Best possible inequalities for the harmonic mean of error function

Abstract

In this paper, we find the least value r and the greatest value p such that the double inequality $erf(M_p(x,y;\lambda ))\le H(erf(x),erf(y);\lambda )\le erf(M_r(x,y;\lambda ))$ holds for all $x,y\ge 1$ (or $0<x,y<1$) with $0<\lambda <1$, where $erf(x)=\frac{2}{\sqrt{\pi }}{\int }_0^x{e}^{-t^2}dt$, and $M_p(x,y;\lambda )={(\lambda x^p+(1-\lambda )y^p)}^{1/p}$ ($p\ne 0$) and $M_0(x,y;\lambda )=x^{\lambda }{y}^{1-\lambda }$ are, respectively, the error function, and weighted power mean.

MSC:33B20, 26D15.

1 Introduction

For $x\in R$, the error function $erf(x)$ is defined as

$$erf(x)=\frac{2}{\sqrt{\pi }}{\int }_0^x{e}^{-t^2}dt.$$

It is well known that the error function $erf(x)$ is odd, strictly increasing on $(-\mathrm{\infty },+\mathrm{\infty })$ with ${lim}_{x\to +\mathrm{\infty }}erf(x)=1$, strictly concave and strictly log-concave on $[0,+\mathrm{\infty })$. For the n th derivation we have the representation

$$\frac{d^n}{d{x}^n}erf(x)={(-1)}^{n-1}\frac{2}{\sqrt{\pi }}{e}^{-x^2}{H}_{n-1}(x),$$

where $H_n(x)={(-1)}^n{e}^{x^2}\frac{d^n}{d{x}^n}(e^{-x^2})$ is a Hermite polynomial.

The error function can be expanded as a power series in the following two ways [1]:

$$erf(x)=\frac{2}{\sqrt{\pi }}\sum _{n=0}^{+\mathrm{\infty }}\frac{{(-1)}^n}{n!(2n+1)}{x}^{2n+1}=e^{-x^2}\sum _{n=0}^{+\mathrm{\infty }}\frac{1}{\mathrm{\Gamma }(n+\frac{3}{2})}{x}^{2n+1}.$$

It also can be expressed in terms of incomplete gamma function and a confluent hypergeometric function:

$$erf(x)=\frac{sgn(x)}{\sqrt{\pi }}\gamma (\frac{1}{2},x^2)=\frac{2x}{\sqrt{\pi }}{1}_{}{F}_1(\frac{1}{2};\frac{3}{2};-x^2).$$

Recently, the error function have been the subject of intensive research. In particular, many remarkable properties and inequalities for the error function can be found in the literature [2–10]. It might be surprising that the error function has applications in heat conduction problems [11, 12].

In [13], Chu proved that the double inequality

$$\sqrt{1-e^{-a{x}^2}}\le erf(x)\le \sqrt{1-e^{-b{x}^2}}$$

holds for all $x\ge 0$ if and only if $0\le a\le 1$ and $b\ge \frac{4}{\pi }$.

Mitrinović and Weinacht [14] established that

$$erf(x)+erf(y)\le erf(x+y)+erf(x)erf(y)$$

for all $x,y\ge 0$, and proved that the inequality become equality if and only if $x=0$ or $y=0$.

In [15, 16] Alzer proved that

$${\alpha }_n=\{\begin{array}{ll}0.90686\cdots ,& \text{if }n=2,\\ 1,& \text{if }n\ge 3\end{array}\text{and}{\beta }_n=n-1$$

(1.1)

are the best possible constants such that the double inequality

$${\alpha }_{n}erf\left(\sum _{i=1}^n{x}_i\right)\le \sum _{i=1}^{n}erf(x_i)-\prod _{i=1}^{n}erf(x_i)\le {\beta }_{n}erf\left(\sum _{i=1}^n{x}_i\right)$$

holds for $n\ge 2$ and all real number $x_i\ge 0$ ($i=1,2,\dots ,n$), and the sharp double inequalities

$$erf(1)<\frac{erf(x+erf(y))}{erf(y+erf(x))}<\frac{2}{\sqrt{\pi }}$$

and

$$0<\frac{erf(xerf(y))}{erf(yerf(x))}\le 1$$

hold for all positive real numbers x, y with $x\ge y$.

