NECESSARY AND SUFFICIENT CONDITIONS FOR A FUNCTION ...

arXiv:0904.1101v2 [math.CA] 17 May 2009

NECESSARY AND SUFFICIENT CONDITIONS FOR A FUNCTION INVOLVING A RATIO OF GAMMA FUNCTIONS TO BE LOGARITHMICALLY COMPLETELY MONOTONIC FENG QI AND BAI-NI GUO Abstract. In the paper, necessary and sufficient conditions are presented for a function involving a ratio of two gamma functions to be logarithmically completely monotonic, and some known monotonicity and inequality results are generalized and extended.

1. Introduction Recall [4, 35] that a function f is said to be logarithmically completely monotonic on an interval I ⊆ R if it has derivatives of all orders on I and its logarithm ln f satisfies 0 ≤ (−1)k [ln f (x)](k) < ∞

(1)

for k ∈ N on I. By looking through “logarithmically completely monotonic function” in the database MathSciNet, it is found that this phrase was first used in [4], but with no a word to explicitly define it. Thereafter, it seems to have been ignored by the mathematical community. In early 2004, this terminology was recovered in [35] and it was immediately referenced in [44], the preprint of [43]. A natural question that one may ask is: Whether is this notion trivial or not? In [35, Theorem 4], it was proved that all logarithmically completely monotonic functions are also completely monotonic, but not conversely. This result was formally published while revising [30]. Hereafter, this conclusion and its proofs were dug in [6, 11, 12] and [50] (the preprint of [33]) once and again. Furthermore, in the paper [6], the logarithmically completely monotonic functions on (0, ∞) were characterized as the infinitely divisible completely monotonic functions studied in [16] and all Stieltjes transforms were proved to be logarithmically completely monotonic on (0, ∞). For more information, please refer to [6]. It is well-known that the classical Euler gamma function Γ(x) may be defined for x > 0 by Z ∞ Γ(x) = tx−1 e−t dt. (2) 0

The logarithmic derivative of Γ(x), denoted by ψ(x) =

Γ′ (x) Γ(x) ,

is called the psi or

digamma function, and ψ (k) (x) for k ∈ N are called the polygamma functions. It is common knowledge that these functions are fundamental and that they have much extensive applications in mathematical sciences. 2000 Mathematics Subject Classification. Primary 26A48, 33B15; Secondary 26A51, 26D07. Key words and phrases. Necessary and sufficient condition, logarithmically completely monotonic function, gamma function, psi function, polygamma function, inequality. The first author was partially supported by the China Scholarship Council. This paper was typeset using AMS-LATEX. 1

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F. QI AND B.-N. GUO

In [13, Theorem 2] and its preprint [42], the following monotonicity was established: The function [Γ(x + y + 1)/Γ(y + 1)]1/x (3) x+y+1 is decreasing with respect to x ≥ 1 for fixed y ≥ 0. Consequently, for positive real numbers x ≥ 1 and y ≥ 0, we have [Γ(x + y + 1)/Γ(y + 1)]1/x x+y+1 . ≤ x+y+2 [Γ(x + y + 2)/Γ(y + 1)]1/(x+1)

(4)

In [32], the above decreasing monotonicity was extended and generalized as follows: The function (3) is logarithmically completely monotonic with respect to x ∈ (0, ∞) if y ≥ 0, so is its reciprocal if −1 < y ≤ − 21 . Consequently, the inequal ity (4) is valid for (x, y) ∈ (0, ∞)×[0, ∞) and reversed for (x, y) ∈ (0, ∞)× −1, − 21 . For (x, y) ∈ (0, ∞) × [0, ∞) and α ∈ [0, ∞), the function [Γ(x + y + 1)/Γ(y + 1)]1/x (x + y + 1)α

(5)

was proved in [54] to be strictly increasing (or decreasing, respectively) with respect to x ∈ (0, ∞) if and only if 0 ≤ α ≤ 21 (or α ≥ 1, respectively), to be strictly increasing with respect to y on [0, ∞) if and only if 0 ≤ α ≤ 1, and to be logarithmically concave with respect to (x, y) ∈ (0, ∞) × (0, ∞) if 0 ≤ α ≤ 41 . For more information on the history, background, motivation and generalizations of this topic, please refer to [1, 5, 14, 25, 26, 37, 38, 42, 46, 53] and related references therein. For given y ∈ (−1, ∞) and α ∈ (−∞, ∞), let  1/x 1 Γ(x + y + 1)    , x ∈ (−y − 1, ∞) \ {0}; α Γ(y + 1) hα,y (x) = (x + y + 1) (6)  1   exp[ψ(y + 1)], x = 0. (y + 1)α

