1 Introduction

For \(r\in (0,1),\) \(a\in (0,1),\) and \(r^{\prime }=\sqrt{1-r^{2}},\) the generalized elliptic integral of the first kind (see [1, Section 5.5]) is defined by

$$\begin{aligned} \left\{ \begin{array}{l} {\mathcal {K}}_{a}\equiv {\mathcal {K}}_{a}(r)=\frac{\pi }{2}F\left( a,1-a;1;r^{2}\right) , \\ {\mathcal {K}}_{a}(0^{+})=\pi /2,\;\; {\mathcal {K}}_{a}(1^{-})=\infty , \end{array} \right. \end{aligned}$$

which can be expressed as a power series

$$\begin{aligned} {\mathcal {K}}_{a}(r)=\frac{\pi }{2}\sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}r^{2n}, \end{aligned}$$

where \(\left( a\right) _{n}\) is the shifted factorial function (or Pochhammer symbol)

$$\begin{aligned} (a)_{n}=a(a+1)(a+2)\ldots (a+n-1). \end{aligned}$$

In 2000, this special function was rediscovered by [2]. In the particular case \(a=1/2,\) the function \({\mathcal {K}}_{a}(r)\) reduces to \({\mathcal {K}}(r),\) the well-known complete elliptic integral of the first kind:

$$\begin{aligned} {\mathcal {K}}(r)=\int _{0}^{\pi /2}\frac{dt}{\sqrt{1-r^{2}\sin ^{2}(t)}}, \end{aligned}$$

which is the particular case of the Gaussian hypergeometric function [3,4,5,6,7,8,9]:

$$\begin{aligned} {\mathcal {K}}(r)=\frac{\pi }{2}F\left( \frac{1}{2},\frac{1}{2};1;r^{2}\right) = \frac{\pi }{2}\sum _{n=0}^{\infty }\frac{\left( \frac{1}{2}\right) _{n}^{2}}{ (n!)^{2}}r^{2n}=\frac{\pi }{2}\sum _{n=0}^{\infty }\left[ \frac{(2n-1)!!}{ (2n)!!}\right] ^{2}r^{2n}=:\frac{\pi }{2}\sum _{n=0}^{\infty }W_{n}^{2}r^{2n}.\nonumber \\ \end{aligned}$$
(1)

Many researchers have obtained some results about this special function \( {\mathcal {K}}_{a}(r)\) ( see [10,11,12,13,14,15,16,17,18,19,20]) . In [21], the upper and lower bounds for \({\mathcal {K}}_{a}(r)\) were shown, and a double inequality was proved as follows:

$$\begin{aligned} 1+\alpha (r^{\prime })^{2}<\frac{{\mathcal {K}}_{a}(r)}{\sin \left( \pi a\right) \log \left( e^{R(a)/2}/r^{\prime }\right) }<1+\beta (r^{\prime })^{2} \end{aligned}$$

holds for all \(a\in (0,1/2]\) and \(r\in (0,1)\) if and only if \(\alpha \le \pi /[R(a)\sin (\pi a)]-1\) and \(\beta \ge a\left( 1-a\right) \), where R(x) is the Ramanujan constant function (see [22]).

The aim of this paper is to provide a new concise bound for \({\mathcal {K}} _{a}(r)\) by a rational function of the arugment \(r^{\prime }\) and obtain the following results.

Theorem 1

Let \(a\in (0,1),\) \(r\in (0,1),\) and \(r^{\prime }=\sqrt{1-r^{2}}\). Then

$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{a}(r)\; \mathbf {>}\frac{\left( 1-3a+3a^{2}\right) r\mathbf {^{\prime }}+\left( 1+3a-3a^{2}\right) }{\left( 1+a-a^{2}\right) r\mathbf {^{\prime }}+\left( 1-a+a^{2}\right) }=:{\mathcal {G}} (r\mathbf {^{\prime }}). \end{aligned}$$
(2)

Letting \(a=1/2,1/3,1/4,\) and 1/6 in the above theorem, respectively, we immediately draw the following corollary.

