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

: 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).


Introduction
It is common knowledge that the classical Euler's gamma function [1,2] is defined by for x > 0 and the digamma function [3] is defined as the logarithmic derivative of the gamma function The functions ψ, ψ , ψ , ψ , ... 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 ψ 2 + ψ . 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 Key Lemmas
In this section, we prove two important lemmas. x 2 + 4x + 12 12(1 + x) 2 α for x > 0 and α ∈ R. Then Proof of Lemma 1. To prove (1) it suffices to show lim x→+∞ ∆(x) = 0 which follows from the double inequality x 2 + 12 12x 4 (1 + x) 2 < ∆(x) < x 2 + 4x + 12 12x 4 (1 + x) 2 (see [5], p. 9). Making use of the previous double inequality yields which is (2). Simple calculation brings To prove (3) it suffices to show The function δ(x) can be rewritten as Making use of well known formulas: for polygamma functions and for values of the Riemann zeta function [7] The equality (4) can be rewritten as Proof of Lemma 2. Consider three cases The case (a).
Because of The proof of the case (a) is complete.
The case (b). Let 1.27 ≤ x < 5.7. Using the following formulas Making use of the inequality (see [5], p. 9) To prove the case (b) it suffices to show It will be done if we prove where Differentiating F yields .
In the paper [8], inequality (2.3) it was established that Using (6) gives By (7), it follows that This implies that The inequality G(x) > 0 may be rearranged as In the paper [10], by using asymptotic expansion, it was deduced that This implies Utilizing (6), Rewriting To show the positivity of the function G 1 (x) it suffices to prove ϕ(x) > 0. By virtue of x ≥ 1.27 we see that ϕ(x) ≥ n 1 (x) ≥ n 2 (x) ≥ n 3 (x) where the functions n 1 (x), n 2 (x), n 3 (x) are derived in Appendix A.

The inequality t(x) ≥ 0 is equivalent to
Making use of the following inequality (see [11], p. 6)  It suffices to show that (1.27) < 0.
Let 0 < x ≤ 0.6. First we show (x) > 0 for 0 < x ≤ 0.8. It is obvious that then the polynomial f (x) 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 (x) be continuous function on a closed interval [a, b] such that f (a) f (b) < 0. Then there is a number x 0 ∈ (a, b) such that f (x 0 ) = 0.
Consider the equation n s (x) = 0. Direct computation gives δ ≈ −2.21 * 10 64 < 0 and This implies that there are only two real roots of n s (x) = 0. Table 1 implies the first root of n s (x) = 0 is in (−330,000, −320,000) and the second root of n s (x) = 0 is in (0.6, 0.7).

Proof of the Main Result
In this section, we prove Qi's Conjecture.

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