Let $\lambda \in (0,1)$, and $A(x,y;\lambda )=\lambda x+(1-\lambda )y$, $G(x,y;\lambda )=x^{\lambda }{y}^{1-\lambda }$, $H(x,y;\lambda )=xy/[\lambda y+(1-\lambda )x]$, and $M_r(x,y;\lambda )={[\lambda x^r+(1-\lambda )y^r]}^{1/r}$ ($r\ne 0$) and $M_0(x,y;\lambda )=x^{\lambda }{y}^{1-\lambda }$ be, respectively, the weighted arithmetic, geometric, harmonic, and power means of two positive numbers x and y. Then it is well known that the inequalities

$$H(x,y;\lambda )=M_{-1}(x,y;\lambda )<G(x,y;\lambda )=M_0(x,y;\lambda )<A(x,y;\lambda )=M_1(x,y;\lambda )$$

hold for all $\lambda \in (0,1)$ and $x,y>0$ with $x\ne y$.

Very recently, Alzer [17] proved that $c_1(\lambda )=[\lambda +(1-\lambda )erf(1)]/[erf(1/(1-\lambda ))]$ and $c_2(\lambda )=1$ are the best possible factors such that the double inequality

$$c_1(\lambda )erf(H(x,y;\lambda ))\le A(erf(x),erf(y);\lambda )\le c_2(\lambda )erf(H(x,y;\lambda ))$$

(1.2)

holds for all $x,y\in [1,+\mathrm{\infty })$ and $\lambda \in (0,1/2)$.

It is natural to ask what are the least value r and the greatest value p such that the double inequality

$$erf(M_p(x,y;\lambda ))\le H(erf(x),erf(y);\lambda )\le erf(M_r(x,y;\lambda ))$$

holds for all $x,y\ge 1$ (or $0<x,y<1$)? The main purpose of this article is to answer this question.

2 Lemmas

In order to prove our main results we need three lemmas, which we present in this section.

Lemma 2.1 Let $r\ne 0$ and $J(x)=\frac{1}{r}[erf(x^{1/r})-\frac{1}{\sqrt{\pi }}{x}^{1/r}{e}^{-x^{2/r}}]$. Then the following statements are true:

• (1) if $-1\le r<0$, then $J(x)<0$ for all $x\in (0,+\mathrm{\infty })$;

• (2) if $0<r<1$, then $J(x)>0$ for all $x\in (0,+\mathrm{\infty })$.

Proof Simple computation leads to

$$J^{\prime }(x)=\frac{1}{r^2}\frac{2}{\sqrt{\pi }}{x}^{1/r-1}{e}^{-x^{2/r}}(\frac{1}{2}+x^{2/r})>0$$

(2.1)

for all $x\in (0,+\mathrm{\infty })$.

• (1) If $-1\le r<0$, then we clearly see that

  $$\underset{x\to +\mathrm{\infty }}{lim}J(x)=0.$$

  (2.2)

Therefore, Lemma 2.1(1) follows easily from (2.1) and (2.2).

• (2) If $0<r<1$, then it is obvious that

  $$\underset{x\to 0^{+}}{lim}J(x)=0.$$

  (2.3)

Therefore, Lemma 2.1(2) follows from (2.1) and (2.3). □

Lemma 2.2 Let $r\ne 0$, $r_0=-1-\frac{4}{e\sqrt{\pi }erf(1)}=-1.9852\cdots$ and $u(x)=\frac{1}{erf(x^{1/r})}$. Then the following statements are true:

• (1) if $r\le r_0$, then $u(x)$ is strictly concave on $[1,+\mathrm{\infty })$;

• (2) if $r_0\le r<-1$, then $u(x)$ is strictly convex on $(0,1]$;

• (3) if $r\ge -1$, then $u(x)$ is strictly convex on $(0,+\mathrm{\infty })$.

Proof Differentiating $u(x)$ leads to

$$u^{\prime }(x)=-\frac{1}{r}\frac{x^{1/r-1}{erf}^{\prime }(x^{1/r})}{{erf}^2(x^{1/r})}$$

(2.4)

and

$$u^{″}(x)=\frac{1}{r^2}\frac{2}{\sqrt{\pi }}\frac{1}{{erf}^2(x^{1/r})}{x}^{1/r-2}{e}^{-x^{2/r}}g(x),$$

(2.5)

where

$$g(x)=(r-1+2x^{2/r})erf\left(x^{1/r}\right)+\frac{4}{\sqrt{\pi }}{x}^{1/r}{e}^{-x^{2/r}}.$$