Corollary 1

Let \(r\in (0,1)\) and \(r^{\prime }=\sqrt{1-r^{2}}\). Then

$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{1/2}(r)\;\equiv & {} \frac{2}{\pi }{\mathcal {K}} (r)\; \mathbf {>}\;\frac{r\mathbf {^{\prime }}+7}{5r\mathbf { ^{\prime }}+3}, \end{aligned}$$
(3)
$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{1/3}(r)\;&\mathbf {>}&\;\frac{3\left( r^{\prime }+5\right) }{11r^{\prime }+7}, \end{aligned}$$
(4)
$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{1/4}(r)\;&\mathbf {>}&\;\frac{7r^{\prime }+25}{19r^{\prime }+13}, \end{aligned}$$
(5)
$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{1/6}(r)\;&\mathbf {>}&\frac{ 3\left( 7r^{\prime }+17\right) }{41r^{\prime }+31}. \end{aligned}$$
(6)

2 Lemmas

In order to prove our main results we need following lemmas.

Lemma 1

Let \(a\in \left( 0,1\right) ,\) \(n\ge 5\). Then

$$\begin{aligned} \frac{n^{2}\left( 1-a+a^{2}\right) ^{2}+4na\left( 1-a\right) -4a^{2}\left( 1-a\right) ^{2}}{\left( n-a\right) (n-1+a)}\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}>\frac{8a\left( 1-a\right) }{2n-1}\frac{1}{n!}\left( \frac{1}{2}\right) _{n} \end{aligned}$$
(7)

holds.

Proof

The inequality required is equivalent to

$$\begin{aligned} T_{n}=: & {} \frac{n^{2}\left( 1-a+a^{2}\right) ^{2}+4na\left( 1-a\right) -4a^{2}\left( 1-a\right) ^{2}}{\left( n-a\right) \left( n-1+a\right) }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}\nonumber \\&- \frac{8a\left( 1-a\right) }{2n-1}\frac{1}{n!}\left( \frac{1}{2}\right) _{n}>0. \end{aligned}$$
(8)

We use mathematical induction to prove \(\left( 7\right) \). We can calculate directly

$$\begin{aligned} T_{5}=\frac{1}{14\,400}a\left( 1-a\right) g(a), \end{aligned}$$

where

$$\begin{aligned} g(a)= & {} 450-1620a+1904a^{2}+1125a^{3}-4405a^{4}+3540a^{5}\\&+542a^{6}-1350a^{7}+180a^{8}+105a^{9}-21a^{10}. \end{aligned}$$

The fact that \(g(a)>0\) for all \(a\in \left( 0,1\right) \) can be proved by the following elementary method:

$$\begin{aligned} g(a)= & {} \frac{92\,925}{1024}+\frac{290\,759}{256}\left( a-\frac{1}{2}\right) ^{2}+\frac{54\,505}{32}\left( a-\frac{1}{2}\right) ^{4} \\&-\frac{16\,769}{8}\left( a-\frac{1}{2}\right) ^{6}+\frac{1665}{ 4}\left( a-\frac{1}{2}\right) ^{8}-21\left( a-\frac{1}{2}\right) ^{10} \\=: & {} \frac{92\,925}{1024}+\frac{290\,759}{256}t+\frac{54\,505}{32} t^{2}-\frac{16\,769}{8}t^{3}+\frac{1665}{4}t^{4}-21t^{5} \\= & {} \frac{6007\,063}{4096}t+\frac{1179\,959}{16\,384}+\frac{60\,521}{32} \left( \frac{152\,609}{242\,084}-t\right) \left( t-\frac{1}{8} \right) ^{2}+21\left( \frac{541}{28}-t\right) \left( t-\frac{1}{8 }\right) ^{4} \\> & {} 0, \end{aligned}$$

where \(t=\left( a-1/2\right) ^{2}\) and \(0\le t<1/4\). So \(T_{5}>0,\) which implies (7) holds for \(n=5\). Assuming that (7) holds for \(n=m>5\), that is,

$$\begin{aligned} \frac{\left( a\right) _{m}\left( 1-a\right) _{m}}{m!}>8a\left( 1-a\right) h(m)\left( \frac{1}{2}\right) _{m}, \end{aligned}$$
(9)

where

$$\begin{aligned} h(m)=\frac{\left( m-a\right) \left( m-1+a\right) }{\left( 2m-1\right) \left[ m^{2}\left( 1-a+a^{2}\right) ^{2}+4ma\left( 1-a\right) -4a^{2}\left( 1-a\right) ^{2}\right] }. \end{aligned}$$

