4.1 Simpson–Spectortype inequality for \(W_{v} ( x ) \)
It is clear that
$$ F_{p} ( x ) < ( > ) c\quad \iff\quad \bigl( W_{v} ( x ) p \bigr) ^{2} ( 2v+2p ) ^{2}< ( > ) cx^{2}, $$
where the latter indeed offers some new Simpson–Spectortype inequalities for \(W_{v} ( x ) \). In fact, by Theorem 1.1 we immediately get the following.
Proposition 4.1
Let
$$\begin{aligned}& E_{1} = \biggl\{ p\geq v+\frac{1}{2},v\geq \frac{3}{2} \biggr\} ,\qquad E_{2}= \biggl\{ p\geq c_{v},2< v< \frac{3}{2} \biggr\} , \\& E_{3} = \{ p\leq v1,v>2 \} ,\qquad E_{4}= \biggl\{ v1< p< v+\frac{1}{2},v>2 \biggr\} , \\& E_{5} = \biggl\{ v+\frac{1}{2}\leq p< c_{v},2< v<  \frac{3}{2} \biggr\} , \end{aligned}$$
where
\(c_{v}\)
is given in (1.7). Then the double inequality
$$ \alpha x^{2}< \bigl( W_{v} ( x ) p \bigr) ^{2} ( 2v+2p ) ^{2}< \beta x^{2} $$
(4.1)
holds for
\(x>0\)
and
\(v>2\)
if and only if
$$\begin{aligned}& \alpha\leq l ( p ) =\min \biggl\{ \frac {2v+2p}{v+2},1 \biggr\} , \\& \beta\geq u ( p ) = \textstyle\begin{cases} 1 & \textit{if } ( p,v ) \in E_{1}\cup E_{2}, \\ \frac{2v+2p}{v+2} & \textit{if } ( p,v ) \in E_{3}, \\ \lambda_{p} & \textit{if } ( p,v ) \in E_{4}, \\ \theta_{p} & \textit{if } ( p,v ) \in E_{5}, \end{cases}\displaystyle \end{aligned}$$
where
\(\theta_{p}=\sup_{x>0}F_{p} ( x ) \)
if
\(( p,v ) \in E_{5}\), and
$$ \lambda_{p}=\frac{ ( W_{v} ( x_{0} ) p ) ^{2} ( 2v+2p ) ^{2}}{x_{0}^{2}}, $$
and here
\(x_{0}\)
is the unique solution of the equation
$$ y^{3} ( p+2v+1 ) y^{2} \bigl( x^{2}2pv \bigr) y+px^{2}+4 ( v+1 ) ( pv1 ) =0 $$
(4.2)
on
\(( 0,\infty ) \)
with
\(y=W_{v} ( x ) \).
Proof
(i) By Theorem 1.1 we see that the lefthand side inequality of (4.1) holds for \(x>0\) if and only if
$$\begin{aligned} \alpha \leq& \textstyle\begin{cases} \frac{2v+2p}{v+2} & \text{if } ( p,v ) \in E_{1}\cup E_{2}, \\ 1 & \text{if } ( p,v ) \in E_{3}, \\ \min \{ \frac{2v+2p}{v+2},1 \} & \text{if } ( p,v ) \in E_{4}, \\ \frac{2v+2p}{v+2} & \text{if } ( p,v ) \in E_{5}. \end{cases}\displaystyle \\ =& \textstyle\begin{cases} \frac{2v+2p}{v+2}, & \text{if } ( p,v ) \in E_{1}\cup E_{2}\cup E_{5}\cup ( E_{4}\cap \{ p\geq v,v>2 \} ) , \\ 1, & \text{if } ( p,v ) \in E_{3}\cup ( E_{4}\cap \{ p\leq v,v>2 \} ) . \end{cases}\displaystyle \end{aligned}$$
It is easy to check that
$$\begin{aligned} &E_{1}\cup E_{2}\cup E_{5}\cup \bigl( E_{4}\cap \{ p\geq v,v>2 \} \bigr) = \{ p\geq v,v>2 \} , \\ &E_{3}\cup \bigl( E_{4}\cap \{ p\leq v,v>2 \} \bigr) = \{ p\leq v,v>2 \} , \end{aligned} $$
which indicate that \(\alpha\leq l ( p ) \).
(ii) The necessary and sufficient conditions for the righthand side inequality of (4.1) to hold are obvious.
(iii) As shown in Simpson and Spector [2], \(W_{v}\) satisfies the Riccati equation
$$ xW_{v}^{\prime} ( x ) =x^{2}+2 ( v+1 ) W_{v} ( x ) W_{v} ( x ) ^{2}. $$
Then
$$\begin{aligned} \frac{x^{3}}{2}F_{p}^{\prime} ( x ) =& \bigl( W_{v} ( x ) p \bigr) xW_{v}^{\prime} ( x )  \bigl( W_{v} ( x ) p \bigr) ^{2}+ ( 2v+2p ) ^{2} \\ =& ( yp ) \bigl( x^{2}+2 ( v+1 ) yy^{2} \bigr)  ( yp ) ^{2}+ ( 2v+2p ) ^{2} \\ =&y^{3}+ ( p+2v+1 ) y^{2}+ \bigl( x^{2}2pv \bigr) ypx^{2}4 ( v+1 ) ( pv1 ) , \end{aligned}$$
where \(y=W_{v} ( x ) \). Clearly, if \(x_{0}\) is the unique solution of the equation \(F_{p}^{\prime} ( x ) =0\) on \(( 0,\infty ) \), then so is equation (4.2) on \(( 0,\infty ) \).
