A Solution to Qi’s Conjecture on a Double Inequality for a Function Involving the Tri- and Tetra-Gamma Functions

Abstract

In the paper, the author gives a solution to a conjecture on a double inequality for a function involving the tri- and tetra-gamma functions, which was first posed in Remark 6 of the paper “Complete monotonicity of a function involving the tri- and tetragamma functions” (2015) and repeated in the seventh open problem of the paper “On complete monotonicity for several classes of functions related to ratios of gamma functions” (2019).

1. Introduction

It is common knowledge that the classical Euler’s gamma function [1,2] is defined by

$$\Gamma \left(x\right)=\underset{0}{\overset{\infty }{\int }}{t}^{x-1}{e}^{-t}dt$$

for $x>0$ and the digamma function [3] is defined as the logarithmic derivative of the gamma function.

$$\psi \left(x\right)=\frac{{\Gamma }^{\prime }\left(x\right)}{\Gamma \left(x\right)}.$$

The functions $\psi ,$ ${\psi }^{\prime },$ ${\psi }^{″},$ ${\psi }^{‴},$ … are known as polygamma functions [4].

Very recently, in the paper [5], F. Qi and R. P. Agarwal surveyed some results related to the function ${\psi }^{\prime 2}+{\psi }^{″}.$ They also posed eight open problems. The goal of the paper is to find a solution of the seventh open problem which was first posed as a conjecture in Remark 6 of the paper [6].

The seventh open problem states that the double inequality

$$\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha }<{\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)<\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\beta }$$

holds on $(0,\infty )$ if and only if $\alpha \ge 6/5$ and $\beta \le 1.$

2. The Key Lemmas

In this section, we prove two important lemmas.

Lemma 1.

Let $\Delta \left(x\right)={\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)$ and

$$\varphi (\alpha ,x)=\Delta \left(x\right)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha }$$

for $x>0$ and $\alpha \in R.$ Then

$$\underset{x\to +\infty }{lim}\varphi (\alpha ,x)=0,$$

(1)

$$\underset{x\to 0^{+}}{lim}{x}^4\varphi (\alpha ,x)=0,$$

(2)

$$\underset{x\to 0^{+}}{lim}\frac{d\left[x^4\varphi (\alpha ,x)\right]}{dx}=-2+\frac{5\alpha }{3}.$$

(3)

Proof of Lemma 1. 

To prove (1) it suffices to show $\underset{x\to +\infty }{lim}\Delta \left(x\right)=0$ which follows from the double inequality

$$\frac{x^2+12}{12x^4{(1+x)}^2}<\Delta \left(x\right)<\frac{x^2+4x+12}{12x^4{(1+x)}^2}$$

(see [5], p. 9). Making use of the previous double inequality yields

$$\underset{x\to 0^{+}}{lim}\left\{x^4\Delta \left(x\right)-{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha }\right\}=0=\underset{x\to 0^{+}}{lim}\left[x^4\varphi (\alpha ,x)\right]$$

which is (2).

Simple calculation brings

$$\begin{array}{c}\frac{d\left[x^4\varphi (\alpha ,x)\right]}{dx}=4x^3\Delta \left(x\right)+x^4\left[2{\psi }^{\prime }\left(x\right){\psi }^{″}\left(x\right)+{\psi }^{‴}\left(x\right)\right]+\alpha {\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha -1}\frac{x+10}{6{(1+x)}^3}.\hfill \end{array}$$

Denote

$$\begin{array}{c}\delta \left(x\right)=4x^3\Delta \left(x\right)+x^4\left[2{\psi }^{\prime }\left(x\right){\psi }^{″}\left(x\right)+{\psi }^{‴}\left(x\right)\right].\hfill \end{array}$$

To prove (3) it suffices to show

$$\underset{x\to 0^{+}}{lim}\delta \left(x\right)=\underset{x\to 0^{+}}{lim}4x^3\left({\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)\right)+x^4\left(2{\psi }^{\prime }\left(x\right){\psi }^{″}\left(x\right)+{\psi }^{‴}\left(x\right)\right)=-2.$$

The function $\delta \left(x\right)$ can be rewritten as $\delta \left(x\right)={\delta }_1\left(x\right)+{\delta }_2\left(x\right)$ where

$${\delta }_1\left(x\right)=2x^3{\psi }^{\prime }\left(x\right)\left(2{\psi }^{\prime }\left(x\right)+x{\psi }^{″}\left(x\right)\right),$$

$${\delta }_2\left(x\right)=x^3\left(4{\psi }^{″}\left(x\right)+x{\psi }^{‴}\left(x\right)\right).$$

(4)