Next, we prove that (7) is valid for \(n=m+1\). By (9) we have

$$\begin{aligned} \frac{\left( a\right) _{m+1}\left( 1-a\right) _{m+1}}{\left( m+1\right) !}= & {} \frac{\left( a+m\right) \left( 1-a+m\right) }{\left( m+1\right) }\frac{ \left( a\right) _{m}\left( 1-a\right) _{m}}{(m!)^{2}} \\> & {} \frac{\left( a+m\right) \left( 1-a+m\right) }{\left( m+1\right) }8a\left( 1-a\right) h(m)\left( \frac{1}{2}\right) _{m},\end{aligned}$$

in order to complete the proof of (7) it suffices to show that

$$\begin{aligned} \frac{\left( a+m\right) \left( 1-a+m\right) }{\left( m+1\right) }8a\left( 1-a\right) h(m)\left( \frac{1}{2}\right) _{m}>8a\left( 1-a\right) h(m+1)\left( \frac{1}{2}\right) _{m+1}, \end{aligned}$$

that is

$$\begin{aligned} \frac{\left( a+m\right) \left( 1-a+m\right) }{\left( m+1\right) }h(m)\left( \frac{1}{2}\right) _{m}>h(m+1)\left( \frac{1}{2}\right) _{m+1}=h(m+1)\left( \frac{1}{2}+m\right) \left( \frac{1}{2}\right) _{m}, \end{aligned}$$

or

$$\begin{aligned} \frac{A}{B}=: & {} \frac{\left( a+m-1\right) \left( m-a\right) }{\left( 2m-1\right) \left( m+1\right) \left[ m^{2}\left( a^{2}-a+1\right) ^{2}+4ma\left( 1-a\right) -4a^{2}\left( a-1\right) ^{2}\right] } \\> & {} \frac{1}{2\left[ \left( m+1\right) ^{2}\left( a^{2}-a+1\right) ^{2}+4\left( m+1\right) a\left( 1-a\right) -4a^{2}\left( a-1\right) ^{2} \right] }=:\frac{C}{D}. \end{aligned}$$

In fact,

$$\begin{aligned} AD-BC= & {} 2a\left( a-1\right) \left( 3a^{2}-6a^{3}+3a^{4}-1\right) \\&-2m\left( a^{2}-a-1\right) \left( 3a-a^{2}-4a^{3}+2a^{4}-1\right) \\&-m^{2}\left( 5a^{2}-12a^{3}+11a^{4}-6a^{5}+2a^{6}+1\right) \\&+m^{3}\left( a^{2}-a+1\right) ^{2} \\= & {} u_{0}(a)+\left( m-3\right) u_{1}(a)+\left( m-3\right) ^{2}u_{2}(a)+\left( m-3\right) ^{3}u_{3}(a), \end{aligned}$$

where

$$\begin{aligned} u_{0}(a)= & {} 12-40a+52a^{2}-60a^{4}+72a^{5}-24a^{6}, \\ u_{1}(a)= & {} 19-50a+57a^{2}+2a^{3}-41a^{4}+48a^{5}-16a^{6}, \\ u_{2}(a)= & {} 8-18a+22a^{2}-6a^{3}-2a^{4}+6a^{5}-2a^{6}, \\ u_{3}(a)= & {} 1-2a+3a^{2}-2a^{3}+a^{4}. \end{aligned}$$

