1 Introduction

Masjed-Jamei [1] obtained the following inequality

$$\begin{aligned} \left( \arctan x\right) ^{2}\le \frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2} }},\left| x\right| <1, \end{aligned}$$
(1.1)

where \(\ln (x+\sqrt{1+x^{2}})\) is the inverse hyperbolic sinefunction \(\mathop {\mathrm{ arcsinh}}\nolimits x=\sinh ^{-1}x\). In [1] the author conjectured that the above inequality is established in a larger interval \((-\infty ,\infty )\). Recently, the authors of this paper [2] first affirmed Masjed-Jamei’s conjecture, obtained some natural generalizations of this inequality, and pose a conjecture about a natural approach of Masjed-Jamei’s inequality inspired by [3,4,5,6,7,8,9].

Proposition 1.1

([2], Theorem 1.1]) The inequality

$$\begin{aligned} \left( \arctan x\right) ^{2}\le \frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2} }} \end{aligned}$$
(1.2)

holds for all \(x\in (-\infty ,\infty )\), and the power number 2 is the best in (1.2).

Proposition 1.2

([2], Theorem 1.3]) Let \(-\infty<\)\(x<\infty \). Then we have

$$\begin{aligned}&-\frac{1}{45}x^{6} \le \left( \arctan x\right) ^{2}-\frac{ x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}}\le -\frac{1}{45}x^{6}+\frac{4}{105} x^{8}, \end{aligned}$$
(1.3)
$$\begin{aligned}&-\frac{1}{45}x^{6}+\frac{4}{105}x^{8}-\frac{11}{225}x^{10} \le \left( \arctan x\right) ^{2}-\frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}} \nonumber \\&\quad \le -\frac{1}{45}x^{6}+\frac{4}{105}x^{8}-\frac{11}{225}x^{10}+\frac{586}{ 10\,395}x^{12}. \end{aligned}$$
(1.4)

Conjecture 1.1

([2], Conjecture 8.1]) Let \(x\in {\mathbf {R}},\)\(m\ge 1,\) and \(v_{n}\) be defined by

$$\begin{aligned} v_{n}=\frac{1}{n}\left( \frac{n!2^{n-1}}{\left( 2n-1\right) !!} -\sum _{i=1}^{n}\frac{1}{2i-1}\right) ,n\ge 3. \end{aligned}$$
(1.5)

Then the double inequality

$$\begin{aligned} \sum _{n=3}^{2m+1}\left( -1\right) ^{n}v_{n}x^{2n}\le \left( \arctan x\right) ^{2}-\frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}}\le \sum _{n=3}^{2m+2}\left( -1\right) ^{n}v_{n}x^{2n} \end{aligned}$$
(1.6)

holds.

Using flexible analysis tools this paper obtains a more general conclusion on the natural approximation of the function \(\left( \arctan x\right) ^{2}-\left( x\mathop {\mathrm{arcsinh}}\nolimits x\right) /\sqrt{1+x^{2}}\), and proves the above conjecture.

Theorem 1.1

Let \(k\ge 3\) and \(v_{n}\) be defined by (1.5). Then the function

$$\begin{aligned} G_{k}(x)=\frac{\sqrt{1+x^{2}}}{x}\left( \left( \arctan x\right) ^{2}-\frac{ x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}}-\sum _{n=3}^{k}\left( -1\right) ^{n}v_{n}x^{2n}\right) \end{aligned}$$
(1.7)

is decreasing and negative on \((0,+\infty )\) when k is an even number, and is increasing and positive on \((0,+\infty )\) when k is an odd number. In particular, for \(m\ge 1\),

  1. (i)

    the inequality

    $$\begin{aligned} \left( \arctan x\right) ^{2}-\frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}} \le \sum _{n=3}^{2m+2}\left( -1\right) ^{n}v_{n}x^{2n} \end{aligned}$$
    (1.8)

    holds for all \(x\in [0,+\infty )\), the constant \(v_{2m+2}\) is best possible in (1.8);

  2. (ii)

    the inequality

    $$\begin{aligned} \sum _{n=3}^{2m+1}\left( -1\right) ^{n}v_{n}x^{2n}\le \left( \arctan x\right) ^{2}-\frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}} \end{aligned}$$
    (1.9)

    holds for all \(x\in [0,+\infty )\), the constant \(-v_{2m+1}\) is best possible in (1.9).