This completes the proof. □
Remark 4.2
Taking \(p=v+1/2\) in Proposition 4.1 gives
$$ \frac{2v+3}{2v+4}x^{2}< W_{v} ( x ) ^{2} ( 2v+1 ) W_{v} ( x ) 2 ( v+1 ) < x^{2}\quad\text{for }x>0\text{ and }v>\frac{3}{2}, $$
where the lefthand side inequality holds for \(x>0\) and \(v>2\), the righthand side one is inequality (1.4).
Setting \(p=v\) in Proposition 4.1 yields
$$ x^{2}< W_{v} ( x ) ^{2}2vW_{v} ( x ) 4 ( v+1 ) < \lambda_{v}x^{2}\quad\text{for }x>0\text{ and }v>2, $$
where the lefthand side inequality is inequality (1.5).
In addition, putting \(p=c_{v}\) with \(2< v<3/2\) in Proposition 4.1, where \(c_{v}\) is given in (1.7), we obtain a new Simpson–Spectortype inequality, which is stated as a corollary.
Corollary 4.3
Let
\(2< v<3/2\). Then the double inequalities
$$\begin{aligned} &\frac{ ( 2v+3 ) ( v+3 ) ( v+4 ) }{ ( v+2 ) ( 2v^{2}+11v+16 ) }x^{2}\\ &\quad < W_{v} ( x ) ^{2}2 \frac{2v^{3}+9v^{2}+9v4}{2v^{2}+11v+16}W_{v} ( x ) 8\frac{ ( 2v+5 ) ( v+1 ) ( v+2 ) }{2v^{2}+11v+16}< x^{2} \end{aligned} $$
hold for
\(x>0\). The lower and upper bounds are sharp.
4.2 Sharp bounds for \(W_{v} ( x ) \) in the form of \(p+r\sqrt{x^{2}+q^{2}}\)
A bound in the form of
$$ A_{p,q} ( x ) =p+\sqrt{x^{2}+q^{2}} $$
(4.3)
for the ratio \(W_{v} ( x ) \) is known as Amostype bound (see [6, 9, 10]). In this subsection, we will give another type of bounds in the form of
$$ B_{p,q,r} ( x ) =p+r\sqrt{x^{2}+q^{2}} $$
(4.4)
for \(W_{v} ( x ) \) by Proposition 4.1. Clearly, \(B_{p,q,1} ( x ) =A_{p,q} ( x ) \).
As mentioned in Introduction, Baricz and Neuman [8, Theorem 2.2] (also see [6, Lemma 4.2.]) have shown that \(W_{v}\) is strictly increasing from \(( 0,\infty ) \) onto \(( 2v+2,\infty ) \) for \(v>2\). This implies that \(W_{v} ( x ) p>0\) for \(p<2v+2\), and then the double inequality of (4.1) is equivalent to
$$ p+\sqrt{\alpha x^{2}+ ( 2v+2p ) ^{2}}< W_{v} ( x ) < p+\sqrt{\beta x^{2}+ ( 2v+2p ) ^{2}} $$
(4.5)
for \(x>0\) and \(p<2v+2\) with \(v>2\). Thus from Proposition 4.1 we derive the following statement.
Proposition 4.4
Let
\(E_{0}= \{ p<2v+2,v>2 \} \). (i) The double inequality (4.5) holds for
\(x>0\)
and
\(( p,v ) \in E_{0}\)
if and only if
$$\begin{aligned}& \alpha\leq l ( p ) =\min \biggl\{ \frac {2v+2p}{v+2},1 \biggr\} , \\& \beta\geq u^{\ast} ( p ) = \textstyle\begin{cases} 1 & \textit{if }p\geq v+\frac{1}{2},v\geq\frac{3}{2}, \\ \frac{2v+2p}{v+2} & \textit{if }p\leq v1,v>2, \\ \lambda_{p} & \textit{if }v1< p< \min \{v+\frac{1}{2},2v+2 \} ,v>2, \end{cases}\displaystyle \end{aligned}$$
where
\(\lambda_{p}\)
is as in Proposition 4.1.
(ii) Furthermore, let
$$\begin{aligned}& L_{p} ( x ) =p+\sqrt{\alpha_{\max}x^{2}+ ( 2v+2p ) ^{2}}, \end{aligned}$$
(4.6)
$$\begin{aligned}& U_{p} ( x ) =p+\sqrt{\beta_{\min}x^{2}+ ( 2v+2p ) ^{2}}. \end{aligned}$$
(4.7)
Then we have
$$\begin{aligned}& \max_{p< 2v+2,v>2}L_{p} ( x ) =v+\sqrt{x^{2}+ ( v+2 ) ^{2}}=L_{v} ( x ) , \end{aligned}$$
(4.8)
$$\begin{aligned}& \min_{p\geq v+1/2,v\geq3/2}U_{p} ( x ) =v+\frac {1}{2}+ \sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}}:=U_{v+1/2}^{ ( 1 ) } ( x ) , \end{aligned}$$
(4.9)
$$\begin{aligned}& \min_{p\leq v1,v>2}U_{p} ( x ) =v1+\sqrt{ \frac {v+3}{v+2}x^{2}+ ( v+3 ) ^{2}}:=U_{v1}^{ ( 2 ) } ( x ) . \end{aligned}$$
(4.10)
Moreover, \(\min_{p\geq v+1/2,v\geq3/2}U_{p} ( x ) \)
and
\(\min_{p\leq v1,v\geq3/2}U_{p} ( x ) \)
are not comparable for
\(x\in ( 0,\infty ) \).
