arXiv:1706.01971v1 [math.CA] 28 May 2017 o all for interval ucin hr am ucini endby defined is deriv function we gamma functions, fun where these 1 some 6, function, of 5, of properties 2, monotonicity monotonic 17, the complete 24, Using 30, the 31, investigate 7, we 22, 14, paper, inequalitie [13, various see ex functions, provide for beta to and articles, appeared many have in articles considered many been have functions beta otermntncpoete 9 0 6.Afunction A 26]. 10, [9, properties Probabi monotonic in their role major to a play functions monotonic Completely Introduction 1 M 20)SbetClassification Subject (2000) AMS function. gamma of inequalities Phrases and Keywords opeeMntnct n nqaiiso Functions of Inequalities and Monotonicity Complete ∗ [email protected] qaiisivliggmaadbt ucin.Sc inequa fun Such these functions. of beta ory. properties and monotonic gamma involving the equalities Using function. gamma x I ∈ nti ae,w netgt h opeemntnct fs of monotonicity complete the investigate we paper, this In eateto ahmtc,YrokUiest,Ibd Jor Irbid, University, Yarmouk Mathematics, of Department ⊂ I R n all and fi a eiaie fayorder any of derivatives has it if n ≥ 0 am ucin eaFnto,cmltl oooi funct monotonic completely Function, Beta Function, Gamma : . eety opeemntnct ffntosivligga involving functions of monotonicity complete Recently, Involving Γ( x = ) ( 33B15. : .Al-Jararha M. − Z 1) 0 ∞ Abstract n f e − ( 1 n f Γ t ) t ( ( x n -function z − ) ) ( 1 z t > x dt, ≥ f ) n , ( 0 ∗ z ) 0 = scle opeeymntnco an on monotonic completely called is ffntosta novn gamma involving that functions of s oeieulte novn gamma involving inequalities some e ,1,4 8 3 2 1 5 ] nthis In 3]. 35, 21, 12, 23, 28, 4, 15, 1, iyadMteaia nlssdue Analysis Mathematical and lity ml 1,2,1,8 4 0.Also, 20]. 34, 8, 18, 29, [19, ample 0 iisaiigi rbblt the- probability in arising lities , . tosivliggmafunction. gamma involving ctions 1 , tos edrvdsm in- some derived we ctions, 2 , 3 m ucin involving functions ome , · · · n if and , a,21163. dan, m and mma ions, Due to the relation between gamma and beta function, we derived some inequalities that are involving beta function. Commonly, beta function is defined by
1 B(x,y)= tx−1(1 − t)y−1dt, x and y > 0, Z0 and its relation with the gamma function is given by
Γ(x + y) B(x,y)= , x,y> 0. Γ(x)Γ(y)
The layout of the paper: In the first section, we prove our main results. In the second section, we apply some of our main results to prove the inequality
Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) ≥ 1, y ≥ x ≥ a + b > b ≥ a> 0. Γ(x − a + 1)Γ(x − b + 1) Γ(y + 1)Γ(y − a − b + 1)
The last section is devoted for the concluding remarks.
2 The Main Results
In this section, we present and prove our main results. First, we present some useful definitions and theorems.
Definition 2.1. A function f(z) is called completely monotonic on an interval I if it has deriva- tives of any order f (n)(z), n = 0, 1, 2, 3, · · ·, and if
(−1)nf (n)(z) ≥ 0 for all x ∈ I and all n ≥ 0. If the above inequality is strict for all x ∈ I and all n ≥ 0, then f(z) is called strictly completely monotonic.
Definition 2.2. A function f(z) is called logarithmically completely monotonic on an interval I if its logarithm has derivatives [ln f(z)](n) of orders n ≥ 1, and if
(−1)n[ln f(z)](n) ≥ 0 for all x ∈ I and all n ≥ 1. If the above inequality is strict for all x ∈ I and all n ≥ 1, then f(z) is called strictly logarithmically completely monotonic.
Theorem 2.1. [29]. Every (strict) logarithmically completely monotonic function is (strict) completely monotonic.