Obviously, the Conjecture 1.1 is from Theorem 1.1 immediately.

2 Lemmas

Lemma 2.1

Let \(\left| x\right| <1\). Then

$$\begin{aligned} \frac{\mathop {\mathrm{arcsinh}}\nolimits x}{\sqrt{1+x^{2}}}= & {} \frac{\ln \left( x+\sqrt{1+x^{2}}\right) }{ \sqrt{1+x^{2}}} \nonumber \\= & {} \sum _{n=1}^{\infty }\left( -1\right) ^{n+1}\frac{n!2^{n-1}}{n\left( 2n-1\right) !!}x^{2n-1}=\sum _{n=1}^{\infty }\frac{\left( -1\right) ^{n+1}\left( 2x\right) ^{2n-1}}{n\left( {\begin{array}{c}2n\\ n\end{array}}\right) }. \end{aligned}$$
(2.1)

Lemma 2.2

Let \(\left| x\right| <1,\) and \(v_{n}\) be defined by (1.5). Then

$$\begin{aligned} \left( \arctan x\right) ^{2}-\frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}} =\sum _{n=3}^{\infty }\left( -1\right) ^{n}v_{n}x^{2n}, \end{aligned}$$
(2.2)

and \(v_{n}>0\) for \(n\ge 3\).

Lemma 2.3

Let \(v_{n}\) be defined by (1.5), and

$$\begin{aligned} \gamma _{3}= & {} 45v_{3}-1, \\ \gamma _{k}= & {} k\left( 2k-1\right) \left( 2k-3\right) v_{k}-2\left( k-1\right) \left( 3k-4\right) \left( 2k-3\right) v_{k-1} \\&+\left( k-2\right) \left( -44k+12k^{2}+41\right) v_{k-2}-2\left( k-3\right) \left( 2k-5\right) \left( k-2\right) v_{k-3}. \end{aligned}$$

Then \(\gamma _{k}=0\) for \(k\ge 3\).

Lemma 2.4

Let \(v_{n}\) be defined by (1.5). For \(k\ge 3\),

$$\begin{aligned} \mu= & {} \left( 1+4k+12k^{2}\right) v_{k}-\left( 4k^{2}-6k+2\right) v_{k-1}>0,\\ \zeta= & {} 2k\left( 3k-1\right) \left( 2k-1\right) v_{k}-\left( k-1\right) \left( 12k^{2}-20k+9\right) v_{k-1}\\&+ 2\left( 2k-3\right) \left( k-1\right) \left( k-2\right) v_{k-2}>0.\end{aligned}$$

3 Proof of Lemma 2.1

Let

$$\begin{aligned} p(x)=\frac{\mathop {\mathrm{arcsinh}}\nolimits x}{\sqrt{1+x^{2}}},\text { }\left| x\right| <1. \end{aligned}$$
(3.1)

Then \(p(0)=0,\) and

$$\begin{aligned} \sqrt{1+x^{2}}p(x)=\mathop {\mathrm{arcsinh}}\nolimits x. \end{aligned}$$
(3.2)

Differentiating (3.2) gives

$$\begin{aligned} \left( 1+x^{2}\right) p^{\prime }(x)+xp(x)=1. \end{aligned}$$
(3.3)

Since p(x) is an odd function, we can express it as a power series

$$\begin{aligned} p(x)=\sum _{n=0}^{\infty }a_{n}x^{2n+1}. \end{aligned}$$
(3.4)

Differentiation of (3.4) yields

$$\begin{aligned} p^{\prime }(x)=\sum _{n=0}^{\infty }\left( 2n+1\right) a_{n}x^{2n}, \end{aligned}$$

Substituting the series of \(p^{\prime }(x)\) and p(x) into (3.3) yields

$$\begin{aligned} a_{0}+\sum _{n=0}^{\infty }\left( \left( 2n+3\right) a_{n+1}+2\left( n+1\right) a_{n}\right) x^{2n+2}=1. \end{aligned}$$