Proof
(i) By Proposition 4.1, the necessary and sufficient condition such that the lefthand side inequality of (4.5) holds for \(x>0\) and \(( p,v ) \in E_{0}\) is clear.
While the righthand side inequality of (4.5) holds for \(x>0\) and \(( p,v ) \in E_{0}\) if and only if
$$ \beta\geq \textstyle\begin{cases} 1 & \text{if } ( p,v ) \in ( E_{1}\cup E_{2} ) \cap E_{0}, \\ \frac{2v+2p}{v+2} & \text{if } ( p,v ) \in E_{3}\cap E_{0}, \\ \lambda_{p} & \text{if } ( p,v ) \in E_{4}\cap E_{0}, \\ \theta_{p} & \text{if } ( p,v ) \in E_{5}\cap E_{0}, \end{cases} $$
where \(E_{i}\) (\(i=15\)) are given in Proposition 4.1. Simplifying yields \(E_{1}\cap E_{0}=E_{1}\), \(E_{2}\cap E_{0}=\emptyset\), \(E_{3}\cap E_{0}=E_{3}\), \(E_{5}\cap E_{0}=\emptyset\), and
$$ E_{4}\cap E_{0}= \biggl\{ v1< p< \min \biggl\{ v+ \frac{1}{2},2v+2 \biggr\} ,v>2 \biggr\} , $$
which imply that \(\beta\geq u^{\ast} ( p ) \).
(ii) To prove the second assertion of this proposition, we first note that the function
$$ p\mapsto p+\sqrt{x^{2}+ ( 2v+2p ) ^{2}} $$
is increasing on \(\mathbb{R}\), and another function
$$ p\mapsto p+\sqrt{\frac{2v+2p}{v+2}x^{2}+ ( 2v+2p ) ^{2}} $$
is decreasing on \((\infty,2v+2]\).
Now, since
$$ L_{p} ( x ) = \textstyle\begin{cases} p+\sqrt{\frac{2v+2p}{v+2}x^{2}+ ( 2v+2p ) ^{2}} & \text{if }v\leq p\leq2v+2, \\ p+\sqrt{x^{2}+ ( 2v+2p ) ^{2}} & \text{if }p\leq v, \end{cases} $$
the function \(p\mapsto L_{p} ( x ) \) is increasing on \((\infty,v]\) and decreasing on \([ v,2v+2 ] \), which implies \(\max_{p\leq 2v+2,v>2}L_{p} ( x ) =L_{v} ( x ) \).
If \(p\geq v+1/2\) with \(v\geq3/2\), then \(\beta_{\min}=1\), and then
$$ U_{p} ( x ) =p+\sqrt{x^{2}+ ( 2v+2p ) ^{2}}:=U_{p}^{ ( 1 ) } ( x ) , $$
(4.11)
which is increasing in p on \([ v+1/2,2v+2 ] \). This gives \(\min_{p\geq v+1/2,v\geq3/2}U_{p} ( x ) =U_{v+1/2}^{ ( 1 ) } ( x ) \).
If \(p\leq v1\) with \(v>2\), then \(\beta_{\min}= ( 2v+2p ) / ( v+2 ) \), and therefore,
$$ U_{p} ( x ) =p+\sqrt{\frac{2v+2p}{v+2}x^{2}+ ( 2v+2p ) ^{2}}:=U_{p}^{ ( 2 ) } ( x ) , $$
(4.12)
which is decreasing in p on \((\infty,2v+2]\). This leads to \(\min_{p\leq v1,v>2}U_{p} ( x ) =U_{v1}^{ ( 2 ) } ( x ) \).
Finally, we show that \(\min_{p\geq v+1/2,v\geq3/2}U_{p} ( x ) \) is not comparable with \(\min_{p\leq v1,v\geq3/2}U_{p} ( x ) \) for \(x\in ( 0,\infty ) \). In fact, we have that for \(v\geq3/2\),
$$\begin{aligned}& U_{v+1/2}^{ ( 1 ) } ( x ) U_{v1}^{ ( 2 ) } ( x ) = \frac{3}{2}+\sqrt{x^{2}+ \biggl( v+\frac {3}{2} \biggr) ^{2}}\sqrt{\frac{v+3}{v+2}x^{2}+ ( v+3 ) ^{2}}, \\& \biggl( \frac{3}{2}+\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}} \biggr) ^{2}\frac{v+3}{v+2}x^{2} ( v+3 ) ^{2} \\& \quad =3\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}} \biggl( 3 \biggl( v+\frac{3}{2} \biggr) + \frac{x^{2}}{v+2} \biggr) , \\& \biggl( 3\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}} \biggr) ^{2} \biggl( 3 \biggl( v+\frac{3}{2} \biggr) +\frac{x^{2}}{v+2} \biggr) ^{2}=x^{2} \frac{x^{2}3 ( v+2 ) ( v+3 ) }{ ( v+2 ) ^{2}}. \end{aligned}$$
From this it is seen that \(U_{v+1/2}^{ ( 1 ) } ( x ) < U_{v1}^{ ( 2 ) } ( x ) \) if \(x>\sqrt{3 ( v+2 ) ( v+3 ) }\) and \(U_{v+1/2}^{ ( 1 ) } ( x ) >U_{v1}^{ ( 2 ) } ( x ) \) if \(0< x<\sqrt{3 ( v+2 ) ( v+3 ) }\).