Now, we turn to prove our main results. Γ(z + 1)Γ(z − a − b + 1) Theorem 2.2. Let a, b ≥ 0. Define f(z)= , z > a + b − 1. Then f(z) Γ(z − a + 1)Γ(z − b + 1) is completely monotonic function.
2 Γ(z + 1)Γ(z − a − b + 1) proof. Let f(z)= . Then f(z) > 0, and Γ(z − a + 1)Γ(z − b + 1)
ln f(z) = ln Γ(z + 1) + ln Γ(z − a − b + 1) + ln Γ(z − a + 1) + ln Γ(z − b + 1). (2.1)
By differentiating Eq. (2.1), we get
d Γ′(z + 1) Γ′(z − a − b + 1) Γ′(z − a + 1) Γ′(z − b + 1) ln f(z) = + − − dz Γ(z + 1) Γ(z − a − b + 1) Γ(z − a + 1) Γ(z − b + 1) = Ψ(z + 1) + Ψ(z − a − b + 1) − Ψ(z − a + 1) − Ψ(z − b + 1), (2.2) where Ψ(z) is the Digamma function (the logarithmic differentiation of the Γ−function). By using the following integral representation of Digamma function
∞ (e−t − e−zt) Ψ(z)= −γ + dt, Re(z) > 0, 1 − e−t Z0 where γ = 0.577218... is Euler’s constant, we get d ln f(z) = Ψ(z + 1) + Ψ(z − a − b + 1) − Ψ(z − a + 1) − Ψ(z − b + 1) dz ∞ e−t(z−a−b+1) e−(a+b)t e−at e−bt dt = − −t − − + 1 0 1 − e Z ∞ e−t(z−a−b+1) = − 1 − e−at 1 − e−bt dt 1 − e−t Z0 ≤ 0, ∀z > a + b − 1. (2.3)
d Therefore, − dz ln f(z) ≥ 0, ∀z > a + b − 1. Inductively, we have
dn ∞ tn−1e−t(z−a−b+1) (−1)n ln f(z)= 1 − e−at 1 − e−bt dt ≥ 0, ∀z > a + b − 1. (2.4) dzn 1 − e−t Z0 Hence, by Theorem 2.1, f(z) is completely monotonic function.
Remark 2.1. Since f(z) is completely monotonic. Then it is decreasing function. Hence, by using the asymptotic relation Γ(z + a) lim z(b−a) = 1, (2.5) z→∞ Γ(z + b) we get , Γ(z + 1)Γ(z − a − b + 1) ≥ 1, z > a + b − 1. (2.6) Γ(z − a + 1)Γ(z − b + 1) For a reference of the above asymptotic relation, see eq. 13 in [16] (see also, [1, 33]).
n−1 n Γ (z + 1)Γ(z − a¯ + 1) Theorem 2.3. Let a1, a2, · · · , an ≥ 0, and let a¯ = i=1 ai. Then f(z)= n , i=1 Γ(z − ai + 1) z > a¯ − 1 is completely monotonic function. P Q
3 Γn−1(z + 1)Γ(z − a¯ + 1) proof. Let f(z)= n , z> a¯ − 1. Then i=1 Γ(z − ai + 1) Q n dnf(z) ∞ t(n−1)e−t(z+1) (−1)n = (n − 1) + eat¯ − eait dt dzn 1 − e−t 0 i ! Z X=1 ≥ 0, z> a¯ − 1.
at¯ n ait This is correct since the function ξ(t) := (n−1)+e − i=1 e , t ≥ 0, is nonnegative function. In fact, ξ(0) = 0 and ξ′(t) =ae ¯ at¯ − n a eait = n a eat¯ − n a eait = n a (eat¯ −eait) > 0, i=1 i i=1 Pi i=1 i i=1 i since the exponential function is increasing function. Hence, ξ′(0) = 0 and ξ′(t) > 0, ∀t > 0. P P P P Therefore, ξ(t) is nonnegative. Hence, by Theorem 2.1, f(z) is completely monotonic.