Making use of well known formulas:

for polygamma functions

$$\begin{array}{c}{\psi }^{\prime }\left(x\right)=\sum _{n=0}^{\infty }\frac{1}{{(n+x)}^2},{\psi }^{″}\left(x\right)=-2\sum _{n=0}^{\infty }\frac{1}{{(n+x)}^3},{\psi }^{‴}\left(x\right)=6\sum _{n=0}^{\infty }\frac{1}{{(n+x)}^4},\hfill \end{array}$$

and for values of the Riemann zeta function [7]

$$\begin{array}{c}\zeta \left(2\right)=\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6},\zeta \left(3\right)=\sum _{n=1}^{\infty }\frac{1}{n^3}=1.20205\dots ,\zeta \left(4\right)=\sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{90}\hfill \end{array}$$

gives

$$2x^3{\psi }^{\prime }\left(x\right)<2x+2x^3\sum _{n=1}^{\infty }\frac{1}{n^2}=2x+\frac{x^3{\pi }^2}{3}$$

and

$$\left|2{\psi }^{\prime }\left(x\right)+x{\psi }^{″}\left(x\right)\right|=\left|2\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^2}-2x\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^3}\right|\le \frac{{\pi }^2}{3}+2x\zeta \left(3\right).$$

$$\underset{x\to 0^{+}}{lim}{\delta }_1\left(x\right)=\underset{x\to 0^{+}}{lim}2x^3\left({\psi }^{\prime }\left(x\right)+x{\psi }^{″}\left(x\right)\right)=0.$$

The equality (4) can be rewritten as

$$\begin{array}{cc}\hfill {\delta }_2\left(x\right)& =4x^3\left(\frac{-2}{x^3}-2\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^3}\right)+x^4\left(\frac{6}{x^4}+6\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^4}\right)\hfill \\ \hfill & =-2-8x^3\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^3}+6x^4\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^4}.\hfill \end{array}$$

Because of

$$\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^3}<\zeta \left(3\right)$$

and

$$\sum _{n=1}^{\infty }\frac{1}{{(n+x)}^4}<\zeta \left(4\right)=\frac{{\pi }^4}{90},$$

we obtain $\underset{x\to 0^{+}}{lim}{\delta }_2\left(x\right)=-2.$ The proof of Lemma 1 is complete. □

Lemma 2.

Let

$$s\left(x\right)=\varphi \left(\frac{6}{5},x\right)=\Delta \left(x\right)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{6/5}$$

for $x>0.$ Then $s\left(x\right)>0$ for $x>0$.

Proof of Lemma 2. 

Consider three cases

(a)

$5.7\le x<\infty ,$

(b)

$1.27\le x<5.7,$

(c)

$0<x\le 1.27.$

The case (a).

In the paper (see [5], p. 9), it was presented that

$$\Delta \left(x\right)>\frac{1}{x^4}\left[\frac{x^2+12}{12{(1+x)}^2}\right]$$

(5)

for $x>0.$ So the case (a) will be done if we show

$$\frac{1}{x^4}\left[\frac{x^2+12}{12{(1+x)}^2}\right]\ge \frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{6/5},$$

which is equivalent to

$$s_1\left(x\right)=log(x^2+12)-\frac{6}{5}log(x^2+4x+12)+\frac{1}{5}log\left(12\right)+\frac{2}{5}log(1+x)>0$$

for $5.7\le x<\infty .$ In order to prove $s_1\left(x\right)>0$ it is sufficient to show $s_1\left(5.7\right)>0$ and $s_1^{\prime }\left(x\right)>0$ for $5.7<x<\infty .$ The inequality $s_1\left(x\right)>0$ is equivalent to

$$\begin{array}{c}g\left(x\right)=12{(x^2+12)}^5{(1+x)}^2-{(x^2+4x+12)}^6>0.\hfill \end{array}$$

Easy computation gives $g\left(5.7\right)\approx 1.06\ast {10}^{09}>0,$ so $s_1\left(5.7\right)>0.$

Differentiation yields

$$s_1^{\prime }\left(x\right)=\frac{x(22x^2+40x-216)}{5(x^2+12)(x+1)(x^2+4x+12)}.$$

Because of

$$\begin{array}{cc}\hfill 22x^2+40x-216=& 2(11x^2+20x-108)=2\left[x-\frac{2}{11}\left(\sqrt{322}-5\right)\right]\left[x+\frac{2}{11}\left(\sqrt{322}+5\right)\right]\hfill \\ \hfill & =2(x+4.1717\dots )(x-2.3535\dots )\hfill \end{array}$$

the function $s_1^{\prime }\left(x\right)$ is positive for $x>\frac{2}{11}\left(\sqrt{322}-5\right).$ The proof of the case (a) is complete.