Equating coefficients of \(x^{2n+2},\) we can obtain

$$\begin{aligned} a_{0}= & {} 1, \\ a_{n+1}= & {} -2\frac{n+1}{2n+3}a_{n},n\ge 0. \end{aligned}$$

Then

$$\begin{aligned} a_{n}=\left( -2\right) ^{n}\frac{n}{2n+1}\frac{n-1}{2n-1}\cdot \cdot \cdot \frac{2}{5}\frac{1}{3}1=\left( -2\right) ^{n}\frac{n!}{\left( 2n+1\right) !!} ,n\ge 0, \end{aligned}$$

and

$$\begin{aligned} p(x)= & {} \frac{\mathop {\mathrm{arcsinh}}\nolimits x}{\sqrt{1+x^{2}}}=\sum _{n=0}^{\infty }\left( -2\right) ^{n}\frac{n!}{\left( 2n+1\right) !!}x^{2n+1}=\sum _{n=1}^{\infty }\left( -2\right) ^{n-1}\frac{\left( n-1\right) !}{\left( 2n-1\right) !!} x^{2n-1} \\= & {} \sum _{n=1}^{\infty }\left( -1\right) ^{n+1}\frac{n!2^{n-1}}{n\left( 2n-1\right) !!}x^{2n-1}=\sum _{n=1}^{\infty }\frac{\left( -1\right) ^{n+1}\left( 2x\right) ^{2n-1}}{n\left( {\begin{array}{c}2n\\ n\end{array}}\right) }. \end{aligned}$$

The last equation is true due to

$$\begin{aligned} \left( {\begin{array}{c}2n\\ n\end{array}}\right) =\frac{2^{n}\left( 2n-1\right) !!}{n!}. \end{aligned}$$

The proof of Lemma 2.1 is complete.

4 Proof of Lemma 2.2

First, by (2.1) we have

$$\begin{aligned} \frac{x\mathop {\mathrm{arcsinh}}\nolimits x}{\sqrt{1+x^{2}}}=\frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{ 1+x^{2}}}=\sum _{n=1}^{\infty }\left( -1\right) ^{n+1}\frac{n!2^{n-1}}{ n\left( 2n-1\right) !!}x^{2n},\text { }\left| x\right| <1. \end{aligned}$$
(4.1)

Second, we have

$$\begin{aligned} \frac{d}{dx}\ \left( \arctan x\right) ^{2}= & {} \frac{2}{1+x^{2}} \arctan x=2\left( \sum _{n=0}^{\infty }\left( -1\right) ^{n}x^{2n}\right) \left( \sum _{n=0}^{\infty }\left( -1\right) ^{n}\frac{x^{2n+1}}{2n+1}\right) \nonumber \\= & {} 2\sum _{n=1}^{\infty }\left( -1\right) ^{n+1}\sum _{i=1}^{n}\frac{1}{2i-1} x^{2n-1}. \end{aligned}$$
(4.2)

Integrating two sides of (4.2) on [0, x] we can obtain

$$\begin{aligned} \ \left( \arctan x\right) ^{2}=\sum _{n=1}^{\infty }\left( -1\right) ^{n+1}\left( \frac{1}{n}\sum _{i=1}^{n}\frac{1}{2i-1}\right) x^{2n}. \end{aligned}$$
(4.3)

From (4.1) and (4.3) we have

$$\begin{aligned} \left( \arctan x\right) ^{2}-\frac{x\mathop {\mathrm{arcsinh}}\nolimits x}{\sqrt{ 1+x^{2}}}= & {} \sum _{n=1}^{\infty }\left( -1\right) ^{n+1}\left( \frac{1}{n} \sum _{i=1}^{n}\frac{1}{2i-1}-\frac{n!2^{n-1}}{n\left( 2n-1\right) !!} x^{2n}\right) x^{2n} \\= & {} \sum _{n=3}^{\infty }\frac{\left( -1\right) ^{n}}{n}\left( \frac{n!2^{n-1} }{\left( 2n-1\right) !!}-\sum _{i=1}^{n}\frac{1}{2i-1}\right) x^{2n} \\= & {} \sum _{n=3}^{\infty }\left( -1\right) ^{n}v_{n}x^{2n}, \end{aligned}$$

where \(v_{n}\) is defined by (1.5). The proof of \(v_{n}>0\)\( \left( n\ge 3\right) \) can be found in \(\left[ 2\right] \).