This completes the proof. □
Remark 4.5
Amos [10, Eq. (11)] offered a lower bound \(A_{v,v+2} ( x ) < W_{v} ( x ) \) for \(x>0\) and \(v\geq0\). Hornik and Grün [9, Theorem 6] showed that the Amostype bound is the sharpest for \(x>0\) and \(v>1\). Yang and Zheng [6, Theorem 4.6] extended the range of v from \(v>1\) to \(v>3/2\). Proposition 4.4 presents another lower bound \(L_{p} ( x ) \) defined in (4.6) for \(W_{v} ( x ) \) for \(x>0\) with \(v>2\) and shows that \(L_{v} ( x ) \) defined by (4.8) is the maximum over all lower bounds \(\{ L_{p} ( x ) :p<2v+2,v>2 \} \). It should be emphasized that our sharpest lower bound \(L_{v} ( x ) \) extends the range of \(A_{v,v+2} ( x ) \) from \(v>3/2\) to \(v>2\) although \(L_{v} ( x ) \) and \(A_{v,v+2} ( x ) \) have the same expression.
Remark 4.6
Amos [10, Eq. (16)] gave an upper bound \(W_{v} ( x ) < A_{v+1/2,v+3/2} ( x ) \) for \(x>0\) and \(v\geq0 \). Hornik and Grün [9, Theorem 3] proved that this Amostype upper bound is the best for \(x>0\) and \(v>1\), where the range of v has been extended from \(v>1\) to \(v>3/2\) in [6, Theorem 4.4] by Yang and Zheng. Since \(U_{v+1/2}^{ ( 1 ) } ( x ) =A_{v+1/2,v+3/2} ( x ) \), our Proposition 4.4 demonstrates the same result in [6, Theorem 4.4] by a slightly different approach.
Remark 4.7
Proposition 4.4 also gives another upper bounds for \(W_{v} ( x ) \) by \(U_{p}^{ ( 2 ) } ( x ) \) defined in (4.10) for \(x>0\) and \(p\leq v1\) with \(v>2\), that is,
$$ W_{v} ( x ) < p+\sqrt{\frac{2v+2p}{v+2}x^{2}+ ( 2v+2p ) ^{2}}=U_{p}^{ ( 2 ) } ( x ) . $$
(4.13)
Not only the above inequalities are valid, but we explain that \(U_{v1}^{ ( 2 ) } ( x ) \) is the minimum over all upper bounds \(\{ U_{p}^{ ( 2 ) } ( x ) :p\leq v1,v>2 \} \), and \(U_{v1}^{ ( 2 ) } ( x ) \) and \(U_{v+1/2}^{ ( 1 ) } ( x ) \) are not comparable in x on \(( 0,\infty ) \) for \(v\geq3/2\). This indicates that \(U_{v1}^{ ( 2 ) } ( x ) \) for \(v>2\) is indeed a new sharpest upper bound for \(W_{v} ( x ) \). Consequently, Proposition 4.4 in fact offers a new type of bounds \(p+r\sqrt{x^{2}+q^{2}}\) (\(r>0\)) for \(W_{v} ( x ) \), which is clearly different from the Amostype bound \(A_{p,q} ( x ) =p+\sqrt {x^{2}+q^{2}}\). Moreover, inequality (4.13) is sharp at \(x=0\) in view of
$$ W_{v} ( x ) U_{p}^{ ( 2 ) } ( x ) \thicksim ( 2v+2 ) p \vert 2v+2p \vert =0 $$
as \(x\rightarrow0\).
As a direct consequence of Proposition 4.4, we have the following.
Corollary 4.8
If
\(v+1/2\leq p<2v+2\)
with
\(v\geq3/2\), then the double inequality
$$ p+\sqrt{\frac{2v+2p}{v+2}x^{2}+ ( 2v+2p ) ^{2}}< W_{v} ( x ) < p+\sqrt{x^{2}+ ( 2v+2p ) ^{2}} $$
(4.14)
holds for all
\(x>0\). Inequalities (4.14) are reversed if
\(p\leq v1\)
with
\(v>2\).
In particular, taking
\(p=v+1/2\), \(v+1\), \(( 2v+2 ) ^{}\)
and
\(p=v1\), −∞, we have
$$\begin{aligned}& \begin{aligned}[b] &v+\frac{1}{2}+\sqrt{\frac{v+3/2}{v+2}x^{2}+ \biggl( v+ \frac {3}{2} \biggr) ^{2}}\\ &\quad < W_{v} ( x ) < v+ \frac{1}{2}+\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}}\quad\textit{for }v\geq\frac{3}{2}, \end{aligned} \end{aligned}$$
(4.15)
$$\begin{aligned}& v+1+\sqrt{\frac{v+1}{v+2}x^{2}+ ( v+1 ) ^{2}}< W_{v} ( x ) < v+1+\sqrt{x^{2}+ ( v+1 ) ^{2}}\quad\textit{for }v>1, \end{aligned}$$
(4.16)
$$\begin{aligned}& 2 ( v+1 ) < W_{v} ( x ) < 2v+2+x\quad\textit{for }v\geq \frac{3}{2}, \end{aligned}$$
(4.17)
$$\begin{aligned}& v1+\sqrt{x^{2}+ ( v+3 ) ^{2}}< W_{v} ( x ) < v1+ \sqrt{\frac{v+3}{v+2}x^{2}+ ( v+3 ) ^{2}}\quad\textit{for }v>2, \\& 2 ( v+1 ) < W_{v} ( x ) < 2 ( v+1 ) +\frac{1}{2} \frac{x^{2}}{v+2}\quad\textit{for }v>2. \end{aligned}$$
(4.18)
Remark 4.9
The righthand side in inequality (4.15) for \(v\geq 0\) was proved in [10, Eq. (16)] by Amos, and for \(v>3/2\) it follows from Neuman’s inequality (1.4). The righthand side one in (4.16) for \(v\geq0\) is also due to Amos [10, Eq. (11)], which for \(v>1\) was proved by Yuan and Kalbfleisch [12, Eq. (A.5)], and Laforgia and Natalini [14, Theorem 1.1]. While the lefthand side inequality in (4.16) for \(v>1\) was showed by Segura [15, Eq. (61)]. Inequalities (4.17) were proved by Yang and Zheng in [6, Remark 4.9]. Moreover, the rational bounds given in (4.18) appeared in [4, Appendix] for \(v>1\), so the righthand side inequality of (4.18) can be viewed as a new one in the sense that the range of v is extended from \(v>1\) to \(v>2\).