The case (b). Let $1.27\le x<5.7.$ Using the following formulas

$$\begin{array}{c}{\psi }^{\prime }\left(x\right)=\frac{1}{x^2}+{\psi }^{\prime }(1+x),{\psi }^{″}\left(x\right)=-\frac{2}{x^3}+{\psi }^{″}(1+x)\hfill \end{array}$$

for $x>0$ yields

$$\begin{array}{c}\Delta \left(x\right)={\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)=\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\Delta (1+x).\hfill \end{array}$$

Making use of the inequality (see [5], p. 9)

$$\begin{array}{c}\Delta \left(x\right)>\frac{1}{x^4}\left[\frac{x^2+12}{12{(x+1)}^2}\right].\hfill \end{array}$$

yields

$$\begin{array}{c}\Delta \left(x\right)>z\left(x\right)=\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\frac{1}{{(x+1)}^4}\left[\frac{{(x+1)}^2+12}{12{(x+2)}^2}\right].\hfill \end{array}$$

To prove the case (b) it suffices to show

$$\begin{array}{c}\alpha \left(x\right)=z\left(x\right)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}>0.\hfill \end{array}$$

It will be done if we prove

• $z\left(x\right)>0for1.27\le x<5.7$,
• $F\left(1.27\right)\ge 0,$
• $F^{\prime }\left(x\right)>0for1.27\le x<5.7$

where

$$\begin{array}{c}F\left(x\right)=log\left(z\left(x\right)\right)+4log\left(x\right)-\frac{6}{5}log\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right].\hfill \end{array}$$

Differentiating F yields

$$\begin{array}{cc}\hfill F^{\prime }\left(x\right)=& \frac{-\frac{4}{x^5}+\frac{6}{x^4}-\frac{4}{x^3}{\psi }^{\prime }(1+x)+\frac{2}{x^2}{\psi }^{″}(1+x)+\left[\frac{-2x^3-7x^2-44x-63}{6{(x+1)}^5{(x+2)}^3}\right]}{\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\frac{1}{{(1+x)}^4}\left[\frac{{(x+1)}^2+12}{12{(x+2)}^2}\right]}\hfill \\ \hfill & +\frac{4}{x}+\frac{6}{5}\left[\frac{2x+20}{(x+1)(x^2+4x+12)}\right].\hfill \end{array}$$

It is clear that, $F\left(1.27\right)\ge 0$ is equivalent to $\alpha \left(1.27\right)\ge 0.$

In the paper [8], inequality (2.3) it was established that

$$\begin{array}{cc}\hfill {\psi }^{\prime }(1+x)& >\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{6x^3}-\frac{1}{30x^5}+\frac{1}{42x^7}-\frac{1}{30x^9}\hfill \\ \hfill & =\frac{210x^8-105x^7+35x^6-7x^4+5x^2-7}{210x^9}\hfill \end{array}$$

(6)

for $x>0.$

Using (6) gives

$$\begin{array}{ccc}\hfill {\psi }^{\prime }\left(2.27\right)& =& \frac{1}{{2.27}^2}+{\psi }^{\prime }\left(3.27\right)>\hfill \\ \hfill & \hfill & \frac{1}{{2.27}^2}+\frac{210\times {2.27}^8-105\times {2.27}^7+35\times {2.27}^6-7\times {2.27}^4+5\times {2.27}^2-7}{210\times {2.27}^9}\hfill \\ \hfill & >& 0.551>0.55.\hfill \end{array}$$

(7)

By (7), it follows that

$$\begin{array}{c}z\left(x\right)>\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}0.55+\frac{1}{{(x+1)}^4}\left[\frac{{(x+1)}^2+12}{12{(x+2)}^2}\right]=\hfill \\ q\left(x\right)=\frac{66x^8+408x^7+821x^6+274x^5-949x^4-960x^3+324x^2+720x+240}{60x^4{(x+1)}^4{(x+2)}^2}.\hfill \end{array}$$

Putting $x=1.27$ in $q\left(x\right)$ yields $z\left(1.27\right)>q\left(1.27\right)=0.0950575080326270>0.095.$

Because of

$$\begin{array}{c}{\left[1.{27}^4\times 0.095\right]}^5-{\left[\frac{1.{27}^2+4\times 1.27+12}{12\times 2.{27}^2}\right]}^6\approx 1.58\ast {10}^{-04}>0\hfill \end{array}$$

we have $\alpha \left(1.27\right)>0.$ So $F\left(1.27\right)>0.$

Next, we show $z\left(x\right)>0$ for $1.27<x<5.7.$ It is well known [9] that

$$\begin{array}{c}{\psi }^{\prime }\left(x\right)>\frac{1}{x}+\frac{1}{2x^2}forx>0.\hfill \end{array}$$

This implies that

$$\begin{array}{cc}\hfill z\left(x\right)>z_1\left(x\right)& =\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}\left(\frac{1}{x+1}+\frac{1}{2{(x+1)}^2}\right)+\frac{1}{{(x+1)}^4}\left[\frac{{(x+1)}^2+12}{12{(x+2)}^2}\right]\hfill \\ \hfill & =\frac{x^6+2x^5+25x^4+72x^3+156x^2+144x+48}{12x^4{(x+1)}^4{(x+2)}^2}>0.\hfill \end{array}$$