n Remark 2.2. Let a1, a2, · · · , an ≥ 0, and let a¯ = i=1 ai. Since
Γn−1(z + 1)Γ(z −Pa¯ + 1) f(z)= n , ∀z > a¯ − 1 i=1 Γ(z − ai + 1) is completely monotonic function, andQ hence, it is decreasing function. Therefore, by using the asymptotic relation (2.5), we get
Γn−1(z + 1)Γ(z − a¯ + 1) f(z)= n ≥ 1, ∀z > a¯ − 1. i=1 Γ(z − ai + 1)
Theorem 2.4. Let 0 ≤ a1 ≤ a2 ≤···≤Q an, 0 ≤ b1 ≤ b2 ≤···≤ bn, and m = max{a1, a2, · · · , an, k k b1, b2, · · · , bn}. Moreover, assume that i=1 bi ≤ i=1 ai, k = 1, 2, · · · ,n. Then n P Γ(zP− ai) f(z)= i=1 , z>m, n Γ(z − b ) Qi=1 i is completely monotonic function. Q n Γ(z − ai) proof. Let f(z)= i=1 , z>m. Then n Γ(z − b ) Qi=1 i Q n n dnf(z) ∞ t(n−1)e−tz (−1)n = eait − ebit dt dzn 1 − e−t 0 i i ! Z X=1 X=1 ≥ 0, z>m.
k k This inequality holds since i=1 bi ≤ i=1 ai, k = 1, 2, · · · ,n and the exponential function ex, x> 0 is convex and increasing function which implies that n eait ≥ n ebit (see page P P i=1 i=1 12 in [25]). Hence, by Theorem 2.1, f(z) is completely monotonic function. P P
Remark 2.3. Let 0 ≤ a1 ≤ a2 ≤···≤ an, 0 ≤ b1 ≤ b2 ≤···≤ bn, and m = max{a1, a2, · · · , an, k k n b1, b2, · · · , bn}. Moreover, assume that i=1 bi ≤ i=1 ai, k = 1, 2, · · · ,n − 1 and i=1 ai = n i bi. Then f(z) is decreasing function as a result of the above theorem. Hence, by the limit =1 n P P P Qi=1 Γ(z−ai) (2.5), we get n ≥ 1, ∀z >m. P Qi=1 Γ(z−bi)
4 n a Remark 2.4. Let 0 ≤ a ≤ a ··· ≤ a and let b =a ¯ := i=1 i , i = 1, 2, · · · , n. Then 1 2 n i n k k k k nP n i=1 bi = ka¯ = n i=1 ai < i=1 ai, k = 1, 2, · · · ,n−1, and i=1 ai = i=1 bi. Hence, f(z)= n n Γ(z − ai) Γ(z − ai) Pi=1 isP completelyP monotonic ∀z > a¯. Moreover, theP inequalityP i=1 ≥ 1 Γ(z − a¯)n Γ(z − a¯)n Q n Q Γ(z − a + 1) holds ∀z > a¯. By letting z = z + 1 in this inequality, we get that i=1 i ≥ 1, ∀z > Γ(z − a¯ + 1)n Q a¯ − 1. For a particular case if we let n = 2, a1 = x > 0, a2 = y >, and z = x + y, then we Γ(x + 1)Γ(y + 1) Γ(x)Γ(y) (x + y)2 get ≥ 1, ∀x,y ≥ 0. Particularly, we get ≥ , ∀x,y > 0. This x+y 2 x+y 2 4xy Γ( 2 + 1) Γ( 2 ) inequality has been proved in different method in [21]. Γ(z + a)Γ(z − a) Theorem 2.5. Let a ≥ 0 and define f(z) = , z > a. Then f(z) is completely Γ(z)2 monotonic function. Γ(x − a)Γ(x + a) proof. Let f(z)= , z > a. Then Γ(z)2
dnf(z) ∞ t(n−1)e−tz (−1)n = eat + e−at − 2 dt dzn 1 − e−t Z0 ∞ t(n−1)e−tz = eat − 1 1 − e−at dt 1 − e−t Z0 ≥ 0, z > a.