5 Proof of Lemma 2.3

The fact \(v_{1}=v_{2}=0\) and \(v_{3}=1/45\) is directly derived from (1.5). Then \(\gamma _{3}=45v_{3}-1=0.\)

Below we assume \(k\ge 4.\) For convenience, we can order

$$\begin{aligned} \sum _{i=1}^{k-3}\frac{1}{2i-1}:=\theta . \end{aligned}$$

Substituting the expression of \(v_{k}\) defined by (1.5) to the left-hand side of the formula (1.6), we can easy verify that

$$\begin{aligned} \gamma _{k}= & {} \left( 2k-1\right) \left( 2k-3\right) \left( \frac{k!2^{k-1}}{ \left( 2k-1\right) !!}-\left( \theta +\frac{1}{2k-5}+\frac{1}{2k-3}+\frac{1}{ 2k-1}\right) \right) \\&-2\left( 3k-4\right) \left( 2k-3\right) \left( \frac{\left( k-1\right) !2^{k-2}}{\left( 2k-3\right) !!}-\left( \theta +\frac{1}{2k-5}+\frac{1}{2k-3} \right) \right) \\&+\left( -44k+12k^{2}+41\right) \left( \frac{\left( k-2\right) !2^{k-3}}{\left( 2k-5\right) !!}-\left( \theta +\frac{1}{2k-5}\right) \right) \\&-2\left( 2k-5\right) \left( k-2\right) \left( \frac{\left( k-3\right) !2^{k-4}}{\left( 2k-7\right) !!}-\theta \right) \\= & {} 0\cdot \theta =0. \end{aligned}$$

6 Proof of Lemma 2.4

We first verify the first inequality, which is equivalent to

$$\begin{aligned}&\left( 4k+12k^{2}+1\right) \frac{1}{k}\left( \frac{k!2^{k-1}}{\left( 2k-1\right) !!}-\sum _{i=1}^{k}\frac{1}{2i-1}\right) \\&\quad >2\left( 2k-1\right) \left( \frac{\left( k-1\right) !2^{k-2}}{\left( 2k-3\right) !!}-\sum _{i=1}^{k-1}\frac{1}{2i-1}\right) , \end{aligned}$$

that is,

$$\begin{aligned} \frac{8\left( k+1\right) !2^{k-1}}{\left( 2k-1\right) !!}>\frac{\left( 2k+1\right) \left( 4k+1\right) }{k}\sum _{i=1}^{k-1} \frac{1}{2i-1}+\frac{4k+12k^{2}+1}{k\left( 2k-1\right) }. \end{aligned}$$
(6.1)

We use mathematical induction to prove (6.1). Obviously, the formula (6.1) holds for \(k=3\). Assuming that (6.1) holds for \(k=m\), we have

$$\begin{aligned} \frac{8\left( m+1\right) !2^{m-1}}{\left( 2m-1\right) !!}>\frac{\left( 2m+1\right) \left( 4m+1\right) }{m}\sum _{i=1}^{m-1} \frac{1}{2i-1}+\frac{4m+12m^{2}+1}{m\left( 2m-1\right) }. \end{aligned}$$
(6.2)

Next, we prove that (6.1) is valid for \(k=m+1\). By (6.2) we have

$$\begin{aligned}&\frac{8\left( m+2\right) !2^{m}}{\left( 2m+1\right) !!}=\frac{2\left( m+2\right) }{2m+1}\frac{8\left( m+1\right) !2^{m-1}}{\left( 2m-1\right) !!}\\&\quad >\frac{2\left( m+2\right) }{2m+1}\left( \frac{\left( 2m+1\right) \left( 4m+1\right) }{m}\sum _{i=1}^{m-1}\frac{1}{2i-1}+\frac{4m+12m^{2}+1}{m\left( 2m-1\right) }\right) . \end{aligned}$$