It is clear that the ranges of x, y and α in the function hα,y (x) extend the corresponding ones in the functions (3) and (5) discussed in [13, 32, 42, 54]. The aim of this paper is to present necessary and sufficient conditions such that the function (6) or its reciprocal are logarithmically completely monotonic. Our main results may be stated as follows. Theorem 1. For y > −1, 1 (1) if and only if α ≥ max 1, y+1 , the function (6) is logarithmically completely monotonic with respect to x ∈ (−y − 1, ∞); 1 , the reciprocal of the function (6) is logarithmically (2) if α ≤ min 1, 2(y+1) completely monotonic with respect to x ∈ (−y − 1, ∞); (3) the necessary condition for the reciprocal of the function (6) to be logarithmically completely monotonic with respect to x ∈ (−y − 1, ∞) is α ≤ 1. By the way, we give an inequality for the difference of the logarithms of gamma functions. Theorem 2. For t ∈ (0, ∞), we have t 1 + 2t − ln Γ(t) < 1 − ψ(t). ln Γ 2t2 1 + 2t

(7)

Finally, as by-product, we prove the following monotonicity result of a function involving the psi function and the logarithms of two gamma functions.

COMPLETE MONOTONICITY OF A FUNCTION INVOLVING GAMMA FUNCTIONS

3

Theorem 3. The function x2 (8) 2(y + 1)(x + y + 1) h 2 is negative and decreasing with respect to x ∈ − 2(y+1) , ∞ for y ∈ −1, − 21 . 1+2y q(x, y) = xψ(x + y + 1) − ln Γ(x + y + 1) + ln Γ(y + 1) −

2. Remarks

Before continuing to verify our theorems, we give several direct consequences of them. Remark 1. As a ready consequence of of the monotonicity results in Theorem 1, the following double inequality may be derived: For t > 0 and y > −1, the inequality a b x+y+1 [Γ(x + y + 1)/Γ(y + 1)]1/x x+y+1 < (9) < x+y+t+1 x+y+t+1 [Γ(x + y + t + 1)/Γ(y + 1)]1/(x+t) 1 1 holds with respect to x ∈ (−y − 1, ∞) if a ≥ max 1, y+1 and b ≤ min 1, 2(y+1) . Accordingly, it is not difficult to see that Theorem 1 extends, refines and generalizes those monotonicity and inequality results obtained in [13, 25, 26, 32, 42, 51, 52, 53, 54]. n 2

for n ∈ N in (9) reveals that r [Γ(n/2 + 1)]2/n n+2 n+2 < < 2/(n+2) n+4 n+4 [Γ((n + 2)/2 + 1)]

Remark 2. Letting t = 1, y = 0 and x =

which is equivalent to r

1/(n+2)

Ω n+2 < n+2 1/n n+4 Ωn

2, this refines √ n/(n+1) 2 √ Ωn/(n+1) , n∈N (14) ≤ Ωn < e Ωn+1 n+1 π in [2, Theorem 1]. Similarly, by setting different values of x, y and t in (9), more similar inequalities as above may be derived immediately. For more information on inequalities for the volume of the unit ball in Rn , please see [2, 3, 36] and related references therein. Remark 3. It is noted that an alternative upper bound in (4) and (9) for t = 1 has been established in [34] and related references therein.

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F. QI AND B.-N. GUO

Remark 4. It is noted that Theorem 1 may be restated as follows: For y ∈ (0, ∞) and α ∈ (−∞, ∞), the function  [Γ(x + y)/Γ(y)]1/x   , x ∈ (−y, ∞) \ {0}  (x + y)α Hy (x) = (15) ψ(y)   e , x = 0 yα

is logarithmically respect completely monotonic with to x ∈ (−y, ∞) if and only if 1 and only if α ≤ 1. α ≥ max 1, y1 , so is its reciprocal if α ≤ min 1, 2y

1 is also Remark 5. We conjecture that when y > − 12 the condition α ≤ 2(y+1) necessary for the reciprocal of the function (6) to be logarithmically completely monotonic with respect to x ∈ (−y − 1, ∞).