(2.6)

It follows from (2.6) that

$$g(1)=(r+1)erf(1)+\frac{4}{e\sqrt{\pi }},$$

(2.7)

$$\begin{array}{c}{g}^{\prime }(x)=4x^{2/r-1}{g}_1(x),\hfill \\ g_1(x)=\frac{1}{r}erf\left(x^{1/r}\right)+\frac{1}{r}\frac{1}{2\sqrt{\pi }}{x}^{1/r}[(1+r)x^{-2/r}-2]e^{-x^{2/r}},\hfill \end{array}$$

(2.8)

$$\begin{array}{c}{g}_1^{\prime }(x)=\frac{1}{r^2}\frac{1}{2\sqrt{\pi }}{x}^{-1/r-1}{e}^{-x^{2/r}}{g}_2(x),\hfill \\ g_2(x)=4x^{4/r}-2r{x}^{2/r}-(1+r).\hfill \end{array}$$

(2.9)

We divide the proof into four cases.

Case 1 $r<-1$. Then from (2.6) and (2.8) together with (2.9) we clearly see that

$$\underset{x\to 0^{+}}{lim}g(x)=+\mathrm{\infty },\underset{x\to +\mathrm{\infty }}{lim}g(x)=0,$$

(2.10)

$$\underset{x\to 0^{+}}{lim}{g}_1(x)=\frac{1}{r}<0,\underset{x\to +\mathrm{\infty }}{lim}{g}_1(x)=+\mathrm{\infty },$$

(2.11)

$$\underset{x\to +\mathrm{\infty }}{lim}{g}_2(x)=-(1+r)>0,$$

(2.12)

and $g_2(x)$ is strictly decreasing on $[0,+\mathrm{\infty })$.

It follows from the monotonicity of $g_2(x)$ and (2.12) that $g_1(x)$ is strictly increasing on $[0,+\mathrm{\infty })$.

The monotonicity of $g_1(x)$ and (2.11) imply that there exists $x_1\in (0,+\mathrm{\infty })$, such that $g_1(x)<0$ for $x\in (0,x_1)$ and $g_1(x)>0$ for $x\in (x_1,+\mathrm{\infty })$. Therefore, $g(x)$ is strictly decreasing on $[0,x_1]$ and strictly increasing on $[x_1,+\mathrm{\infty })$.

From the piecewise monotonicity of $g(x)$ and (2.10) we clearly see that there exists $x_2\in (0,+\mathrm{\infty })$, such that $g(x)>0$ for $x\in (0,x_2)$ and $g(x)<0$ for $x\in (x_2,+\mathrm{\infty })$.

If $r\le r_0$, then (2.7) leads to $g(1)\le 0$, this implies that $g(x)<0$ for $x\in (1,+\mathrm{\infty })$. Therefore, (2.5) leads to the conclusion that $u(x)$ is strictly concave on $[1,+\mathrm{\infty })$.

If $r_0\le r<-1$, then (2.7) leads to $g(1)\ge 0$, this implies that $g(x)>0$ for $x\in (0,1)$. Therefore, (2.5) leads to the conclusion that $u(x)$ is strictly convex on $(0,1)$.

Case 2 $-1\le r<0$. Then we clearly see that the function $(1+r)x^{-2/r}-2$ is strictly increasing on $(0,+\mathrm{\infty })$ with ${lim}_{x\to 0^{+}}[(1+r)x^{-2/r}-2]=-2$, and

$$g_1(x)<\frac{1}{r}[erf\left(x^{1/r}\right)-\frac{1}{\sqrt{\pi }}{x}^{1/r}{e}^{-x^{2/r}}].$$

(2.13)

Therefore, Lemma 2.1(1) and (2.13) imply that $g_1(x)<0$ for $x\in (0,+\mathrm{\infty })$. This leads to the conclusion that $g(x)$ is strictly decreasing on $(0,+\mathrm{\infty })$.

From (2.6) we get

$$\underset{x\to +\mathrm{\infty }}{lim}g(x)=0$$

(2.14)

for $-1\le r<0$.

It follows from the monotonicity of $g(x)$ and (2.14) that $g(x)>0$ for $x\in (0,+\mathrm{\infty })$. Therefore, (2.5) leads to the conclusion that $u(x)$ is strictly convex on $(0,+\mathrm{\infty })$.