In order to complete the proof of (6.1) it suffices to show that

$$\begin{aligned}&\frac{2\left( m+2\right) }{2m+1}\left( \frac{\left( 2m+1\right) \left( 4m+1\right) }{m}\sum _{i=1}^{m-1}\frac{1}{2i-1}+\frac{4m+12m^{2}+1}{m\left( 2m-1\right) }\right) \\&\quad >\frac{\left( 2m+3\right) \left( 4m+5\right) }{m+1}\sum _{i=1}^{m}\frac{1}{ 2i-1}+\frac{28m+12m^{2}+17}{\left( 2m+1\right) \left( m+1\right) }, \end{aligned}$$

which is

$$\begin{aligned} \sum _{i=1}^{m-1}\frac{1}{2i-1}>\frac{4\left( 4m^{4}+4m^{3}-4m^{2}-6m-1\right) }{\left( 2m-1\right) \left( 2m+1\right) \left( 7m+4m^{2}+4\right) }. \end{aligned}$$
(6.3)

Using mathematical induction again to prove (6.3). First, we can see that (6.3) holds for \(m=3\). Second, assuming that (6.3) holds for \(m=k\), we have

$$\begin{aligned} \sum _{i=1}^{k-1}\frac{1}{2i-1}>\frac{4\left( 4k^{4}+4k^{3}-4k^{2}-6k-1\right) }{\left( 2k-1\right) \left( 2k+1\right) \left( 7k+4k^{2}+4\right) }. \end{aligned}$$

Via the inequality above, we have

$$\begin{aligned} \sum _{i=1}^{k}\frac{1}{2i-1}= & {} \sum _{i=1}^{k-1}\frac{1}{2i-1}+\frac{1}{2k-1}\\> & {} \frac{4\left( 4k^{4}+4k^{3}-4k^{2}-6k-1\right) }{\left( 2k-1\right) \left( 2k+1\right) \left( 7k+4k^{2}+4\right) }+\frac{1}{2k-1}. \end{aligned}$$

In order to prove that (6.3) is true for \(m=k+1\) it suffices to show that

$$\begin{aligned}&\frac{4\left( 4k^{4}+4k^{3}-4k^{2}-6k-1\right) }{\left( 2k-1\right) \left( 2k+1\right) \left( 7k+4k^{2}+4\right) }+\frac{1}{2k-1} \\&\quad >\frac{4\left( 14k+32k^{2}+20k^{3}+4k^{4}-3\right) }{\left( 2k+1\right) \left( 2k+3\right) \left( 15k+4k^{2}+15\right) }, \end{aligned}$$

that is,

$$\begin{aligned} \frac{1}{2k-1}>\frac{4\left( 172k+184k^{2}+80k^{3}+12k^{4}+57\right) }{\left( 2k+3\right) \left( 2k-1\right) \left( 7k+4k^{2}+4\right) \left( 15k+4k^{2}+15\right) }. \end{aligned}$$

The last inequality holds due to

$$\begin{aligned}&\left( 2k+3\right) \left( 2k-1\right) \left( 7k+4k^{2}+4\right) \left( 15k+4k^{2}+15\right) \\&\qquad -4\left( 2k-1\right) \left( 172k+184k^{2}+80k^{3}+12k^{4}+57\right) \\&\quad =\left( k+1\right) \left( 121k+80k^{2}+16k^{3}+48\right) \left( 2k-1\right) ^{2}>0. \end{aligned}$$