We can prove \(u_{i}(a)>0\) \(\left( i=0,1,2,3\right) \) for all \(a\in \left( 0,1\right) \) as follows:

$$\begin{aligned} u_{0}(a)= & {} \frac{25}{8}+\frac{59}{2}\left( a-\frac{1}{2}\right) ^{2}+30\left( a-\frac{1}{2}\right) ^{4}-24\left( a-\frac{1}{2}\right) ^{6} \\=: & {} \frac{25}{8}+\frac{59}{2}t+30t^{2}-24t^{3}=\frac{287}{8}t+\frac{11}{4}+24\left( 1-t\right) \left( t-\frac{1}{8}\right) ^{2}>0, \\ u_{1}(a)= & {} \frac{115}{16}+\frac{87}{2}\left( a-\frac{1}{2} \right) ^{2}+19\left( a-\frac{1}{2}\right) ^{4}-16\left( a-\frac{1}{2} \right) ^{6} \\=: & {} \frac{115}{16}+\frac{87}{2}t+19t^{2}-16t^{3}=\frac{95}{2}t+\frac{445}{64} +16\left( \frac{15}{16}-t\right) \left( t-\frac{1}{8}\right) ^{2}>0, \\ u_{2}(a)= & {} \frac{121}{32}+\frac{125}{8}\left( a-\frac{1}{2} \right) ^{2}+\frac{11}{2}\left( a-\frac{1}{2}\right) ^{4}-2\left( a-\frac{1}{ 2}\right) ^{6} \\=: & {} \frac{121}{32}+\frac{125}{8}t+\frac{11}{2}t^{2}-2t^{3}=\frac{541}{32}t+ \frac{237}{64}+2\left( \frac{5}{2}-t\right) \left( t-\frac{1}{8}\right) ^{2}>0, \\ u_{3}(a)= & {} \frac{9}{16}+\frac{3}{2}\left( a-\frac{1}{2}\right) ^{2}+\left( a-\frac{1}{2}\right) ^{4}=:\frac{9}{16}+\frac{3}{2}t+t^{2}>0, \end{aligned}$$

where \(t=\left( a-1/2\right) ^{2}\) and \(0\le t<1/4\). So \(AD-BC>0\) for all \( m>5\).

This completes the proof of Lemma 1. \(\square \)

Lemma 2

[26,27,28,29,30,31] Let \(\{a_{k}\}_{k=0}^{\infty }\) be a nonnegative real sequence with \(a_{m}>0\) and \(\sum _{k=m+1}^{\infty }a_{k}>0\), and

$$\begin{aligned} S(t)=-\sum _{k=0}^{m}a_{k}t^{k}+\sum _{k=m+1}^{\infty }a_{k}t^{k} \end{aligned}$$

be a convergent power series on the interval (0, r) \((r>0)\). Then the following statements are true:

1.:

If \(S(r^{-})\le 0\), then \(S(t)<0\) for all \(t\in (0,r)\);

2.:

If \(S(r^{-})>0\), then there exists \(t_{0}\in (0,r)\) such that \(S(t)<0\) for \(t\in (0,t_{0})\) and \(S(t)>0\) for \(t\in (t_{0},r)\).

Lemma 2 appeared first in [26], then was proven in [27]. In a recent paper [28], the above power series S(t) in Lemma 2 has be called “NP type power series”.

3 Proof of Theorem 1

By

$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{a}(r)=\sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}r^{2n},\;\sqrt{1-r^{2}} =\sum _{n=0}^{\infty }\frac{\left( -\frac{1}{2}\right) _{n}}{n!}r^{2n} \end{aligned}$$

