key: cord-0781958-ahh1rraj authors: Qi, Feng; Guo, Bai-Ni title: From inequalities involving exponential functions and sums to logarithmically complete monotonicity of ratios of gamma functions date: 2020-08-10 journal: J Math Anal Appl DOI: 10.1016/j.jmaa.2020.124478 sha: 5efea6719aa7aad67f4244266b8dd49931bb9162 doc_id: 781958 cord_uid: ahh1rraj In the paper, the authors review origins, motivations, and generalizations of a series of inequalities involving finitely many exponential functions and sums, establish three new inequalities involving finitely many exponential functions and sums by finding convexity of a function related to the generating function of the Bernoulli numbers, survey the history, backgrounds, generalizations, logarithmically complete monotonicity, and applications of a series of ratios of finitely many gamma functions, present complete monotonicity of a linear combination of finitely many trigamma functions, construct a new ratio of finitely many gamma functions, derive monotonicity, logarithmic convexity, concavity, complete monotonicity, and the Bernstein function property of the newly constructed ratio of finitely many gamma functions, and suggest two linear combinations of finitely many trigamma functions and two ratios of finitely many gamma functions to be investigated. was proved to be true. In [20, Lemma 1.4] , the inequality (1.1) was generalized as one which can be reformulated as 1 y − 1 − n k=1 1 y 1/λ k − 1 > 0, (1.2) where y > 1 and λ 1 , λ 2 , . . . , λ n ∈ (0, 1) such that where x = ln y > 0 and λ ∈ (0, 1); (2) inequality (1.2) can be rewritten as where x = ln y > 0 and λ 1 , λ 2 , . . . , λ n ∈ (0, 1) such that n k=1 λ k = 1; (3) inequality (1.3) can be reformulated as (1.8) without restrictions n k=1 λ k = 1 and λ 1 , λ 2 , . . . , λ n < 1; (4) inequality (1.4) can be rewritten as where λ k > 0 and x > 0. (5) inequality (1.5) can be reformulated as (9) when taking m = n and λ 1i = λ i1 > 0 for 1 ≤ i ≤ n and letting λ ij → 0 + for 2 ≤ i, j ≤ n, the inequality (1.6) becomes 1 e x/ n j=1 λ1j − 1 this inequality is equivalent to (1.8) without restrictions n k=1 λ k = 1 and λ 1 , λ 2 , . . . , λ n < 1. In a word, the inequality (1.6) established in [29, is valid for x > 0 and λ ij > 0? where ν i = n j=1 λ ij and τ j = m i=1 λ ij . Motivated by proofs of inequalities (1.3), (1.4), and (1.6) in the papers [22, 29, 31] , we would like to ask a question: what is the largest range of α such that the function t α H 1 t is convex on (0, ∞)? The following lemmas are useful in this paper. are both decreasing on (a, b). A function ϕ : [0, ∞) → R is said to be star-shaped if ϕ(νt) ≤ νϕ(t) for all ν ∈ [0, 1] and t ≥ 0. A real function ϕ defined on a set S ⊂ R n is said to be super-additive if s, t ∈ S implies s + t ∈ S and ϕ(s + t) ≥ ϕ(s) + ϕ(t). See [15, Chapter 16] and [17, Section 3.4] . (1) if ϕ is convex on [0, ∞) with ϕ(0) ≤ 0, then ϕ is star-shaped; (2) if ϕ : [0, ∞) → R is star-shaped, then ϕ is super-additive. Now we give an answer to the third question above and find something more. Then (2) if 0 ≤ α < 1, the function H α (t) has a unique inflection point on (0, ∞); (3) if α < 0, the function H α (t) has only two inflection points on (0, ∞); (4) the function H α (t) has the limits Proof. By direct computation, we have where which can be regarded as a quadratic in the variable α, and (4t + 1)e 2t − 2(2t 2 + 2t + 1)e t + 1 = t 2 + ∞ k=3 Since 2te t + (4t + 1)e 2t − 2(2t 2 + 2t + 1)e t + 1 (e t − 1) and d dt making use of Lemma 2.1 twice, we can deduce that the function H 1 (t) is increasing on (0, ∞). It is straightforward that on (0, ∞). When k ≥ 6, it is standard argument to verify that the terms 4 k − 9k 3 and 2 k+2 − k 3 are positive. It is easy to see that the first derivative of the function with respect to k is Accordingly, when k ≥ 11, the terms are all positive. Then, for k ≥ 11, all coefficients of t k in the infinite series are positive. When 6 ≤ k ≤ 10, numerical computation shows that the coefficients of Consequently, we can conclude that ∆(t) > 0 on (0, ∞). This means that the function