Case 3 $0<r<1$. Then we clearly see that the function $(1+r)x^{-2/r}-2$ is strictly decreasing on $(0,+\mathrm{\infty })$ with ${lim}_{x\to +\mathrm{\infty }}[(1+r)x^{-2/r}-2]=-2$, and

$$g_1(x)>\frac{1}{r}[erf\left(x^{1/r}\right)-\frac{1}{\sqrt{\pi }}{x}^{1/r}{e}^{-x^{2/r}}].$$

(2.15)

It follows from Lemma 2.1(2) and (2.15) that $g_1(x)>0$ for $x\in (0,+\mathrm{\infty })$. This leads to $g(x)$ being strictly increasing on $(0,+\mathrm{\infty })$.

It follows from (2.6) that

$$\underset{x\to 0^{+}}{lim}g(x)=0$$

(2.16)

for $0<r<1$.

From the monotonicity of $g(x)$ and (2.16) we know that $g(x)>0$ for $x\in (0,+\mathrm{\infty })$. Therefore, (2.5) leads to the conclusion that $u(x)$ is strictly convex on $(0,+\mathrm{\infty })$.

Case 4 $r\ge 1$. Then from (2.6) we clearly see that $g(x)>0$ for $x\in (0,+\mathrm{\infty })$. Therefore, $u(x)$ is strictly convex on $(0,+\mathrm{\infty })$ follows easily from (2.5). □

Lemma 2.3 The function $h(x)=x^2+\frac{x{e}^{-x^2}}{{\int }_0^x{e}^{-t^2}dt}$ is strictly increasing on $(0,+\mathrm{\infty })$.

Proof Simple computations lead to

$$h^{\prime }(x)=\frac{h_1(x)}{{({\int }_0^x{e}^{-t^2}dt)}^2},$$

(2.17)

where

$$\begin{array}{c}{h}_1(x)=2x{\left({\int }_0^x{e}^{-t^2}dt\right)}^2+(1-2x^2)e^{-x^2}{\int }_0^x{e}^{-t^2}dt-x{e}^{-2x^2},\hfill \\ h_1(0)=0,h_1(1)=0.7054\cdots >0,\hfill \end{array}$$

(2.18)

$$h_1^{\prime }(x)=2{\left({\int }_0^x{e}^{-t^2}dt\right)}^2+2x(2x^2-1)e^{-x^2}{\int }_0^x{e}^{-t^2}dt+2x^2{e}^{-2x^2},$$

(2.19)

$$h_1^{\prime }(0)=0,$$

(2.20)

$$h_1^{″}(x)=e^{-x^2}{h}_2(x),$$

(2.21)

$$\begin{array}{c}{h}_2(x)=(-8x^4+16x^2+2){\int }_0^x{e}^{-t^2}dt+(-4x^3+2x)e^{-x^2},\hfill \\ h_2(0)=0,\hfill \end{array}$$

(2.22)

$$h_2^{\prime }(x)=32x(1-x^2){\int }_0^x{e}^{-t^2}dt+4e^{-x^2}.$$

(2.23)

We divide the proof into two cases.

Case 1 $x\ge 1$. Then (2.19) leads to $h_1^{\prime }(x)>0$. Therefore, $h^{\prime }(x)>0$ follows from (2.18) and (2.17).

Case 2 $0<x<1$. Then from (2.23) we clearly see that $h_2^{\prime }(x)>0$. Therefore, $h^{\prime }(x)>0$ follows from (2.17) and (2.18) together with (2.20)-(2.22). □

3 Main results

Theorem 3.1 Let $\lambda \in (0,1)$ and $r_0=-1-\frac{4}{e\sqrt{\pi }erf(1)}=-1.9852\cdots$. Then the double inequality

$$erf(M_{\mu }(x,y;\lambda ))\le H(erf(x),erf(y);\lambda )\le erf(M_{\nu }(x,y;\lambda ))$$

(3.1)

holds for all $0<x,y<1$ if and only if $\mu \le r_0$ and $\nu \ge -1$.

Proof Firstly, we prove that (3.1) holds if $\mu \le r_0$ and $\nu \ge -1$.

If $\mu \le r_0$, $u(z)=\frac{1}{erf(z^{1/\mu })}$, then Lemma 2.2(1) leads to

$$\lambda u(s)+(1-\lambda )u(t)\le u(\lambda s+(1-\lambda )t)$$

(3.2)

for $\lambda \in (0,1)$ and $s,t>1$.