Hence, by Theorem 2.1, f(z) is completely monotonic function. Γ(z + a)Γ(z − a) Remark 2.5. Let a ≥ 0. Then f(z)= , z > a is decreasing function. By using Γ(z)2 Γ(z + a) Γ(z + a)Γ(z − a) the asymptotic relation lim z(b−a) = 1, we get ≥ 1, ∀z > a. This z→∞ Γ(z + b) Γ(z)2 inequality has been proved in [15] by using classical integral inequalities. Remark 2.6. Using the same argument above, we can show that Γ(z + a + 1)Γ(z − a + 1) f(z)= , z > a − 1 Γ(z + 1)2 is completely monotonic, and so it is decreasing function. Consequently, the inequality Γ(z + a + 1)Γ(z − a + 1) ≥ 1 (2.7) Γ(z + 1)2 holds ∀z > a − 1. Moreover, since f(z) is decreasing function, then f(z) ≤ f(0). Hence, we get Γ(z + a + 1)Γ(z − a + 1) 1 ≤ ≤ Γ(1 − a)Γ(1 + a), 0 ≤ a< 1, z ≥ 0. (2.8) Γ(z + 1)2 Moreover, by using the formula Γ(z +1) = zΓ(z), we get z2 Γ(z + a)Γ(z − a) z2 ≤ ≤ (Γ(1 − a)Γ(1 + a)), 0 ≤ a< 1, z > a. z2 − a2 Γ(z)2 z2 − a2
5 πa By using the formula Γ(1 − a)Γ(1 + a)= sin πa , 0
In this section, we prove the following inequality: Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) ≥ 1, 0 < a ≤ b < a + b ≤ x ≤ y. Γ(x − a + 1)Γ(x − b + 1) Γ(y + 1)Γ(y − a − b + 1) This inequality arises in the Probability in the problems dealing with the general binomial coef- α Γ(α+1) ficients β = Γ(β+1)Γ(α−β+1) . To prove this inequality, we present some useful remarks. Remark 3.1. Let a, b > 0 and let M ≥ a + b. Then ∀z such that 0 < a ≤ b < a + b ≤ z < e−t(z−a−b+1) M, the function δ(z) = (1 − e−at)(1 − e−bt) is nonnegative and is not identically 1 − e−t zero. Hence, as a consequence of the proof of Theorem 2.2, we have f ′(z) < 0. Therefore, Γ(z + 1)Γ(z − a − b + 1) f(z)= is strictly decreasing function on [a + b, M). Hence, f(a + b)= Γ(z − a + 1)Γ(z − b + 1) Γ(a + b + 1) Γ(z − a + 1)Γ(z − b + 1) > 1. Let g(z) = , z ≥ a + b > b ≥ a > 0. Then Γ(a + 1)Γ(b + 1) Γ(z + 1)Γ(z − a − b + 1) 1 Γ(z − a + 1)Γ(z − b + 1) g(z) = , and so g(z) = ≤ 1, ∀z ≥ a + b > b ≥ a > 0. Also, for f(z) Γ(z + 1)Γ(z − a − b + 1) z ∈ [a + b, M), we have g′(z) > 0. Remark 3.2. Let 0 < a ≤ b. Define h(x,y) = f(x)g(y), x,y ≥ a + b, where f(x) = Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) and g(y) = . Then h(x,y) is positive and Γ(x − a + 1)Γ(x − b + 1), Γ(y + 1)Γ(y − a − b + 1) continuous on the rectangular domain D = [a + b, ∞) × [a + b, ∞). Using the asymptotic relation (b−a) Γ(z + a) lim z = 1, we have limk(x,y)k→∞ h(x,y) → 1. z→∞ Γ(z + b)
6 In the following theorem, we prove that h(x,y) ≥ 1 on the triangular domain
Ω= {(x,y) | 0 < a ≤ b < a + b ≤ x ≤ y} ⊂ D. Γ(x + 1)Γ(x − a − b + 1) Γ(y − a + 1)Γ(y − b + 1) Theorem 3.1. Let h(x,y) = , (x,y) ∈ Ω. Γ(x − a + 1)Γ(x − b + 1) Γ(y + 1)Γ(y − a − b + 1) Then h(x,y) ≥ 1, ∀(x,y) ∈ Ω. proof. Since h(x,y) is positive, continuous, and limk(x,y)k→∞ h(x,y) → 1. Then, by using Remark 3.1 and Remark 3.2, there exists a sufficiently large constant M > 0, such that Γ(a + b + 1) Γ(M − a + 1)Γ(M − b + 1) h(a + b, M)= ≥ 1, Γ(a + 1)Γ(b + 1) Γ(M + 1)Γ(M − a − b + 1) Now, define the closed and bounded rectangular region:
R = {(x,y)| 0 < a ≤ b < a + b ≤ x ≤ y ≤ M} .