Finally, we show $F^{\prime }\left(x\right)>0$ for for $1.27\le x<5.7.$ The inequality $F^{\prime }\left(x\right)>0$ is equivalent to

$$\begin{array}{cc}\hfill G\left(x\right)=& -\frac{4}{x^5}+\frac{6}{x^4}-\frac{4}{x^3}{\psi }^{\prime }(1+x)+\frac{2}{x^2}{\psi }^{″}(1+x)-\frac{2x^3+7x^2+44x+63}{6{(x+1)}^5{(x+2)}^3}+\hfill \\ \hfill & \left[\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\frac{{(x+1)}^2+12}{12{(1+x)}^4{(x+2)}^2}\right]\times \hfill \\ \hfill & \left[\frac{4}{x}+\frac{6}{5}\left(\frac{2x+20}{(x+1)(x^2+4x+12)}\right)\right]>0.\hfill \end{array}$$

The inequality $G\left(x\right)>0$ may be rearranged as

$$\begin{array}{cc}\hfill G\left(x\right)=& {\psi }^{\prime }(1+x)\left[\frac{20x^3+124x^2+560x+240}{5x^3(x+1)(x^2+4x+12)}\right]+\frac{2{\psi }^{″}(1+x)}{x^2}-\hfill \\ \hfill & \frac{1}{30x^3{(x+1)}^5{(x+2)}^3(x^2+4x+12)}\times \left(60x^9+1044x^8+9267x^7+47554x^6+\right.\hfill \\ \hfill & \left.147543x^5+285108x^4+344260x^3+254352x^2+105984x+18624\right)>0.\hfill \end{array}$$

In the paper [10], by using asymptotic expansion, it was deduced that

$$\begin{array}{c}{\psi }^{″}\left(x\right)=-\frac{1}{x^2}-\frac{1}{x^3}-\frac{1}{2x^4}+\frac{1}{6x^6}-\frac{\theta }{6x^8}forx>0,0\le \theta \le 1.\hfill \end{array}$$

This implies

$${\psi }^{″}\left(x\right)\ge -\frac{1}{x^2}-\frac{1}{x^3}-\frac{1}{2x^4}+\frac{1}{6x^6}-\frac{1}{6x^8}forx>0.$$

(8)

Utilizing (6), (8) we obtain that $G\left(x\right)\ge G_1\left(x\right)$, where

$$\begin{array}{cc}\hfill G_1\left(x\right)=& \left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{6x^3}-\frac{1}{30x^5}+\frac{1}{42x^7}-\frac{1}{30x^9}\right)\left[\frac{20x^3+124x^2+560x+240}{5x^3(x+1)(x^2+4x+12)}\right]\hfill \\ \hfill & -\frac{2}{x^2}\left[\frac{1}{{(1+x)}^2}+\frac{1}{{(1+x)}^3}+\frac{1}{2{(1+x)}^4}-\frac{1}{6{(1+x)}^6}+\frac{1}{6{(1+x)}^8}\right]-\hfill \\ \hfill & \frac{1}{30x^3{(x+1)}^5{(x+2)}^3(x^2+4x+12)}\times \left(60x^9+1044x^8+9267x^7+\right.\hfill \\ \hfill & 47554x^6+147543x^5+285108x^4+344260x^3+254352x^2+105984x+\hfill \\ \hfill & \left.18624\right).\hfill \end{array}$$

Rewriting $G_1\left(x\right)$ yields $G_1\left(x\right)=\phi \left(x\right)/\varphi \left(x\right)$, where

$$\begin{array}{cc}\phi \left(x\right)=\hfill & 385x^{19}+1435x^{18}+2100x^{17}+18662x^{16}+153421x^{15}+\hfill \\ \hfill & 477255x^{14}+718276x^{13}+568208x^{12}+232044x^{11}+38256x^{10}-\hfill \\ \hfill & 46764x^9-314648x^8-1233916x^7-2806448x^6-3866740x^5-\hfill \\ \hfill & 3395384x^4-1933504x^3-697344x^2-145600x-13440,\hfill \\ \varphi \left(x\right)=\hfill & 1050x^{12}{(x+1)}^8{(x+2)}^3(x^2+4x+12).\hfill \end{array}$$

To show the positivity of the function $G_1\left(x\right)$ it suffices to prove $\phi \left(x\right)>0.$ By virtue of $x\ge 1.27$ we see that $\phi \left(x\right)\ge n_1\left(x\right)\ge$ $n_2\left(x\right)\ge$ $n_3\left(x\right)$ where the functions $n_1\left(x\right),$ $n_2\left(x\right),$ $n_3\left(x\right)$ are derived in Appendix A.

