1 $, the expectation of $ X $ is $ EX=\frac{\theta a}{\theta -1} $. Then, for any given real number $ \kappa > 0 $, we have

$ \begin{eqnarray*} P\left ( X\leq \kappa EX \right )&=&\int_{a}^{\frac{\kappa \theta a}{\theta -1}}\theta a^{\theta }t^{-\left ( \theta +1 \right )}dt =-\left ( \frac{a}{t} \right )^{\theta }\big|_{a}^{\frac{\kappa \theta a}{\theta -1}} =1-\left ( \frac{\theta -1}{\kappa \theta }\right )^{\theta }, \end{eqnarray*} $

which shows that $ P\left (X\leq \kappa EX \right) $ is independent of $ a $. Note that, in order to make sense of the above equality, if $ \kappa < 1 $, the parameter $ \theta $ should satisfy that $ 1 < \theta\le \frac{1}{1-\kappa}; $ and if $ \kappa\ge 1 $, the parameter $ \theta $ should satisfy that $ \theta > 1. $

Define a function

$ g_{\kappa}(\theta):= 1-\left(\frac{\theta -1}{\kappa \theta }\right )^{\theta }, \ 1<\theta \leq \frac{1}{1-\kappa }, \kappa <1\ \mbox{or}\ \theta > 1, \kappa \geq 1. $

The main result of this section is

Proposition 3.1 (ⅰ) If $ \kappa < 1 $, then

$ \min\limits_{\theta \in \left ( 1, \frac{1}{1-\kappa } \right ] }g_{\kappa }\left ( \theta \right )=g_{\kappa }\left ( \frac{1}{1-\kappa}\right )=0. $

(ⅱ) If $ \kappa=1 $, then $ \inf\limits_{\theta \in (1, \infty)}g_1(\theta)=\lim\limits_{\theta\to \infty}g_1(\theta)=1-e^{-1} $.

(ⅲ) If $ \kappa > 1 $, then $ \min\limits_{\theta \in \left (1, \infty \right) }g_{\kappa }\left (\theta \right)= g_{\kappa}\left (\theta_{0}\left (\kappa \right) \right), $ where $ \theta _{0}\left (\kappa \right)=\frac{1}{1-x_{0}\left (\kappa \right) } $, and $ x_{0}\left (\kappa \right) $ is the unique null point of function $ \varphi _{\kappa }\left (x \right): = 1-\frac{1}{x}-\ln\frac{x}{\kappa } $ on the interval $ \left (0, 1 \right) $.

Note that $ \left (\frac{\theta -1}{\kappa \theta }\right)^{\theta }=e^{\theta \ln\frac{\theta -1}{\kappa \theta }} $. Let $ x=1-\frac{1}{\theta } $ and define function

$ \begin{eqnarray} h_{\kappa }\left ( x \right ):=\frac{\ln \frac{x}{\kappa }}{x-1}, \ 0<x \leq \kappa , \kappa < 1, \ \mbox{or}\ 0<x<1, \kappa \geq 1. \end{eqnarray} $

(3.1)

Then

$ \begin{eqnarray} g_{\kappa}\left ( \theta \right )=1-e^{-h_{\kappa }\left ( x \right )}, \end{eqnarray} $

(3.2)

and in order to finish the proof of Proposition 3.1, it is enough to prove the following lemma.

Lemma 3.2

(ⅰ) If $ \kappa < 1 $, then $ \min\limits_{x\in (0, \kappa] }h_{\kappa}(x)=h_{\kappa}(\kappa)=0. $

(ⅱ) If $ \kappa=1 $, then $ \inf\limits_{x\in (0, 1) }h_1(x)=\lim\limits_{x\to 1^-}h_1(x)=1 $.

(ⅲ) If $ \kappa > 1 $, then $ \min\limits_{x\in (0, 1) }h_{\kappa}(x)=h_{\kappa}\left (x_{0}\left (\kappa \right) \right), $ where $ x_{0}\left (\kappa \right) $ is the unique null point of function $ \varphi _{\kappa }\left (x \right): = 1-\frac{1}{x}-\ln\frac{x}{\kappa } $ on the interval $ \left (0, 1 \right) $.

Proof (ⅰ) If $ \kappa < 1 $, by (3.1) we have

$ \begin{eqnarray} {h}'_{\kappa }\left ( x \right )=\frac{1-\frac{1}{x}-\ln\frac{x}{\kappa }}{\left ( x-1 \right )^{2}}=\frac{\varphi _{\kappa }\left ( x \right )}{\left ( x-1 \right )^{2}}, \quad 0<x \leq \kappa. \end{eqnarray} $

(3.3)

By the definition of $ \varphi _{\kappa }\left (x \right) $, we get that

$ \begin{eqnarray*} \label{4.4} {\varphi}'_{\kappa }\left ( x \right )=\frac{1-x}{x^{2}}> 0, \quad \forall 0<x \leq \kappa. \end{eqnarray*} $