Since h(x,y) is continuous function on R. Then it takes its absolute values on the boundaries of R, or at the points in R where ∇h(x,y) = 0 (∇h(x,y) is the gradient of h(x,y)). Clearly, ∇h(x,y) = f ′(x)g(y)i + f(x)g′(y)j, where f(x) and g(y) are defined in Remark 3.2. Moreover, by Remark 3.1, we have f(x) > 0, g(y) > 0, g′(y) > 0, and f ′(x) < 0 in R. Hence, ∇h(x,y) 6= 0 in R. Hence, the absolute values of h(x,y) must occur at the boundaries of R. In fact, the boundaries of R are the line segments
1. l1 = {(x, M) | a + b ≤ x ≤ M} ,
2. l2 = {(a + b, y) | a + b ≤ x ≤ M} , and
3. l3 = {(x,x) | a + b ≤ x ≤ M} .
Obviously, h(x,y) = h(x,x) = 1 on the line segment l3. Moreover, on the line segment l1, we have Γ(x + 1)Γ(x − a − b + 1) Γ(M − a + 1)Γ(M − b + 1) h(x, M)= , a + b ≤ x ≤ M, Γ(x − a + 1)Γ(x − b + 1) Γ(M + 1)Γ(M − a − b + 1) which is decreasing function in x with a negative derivative. Similarly, on the line segment l1, we have Γ(a + b + 1) Γ(y − a + 1)Γ(y − b + 1) h(a + b, y)= , a + b ≤ y ≤ M, Γ(a + 1)Γ(b + 1) Γ(y + 1)Γ(y − a − b + 1) which is increasing function in y with a positive derivative. Hence, h(x,y) takes its absolut values at the points (a + b, a + b), (M, M), and (a + b, M). Clearly, h(a + b, a + b)= h(M, M) = 1. At the point (a + b, M), we have Γ(a + b + 1) Γ(M − a + 1)Γ(M − b + 1) h(a + b, M)= ≥ 1. Γ(a + 1)Γ(b + 1) Γ(M + 1)Γ(M − a − b + 1) This implies that h(x,y) ≥ 1, ∀(x,y) ∈ Ω. This completes the proof. Finally, we have the following remark:
7 Γ(x + 1)Γ(y − a + 1) Remark 3.3. Let ψ(x,y) = , y>x>a> 0. Let x ∈ (a, y). Then there Γ(y + 1)Γ(x − a + 1) exists αx > 0, such that x = y − αx. Hence, we define
Γ(y − αx + 1)Γ(y − a + 1) ξ(y)= ψ(y − αx,y)= , y>αx + a>a> 0. Γ(y + 1)Γ(y − αx − a + 1) Then, by using the above Remark 3.2, we have
Γ(y − αx + 1)Γ(y − a + 1) ψ(y − αx,y)= ≤ 1, ∀y > αx + a>a> 0. Γ(y + 1)Γ(y − αx − a + 1) Therefore, Γ(x + 1)Γ(y − a + 1) ψ(x,y)= ≤ 1, y>x>a> 0. Γ(y + 1)Γ(x − a + 1) Consequently, we also have Γ(x + 1)Γ(y − a − b + 1) ≤ 1, y > x > a + b > b ≥ a> 0. Γ(y + 1)Γ(x − a − b + 1)
4 Concluding Remarks
We have seen in Remark 2.1 that Γ(z + 1)Γ(z − a − b + 1) ≥ 1, ∀z ≥ a + b − 1. (4.1) Γ(z − a + 1)Γ(z − b + 1)
Let z = a + b, a, b ≥ 0 in (4.1), then we get
Γ(a + b + 1) ≥ 1, ∀a, b ≥ 0. (4.2) Γ(a + 1)Γ(b + 1)
By using the fact that Γ(z +1) = zΓ(z), we have