Now let us return to Proposition 4.1. First, if \(( p,v ) \in E_{2}= \{ p\geq c_{v},2< v<3/2 \} \), then the righthand side inequality of (4.1) holds for \(x>0\) if and only if \(\beta\geq1\), which implies that the double inequality
$$ p\sqrt{\beta x^{2}+ ( 2v+2p ) ^{2}}< W_{v} ( x ) < p+\sqrt{\beta x^{2}+ ( 2v+2p ) ^{2}} $$
(4.19)
holds for \(x>0\). Second, according to the guess presented in Remark 3.1, \(\theta_{p}=\sup_{x>0}F_{p} ( x ) \) may equal \(F_{p} ( \infty ) =1\) for certain \(p\in{}[ v+1/2,c_{v})\) with \(2< v<3/2\). If so, then the righthand side inequality of (4.1) holds for \(x>0\) if and only if \(\beta\geq\theta_{p}=1\), which implies that the double inequality (4.19) also holds for \(x>0\) and certain \(p\in{}[ v+1/2,c_{v})\) with \(2< v<3/2\), where \(\beta=1\) is the best possible constant. In fact, this claim is valid.
Proposition 4.10
Let
\(2< v<3/2\). Then the double inequality (4.19) holds for
\(x>0\)
and
\(p\geq c_{v}^{\ast}=v+1/ ( 2v+5 ) \)
with the best constant
\(\beta=1\).
Proof
(i) For \(p\geq c_{v}\), the desired result is evidently valid by Proposition 4.1.
(ii) For \(c_{v}^{\ast}\leq p< c_{v}\), where \(c_{v}\) is defined by (1.7), it is easy to check that
$$\begin{aligned}& c_{v}^{\ast} \biggl( v+\frac{1}{2} \biggr) =  \frac{1}{2}\frac {2v+3}{2v+5}>0, \\& c_{v}^{\ast}c_{v} = 2\frac{ ( 2v+3 ) ( v+2 ) ( v+3 ) }{ ( 2v+5 ) ( 2v^{2}+11v+16 ) }< 0, \end{aligned}$$
which imply \(c_{v}^{\ast}\in{}[ c+1/2,c_{v})\). To prove the desired assertion, it suffices to prove that
$$ F_{p} ( x ) =\frac{f_{1} ( x ) }{f_{2} ( x ) }=\frac{\sum_{n=0}^{\infty}a_{n+1} ( x^{2}/4 ) ^{n}}{\sum_{n=0}^{\infty}b_{n+1} ( x^{2}/4 ) ^{n}}< 1 $$
(4.20)
for \(x>0\) and \(c_{v}^{\ast}\leq p< c_{v}\) with \(2< v<3/2\). Indeed, we have
$$ \frac{a_{1}}{b_{1}}1=\frac{pv}{v+2}< 0, $$
and for \(n\geq2\),
$$\begin{aligned} \frac{a_{n}}{b_{n}}1 =&\frac{p ( 2n+2v+1 ) +n ( 2v+1 ) + ( v+1 ) ( 2v1 ) }{ ( n+v+1 ) ( n+2v+2 ) } \\ \leq&\frac{c_{v}^{\ast} ( 2n+2v+1 ) +n ( 2v+1 ) + ( v+1 ) ( 2v1 ) }{ ( n+v+1 ) ( n+2v+2 ) } \\ =&\frac{ ( n2 ) ( 2v+3 ) }{ ( n+2v+2 ) ( 2v+5 ) ( n+v+1 ) }\leq0, \end{aligned}$$
which yield
$$ F_{p} ( x ) =\frac{\sum_{n=0}^{\infty}a_{n+1} ( x^{2}/4 ) ^{n}}{\sum_{n=0}^{\infty}b_{n+1} ( x^{2}/4 ) ^{n}}< \frac {\sum_{n=0}^{\infty}b_{n+1} ( x^{2}/4 ) ^{n}}{\sum_{n=0}^{\infty }b_{n+1} ( x^{2}/4 ) ^{n}}=1. $$
In view of \(F_{p} ( \infty ) =1\), the upper bound given in (4.20) is sharp, and by Proposition 4.1 the desired assertion follows. Thus we complete the proof. □
Remark 4.11
It is easy to check that the lower bound for \(W_{v} ( x ) \) given in (4.19) is weaker than \(2v+2\), but the upper bound for \(\beta=1\) is clearly a new Amostype bound for \(p\geq c_{v}^{\ast}\) with \(2< v<3/2\) since it is not comparable with the sharpest upper bound \(U_{v1}^{ ( 2 ) } ( x ) \) for \(p\geq c_{v}^{\ast}\) with \(2< v<3/2\), while another one \(U_{v+1/2}^{ ( 1 ) } ( x ) \) is restricted in \(v\geq3/2\).