Remark 6. This paper is a corrected version of the preprint [39]. 3. Lemmas In order to prove our main result, the following lemmas are needed. Lemma 1. For x ∈ (0, ∞) and k ∈ N, we have ln x − and

1 1 < ψ(x) < ln x − x 2x

k! (k − 1)! k! (k − 1)! + k+1 < (−1)k+1 ψ (k) (x) < + k+1 . xk 2x xk x

(16)

(17)

Proof. In [17, Theorem 2.1], [28, Lemma 1.3] and [29, Lemma 3], the function ψ(x) − ln x + αx was proved to be completely monotonic on (0, ∞), i.e., h α i(i) (−1)i ψ(x) − ln x + ≥0 (18) x for i ≥ 0, if and only if α ≥ 1, so is its negative, i.e., the inequality (18) is reversed, x Γ(x) was proved if and only if α ≤ 21 . In [8] and [20, Theorem 2.1], the function exx−α to be logarithmically completely monotonic on (0, ∞), i.e., (k) ex Γ(x) (−1)k ln x−α ≥0 (19) x for k ∈ N, if and only if α ≥ 1, so is its reciprocal, i.e., the inequality (19) is reversed, if and only if α ≤ 12 . Rearranging either (18) or (19) and considering the fact [43, p. 82] that a completely monotonic function which is non-identically zero cannot vanish at any point on (0, ∞) leads to the double inequalities (16) and (17). The proof of Lemma 1 is complete. Lemma 2. For x ∈ (0, ∞) and k ∈ N, we have

k! (k − 1)! k! (k − 1)! + k+1 < (−1)k+1 ψ (k) (x) < + k+1 . (x + 1)k x (x + 1/2)k x

(20)

Proof. In [15, Theorem 1], the following necessary and sufficient conditions are obtained: For real numbers α 6= 0 and β, the function α x e Γ(x + 1) , x ∈ (max{0, −β}, ∞) gα,β (x) = (x + β)x+β is logarithmically completely monotonic if and only if either α > 0 and β ≥ 1 or α < 0 and β ≤ 21 . Further considering the fact [43, p. 82] that a completely


5

monotonic function which is non-identically zero cannot vanish at any point on (0, ∞) gives (−1)k [ln gα,β (x)](k) = (−1)k α[x + ln Γ(x) + ln x − (x + β) ln(x + β)](k) > 0

for k ∈ N and x ∈ (0, ∞) if and only if either α > 0 and β ≥ 1 or α < 0 and β ≤ 12 . As a result, from straightforward calculation and standard arrangement, the inequality 1 1 1 − < ψ(x) < ln(x + 1) − (21) ln x + 2 x x and (20) follow readily. Lemma 2 is thus proved. Remark 7. In [9, Corollary 3] and [40, Theorem 1], it was proved that the double inequality 1 1 1 ln x + − < ψ(x) < ln(x + e−γ ) − (22) 2 x x holds on (0, ∞). In [40, Theorem 1], it was also shown that the scalars 12 and e−γ = 0.56 · · · in (22) are the best possible. It is obvious that the inequality (22) refines and sharpens (21). In [41], the inequality (20) was further refined and sharpened. 4. Proofs of theorems Now we are in a position to prove our theorems. Proof of Theorem 1. For x 6= 0, taking the logarithm of hα,y (x) gives

ln Γ(x + y + 1) − ln Γ(y + 1) − α ln(x + y + 1). x Direct differentiation yields ln hα,y (x) =

[ln hα,y (x)](k) =

k k! X (−1)k−i xi ψ (i−1) (x + y + 1) xk+1 i=0 i!

(−1)k k! ln Γ(y + 1) (−1)k−1 (k − 1)!α − − xk+1 (x + y + 1)k

(23)

for k ∈ N, where ψ (−1) (x + y + 1) and ψ (0) (x + y + 1) stand for ln Γ(x + y + 1) and ψ(x + y + 1) respectively. Furthermore, a simple calculation gives ′ k+1 x [ln hα,y (x)](k) = (−1)k−1 xk (−1)k−1 ψ (k) (x + y + 1) (k − 1)!α k!(y + 1)α − . − (x + y + 1)k (x + y + 1)k+1 Utilizing (17) in the above equation leads to ′ (−1)k−1 k+1 (k − 1)!(1 − α) k![1/2 − (y + 1)α] + ≤ x [ln hα,y (x)](k) (x + y + 1)k (x + y + 1)k+1 xk (k − 1)!(1 − α) k![1 − (y + 1)α] ≤ + (x + y + 1)k (x + y + 1)k+1 for k ∈ N, x 6= 0, y ∈ (−1, ∞) and α ∈ (−∞, ∞). Therefore, ( (−1)k−1 k+1 (k) ′ ≤ 0, if α ≥ 1 and α ≥ x [ln hα,y (x)] xk ≥ 0, if α ≤ 1 and α ≤