Then we turn to the proof of second inequality. The desired one is equivalent to that

$$\begin{aligned}&2k\left( 3k-1\right) \left( 2k-1\right) \frac{1}{k}\left( \frac{k!2^{k-1}}{\left( 2k-1\right) !!}-\sum _{i=1}^{k}\frac{1}{2i-1}\right) \\&\qquad -\left( k-1\right) \left( 12k^{2}-20k+9\right) \frac{1}{\left( k-1\right) } \left( \frac{\left( k-1\right) !2^{k-2}}{\left( 2k-3\right) !!} -\sum _{i=1}^{k-1}\frac{1}{2i-1}\right) \\&\qquad +2\left( 2k-3\right) \left( k-1\right) \left( k-2\right) \frac{ 1}{\left( k-2\right) }\left( \frac{\left( k-2\right) !2^{k-3}}{\left( 2k-5\right) !!}-\sum _{i=1}^{k-2}\frac{1}{2i-1}\right) \\&\quad >0, \end{aligned}$$

that is,

$$\begin{aligned} \frac{\left( k+1\right) !2^{k}}{\left( 2k-3\right) !!}>\left( 2k-1\right) \left( 2k+1\right) \sum _{i=1}^{k-2}\frac{1}{ 2i-1}+\frac{12k^{2}-12k-1}{2k-3}, \end{aligned}$$

or

$$\begin{aligned} \frac{\left( k+2\right) !2^{k+1}}{\left( 2k-1\right) !!}>\left( 2k+1\right) \left( 2k+3\right) \sum _{i=1}^{k-1}\frac{1}{2i-1}+\frac{12k+12k^{2}-1}{ 2k-1}. \end{aligned}$$
(6.4)

By (6.1) we have

$$\begin{aligned}&\frac{\left( k+2\right) !2^{k+1}}{\left( 2k-1\right) !!}=\frac{ k+2}{2}\frac{8\left( k+1\right) !2^{k-1}}{\left( 2k-1\right) !!} \\&\quad >\frac{k+2}{2}\left( \frac{\left( 2k+1\right) \left( 4k+1\right) }{k} \sum _{i=1}^{k-1}\frac{1}{2i-1}++\frac{4k+12k^{2}+1}{k\left( 2k-1\right) } \right) . \end{aligned}$$

In order to complete the proof of (6.4) it suffices to show that

$$\begin{aligned} \sum _{i=1}^{k-1}\frac{1}{2i-1}>1-\frac{4k}{4k^{2}-1}, \end{aligned}$$

which is true due to

$$\begin{aligned} \sum _{i=1}^{k-1}\frac{1}{2i-1}>1. \end{aligned}$$

7 Proof of Theorem 1.1

Let \(\arctan x=t,x\in (-\infty ,\infty )\). Then \(x=\tan t,t\in (-\pi /2,\pi /2)\). Because the functions involved in this section are all even functions, we only assume \(x>0\), which is corresponding to \(t\in (0,\pi /2)\). Since

$$\begin{aligned} G_{k}(x)= & {} \frac{\sqrt{1+x^{2}}}{x}\left( \left( \arctan x\right) ^{2}- \frac{x\ln \left( x+\sqrt{1+x^{2}}\right) }{\sqrt{1+x^{2}}}-\sum _{n=3}^{k}\left( -1\right) ^{n}v_{n}x^{2n}\right) \\= & {} \frac{t^{2}-\left( \sin t\right) \ln (\tan t+\sec t)-\sum _{n=3}^{k}\left( -1\right) ^{n}v_{n}\tan ^{2n}t}{\sin t} \\= & {} \frac{t^{2}}{\sin t}-\ln \frac{1+\sin t}{\cos t}-\frac{1}{\sin t} \sum _{n=3}^{k}\left( -1\right) ^{n}v_{n}\tan ^{2n}t \\:= & {} F_{k}(t). \end{aligned}$$

we examine the properties of the function \(F_{k}(t)\)\((k\ge 3)\) as follows.

Computing to give

$$\begin{aligned} F_{k}^{\prime }(t)= & {} -\frac{1}{\sin ^{2}t}\left( t^{2}\cos t-2t\sin t\right) -\frac{1}{\cos t}+\frac{\cos t}{\sin ^{2}t}\sum _{n=3}^{k}\left( -1\right) ^{n}v_{n}\tan ^{2n}t \\&-\frac{1}{\sin t}\sum _{n=3}^{k}\left( -1\right) ^{n}2nv_{n}\left( \tan ^{2n-1}t+\tan ^{2n+1}t\right) \\:= & {} \frac{\cos t}{\sin ^{2}t}g_{k}(t), \end{aligned}$$