4.3 Some computable bounds for \(W_{v} ( x ) \)
From Proposition 4.4 we see that the minimum \(\beta_{\min }=\lambda _{p}\) for \(v1< p<\min \{ v+1/2,2v+2 \} \) with \(v>2\) such that the inequality
$$ W_{v} ( x ) < p+\sqrt{\lambda_{p}x^{2}+ ( 2v+2p ) ^{2}}=U_{p}^{ ( 3 ) } ( x ) $$
holds for \(x>0\). Since \(\lambda_{p}=f_{1} ( x_{0} ) /f_{2} ( x_{0} ) \), where \(x_{0}\) is the unique solution of equation (4.2) on \(( 0,\infty ) \), the number \(\lambda_{p}\) is usually not computable, and so is \(U_{p}^{ ( 3 ) } ( x ) \). Therefore, it is interesting and useful to find some upper bounds for \(\lambda_{p}\) by elementary functions.
In this subsection, we will find some upper bounds for \(\lambda_{p}\) in terms of elementary functions to obtain some computable upper bounds for \(W_{v} ( x ) \) by using relation (3.7), that is,
$$ \lambda_{p}=\frac{f_{1} ( x_{0} ) }{f_{2} ( x_{0} ) }>\min \biggl\{ \frac{2v+2p}{v+2},1 \biggr\} , $$
and an analogous technique used in the proof of Subcase 2.4 of Theorem 1.1.
Proposition 4.12
Let
\(v1< p<\min \{ v+1/2,2v+2 \} \)
with
\(v>2\). Then the inequality
$$ W_{v} ( x ) < p+\sqrt{\lambda_{p}^{\ast}x^{2}+ ( 2v+2p ) ^{2}} $$
holds for
\(x>0\), where
$$ \lambda_{p}^{\ast}=\min \biggl\{ \frac{4v+52p}{2 ( v+2 ) }, \frac{v+3}{v+2} \biggr\} . $$
(4.21)
Proof
It suffices to prove \(a_{n}/b_{n}\leq\lambda_{p}^{\ast}\). We first prove that
$$ \frac{a_{n}}{b_{n}}\leq\frac{4v+52p}{2 ( v+2 ) }=1+\frac {v+1/2p}{v+2}= \frac{2v+2p}{v+2}+\frac{1}{2}\frac{1}{v+2} $$
holds for all \(n\geq1\) by dividing the proof into two cases.
Case 1.
\(\min \{ v+1/2,2v+2 \} =v+1/2\), namely \(v\geq 3/2\). For this, we write \(a_{n}/b_{n}\) as
$$ \frac{a_{n}}{b_{n}}=\frac{2n+2v+1}{ ( n+2v+2 ) ( n+v+1 ) } \biggl( v+\frac{1}{2}p \biggr) + \frac{n^{2}+ ( 3v+3 ) n+2v^{2}+3v+1/2}{ ( n+2v+2 ) ( n+v+1 ) }. $$
Then we have
$$\begin{aligned} \frac{a_{n}}{b_{n}}\frac{4v+52p}{2 ( v+2 ) } =&\frac {2n+2v+1}{ ( n+2v+2 ) ( n+v+1 ) } \biggl( v+ \frac {1}{2}p \biggr) \frac{v+1/2p}{v+2} \\ &{}+\frac{n^{2}+ ( 3v+3 ) n+2v^{2}+3v+1/2}{ ( n+2v+2 ) ( n+v+1 ) }1 \\ =&\frac{ ( n1 ) ( n+v ) ( v+1/2p ) }{ ( v+2 ) ( n+v+1 ) ( n+2v+2 ) }\frac {v+3/2}{ ( n+2v+2 ) ( n+v+1 ) }< 0 \end{aligned}$$
for \(n\geq1\).
Case 2.
\(\min \{ v+1/2,2v+2 \} =2v+2\), namely \(2< v<3/2\). Similarly, we write \(a_{n}/b_{n}\) as
$$ \frac{a_{n}}{b_{n}}=\frac{2n+2v+1}{ ( n+2v+2 ) ( n+v+1 ) } ( 2v+2p ) +\frac{n1}{n+2v+2}. $$
Then we get
$$\begin{aligned} \frac{a_{n}}{b_{n}}\frac{4v+52p}{2 ( v+2 ) } &=\frac {2n+2v+1}{ ( n+2v+2 ) ( n+v+1 ) } ( 2v+2p )  \frac{2v+2p}{v+2} \\ &\quad {}+\frac{n1}{n+2v+2}\frac{1}{2}\frac{1}{v+2} \\ &=\frac{ ( n1 ) ( n+v ) ( 2v+2p ) }{ ( v+2 ) ( n+v+1 ) ( n+2v+2 ) }+\frac {1}{2}\frac{ ( n2 ) ( 2v+3 ) }{ ( v+2 ) ( n+2v+2 ) }< 0 \end{aligned} $$
for \(n\geq1\).