1 y+1 1 2(y+1)

(24)

(25)

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F. QI AND B.-N. GUO

for k ∈ N, y > −1 and x 6= 0. For x > 0, the equation (25) means ( 1 2k ≤ 0, if α ≥ 1 and α ≥ y+1 ′ x [ln hα,y (x)](2k−1) 1 ≥ 0, if α ≤ 1 and α ≤ 2(y+1) and

2k+1 ′ x [ln hα,y (x)](2k)

(

≥ 0, if α ≥ 1 and α ≥ ≤ 0, if α ≤ 1 and α ≤

1 y+1 1 2(y+1)

for k ∈ N. From (23), it is easy to see that lim xk+1 [ln hα,y (x)](k) = 0

(26)

x→0

for k ∈ N and any given y > −1. As a result, ( (2k−1) < 0, if α ≥ 1 and α ≥ [ln hα,y (x)] > 0, if α ≤ 1 and α ≤

and (2k)

[ln hα,y (x)]

(

for k ∈ N and x ∈ (0, ∞), that is,

> 0, if α ≥ 1 and α ≥ < 0, if α ≤ 1 and α ≤

(−1)k [ln hα,y (x)](k)

(

1 y+1 1 2(y+1)

1 y+1 1 2(y+1)

> 0, if α ≥ 1 and α ≥ < 0, if α ≤ 1 and α ≤

1 y+1 1 2(y+1)

(27)

(28)

(29)

for k ∈ N and x ∈ (0, ∞). Hence, the function (6) is logarithmically completely 1 monotonic with respect to x on (0, ∞) if α ≥ 1 and α ≥ y+1 , so is the reciprocal 1 of the function (6) if either 0 < α ≤ 1 and α ≤ 2(y+1) or α ≤ 0 and y > −1. If x ∈ (−y − 1, 0), the equation (25) means ( 1 k+1 (k) ′ ≥ 0, if α ≥ 1 and α ≥ y+1 x [ln hα,y (x)] 1 ≤ 0, if α ≤ 1 and α ≤ 2(y+1) for k ∈ N. By virtue of (26), it follows that ( k+1 (k) ≤ 0, if α ≥ 1 and α ≥ x [ln hα,y (x)] ≥ 0, if α ≤ 1 and α ≤

1 y+1 1 2(y+1)

for k ∈ N, which is equivalent to that the equations (27) and (28) hold for x ∈ (−y − 1, 0). As a result, the equation (29) is valid for k ∈ N and x ∈ (−y − 1, 0). Therefore, the function hα,y (x) has the same logarithmically complete monotonicity properties on (−y − 1, 0) as on (0, ∞). Conversely, if hα,y (x) is logarithmically completely monotonic on (−y − 1, ∞), then [ln hα,y (x)]′ < 0 on (−y − 1, ∞), which can be simplified as X 1 (−1)1−i xi ψ (i−1) (x + y + 1) ln Γ(y + 1) 1 (30) + α ≥ (x + y + 1) 2 x i=0 i! x2 1 [(x + y + 1) ln Γ(y + 1) − (y + 1)(x + y + 1)ψ(x + y + 1) x2 + (x + y + 1)2 ψ(x + y + 1) − (x + y + 1) ln Γ(x + y + 1)] x + y + 1 xψ(x + y + 1) − ln Γ(x + y + 1) ln Γ(y + 1) . + = x x x

=

(31) (32)

From (16), it is easy to see that

lim [x2 ψ(x)] = 0.

x→0+

(33)


7

It is common knowledge that Γ(x + 1) = xΓ(x)