It is obvious that the proof of $\phi \left(x\right)>0$ will be done if we show $n_3\left(1.27\right)>0,$ $n_3^{\prime }\left(1.27\right)>0,$ $n_3^{″}\left(x\right)>0$ on $(1.27,5.7).$

A direct differentiation yields

$$\begin{array}{cc}\hfill n_3^{\prime }\left(x\right)=& \frac{14197117854302775}{274877906944}{x}^5+\frac{44097108486849535}{274877906944}{x}^4+\hfill \\ \hfill & \frac{1613308847079861}{8589934592}{x}^3+\frac{1424065194179289}{2147483648}{x}^2-\hfill \\ \hfill & \frac{328178783611353}{536870912}x+\frac{7951559879725953}{8589934592}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill n_3^{″}\left(x\right)=& \frac{70985589271513875}{274877906944}{x}^4+\frac{44097108486849535}{68719476736}{x}^3+\hfill \\ \hfill & \frac{4839926541239583}{8589934592}{x}^2+\frac{1424065194179289}{1073741824}x-\hfill \\ \hfill & \frac{328178783611353}{536870912}.\hfill \end{array}$$

From $n_3^{\prime \prime \prime }\left(x\right)>0,$ $n_3^{″}\left(1.27\right)\approx 3.96\ast {10}^{06}$ we deduce $n_3\left(x\right)$ is a convex function on $(1.27,5.7).$ The proof of the case (b) follows from $n_3\left(1.27\right)\approx 2.12\ast {10}^{06}>0,$ $n_3^{\prime }\left(1.27\right)\approx 2.19\ast {10}^{06}>0.$

The case (c). We have $0<x\le 1.27.$ Using

$$\begin{array}{c}{\left[{\psi }^{\prime }\left(x\right)\right]}^2=\frac{1}{x^4}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+{\left[{\psi }^{\prime }(1+x)\right]}^2,\hfill \end{array}$$

$$\begin{array}{c}{\psi }^{″}\left(x\right)=\frac{-2}{x^3}+{\psi }^{″}(1+x)\hfill \end{array}$$

gives

$$\begin{array}{cc}\hfill s\left(x\right)=\varphi \left(\frac{6}{5},x\right)& =\Delta \left(x\right)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}\hfill \\ \hfill & ={\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}\hfill \\ \hfill & =\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+{\left[{\psi }^{\prime }(1+x)\right]}^2+{\psi }^{″}(1+x)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}\hfill \\ \hfill & =\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\Delta (1+x)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}.\hfill \end{array}$$

Replacing x by $1+x$ in (5) yields

$$\Delta (1+x)\ge \frac{1}{{(1+x)}^4}\left[\frac{{(1+x)}^2+12}{12{(2+x)}^2}\right].$$

So

$$s\left(x\right)=\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\Delta (1+x)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}\ge t\left(x\right),$$

where

$$t\left(x\right)=\frac{1}{x^4}-\frac{2}{x^3}+\frac{2}{x^2}{\psi }^{\prime }(1+x)+\frac{{(1+x)}^2+12}{12{(1+x)}^4{(2+x)}^2}-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(x+1)}^2}\right]}^{6/5}.$$

The inequality $t\left(x\right)\ge 0$ is equivalent to

$$\begin{array}{cc}\hfill r\left(x\right)=& 12{(1+x)}^4{(2+x)}^2(1-2x)+x^4({(1+x)}^2+12)+\hfill \\ \hfill & 24x^2{(1+x)}^4{(2+x)}^2{\psi }^{\prime }(1+x)-12{(1+x)}^4{(2+x)}^2\times \hfill \\ \hfill & {\left((x^2+4x+12)/\left(12{(1+x)}^2\right)\right)}^{6/5}\ge 0.\hfill \end{array}$$

Making use of the following inequality (see [11], p. 6)

$$\begin{array}{c}{\psi }^{\prime }\left(x\right)>\frac{1}{x^2}+\frac{1}{\left(x+\frac{1}{|{\psi }^{\prime }\left(1\right)|}\right)}\hfill \end{array}$$

we come to the conclusion that $r\left(x\right)\ge h\left(x\right)$, where

$$\begin{array}{cc}\hfill h\left(x\right)=& 12{(1+x)}^4{(2+x)}^2(1-2x)+24x^2{(1+x)}^4{(2+x)}^2{(1/(1+x)}^2+\hfill \\ \hfill & 1/{(2+x)}^2+1/{(3+x)}^2+1/(3+x+1/1.64))+x^4({(1+x)}^2+12)-\hfill \\ \hfill & 12{(1+x)}^4{(2+x)}^2{\left((x^2+4x+12)/\left(12{(1+x)}^2\right)\right)}^{6/5}.\hfill \end{array}$$

(We used ${\psi }^{\prime }\left(1\right)={\pi }^2/6>1.64.$)