Let $s=x^{\mu }$, $t=y^{\mu }$, and $0<x,y<1$. Then (3.2) leads to the first inequality in (3.1).

Since the function $t\mapsto erf(M_t(x,y;\lambda ))$ is strictly increasing on R if $\nu \ge -1$, it is enough to prove the second inequality in (3.1) is true for $-1\le \nu <0$.

Let $-1\le \nu <0$ and $u(z)=\frac{1}{erf(z^{1/\nu })}$. Then Lemma 2.2(3) leads to

$$u(\lambda s+(1-\lambda )t)\le \lambda u(s)+(1-\lambda )u(t)$$

(3.3)

for $\lambda \in (0,1)$ and $s,t>1$.

Therefore, the second inequality in (3.1) follows from $s=x^{\nu }$ and $t=y^{\nu }$ together with (3.3).

Secondly, we prove that the second inequality in (3.1) implies $\nu \ge -1$.

Let $0<x,y<1$. Then the second inequality in (3.1) leads to

$$D(x,y):=erf(M_{\nu }(x,y;\lambda ))-H(erf(x),erf(y);\lambda )\ge 0.$$

(3.4)

It follows from (3.4) that

$$D(y,y)=\frac{\partial }{\partial x}D(x,y){|}_{x=y}=0$$

and

$$\frac{{\partial }^2}{\partial x^2}D(x,y){|}_{x=y}=\lambda (1-\lambda )y^{-1}{erf}^{\prime }(y)[\nu -1+2(y^2+\frac{y{e}^{-y^2}}{{\int }_0^y{e}^{-t^2}dt})].$$

(3.5)

Therefore,

$$\nu \ge \underset{y\to 0^{+}}{lim}[1-2(y^2+\frac{y{e}^{-y^2}}{{\int }_0^y{e}^{-t^2}dt})]=-1$$

follows from (3.4) and (3.5) together with Lemma 2.3.

Finally, we prove that the first inequality in (3.1) implies $\mu \le r_0$.

Let $y\to 1$. Then the first inequality in (3.1) leads to

$$L(x)=:H(erf(x),erf(1);\lambda )-erf(m_{\mu }(x,1;\lambda ))\ge 0$$

(3.6)

for $0<x<1$.

It follows from (3.6) that

$$L(1)=0,$$

(3.7)

$${[\lambda erf(1)+(1-\lambda )erf(x)]}^2{L}^{\prime }(x)=\frac{2\lambda }{\sqrt{\pi }}{e}^{-x^2}{L}_1(x),$$

(3.8)

where

$$\begin{array}{c}{L}_1(x)=erf{(1)}^2-x^{\mu -1}{(\lambda x^{\mu }+1-\lambda )}^{\frac{1}{\mu }-1}{[\lambda erf(1)+(1-\lambda )erf(x)]}^2{e}^{x^2-{(\lambda x^{\mu }+1-\lambda )}^{\frac{2}{\mu }}},\hfill \\ \underset{x\to 1^{-}}{lim}{L}_1(x)=0,\hfill \end{array}$$

(3.9)

$$\underset{x\to 1^{-}}{lim}{L}_1^{\prime }(x)=(1-\lambda )erf{(1)}^2[-1-\mu -\frac{4}{e\sqrt{\pi }erf(1)}].$$

(3.10)

If $\mu >r_0$, then from (3.10) we know that there exists a small ${\delta }_1>0$, such that $L_1^{\prime }(x)<0$ for $x\in (1-{\delta }_1,1)$. Therefore, $L_1(x)$ is strictly decreasing on $[1-{\delta }_1,1]$.

The monotonicity of $L_1(x)$ together with (3.8) and (3.9) imply that there exists ${\delta }_2>0$, such that $L(x)$ is strictly increasing on $(1-{\delta }_2,1)$

It follows from the monotonicity of $L(x)$ and (3.7) that there exists ${\delta }_3>0$, such that $L(x)<0$ for $x\in (1-{\delta }_3,1)$, this contradicts with (3.6). □

Theorem 3.2 Let $\lambda \in (0,1)$ and $r_0=-1-\frac{4}{e\sqrt{\pi }erf(1)}=-1.9852\cdots$. Then the double inequality

$$erf(M_p(x,y;\lambda ))\le H(erf(x),erf(y);\lambda )\le erf(M_r(x,y;\lambda ))$$

(3.11)

holds for all $x,y\ge 1$ if and only if $p=-\mathrm{\infty }$ and $r\ge r_0$.