(34)

for x > 0. Taking the logarithm on both sides of (34), rearranging, and taking limit leads to lim+ [x ln Γ(x)] = lim+ [x ln Γ(x + 1)] − lim+ [x ln x] = 0. (35) x→0

x→0

x→0

Taking logarithmic derivatives on both sides of (34) yields ψ(x + 1) = x > 0, and so lim+ [xψ(x)] = 1 + lim+ [xψ(x + 1)] = 1. x→0

1 x

+ ψ(x) for

x→0

(36)

Thus, by utilizing (33), (35) and (36), it is revealed that the limit of the function (31) 1 as x → (−y − 1)+ , that is, as x + y + 1 → 0+ , equals y+1 . By L’Hˆ ospital’s rule and the double inequality (17) for k = 1, we have xψ(x + y + 1) − ln Γ(x + y + 1) = lim [xψ ′ (x + y + 1)] = 1. lim x→∞ x→∞ x

Hence, the limit of the function (32) as x → ∞ equals 1. In a word, the necessary condition for hα,y (x) to be logarithmically completely monotonic is α ≥ 1 and 1 α ≥ y+1 . If the reciprocal of hα,y (x) is logarithmically completely monotonic, then the inequality (30) is reversed. Since the limit of the function (32) equals 1 as x → ∞, as showed above, then the necessary condition α ≤ 1 is obtained. The proof of Theorem 1 is complete. Proof of Theorem 2. It is clear that Z 1 + 2t t t 1 + 2t − ln Γ(t) = − ln Γ ψ(u) du, 2t2 1 + 2t 2t2 t/(1+2t) so the required inequality (7) can be rewritten as Z 1 + 2t t p(t) , ψ(u) du − ψ(t) + 1 > 0, 2t2 t/(1+2t)

t > 0.

In [31], it was obtained that 1 1 1 < ψ(x + 1) − ln x < − , x > 0, 2x 12x2 2x which is equivalent to 1 1 1 ln x − − < ψ(x) < ln x − , x > 0. 2x 12x2 2x From (38), it follows that Z 1 + 2t t 1 1 1 p(t) > +1 du − ln t − ln u − − 2t2 2u 12u2 2t t/(1+2t)

(37)

(38)

4t − 3 ln(2t + 1) − 1 12t2 q(t) . , 12t2 =

1 8 Since q ′ (t) = 2(4t−1) 2t+1 , the function q(t) increases for t > 4 . By q 7 = 0.002 · · · > 0, it is easy to see that the inequality (7) holds for t ≥ 78 = 1.1428 · · · . In [10, Lemma 1], [21, Theorem 1] and [22, Theorem 1], it was proved that the inequality (a−b) Γ(a) ψ(L(a,b)) e < (39) Γ(b)

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F. QI AND B.-N. GUO

is valid for positive numbers a and b with a 6= b, where b−a L(a, b) = (40) ln b − ln a t in (39) and (40) stands for the logarithmic mean. Hence, letting a = t and b = 1+2t gives 2t2 ln Γ(t/(1 + 2t)) − ln Γ(t) . >ψ t/(1 + 2t) − t (1 + 2t) ln(1 + 2t) Since the inequality (7) can be rearranged as ln Γ(t/(1 + 2t)) − ln Γ(t) > ψ(t) − 1, t > 0, t/(1 + 2t) − t it is sufficient to show that Z t 2t2 = ψ ′ (u) du < 1 (41) ψ(t) − ψ (1 + 2t) ln(1 + 2t) 2t2 /(1+2t) ln(1+2t) holds on 0, 78 . In [27, Theorem 1] and [49, Theorem 1], the following double inequalities were obtained: For real numbers s > 0 and t > 0 with s 6= t and an integer i ≥ 0, the inequality Z (−1)i t (i) (−1)i ψ (i) (Lp (s, t)) ≤ ψ (u) du ≤ (−1)i ψ (i) (Lq (s, t)) (42) t−s s

holds if p ≤ −i − 1 and q ≥ −i, where Lp (a, b) is the generalized logarithmic mean of order p ∈ R for positive numbers a and b with a 6= b, which can be defined [7, p. 385] by  1/p  bp+1 − ap+1   , p 6= −1, 0;   (p + 1)(b − a)     b−a , p = −1; Lp (a, b) = (43) ln b − ln a       1 bb 1/(b−a)    , p = 0. e aa Taking s = leads to Z t

2t2 (1+2t) ln(1+2t) ,

i = 1 and p = −2 in the left-hand side inequality of (42)