The inequality $h\left(x\right)\ge 0$ is equivalent to

$$\begin{array}{cc}\hfill T\left(x\right)=& log\left(12{(1+x)}^4{(2+x)}^2(1-2x)+24x^2{(1+x)}^4{(2+x)}^2\left(1/\right.\right.\hfill \\ \hfill & {(1+x)}^2+\left.1/{(2+x)}^2+1/{(3+x)}^2+1/(3+x+1/1.64)\right)+x^4\left({(1+x)}^2+\right.\hfill \\ \hfill & \left.\left.12\right)\right)-log\left(12{(1+x)}^4{(2+x)}^2\right)-log{\left((x^2+4x+12)/\left(12{(1+x)}^2\right)\right)}^{6/5}\ge 0.\hfill \end{array}$$

Differentiating $T\left(x\right)$ yields $T^{\prime }\left(x\right)=\varrho \left(x\right)/\tau \left(x\right),$ where

$$\begin{array}{cc}\hfill \varrho \left(x\right)=& -190374x^{13}-4286906x^{12}-45015548x^{11}-282667076x^{10}-\hfill \\ \hfill & 1146875462x^9-3046852634x^8-5055591704x^7-4276245096x^6+\hfill \\ \hfill & 386216448x^5+4183878144x^4+3096950400x^3+875102976x^2+\hfill \\ \hfill & 292571136x\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \tau \left(x\right)=& 5(41x+148)(x+1)(x+2)(x+3)(x^2+4x+12)\times \hfill \\ \hfill & \left(-67x^9-412x^8+1054x^7+20728x^6+98457x^5+\right.\hfill \\ \hfill & \left.251532x^4+404436x^3+425376x^2+252144x+63936\right).\hfill \end{array}$$

To prove $T\left(x\right)\ge 0$ it suffices to show $T\left(0\right)\ge 0,$ $T\left(1.27\right)\ge 0,$ $\tau \left(x\right)>0,$ $\varrho \left(0\right)=0,$ ${\varrho }^{\prime }\left(0\right)>0,$ $\varrho \left(1.27\right)<0,$ $\varrho \left(x\right)=0$ has only one real root in $(0,1.27).$ It is evident that if $h\left(1.27\right)>0$ then $T\left(1.27\right)>0.$ Some calculations give $T\left(0\right)=0,$ $h\left(1.27\right)\approx 2.70.$ So $T\left(1.27\right)>0.$ The inequality $\tau \left(x\right)>0$ is equivalent to

$$\begin{array}{cc}\hfill d_1\left(x\right)=& -67x^9-412x^8+1054x^7+20728x^6+\hfill \\ \hfill & 98457x^5+251532x^4+404436x^3+425376x^2+252144x+63936>0.\hfill \end{array}$$

Because of

$$\begin{array}{cc}\hfill d_1\left(x\right)>d_2\left(x\right)=& (-67\times 1.{27}^2-412\times 1.27+1054)x^7+20728x^6+\hfill \\ \hfill & 98457x^5+251532x^4+404436x^3+425376x^2+252144x+63936,\hfill \end{array}$$

and $-67\times 1.{27}^2-412\times 1.27+1054=422.6957$ we obtain $\tau \left(x\right)>0.$ Direct computation yields $\varrho \left(0\right)=0,$ $\varrho \left(1.27\right)\approx -5.88\ast {10}^{10},$ ${\varrho }^{\prime }\left(0\right)=$ 292,571,136. Further, we prove that $\varrho \left(x\right)$ has only one root in $(0,1.27).$ Denote $n_1\left(x\right)=\varrho \left(x\right)/x.$ Because of $\varrho \left(x\right)=0$ in $(0,1.27)$ if and only if $n_1\left(x\right)=0$ in $(0,1.27).$

It suffices to show that

• $\varrho \left(x\right)>0$ for $0<x\le 0.8,$
• $n_1^{\prime }\left(x\right)<0$ for $0.8<x<1.27,$
• $\varrho \left(1.27\right)<0.$

Let $0<x\le 0.6.$ First we show $\varrho \left(x\right)>0$ for $0<x\le 0.8.$ It is obvious that

$$\begin{array}{cc}\hfill n_1\left(x\right)\ge n_s\left(x\right)=& 0.6^6\left(-190374x^6-4286906x^5-45015548x^4-\right.\hfill \\ \hfill & \left.282667076x^3-1146875462x^2-3046852634x-5055591704\right)-\hfill \\ \hfill & 4276245096x^5+386216448x^4+4183878144x^3+3096950400x^2+\hfill \\ \hfill & 875102976x+292571136.\hfill \end{array}$$

Differentiation yields

$$\begin{array}{cc}\hfill n_s^{\prime }\left(x\right)=& -832695876x^5/15625-66819454779474x^4/3125+\hfill \\ \hfill & 24007262662032x^3/15625+195501095104788x^2/15625+\hfill \\ \hfill & 95107555576404x/15625+11452328429814/15625,\hfill \end{array}$$

$$\begin{array}{cc}\hfill n_s^{″}\left(x\right)=& -832695876x^4/3125-267277819117896x^3/3125+\hfill \\ \hfill & 72021787986096x^2/15625+\hfill \\ \hfill & 391002190209576x/15625+95107555576404/15625.\hfill \end{array}$$