where

$$\begin{aligned} g_{k}(t)= & {} -\left( t-\tan t\right) ^{2}-2\sum _{n=3}^{k}\left( -1\right) ^{n}nv_{n}\tan ^{2n+2}t-\sum _{n=3}^{k}\left( -1\right) ^{n}v_{n}\left( 2n-1\right) \tan ^{2n}t \\= & {} -\left( t-\tan t\right) ^{2}-2\left( \left( -1\right) ^{k}kv_{k}\tan ^{2k+2}t+\sum _{n=3}^{k-1}\left( -1\right) ^{n}nv_{n}\tan ^{2n+2}t\right) \\&-\left( \left( -1\right) ^{3}5v_{3}\tan ^{6}t+\sum _{n=4}^{k}\left( -1\right) ^{n}v_{n}\left( 2n-1\right) \tan ^{2n}t\right) \\= & {} -\left( t-\tan t\right) ^{2}-2\left( -1\right) ^{k}kv_{k}\tan ^{2k+2}t-2\sum _{n=3}^{k-1}\left( -1\right) ^{n}nv_{n}\tan ^{2n+2}t\\&+5v_{3}\tan ^{6}t-\sum _{n=4}^{k}\left( -1\right) ^{n}v_{n}\left( 2n-1\right) \tan ^{2n}t\\= & {} -\left( t-\tan t\right) ^{2}+5v_{3}\tan ^{6}t-2\sum _{n=3}^{k-1}\left( -1\right) ^{n}nv_{n}\tan ^{2n+2}t \\&-\sum _{n=4}^{k}\left( -1\right) ^{n}v_{n}\left( 2n-1\right) \tan ^{2n}t-2\left( -1\right) ^{k}kv_{k}\tan ^{2k+2}t \\= & {} -\left( t-\tan t\right) ^{2}+5v_{3}\tan ^{6}t-2\sum _{n=4}^{k}\left( -1\right) ^{n-1}\left( n-1\right) v_{n-1}\tan ^{2n}t \\&-\sum _{n=4}^{k}\left( -1\right) ^{n}v_{n}\left( 2n-1\right) \tan ^{2n}t-2\left( -1\right) ^{k}kv_{k}\tan ^{2k+2}t \\= & {} -\left( t-\tan t\right) ^{2}+5v_{3}\tan ^{6}t\\&-\sum _{n=4}^{k}\left( -1\right) ^{n-1}\left( 2\left( n-1\right) v_{n-1}-\left( 2n-1\right) v_{n}\right) \tan ^{2n}t \\&-2\left( -1\right) ^{k}kv_{k}\tan ^{2k+2}t. \end{aligned}$$

Then

$$\begin{aligned} g_{k}^{\prime }(t)= & {} 2\left( t\tan ^{2}t-\tan ^{3}t\right) +30v_{3}\left( \tan ^{7}t+\tan ^{5}t\right) \\&-\sum _{n=4}^{k}\left( -1\right) ^{n-1}2n\left( 2\left( n-1\right) v_{n-1}-\left( 2n-1\right) v_{n}\right) \left( \tan ^{2n+1}t+\tan ^{2n-1}t\right) \\&-2\left( -1\right) ^{k}k\left( 2k+2\right) v_{k}\left( \tan ^{2k+3}t+\tan ^{2k+1}t\right) \\:= & {} \left( \tan ^{2}t\right) f_{k}(t), \end{aligned}$$

where

$$\begin{aligned} f_{k}(t)= & {} 2\left( t-\tan t\right) \\&-\sum _{i=3}^{k}\left( -1\right) ^{i-1}2i\left[ 2\left( i-1\right) v_{i-1}-\left( 2i-1\right) v_{i}\right] \left( \tan ^{2i-1}t+\tan ^{2i-3}t\right) \\&-2\left( -1\right) ^{k}k\left( 2k+2\right) v_{k}\left( \tan ^{2k+1}t+\tan ^{2k-1}t\right) . \end{aligned}$$