Γ(a + b + 1) (a + b) Γ(a + b) = ≥ 1, ∀a, b > 0. (4.3) Γ(a + 1)Γ(b + 1) ab Γ(a)Γ(b) Hence, Γ(a + b) ab ≥ , ∀a, b > 0. (4.4) Γ(a)Γ(b) (a + b) Γ(a + b) Since B(a, b)= , we get Γ(a)Γ(b)
(a + b) B(a, b) ≤ , ∀a, b > 0 (4.5) ab y y Moreover, by using the fact B(x,y +1) = B(x + 1,y)= B(x,y), we get x x + y b b 1 B(a, b +1) = B(a + 1, b)= B(a, b) ≤ , a, b > 0. a a + b a
8 1 1 Hence, B(a, b + 1) ≤ and B(a + 1, b) ≤ for a, b > 0. In [15], the authors proved the following a b inequality: 1 B(a, b) ≤ , 0 < a, b ≤ 1. (4.6) ab Clearly, if 0 < a, b ≤ 1 and a + b ≤ 1, the upper bound of ⌊(a, b) given in (4.5) is better than the one given in (4.6). Moreover, the inequality (4.5) is valid for a, b > 0 while the inequality (4.6) is restricted on a, b ∈ (0, 1]. Let a = b in Eq. (2.6) and Eq. (2.8). Then we get Γ(2a + 1) ≥ 1, a ≥ 0, (4.7) Γ(a + 1)2 and Γ(2a) a ≥ , a> 0, (4.8) Γ(a)2 2 respectively. Assume that 0 < a ≤ 1. then we have the following inequality (see eq. 2.8 in [21]:
Γ(a)2 2a − a2 ≥ , (4.9) Γ(2a) a2 By combining (4.8) with (4.9), we get
2a − a2 Γ(a)2 2 ≤ ≤ , 0 < a ≤ 1. a2 Γ(2a) a Assume that a, b > 0 and 0 < z ≤ 1. Then inequality (4.1) implies that Γ(z − a − b + 1) 1 ≥ ≥ 1, 0 < a ≤ b < a + b ≤ z ≤ 1. (4.10) Γ(z − a + 1)Γ(z − b + 1) Γ(z + 1) This is correct since Γ(x + 1) ≤ 1, ∀z ∈ [0, 1]. Moreover, we have Γ(z + 1)Γ(z − a − b + 1) Γ(a + b + 1) 1 ≤ ≤ , z ≥ a + b > b ≥ a> 0. (4.11) Γ(z − a + 1)Γ(z − b + 1) Γ(a + 1)Γ(b + 1) Hence, for z ≥ a + b > b ≥ a> 0, we have 1 Γ(z − a − b + 1) 1 Γ(a + b + 1) ≤ ≤ . (4.12) Γ(z + 1) Γ(z − a + 1)Γ(z − b + 1) Γ(z + 1) Γ(a + 1)Γ(b + 1) Therefore, Γ(z − a − b + 1) lim = 0. z→∞ Γ(z − a + 1)Γ(z − b + 1) Now, let a = b in (4.1), then we get Γ(z + 1)Γ(z − 2a + 1) ≥ 1, z> 2a − 1. Γ(z − a + 1)2 Equivalently, Γ(z)Γ(z − 2a) (z − a)2 a2 ≥ =1+ , z> 2a. (4.13) Γ(z − a)2 z(z − 2a) z(z − 2a)
9 Clearly, Γ(z)Γ(z − 2a) ≥ 1, z> 2a. Γ(z − a)2 Γ(z)Γ(z−2a) In fact, in [27] many lower bounds of Γ(z−a)2 were presented. Some of these lower bounds are given in the following inequalities:
Γ(z)Γ(z − 2a) a2 + z ≥ , z> 0 and z − 2a> 0, (4.14) Γ(z − a)2 z and Γ(z)Γ(z − 2a) a2(z − 2) ≥ 1+ , z> 2, z − 2a> 0, and a 6= 0, −1. (4.15) Γ(z − a)2 (z − a − 1)2
References
[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with formulas, Graphs, and Mathematical Tables, Dover, new Yourk, 1965.