Now recall from [12,13] that if

$$\begin{array}{c}f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \delta =& 256e^3{a}^3-192a^{2}bd{e}^2-128a^2{c}^2{e}^2+144a^{2}c{d}^{2}e-27a^2{d}^4+\hfill \\ \hfill & 144a{b}^{2}c{e}^2-6b^{2}ae{d}^2-80ab{c}^{2}de+18abc{d}^3+16c^{4}ae-\hfill \\ \hfill & 4a{c}^3{d}^2-27b^4{e}^2+18cde{b}^3-4d^3{b}^3-4b^2{c}^{3}e+b^2{c}^2{d}^2<0\hfill \end{array}$$

then the polynomial $f\left(x\right)$ has two real distinct roots and two complex but not real roots.

Recall [14] the Bolzano Theorem which states: Let $a<b$ be two real numbers, let $f\left(x\right)$ be continuous function on a closed interval $[a,b]$ such that $f\left(a\right)f\left(b\right)<0.$ Then there is a number $x_0\in (a,b)$ such that $f\left(x_0\right)=0.$

Consider the equation $n_s^{″}\left(x\right)=0.$ Direct computation gives

$$\begin{array}{c}\delta \approx -2.21\ast {10}^{64}<0\hfill \end{array}$$

and

This implies that there are only two real roots of $n_s^{″}\left(x\right)=0.$ Table 1 implies the first root of $n_s^{″}\left(x\right)=0$ is in (−330,000, −320,000) and the second root of $n_s^{″}\left(x\right)=0$ is in $(0.6,0.7).$

Table 1. Values of $n_s^{″}\left(x\right)$.

Points	Values of ${\mathit{n}}_{\mathit{s}}^{″}\left(\mathit{x}\right)$
−330,000	≈$-8.63\ast {10}^{25}$
−320,000	≈$8.54\ast {10}^{24}$
$0.6$	≈$4.28\ast {10}^{09}$
$0.7$	≈$-3.47\ast {10}^{09}$

Because of $n_s^{″}\left(0\right)>0,$ $n_s^{\prime }\left(0\right)>0,$ $n_s\left(0\right)>0$ we obtain $\varrho \left(x\right)>0$ for $0<x\le 0.6.$

Similarly, for $0.6<x\le 0.8$ we deduce

$$\begin{array}{cc}{n}_1\left(x\right)\ge n_r\left(x\right)=\hfill & 0.8{.}^6\left(-190374\times 0.8^6-4286906\times 0.8^5-\right.\hfill \\ \hfill & 45015548\times 0.8^4-282667076\times 0.8^3-1146875462\times 0.8^2-3046852634\times \hfill \\ \hfill & \left.0.8-5055591704\right)-4276245096x^5+386216448x^4+4183878144x^3+\hfill \\ \hfill & 3096950400x^2+875102976x+292571136.\hfill \end{array}$$

Differentiation yields

$$\begin{array}{cc}\hfill n_r^{\prime }\left(x\right)=& -21381225480x^4+1544865792x^3+12551634432x^2+\hfill \\ \hfill & 6193900800x+875102976.\hfill \end{array}$$

Consider the equation $n_r^{\prime }\left(x\right)=0.$ Straightforwardly computing acquires

$$\begin{array}{c}\delta \approx -3.27\ast {10}^{59}<0\hfill \end{array}$$

and

This implies that there are only two real roots of $n_r^{\prime }\left(x\right)=0.$ Table 2 implies that the first root of $n_r^{\prime }\left(x\right)=0$ is in $(-0.2,-0.3)$ and the second root of $n_r^{\prime }\left(x\right)=0$ is in $(0.9,1).$

Table 2. Values of $n_r^{\prime }\left(x\right)$.

Points	Values of ${\mathit{n}}_{\mathit{r}}^{\prime }\left(\mathit{x}\right)$
$-0.2$	≈$9.18\ast {10}^{07}$
$-0.3$	≈$-6.83\ast {10}^{07}$
1	−215,721,480
$0.9$	≈$3.71\ast {10}^{09}$

It brings $n_r^{\prime }\left(x\right)>0$ on $0.6<x\le 0.8$ (evidently $n_r^{\prime }\left(0\right)>0$). Because of $n_r\left(0.6\right)\approx 3.53\ast {10}^{08}>0$ we obtain $n_r\left(x\right)>0$ on $0.6<x\le 0.8.$ So $\rho \left(x\right)>0$ for $0.6<x\le 0.8.$ Now we prove $n_1^{\prime }\left(x\right)<0$ for $0.8<x\le 1.27.$

$$\begin{array}{cc}\hfill n_1^{\prime }\left(x\right)=& -2284488x^{11}-47155966x^{10}-450155480x^9-2544003684x^8-\hfill \\ \hfill & 9175003696x^7-21327968438x^6-30333550224x^5-21381225480x^4+\hfill \\ \hfill & 1544865792x^3+12551634432x^2+6193900800x+875102976.\hfill \end{array}$$