It follows that function $ \varphi _{\kappa }\left (x \right) $ is strictly increasing on $ \left (0, \kappa \right] $ and thus

$ \begin{eqnarray} \varphi _{\kappa }\left ( x \right ) \leq \varphi _{\kappa }\left ( \kappa \right )=\frac{\kappa -1}{\kappa }< 0, \quad \forall 0<x\leq \kappa. \end{eqnarray} $

(3.4)

Then, by (3.3) and (3.4) we get $ {h}'_{\kappa }\left (x \right) < 0, \quad \forall 0 < x\leq \kappa, $ which implies that function $ h_{\kappa}\left (x \right) $ is strictly decreasing on $ \left (0, \kappa \right] $. Thus $ \underset{x\in \left (0, \kappa \right]}{\min}h_{\kappa}\left (x \right)=h_{\kappa}\left (\kappa \right)=0. $

If $ \kappa \geq 1 $, by (3.1) and the definition of $ \varphi_{\kappa}(x) $ again, we also have

$ \begin{eqnarray} {h}'_{\kappa }\left ( x \right )=\frac{\varphi _{\kappa }\left ( x \right )}{\left ( x-1 \right )^{2}}, \quad 0< x <1, \end{eqnarray} $

(3.5)

and

$ \begin{eqnarray*} \label{4.6} {\varphi}'_{\kappa }\left ( x \right )=\frac{1-x}{x^{2}}> 0, \quad \forall 0 <x <1. \end{eqnarray*} $

It follows that function $ \varphi _{\kappa }\left (x \right) $ is strictly increasing on $ \left (0, 1 \right) $.

(ⅱ) If $ \kappa = 1 $, then

$ \varphi _{\kappa }\left ( x \right ) < \varphi _{\kappa }\left ( 1 \right )=-\ln \kappa= 0, \quad \forall 0<x<1. $

By (3.5), we get that $ {h}'_{\kappa }\left (x \right) < 0, \quad \forall 0 < x < 1, $ which implies that function $ h_{\kappa}\left (x \right) $ is strictly decreasing on $ \left (0, 1 \right) $. Thus,

$ \inf\limits_{x\in \left ( 0, 1 \right) }h_{\kappa}\left ( x \right )=\lim\limits_{x\rightarrow 1^-}h_{\kappa}\left ( x \right )=\lim\limits_{x\rightarrow 1^-}\frac{\ln x}{x-1}=1. $

(ⅲ) If $ \kappa > 1 $, then $ \varphi _{\kappa }\left (1 \right)=\ln \kappa > 0 $. Moreover,

$ \begin{eqnarray*} \lim\limits_{x\rightarrow 0^+ }\varphi _{\kappa } \left ( x \right )=\lim\limits_{x\rightarrow 0^+ }\left(1-\frac{1}{x}-\ln x+\ln\kappa \right) =\lim\limits_{x\rightarrow 0^+ }\left ( 1+\ln\kappa -\frac{x\ln x+1}{x} \right ) =-\infty. \end{eqnarray*} $

Since the function $ \varphi _{\kappa } \left (x \right) $ is continuous on $ (0, 1) $, by the zero point theorem, there exists $ x_{0}\left (\kappa \right) \in (0, 1) $ depending on parameter $ \kappa $ fulfills that $ \varphi _{\kappa } \left (x_{0}\left (\kappa \right) \right)=0. $ By the monotonicity of function $ \varphi _{\kappa } \left (x \right) $ on $ \left (0, 1 \right) $, we know that $ x_{0}\left (\kappa \right) $ is the unique null point of $ \varphi _{\kappa } \left (x \right) $ and

$ \begin{eqnarray*} \varphi _{\kappa } \left ( x \right )&<&0, \quad \forall x\in \left ( 0, x_{0}\left ( \kappa \right ) \right );\\ \varphi _{\kappa } \left ( x \right )&>&0, \quad \forall x\in \left ( x_{0}\left ( \kappa \right ), 1 \right ). \end{eqnarray*} $

Then, by (3.5) we have

$ \begin{eqnarray*} {h}'_{\kappa }\left ( x \right )&<&0, \quad \forall x\in \left ( 0, x_{0}\left ( \kappa \right ) \right );\\ {h}'_{\kappa }\left ( x \right )&>&0, \quad \forall x\in \left ( x_{0}\left ( \kappa \right ), 1 \right ). \end{eqnarray*} $

Therefore, the function $ h_{\kappa }\left (x \right) $ is strictly decreasing on $ \left (0, x_{0}\left (\kappa \right) \right) $ and is strictly increasing on $ \left (x_{0}\left (\kappa \right), 1 \right) $. Thus

$ \underset{x\in \left ( 0, 1\right )}{\min}h_{\kappa}\left ( x \right )=h_{\kappa}\left ( x_{0}\left ( \kappa \right ) \right ). $

The proof is complete.