Second, to prove that for all \(n\geq1\),
$$ \frac{a_{n}}{b_{n}}\leq\frac{v+3}{v+2}, $$
we write \(a_{n}/b_{n}\) in the form of
$$ \frac{a_{n}}{b_{n}}=\frac{2n+2v+1}{ ( n+2v+2 ) ( n+v+1 ) } ( v1p ) +\frac{n^{2}+3 ( v+2 ) n+2v^{2}+6v+2}{ ( n+2v+2 ) ( n+v+1 ) }. $$
Then, for \(n\geq1\), we have
$$\begin{aligned} \frac{a_{n}}{b_{n}}\frac{v+3}{v+2} =&\frac{ ( v1p ) ( 2n+2v+1 ) }{ ( n+2v+2 ) ( n+v+1 ) }+ \frac {n^{2}+3 ( v+2 ) n+2v^{2}+6v+2}{ ( n+2v+2 ) ( n+v+1 ) }\frac{v+3}{v+2} \\ =&\frac{ ( v1p ) ( 2n+2v+1 ) }{ ( n+2v+2 ) ( n+v+1 ) }\frac{ ( n1 ) ( n2 ) }{ ( v+2 ) ( n+v+1 ) ( n+2v+2 ) }< 0. \end{aligned}$$
Finally, it is obtained that
$$ \lambda_{p}=\frac{f_{1} ( x_{0} ) }{f_{2} ( x_{0} ) }=\frac{\sum_{n=0}^{\infty}a_{n} ( x_{0}^{2}/4 ) ^{n}}{\sum_{n=0}^{\infty}b_{n} ( x_{0}^{2}/4 ) ^{n}}< \frac{\sum_{n=0}^{\infty}\lambda _{p}^{\ast}b_{n} ( x_{0}^{2}/4 ) ^{n}}{\sum_{n=0}^{\infty }b_{n} ( x_{0}^{2}/4 ) ^{n}}=\lambda_{p}^{\ast }, $$
which completes the proof. □
Now by Proposition 4.12 we have the following.
Corollary 4.13
Let
\(v1< p<\min \{ v+1/2,2v+2 \} \)
with
\(v>2\).

(i)
For
\(v1< p\leq v1/2\), the inequality
$$ W_{v} ( x ) < p+\sqrt{\frac{v+3}{v+2}x^{2}+ ( 2v+2p ) ^{2}}=U_{p}^{\ast\ast} ( x ) $$
(4.22)
holds for
\(x>0\).

(ii)
For
\(v1/2< p<\min \{ v+1/2,2v+2 \} \), the inequality
$$ W_{v} ( x ) < p+\sqrt{\frac{4v+52p}{2 ( v+2 ) }x^{2}+ ( 2v+2p ) ^{2}}=U_{p}^{\ast} ( x ) $$
(4.23)
holds for
\(x>0\). In particular, taking
\(p=v\)
and letting
\(p\rightarrow v+1/2 \)
with
\(v\geq3/2\)
and
\(p\rightarrow2v+2\)
with
\(2< v<3/2\), the following inequalities hold for
\(x>0\):
$$\begin{aligned}& W_{v} ( x ) < v+\sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}}=U_{v}^{\ast} ( x )\quad \textit{for }v>2, \end{aligned}$$
(4.24)
$$\begin{aligned}& W_{v} ( x ) < v+\frac{1}{2}+\sqrt{x^{2}+ \biggl( v+ \frac {3}{2} \biggr) ^{2}}=U_{v+1/2}^{\ast} ( x ) \quad \textit{for }v\geq\frac{3}{2}, \end{aligned}$$
(4.25)
$$\begin{aligned}& W_{v} ( x ) < 2v+2+\frac{x}{\sqrt{2v+4}}=U_{2v+2}^{\ast } (x )\quad \textit{for }2< v< \frac{3}{2}. \end{aligned}$$
(4.26)
Remark 4.14
It is easy to check that the function \(p\mapsto U_{p}^{\ast \ast} ( x ) \) defined in (4.22) is increasing on \(\mathbb{R}\), which yields
$$ U_{p}^{\ast\ast} ( x ) >U_{v1}^{\ast\ast} ( x ) =v1+\sqrt{\frac{v+3}{v+2}x^{2}+ ( v+3 ) ^{2}}=U_{v1}^{ ( 2 ) } ( x ) $$
for \(v1< p\leq v1/2\) with \(v>2\). This shows that the upper bound \(U_{p}^{\ast\ast} ( x ) \) for \(W_{v} ( x ) \) is weaker than \(U_{v1}^{ ( 2 ) } ( x ) \) as the sharpest one given in Proposition 4.4. Inequality (4.26) seems to be a new one.
Remark 4.15
Clearly, \(U_{v+1/2}^{\ast} ( x ) =U_{v+1/2}^{ ( 1 ) } ( x ) \) for \(v\geq3/2\). In general, the upper bound \(U_{p}^{\ast} ( x ) \) for \(v1/2< p<\min \{ v+1/2,2v+2 \} \) with \(v>2\) given in (4.23) is not comparable with other two sharpest upper bounds \(U_{v+1/2}^{ ( 1 ) }\) for \(v\geq3/2\) and \(U_{v1}^{ ( 2 ) } ( x ) \) for \(v>2\) given in Proposition 4.4. For example, for \(v\geq3/2\), \(U_{v}^{\ast } ( x ) < U_{v+1/2}^{ ( 1 ) } ( x ) \) if \(0< x<\sqrt {2 ( v+2 ) }\) and \(U_{v}^{\ast} ( x ) >U_{v+1/2}^{ ( 1 ) } ( x ) \) if \(x>\sqrt{2 ( v+2 ) }\), since
$$\begin{aligned}& \begin{aligned} U_{v}^{\ast} ( x ) U_{v+1/2} ( x ) &=v+\sqrt {\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}} \biggl( v+\frac{1}{2}+\sqrt{x^{2}+ \biggl( v+ \frac{3}{2} \biggr) ^{2}} \biggr) \\ &=\sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}} \biggl( \frac{1}{2}+\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}} \biggr) , \end{aligned} \\& \begin{aligned} &\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2} \biggl( \frac{1}{2}+\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}} \biggr) ^{2}\\ &\quad =\frac{ ( v+2 ) ( 2v+3 ) +x^{2}}{2 ( v+2 ) }\sqrt{x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2}}, \end{aligned} \\& \biggl( \frac{ ( v+2 ) ( 2v+3 ) +x^{2}}{2 ( v+2 ) } \biggr) ^{2} \biggl( x^{2}+ \biggl( v+\frac{3}{2} \biggr) ^{2} \biggr) =\frac{1}{4}x^{2} \frac{x^{2}2 ( v+2 ) }{ ( v+2 ) ^{2}}. \end{aligned}$$