We first show $n_1^{\prime }\left(x\right)<0$ for $0.85\le x\le 1.27.$ It is obvious that

$$\begin{array}{cc}\hfill n_1^{\prime }\left(x\right)<v\left(x\right)=& -2284488x^{11}-47155966x^{10}-450155480x^9-\hfill \\ \hfill & 2544003684x^8-9175003696x^7-21327968438x^6-30333550224x^5-\hfill \\ \hfill & 21381225480x^4+1544865792\times 1.{27}^3+12551634432\times 1.{27}^2+\hfill \\ \hfill & 6193900800\times 1.27+875102976.\hfill \end{array}$$

Because of $v^{\prime }\left(x\right)<0,$ $v\left(0.85\right)\approx -4.26\ast {10}^{09}$ we obtain $n_1^{\prime }\left(x\right)<0$ for $0.85\le x\le 1.27.$

Next, we show $n_1^{\prime }\left(x\right)<0$ for $0.8\le x\le 0.85.$ Easy to see that

$$\begin{array}{cc}\hfill n_1^{\prime }\left(x\right)<v\left(x\right)=& -2284488x^{11}-47155966x^{10}-450155480x^9-\hfill \\ \hfill & 2544003684x^8-9175003696x^7-21327968438x^6-30333550224x^5-\hfill \\ \hfill & 21381225480x^4+1544865792\times 0.{85}^3+12551634432\times 0.{85}^2+\hfill \\ \hfill & 6193900800\times 0.85+875102976.\hfill \end{array}$$

Because of $v^{\prime }\left(x\right)<0,$ $v\left(0.8\right)\approx -1.05\ast {10}^{10}$ we get $n_1^{\prime }\left(x\right)<0$ for $0.8\le x\le 0.85.$

From $n_1\left(0.8\right)\approx 1.67\ast {10}^{09}$ and $n_1\left(1.27\right)\approx -4.63\ast {10}^{10}$ we can conclude that $\varrho \left(x\right)$ has only one root in $(0,1.27)$ which completes the proof of Lemma 2. □

3. Proof of the Main Result

In this section, we prove Qi’s Conjecture.

Theorem 1.

Let $\Delta \left(x\right)={\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)$ for $x>0.$ Then

$$\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha }<\Delta \left(x\right)<\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\beta }$$

(9)

holds on $(0,\infty )$ if and only if $\alpha \ge 6/5$ and $\beta \le 1.$

Proof of Theorem 1. 

The upper bound of (9) follows from the following conclusion. Le x be a fixed positive real number. Denote

$$F(\beta ,x)=\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\beta }.$$

Then $F_{\beta }^{\prime }(\beta ,x)<0.$ So, $F(\beta ,x)$ is a decreasing function in $\beta$ for each fixed $x>0.$ In the paper [5], it was obtained that

$$\frac{1}{x^4}\frac{x^2+12}{12{(1+x)}^2}<\Delta \left(x\right)<\frac{1}{x^4}\frac{x^2+4x+12}{12{(1+x)}^2}$$

for $x>0.$ From $F\left(1\right)\ge \Delta \left(x\right)$ and

$$\underset{x\to +\infty }{lim}{x}^4\Delta \left(x\right)-{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\beta }=\frac{1}{12}-\frac{1}{{12}^{\beta }}$$

we can derive that $\beta =1$ is an optimal constant.

Now we show the lower bound of (9). Lemma 2 implies

$$\Delta \left(x\right)-\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha }>0$$

on $(0,\infty )$ for $\alpha \ge 6/5.$ The cases (2), (3) of Lemma 2 imply that $\alpha =6/5$ is the best constant.

This completes the proof. □

4. Materials and Methods

In this paper, MATLAB software and methods of mathematical analysis were used.

5. Conclusions

The main result of this paper is the Theorem 1. The Theorem 1 says that the double inequality

$$\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\alpha }<{\left[{\psi }^{\prime }\left(x\right)\right]}^2+{\psi }^{″}\left(x\right)<\frac{1}{x^4}{\left[\frac{x^2+4x+12}{12{(1+x)}^2}\right]}^{\beta }$$

holds on $(0,\infty )$ if and only if $\alpha \ge 6/5$ and $\beta \le 1$ where ${\psi }^{\prime }\left(x\right)$ and ${\psi }^{″}\left(x\right)$ are the tri- and tetra-gamma functions respectively. The double inequality was posed by F. Qi and R. P. Agarwal as the seventh open problem in Remark 6 of the paper [5].