we have

$$\begin{aligned} {\mathcal {F}}(r)=: & {} \left[ 4a\left( 1-a\right) -r^{2}\left( -1-a+a^{2}\right) ^{2}\right] \frac{2}{\pi }{\mathcal {K}}_{a}(r) \\&-\left( 4a\left( a-1\right) -r^{2}\left( 1+a-a^{2}\right) \left( 1-3a+3a^{2}\right) +8a\left( 1-a\right) \sqrt{1-r^{2}} \right) \\= & {} 4a\left( 1-a\right) \frac{2}{\pi }{\mathcal {K}}_{a}(r)-r^{2}\left( -1-a+a^{2}\right) ^{2}\frac{2}{\pi }{\mathcal {K}}_{a}(r) \\&-4a\left( a-1\right) +r^{2}\left( 1+a-a^{2}\right) \left( 1-3a+3a^{2}\right) -8a\left( 1-a\right) \sqrt{1-r^{2}} \end{aligned}$$
$$\begin{aligned}= & {} 4a\left( 1-a\right) \sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}r^{2n}-\left( 1+a-a^{2}\right) ^{2}\sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{ (n!)^{2}}r^{2n+2} \\&-4a\left( a-1\right) +r^{2}\left( 1+a-a^{2}\right) \left( 1-3a+3a^{2}\right) -8a\left( 1-a\right) \sum _{n=0}^{\infty }\frac{\left( - \frac{1}{2}\right) _{n}}{n!}r^{2n} \\= & {} 4a\left( 1-a\right) \sum _{n=3}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}r^{2n}-\left( 1+a-a^{2}\right) ^{2}\sum _{n=2}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{ (n!)^{2}}r^{2n+2} \\&-8a\left( 1-a\right) \sum _{n=3}^{\infty }\frac{\left( -\frac{1}{2}\right) _{n}}{n!}r^{2n} \\= & {} 4a\left( 1-a\right) \sum _{n=3}^{\infty }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{(n!)^{2}}r^{2n}-\left( 1+a-a^{2}\right) ^{2}\sum _{n=3}^{\infty }\frac{\left( a\right) _{n-1}\left( 1-a\right) _{n-1} }{(\left( n-1\right) !)^{2}}r^{2n} \\&+4a\left( 1-a\right) \sum _{n=3}^{\infty }\frac{\left( \frac{1}{2}\right) _{n-1}}{n!}r^{2n} \\=: & {} \sum _{n=3}^{\infty }a_{n}r^{2n}, \end{aligned}$$

where

$$\begin{aligned} a_{n}= & {} 4a\left( 1-a\right) \frac{\left( a\right) _{n}\left( 1-a\right) _{n} }{(n!)^{2}}-\left( 1+a-a^{2}\right) ^{2}\frac{\left( a\right) _{n-1}\left( 1-a\right) _{n-1}}{(\left( n-1\right) !)^{2}}+4a\left( 1-a\right) \frac{ \left( \frac{1}{2}\right) _{n-1}}{n!} \\= & {} 4a\left( 1-a\right) \frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{ (n!)^{2}}-\left( 1+a-a^{2}\right) ^{2}\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{\left[ \left( n-1\right) !\right] ^{2}\left[ a+(n-1)\right] \left[ 1-a+(n-1)\right] } \\&+4a\left( 1-a\right) \frac{\left( \frac{1}{2}\right) _{n}}{n!\left[ \frac{1 }{2}+(n-1)\right] } \\= & {} -\frac{n^{2}\left( a^{2}-a+1\right) ^{2}+4na\left( 1-a\right) -4a^{2}\left( a-1\right) ^{2}}{\left( n-a\right) \left( n+a-1\right) }\frac{\left( a\right) _{n}\left( 1-a\right) _{n}}{ (n!)^{2}}+\frac{8a\left( 1-a\right) }{n!\left( 2n-1\right) }\left( \frac{1}{2 }\right) _{n} \\= & {} -T_{n}. \end{aligned}$$

A direct verification yields

$$\begin{aligned} a_{3}= & {} \frac{a^{2}\left( 1-a\right) ^{2}}{36}\left( 3-4a-a^{2}+10a^{3}-5a^{4}\right) \\= & {} \frac{a^{2}\left( 1-a\right) ^{2}}{36}\left[ 5\left( a-\frac{1}{2}\right) ^{2}\left( a+\frac{1}{2}\right) \left( \frac{3}{2}-a\right) +\frac{3}{2} \left( a-\frac{1}{2}\right) ^{2}+\frac{27}{16}\right] >0 \end{aligned}$$

for all \(a\in \left( 0,1\right) ,\) and by Lemma 1 we have \(a_{n}<0\) for all \(n\ge 5\).