Similarly, for \(v>2\), \(U_{v}^{\ast} ( x ) < U_{v1}^{ ( 2 ) } ( x ) \) if \(x>\sqrt{8 ( v+2 ) ( v+3 ) }\) and \(U_{v}^{\ast} ( x ) >U_{v1}^{ ( 2 ) } ( x ) \) if \(0< x<\sqrt{8 ( v+2 ) ( v+3 ) }\). In fact, some elementary computations give
$$\begin{aligned}& \begin{aligned} U_{v}^{\ast} ( x ) U_{v1}^{ ( 2 ) } ( x ) &=v+\sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}} \biggl( v1+\sqrt{\frac{v+3}{v+2}x^{2}+ ( v+3 ) ^{2}} \biggr) \\ &=\sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}}+1\sqrt{ \frac{v+3}{v+2}x^{2}+ ( v+3 ) ^{2}}, \end{aligned} \\& \begin{aligned} & \biggl( \sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}}+1 \biggr) ^{2} \biggl( \frac{v+3}{v+2}x^{2}+ ( v+3 ) ^{2} \biggr) \\ &\quad=2\sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}} \frac{4 ( v+2 ) ^{2}+x^{2}}{2 ( v+2 ) }, \end{aligned} \\& \biggl( 2\sqrt{\frac{2v+5}{2 ( v+2 ) }x^{2}+ ( v+2 ) ^{2}} \biggr) ^{2} \biggl( \frac{4 ( v+2 ) ^{2}+x^{2}}{2 ( v+2 ) } \biggr) ^{2}= \frac{1}{4}x^{2}\frac {x^{2}8 ( v+2 ) ( v+3 ) }{ ( v+2 ) ^{2}}. \end{aligned}$$
It thus can be seen that the upper bound \(U_{p}^{\ast} ( x ) \) for \(v1/2< p<\min \{ v+1/2,2v+2 \} \) with \(v>2\) belongs to the new type of bounds \(p+r\sqrt{x^{2}+q^{2}}\) (\(r>0\)) for \(W_{v} ( x ) \).
Let us return to Proposition 4.1 again. We note that the number \(\lambda_{p}\) is also not computable in the case of
$$ ( p,v ) \in E_{4}\cap \{ p>2v+2,v>2 \} = \biggl\{ 2v+2< p< v+ \frac{1}{2},2< v< \frac{3}{2} \biggr\} . $$
If a better upper estimation \(\lambda_{p}^{\ast\ast}>0\) holds for \(\lambda_{p}\), then by Proposition 4.1 we can obtain some bounds \(W_{v} ( x ) \) similar to the double inequality (4.19), which also implies the new type of bounds. In fact, by the same technique as the proof of Proposition 4.12, we can prove the following proposition, but omit all the details of the proof.
Proposition 4.16
Let
\(2v+2< p< v+1/2\)
with
\(2< v<3/2\). Then the double inequality
$$ p\sqrt{\lambda_{p}^{\ast\ast}x^{2}+ ( 2v+2p ) ^{2}}< W_{v} ( x ) < p+\sqrt{\lambda_{p}^{\ast\ast }x^{2}+ ( 2v+2p ) ^{2}} $$
(4.27)
holds for
\(x>0\), where
$$ \lambda_{p}^{\ast\ast}=\min \biggl\{ \frac{v+1/2p}{v+2}+ \frac {4v^{2}+18v+21}{4 ( v+2 ) ( v+3 ) },\frac {1}{2 ( v+2 ) } \biggr\} . $$
From Proposition 4.10, we know the number \(\theta_{p}=1\) for \(p\geq c_{v}^{\ast}=v+1/ ( 2v+5 ) \) with \(2< v<3/2\). It remains to estimate \(\theta_{p}\) for \(p\in{}[ v+1/2,c_{v}^{\ast})\) with \(2< v<3/2\). By a similar technique as the proof of Proposition 4.12, we have
$$ \theta_{p}< \theta_{p}^{\ast}=\min \biggl\{ \frac {4v^{2}+18v+21}{4 ( v+2 ) ( v+3 ) },\frac{c_{v}^{\ast}p}{v+2}+1 \biggr\} $$
(4.28)
for \(p\in{}[ v+1/2,c_{v}^{\ast})\) with \(2< v<3/2\). Thus, by Proposition 4.1 we conclude the following proposition.
Proposition 4.17
Let
\(v+1/2\leq p< c_{v}^{\ast}=v+1/ ( 2v+5 ) \)
with
\(2< v<3/2\). Then the double inequality
$$ p\sqrt{\theta_{p}^{\ast}x^{2}+ ( 2v+2p ) ^{2}}< W_{v} ( x ) < p+\sqrt{\theta_{p}^{\ast}x^{2}+ ( 2v+2p ) ^{2}} $$
(4.29)
holds for
\(x>0\), where
\(\theta_{p}^{\ast}\)
is given in (4.28).
Remark 4.18
Similarly, the lower bounds given by (4.27) and (4.29) are trivial due to the fact that they are weaker than \(2v+2\). However, the upper bounds are new ones which belong to the type of bounds \(p+r\sqrt{x^{2}+q^{2}}\) (\(r>0\)).