1/2 2t3 2t2 ′ ψ . ψ (u) du ≤ t− (2t + 1) ln(2t + 1) (2t + 1) ln(2t + 1) 2t2 /(1+2t) ln(1+2t) ′

Combining this with (41) reveals that it suffices to prove 1/2 2t3 (2t + 1) ln(2t + 1) ψ′ ≤ (2t + 1) ln(2t + 1) t[(2t + 1) ln(2t + 1) − 2t] for t ∈ 0, 87 . The right-hand side inequality in (20) for k = 1 results in 1/2 (2t + 1) ln(2t + 1) 1 2t3 ≤ +q ψ′ (2t + 1) ln(2t + 1) 2t3 2t3

(2t+1) ln(2t+1)

(44)

+

1 2

.

Then, in order to prove (44), it is enough to show (2t + 1) ln(2t + 1) +q 2t3

for t ∈ 0, 87 .

1 2t3 (2t+1) ln(2t+1)

+

1 2

≤

(2t + 1) ln(2t + 1) t[(2t + 1) ln(2t + 1) − 2t]

(45)


9

In [18, p. 296] and [19, p. 274, (3.6.19)], the following inequality was collected: 1 2 1 1 ln 1 + < 1+ , x > 0, (46) − x 2x + 1 12x 12(x + 1)

which can be rewritten as

t t2 + 12t + 12 , ln(1 + t) < 6(t + 1)(t + 2)

t > 0.

Therefore, to verify (45), it is sufficient to prove (2t + 1) 2t 4t2 + 24t + 12 1 · +q 6(2t+1)(2t+2) 2t3 6(2t + 1)(2t + 2) 2t3 (2t+1) · 2t(4t2 +24t+12) +

for t ∈ 0, 78 , which can be simplified as r

≤ h t 1−

2t (2t+1)

(47)

1 2

1 ·

6(2t+1)(2t+2) 2t(4t2 +24t+12)

i

(48)

t2 + 6t + 3 t2 + 6t + 3 2 q ≤ + , 2 (t+1) 3t2 (t + 1) t2 (t + 3) 1 + 12t 2 t +6t+3

12t2 (t + 1) 3t3 + 11t2 + 3t − 3 q(t) . ≥ , 2 t + 6t + 3 t2 + 6t + 3 2 t2 + 6t + 3

(49)

Since q(t) is increasing on (0, ∞) with q(0) = −3 and q(1) = 14, the function q(t) 2 1 has a unique zero t0 ∈ (0, 1). From q 3 = − 3 , we can locate more accurately that 1 t0 ∈ 3 , 1 . When 0 < t ≤ t0 , the function q(t) is non-positive, so the inequality (49) is clearly valid. When t ≥ t0 , the function q(t) is non-negative, so squaring both sides of (49) and simplifying gives h(t) , 9t6 + 54t5 + 55t4 − 60t3 − 93t2 − 18t + 9 ≤ 0.

Direct differentiation yields

h′ (t) = 54t5 + 270t4 + 220t3 − 180t2 − 186t − 18,

h′′ (t) = 270t4 + 1080t3 + 660t2 − 360t − 186,

h(3) (t) = 1080t3 + 3240t2 + 1320t − 360.

It is clear that the function h(3) (t) is increasing with limt→∞ h(3) (t) = ∞ and h(3) (0) = −360, so the function h(3) (t) has a unique zero which is the unique minimum point of the function h′′ (t). Since h′′ (0) = −186 and limt→∞ h′′ (t) = ∞, the function h′′ (t) has a unique zero which is the unique minimum point of the function h′ (t). From h′ (0) = −18 and limt→∞ h′ (t) = ∞, we conclude that the function h′ (t) has a unique zero which is the point of the unique minimum 404759 8 function h(t) on (0, ∞). Due to h(0) = 9, h 31 = − 700 , h = − 81 7 117649 and 1 8 limt→∞ h(t) = ∞, it is not difficult to see that the function h(t) < 0 on 3 , 7 . As 8 a result, the inequalities (49), and so (48), holds on 0, 7 . The proof of Theorem 2 is complete. Proof of Theorem 3. Differentiating directly and using the right-hand side inequality in (17) yields ∂q(x, y) (x + 2y + 2) ′ = x ψ (x + y + 1) − ∂x 2(y + 1)(x + y + 1)2 1 (x + 2y + 2) 1