H 2 (t) is increasing on (0, ∞). From the above increasing monotonicity of H 1 (t) and H 2 (t) on (0, ∞), it follows that, if and only if α ≥ 1, the function H α (t) is positive on (0, ∞). Therefore, if and only if α ≥ 1, the second derivative d 2 Hα(t) The proof of the existence of inflection points of the function H α (t) on (0, ∞) is straightforward. It is easy to see and , t ∈ (0, ∞), the limits in (3.2) follow immediately. The proof of Theorem 3.1 is complete. has a unique maximum on (0, ∞). Proof. By standard computation and by virtue of (3.3), we have where By standard calculation, we have Employing Lemma 2.1 twice we conclude that the function H (t) is decreasing on (0, ∞). Then the function H 1 t is increasing on (0, ∞), with the limits It is not difficult to see that, if α < 0, by the limits in (3.4) and (3.5), the second derivative d 2 ln Hα(t) dt 2 has a zero, the first derivative d ln Hα(t) dt has only one zero, and the function ln H α (t) has only one inflection point and has only one maximum point on (0, ∞). The proof of Theorem 3.2 is complete. Remark 3.1. It is well known [17, Section 1.3] that a logarithmically convex function must be convex, but not conversely. It is also well known [17, Section 1.3] that a concave function must be logarithmically concave, but not conversely. The function H α (t) is an example that a logarithmically concave function may not be concave, that a convex function may not be logarithmically convex, and so on. Making use of some conclusions in Theorems 3.1 and 3.2, we now start out to derive several inequalities involving exponential functions and sums and to answer the first and second questions related to inequalities (1.11) and (1.12). Theorem 4.1. For α ≥ 1, x > 0, and λ ij > 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ n, Proof. Combining the first conclusion and the limit (3.1) in Theorem 3.1 with Lemma 2.2 yields that, if α ≥ 1, the function H α (t) with redefining H α (0) = 0 is convex, then star-shaped, and then supper-additive on [0, ∞). Consequently, it follows that Accordingly, we obtain which can be rearranged as The proof of Theorem 4.1 is complete. Remark 4.1. Setting m = n and λ 1k = λ k1 = λ k > 0 for 1 ≤ k ≤ n and letting λ ij → 0 + for 2 ≤ i, j ≤ n in the inequality (4.1) result in inequality (1.11) for α ≥ 0. The inequality (4.1) is equivalent to (1.12) for α ≥ 0 and ρ ≤ 2. Proof. Similarly to what has been done in Theorem 4.1, we obtain Consequently, it follows that Rearranging and simplifying the above two inequalities lead to (4.2) and (4.3). The proof of Theorem 4.2 is complete. A new ratio of many gamma functions and its properties A nonnegative function F (x) defined on a finite or infinity interval I is called a Bernstein function if its derivative f (x) is completely monotonic on I. See the monograph [32] . Among these three concepts, there are the following relations: (1) A logarithmically completely monotonic function is completely monotonic, but not conversely. See [5, 8, 24, 25] Let p ∈ (0, 1) and k, n be nonnegative integers such that 0 ≤ k ≤ n. In [2, Theorem] , motivated by inequalities related to binomial probability studied in [13, 14] , with the help of the inequality (1.1), Alzer proved [2] that the function is completely monotonic on (0, ∞). Indeed, Alzer implicitly proved logarithmically complete monotonicity of G(x) on (0, ∞). In [20, Theorem 2.1] and [31] , with the aid of inequalities (1.2) and (1.3), the function G(x) defined in (5.1) and its logarithmically complete monotonicity were generalized as follows. Let m ≥ 2, λ i > 0 for 1 ≤ i ≤ m, p i ∈ (0, 1) for 1 ≤ i ≤ m, and m i=1 p i = 1. Then the function is logarithmically completely monotonic on (0, ∞ With the help of the inequality (1.4), for q ∈ (0, 1) and m ≥ 2, the q-analogue of the function Q(x) in (5.2) was proved in [22] to be logarithmically completely monotonic on (0, ∞), where λ i > 0 for 1 ≤ i ≤ m and p i ∈ (0, 1) for 1 ≤ i ≤ m with and proved that the function g m,n (t) is logarithmically complete monotonicity on (0, ∞). Let λ i,j = e −(i+j/2) for 1 ≤ i ≤ m = 3 and 