Computing to get

$$\begin{aligned}&f_{k}^{\prime }(t) =-2\tan ^{2}t \\&\qquad +\sum _{i=3}^{k}\left( -1\right) ^{i-1}2i\left[ 2\left( i-1\right) v_{i-1}-\left( 2i-1\right) v_{i}\right] \cdot \\&\qquad \left[ \left( 1-2i\right) \tan ^{2i}t+4\left( 1-i\right) \tan ^{2i-2}t+\left( 3-2i\right) \tan ^{2i-4}t\right] \\&\qquad -2\left( -1\right) ^{k}k\left( 2k+2\right) v_{k}\left[ \left( 2k+1\right) \tan ^{2k+2}t+4k\tan ^{2k}t+\left( 2k-1\right) \tan ^{2k-2}t\right] \\&\quad :=-2\left( 1-45v_{3}\right) \tan ^{2}t+\sum _{i=4}^{k}\left( -1\right) ^{i-1}2\gamma _{i}\tan ^{2i-4}t \\&\qquad +2\left( -1\right) ^{k+1}\left( \tan ^{2k-2}t\right) \left( \lambda \tan ^{4}t+k\mu \tan ^{2}t+\zeta \right) \\&\quad =\sum _{i=3}^{k}\left( -1\right) ^{i-1}2\gamma _{i}\tan ^{2i-4}t+2\left( -1\right) ^{k+1}\left( \tan ^{2k-2}t\right) \left( \lambda \tan ^{4}t+k\mu \tan ^{2}t+\zeta \right) , \\&\quad =2\left( -1\right) ^{k+1}\left( \tan ^{2k-2}t\right) \left( \lambda \tan ^{4}t+k\mu \tan ^{2}t+\zeta \right) \end{aligned}$$

by Lemm 2.3, where

$$\begin{aligned} \lambda= & {} k\left( 2k+2\right) \left( 2k+1\right) v_{k}, \\ \mu= & {} \left( 1+4k+12k^{2}\right) v_{k}-\left( 4k^{2}-6k+2\right) v_{k-1}, \\ \zeta= & {} 2k\left( 3k-1\right) \left( 2k-1\right) v_{k}-\left( k-1\right) \left( 12k^{2}-20k+9\right) v_{k-1}\\&+ 2\left( 2k-3\right) \left( k-1\right) \left( k-2\right) v_{k-2}. \end{aligned}$$

By Lemmas 2.2 and 2.4 respectively, we have \(\lambda >0\), and \(\mu ,\zeta >0\). So we have

$$\begin{aligned} \lambda \tan ^{4}t+k\mu \tan ^{2}t+\zeta >0, \end{aligned}$$

which leads to that \(f_{k}^{\prime }(t)<0\) for all \(t\in (0,\pi /2)\) when k is an even number and \(f_{k}^{\prime }(t)>0\) for all \(t\in (0,\pi /2)\) when k is an odd number. Noticing these facts \(F_{k}(0^{+})=g_{k}(0)=f_{k}(0)=0,\) by differential method we get the conclusion that \(F_{k}(t)\) is decreasing and negative on \((0,\pi /2)\) when k is an even number, and is increasing and positive on \((0,\pi /2)\) when k is an odd number. Because the transformation \(x=\tan t\) increases monotonically on the interval \((0,\pi /2) \), the function \(G_{k}(x)\) involved in Theorem 1.1 has the same properties as \(F_{k}(t)\), that is to say, \(G_{k}(x)\) is decreasing and negative on \((0,+\infty )\) when k is an even number and is increasing and positive on \((0,+\infty )\) when k is an odd number.

In view of

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{\left( \arctan x\right) ^{2}-\frac{x\ln (x+\sqrt{ 1+x^{2}})}{\sqrt{1+x^{2}}}-\sum _{n=3}^{k-1}\left( -1\right) ^{n}v_{n}x^{2n}}{ x^{2k}}=\left( -1\right) ^{k}v_{k} \end{aligned}$$
(7.1)

by Lemma 2.2, the proof of Theorem 1.1 is complete.

Remark 7.1

We know that the inequality (1.1) can be traced back to the generalization of the famous Cauchy-Schwarz inequality, which can be found in [10] and the references cited therein.