Proof Firstly, we prove that inequality (3.11) holds if $p=-\mathrm{\infty }$ and $r\ge r_0$. Since the first inequality in (3.11) is true if $p=-\mathrm{\infty }$, thus we only need to prove the second inequality in (3.11).

It follows from the monotonicity of the function $erf(M_t(x,y;\lambda ))$ with respect to t that we only need to prove the second inequality in (3.11) holds for $r_0\le r<-1$.

Let $r_0\le r<-1$ and $u(z)=\frac{1}{erf(z^{1/r})}$. Then Lemma 2.2(2) leads to

$$u(\lambda s+(1-\lambda )t)\le \lambda u(s)+(1-\lambda )u(t)$$

(3.12)

for $\lambda \in (0,1)$ and $s,t\in (0,1]$.

Therefore, the second inequality in (3.11) follows from $s=x^r$ and $t=y^r$ together with (3.12).

Secondly, we prove that the second inequality in (3.11) implies $r\ge r_0$.

Let $x\ge 1$ and $y\ge 1$. Then the second inequality in (3.11) leads to

$$K(x,y)=:erf(M_r(x,y;\lambda ))-H(erf(x),erf(y);\lambda )\ge 0.$$

(3.13)

It follows from (3.13) that

$$K(y,y)=\frac{\partial }{\partial x}K(x,y){|}_{x=y}=0$$

and

$$\frac{{\partial }^2}{\partial x^2}K(x,y){|}_{x=y}=\lambda (1-\lambda )y^{-1}{erf}^{\prime }(y)[r-1+2(y^2+\frac{y{e}^{-y^2}}{{\int }_0^y{e}^{-t^2}dt})].$$

(3.14)

Therefore,

$$r\ge \underset{y\to 1^{+}}{lim}[1-2(y^2+\frac{y{e}^{-y^2}}{{\int }_0^y{e}^{-t^2}dt})]=r_0$$

follows from (3.13) and (3.14) together with Lemma 2.3.

Finally, we prove that the first inequality in (3.11) implies $p=-\mathrm{\infty }$. We divide the proof into two cases.

Case 1 $p\ge 0$. Then for any fixed $y\in (1,+\mathrm{\infty })$ one has

$$\underset{x\to +\mathrm{\infty }}{lim}erf(M_p(x,y;\lambda ))=1$$

and

$$\underset{x\to +\mathrm{\infty }}{lim}H(erf(x),erf(y);\lambda )=\frac{erf(y)}{\lambda erf(y)+1-\lambda }<1,$$

which contradicts with the first inequality in (3.11).

Case 2 $-\mathrm{\infty }<p<0$. Let $x\ge 1$, $\theta ={\lambda }^{1/p}$, and $y\to +\mathrm{\infty }$. Then the first inequality in (3.11) leads to

$$T(x)=:H(erf(x),1;\lambda )-erf(\theta x)\ge 0.$$

(3.15)

It follows from (3.15) that

$$\underset{x\to +\mathrm{\infty }}{lim}T(x)=0$$

(3.16)

and

$${[\lambda +(1-\lambda )erf(x)]}^2{T}^{\prime }(x)=\frac{2}{\sqrt{\pi }}{e}^{-x^2}[\lambda -{(\lambda +(1-\lambda )erf(x))}^2\theta e^{(1-{\theta }^2)x^2}].$$

(3.17)

Note that

$$\underset{x\to +\mathrm{\infty }}{lim}[\lambda -{(\lambda +(1-\lambda )erf(x))}^2\theta e^{(1-{\theta }^2)x^2}]=\lambda >0.$$

(3.18)

It follows from (3.17) and (3.18) that there exists a large enough ${\eta }_1\in (0,+\mathrm{\infty })$, such that $T^{\prime }(x)>0$ for $x\in ({\eta }_1,+\mathrm{\infty })$, hence $T(x)$ is strictly increasing on $[{\eta }_1,+\mathrm{\infty })$.

From the monotonicity of $T(x)$ and (3.16) we conclude that there exists a large enough ${\eta }_2\in (0,+\mathrm{\infty })$, such that $T(x)<0$ for $x\in ({\eta }_2,+\mathrm{\infty })$, this contradicts with (3.15). □