1 ≤ j ≤ n = 5. Then and the graph of the function g 3,5 (t) on the interval (0, 2) is showed in Figure 1 . The graph showed in Figure 1 implies that the function g 3,5 (t) is not logarithmically completely monotonic on (0, ∞). Consequently, the conclusions in [19, Theorem 2.1] are thoroughly wrong. Let λ ij > 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ n, let ν i = n j=1 λ ij and τ j = m i=1 λ ij for 1 ≤ i ≤ m and 1 ≤ j ≤ n, and let for ρ ∈ R. This function f (t) is a generalization of the function g(t) in (5.3). In [29, Theorem 4.1] , with the help of the inequality (1.6), the following conclusions were obtained: (1) when ρ ≤ 2, the second derivative [ln f (t)] is a completely monotonic function of t ∈ (0, ∞) and maps from (0, ∞) onto the open interval 0, (2) when ρ = 2, the logarithmic derivative [ln These results correct mistakes appeared in [19, Theorem 2.1]. Some of the above results have been applied in [2, 20, 22, 29] to multinomial probability, to the Bernstein estimators on the simplex, to constructing combinatorial inequalities for multinomial coefficients, to constructing inequalities for multivariate beta functions, and the like. Complete monotonicity of a linear combination of finitely many trigamma functions. In the paper [9] , the authors discussed complete monotonicity of the linear combination m k=1 a k ψ(b k x + δ) for δ ≥ 0 and a k , b k > 0. Now we discuss complete monotonicity of a linear combination of finitely many trigamma functions. Theorem 5.1. Let λ ij > 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ n, let ν i = n j=1 λ ij and τ j = m i=1 λ ij for 1 ≤ i ≤ m and 1 ≤ j ≤ n, and let ρ, θ ∈ R. If ρ ≤ 2 and θ ≥ 0, then the linear combination is completely monotonic on (0, ∞). Proof. Employing the integral representation Consequently, if θ ≥ 0 and ρ ≤ 2, the function P (t) is completely monotonic on (0, ∞). The proof of Theorem 5.1 is complete. A new ratio of many gamma functions and its properties. Let λ ij > 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ n, let ν i = n j=1 λ ij and τ j = m i=1 λ ij for 1 ≤ i ≤ m and 1 ≤ j ≤ n, and let for ρ, θ ∈ R. It is clear that, when θ = 0, the function F (t) becomes f (t) defined in (5.4). Theorem 5.2. The function F (t) has the following properties: (1) If ρ ≤ 2 and θ ≥ 0, the second derivative [ln F (t)] is completely monotonic and maps from (0, ∞) onto the interval (2) If ρ ≤ 2 and θ ≥ 0, the logarithmic derivative [ln F (t)] = F (t) F (t) is increasing and concave on (0, ∞). Proof. Taking the logarithm of F (t) in (5.6) and differentiating give ln and [ln F (t)] = P (t), where P (t) is defined in (5.5) . From Theorem 5.1, it follows immediately that, if ρ ≤ 2 and θ ≥ 0, the second derivative [ln F (t)] is completely monotonic on (0, ∞) and, consequently, that the function F (t) is logarithmically convex on (0, ∞). It is easy to obtain that lim t→0 + F (t) = 1, Since P (t) = [ln F (t)] is completely monotonic on (0, ∞), the logarithmic derivative [ln F (t)] is increasing and concave on (0, ∞). Utilizing lim t→∞ [ψ(t)−ln t] = 0 in [10, Theorem 1] and [11, Section 1.4 where the last term is equal to ∞, θ > 0 or ρ < 2. λ θ ij e −λij ts ds where, when ρ = 2 and θ = 0, we used the fact [29] that The proof of Theorem 5.2 is complete. Finally, basing on inequalities (4.2) and (4.3) in Theorem 4.2, motivated by Theorems 5.1 and 5.2, we would like to suggest to consider two linear combinations L 1 (t), L 2 (t) and two ratios R 1 (t), R 2 (t) defined by λ θ ijk ψ 1 + λ ijk t , where t > 0, θ ≥ 0, and λ ijk > 0 for 1 ≤ i ≤ , 1 ≤ j ≤ m, and 1 ≤ k ≤ n. For the sake of saving the space and shortening the length of this paper, we would not like to write down our guesses on possible conclusions and their detailed proofs for the functions L 1 (t), L 2 (t), R 1 (t), and R 2 (t) on (0, ∞). Remark 6.1. An anonymous referee recommends three papers [7, 12, 35] which are highly relevant to the topic of the paper. Remark 6.2. This paper is a revised version of the preprint [26] and a companion of the paper [30] . 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