So no matter what the sign of \(a_{4}\) is, those coefficients in power series of \(-{\mathcal {F}}(r)\) satisfy the conditions of Lemma 2, and \(-{\mathcal {F}} (1^{-})=\infty \). From Lemma 2, it follows that there is a unique \( r_{0}\in \left( 0,1\right) \) such that \(-{\mathcal {F}}(r)<0\) for \(r\in (0,r_{0})\) and \(-{\mathcal {F}}(r)>0\) for \(r\in (r_{0},1)\), that is, \(\mathcal {F }(r)>0\) for \(r\in (0,r_{0})\) and \({\mathcal {F}}(r)<0\) for \(r\in (r_{0},1),\) and \(r_{0}\) is the unique zero of \({\mathcal {F}}(r)\) on \(\left( 0,1\right) \). At the same time, since

$$\begin{aligned}&{\mathcal {F}}(r)=\left[ \left( a-a^{2}+1\right) r\mathbf {^{\prime }}-\left( -a+a^{2}+1\right) \right] \\&\quad \times \left\{ \begin{array}{c} \left[ \left( a-a^{2}+1\right) r\mathbf {^{\prime }}+\left( -a+a^{2}+1\right) \right] \frac{2}{\pi }{\mathcal {K}}_{a}(r) \\ -\left[ \left( -3a+3a^{2}+1\right) r\mathbf {^{\prime }}+\left( 3a-3a^{2}+1\right) \right] \end{array} \right\} , \end{aligned}$$

and the function

$$\begin{aligned} {\mathcal {S}}(r)=: & {} \left( a-a^{2}+1\right) r\mathbf {^{\prime }}-\left( -a+a^{2}+1\right) \\= & {} \frac{\left[ \left( a-a^{2}+1\right) r\mathbf {^{\prime }}-\left( -a+a^{2}+1\right) \right] \left[ \left( a-a^{2}+1\right) r\mathbf {^{\prime }} +\left( -a+a^{2}+1\right) \right] }{\left( a-a^{2}+1\right) r\mathbf { ^{\prime }}+\left( -a+a^{2}+1\right) } \\= & {} -\frac{\left( 1+a-a^{2}\right) ^{2}}{\left( a-a^{2}+1\right) r\mathbf { ^{\prime }}+\left( -a+a^{2}+1\right) }\left[ r^{2}-\frac{4a\left( 1-a\right) }{\left( 1+a-a^{2}\right) ^{2}}\right] \\= & {} -\frac{\left( 1+a-a^{2}\right) ^{2}}{\left( a-a^{2}+1\right) r\mathbf { ^{\prime }}+\left( -a+a^{2}+1\right) }\left[ r-\frac{2\sqrt{a\left( 1-a\right) }}{1+a-a^{2}}\right] \left[ r+\frac{2\sqrt{a\left( 1-a\right) }}{1+a-a^{2}}\right] \end{aligned}$$

has a unique zero on \(\left( 0,1\right) \), which leads to \(r_{0}=2\sqrt{ a\left( 1-a\right) }/\left( 1+a-a^{2}\right) \). The fact that \(r_{0}\in \left( 0,1\right) \) is true because

$$\begin{aligned} r_{0}>0\Longleftrightarrow 1+a-a^{2}>0 \end{aligned}$$

and

$$\begin{aligned} r_{0}< 1\Longleftrightarrow & {} 2\sqrt{a\left( 1-a\right) }<\left( 1+a-a^{2}\right) \Longleftrightarrow 4a\left( 1-a\right) <\left( 1+a-a^{2}\right) ^{2} \\\Longleftrightarrow & {} \left( 1+a-a^{2}\right) ^{2}-4a\left( 1-a\right) =\left( 1-a+a^{2}\right) ^{2}>0 \end{aligned}$$

hold for all \(a\in \left( 0,1\right) \). Clearly, \({\mathcal {S}}(r)>0\) for \( r\in (0,r_{0})\) and \({\mathcal {S}}(r)<0\) for \(r\in (r_{0},1)\). In a word,

$$\begin{aligned} \frac{2}{\pi }{\mathcal {K}}_{a}(r)-{\mathcal {G}}(r\mathbf {^{\prime }})=\frac{ {\mathcal {F}}(r)}{{\mathcal {S}}(r)\left[ \left( a-a^{2}+1\right) r\mathbf { ^{\prime }}+\left( -a+a^{2}+1\right) \right] }>0 \end{aligned}$$

holds for all \(r\in \left( 0,1\right) \) with \(r\ne r_{0}\). Because these two functions \({\mathcal {K}}_{a}(r)\) and \({\mathcal {G}}(r\mathbf {^{\prime }})\) are continuous at the point \(r_{0}\), the inequality (2) still holds for \( r=r_{0}\).

The proof of Theorem 1 is complete.