[2] R. P. Agarwal, N. Elezovi´c, and J. Peˇcari´c, On Some Inequalities for Beta and Gamma Functions Via Some Classical Inequalities, Journal of Inequalities and Applications, 5(2005), 593–613.
[3] H. Alzer, Some Gamma Function Inequalities, Mathematics of Computations, 60(1993), No. 201,337–346.
[4] H. Alzer, On Some Inequalities for the Gamma and Psi functions, Mathematics of Compu- tation, 66(1997), No. 217, 373–389.
[5] H. Alzer, A Power Mean Inequality for the Gamma Function, Monatsh. Math., 131 (2000), 179-188.
[6] H. Alzer, Sharp Inequalities for the beta function, Indag. Mathem., 12 (2001), No. 1, 15–21.
1 [7] H. Alzer, Inequalities Involving Γ(x) and Γ( x ), Journal of Computational and Applied Math- ematics, 192 (2006), 460–480.
[8] H. Alzer and N. Batir, Monotonicity Properties of the Gamma Function, Applied Mathe- matical Letters, 20 (2007), 778–781.
[9] H. Alzer and C. Berg, Some Classes of Completely Monotonic Functions, Annales Academiae Scientiarum Fennicae, 27(2002), 445–460.
[10] H. Alzer and C. Berg, Some Classes of Completely Monotonic Functions, II, The Ramanujan Journal, 11(2006), No. 2, 225–248.
[11] H. Alzer and A. Z. Grinshpan, Inequalities for the Gamma and q−gamma Functions, Journal of Approximation Theory, 144 (2007), 67–83.
10 [12] N. Batir, Inequalities for the double Gamma Functions, J. Math. Anal. Appl., 351(2009), 182–185.
[13] J. Bustoz and M. E. H. Ismail, On Gamma Function Inequality, Mathematics of Computa- tion, 47(1986), No. 176, 659–667.
[14] C. P. Chen and F. Qi, Inequalities Related to the gamma Function, The Australian Journal of Mathematical Analysis and Applications, 1 (2004), No. 1, 1–18.
[15] S. S. Dragomir, R. P. Agarwal, and N. S. Barnett, Inequalities for Beta and Gamma Func- tions Via Some Classical and New Integral Inequalities, J. of Inequal. & Appl., 5 (2000), no. 2, pp. 103-165
[16] A. Erd´ely, W. Magnus, F. Oberhettinger, F. G. Tricomi, Higher Transcendental Functions, Volume I, McGraw Hill, New York, 1953.
[17] M. Garg, A.Sharma, and P. Manohar , Some Inequalities Involving Ratios and Products of Gamma Function, Le Matematiche, LXVIII (2003), Fasc. II, pp. 219-225.
[18] C. Giordano and A. Laforgia , Inequalities and Monotonicity Properties for the Gamma Function, Journal of Computational and Applied Mathematics, 133(2001), 387–396.
[19] B. N. Guo and F.Qi, Inequalities and Monotonicity for the Ratio of Gamma Functions, 7(2003), No. 2, 239–247.
[20] M. E. H. Ismail, L. Lorch, and M. E. Muldoon, Completely Monotonic Functions Associated with the Gamma Function and its q−Analogues, Journal of Mathematical Analysis and Applications, 116(1986), 1–9.
[21] P. Iv´ady, On the Beta Function Inequality, Journal of Mathematical Inequalities, 6(2012), No. 3, 333–341.
[22] A. Laforgia , Further Inequalities for the Gamma Function, Mathematics of Computation, 42(1984), 597–600.
[23] L. Maligranda, J. E. Pecaric, and L. E. Persson, Stolarsky’s Inequality with General Weights, Proceeding of the American Mathematical Society, 123, No. 7, 2113–2118.
[24] M. Marcer, Some New Inequalities for the Gamma, Beta and Zeta Function, Journal of Inequalities in Pure and Applied Mathematics, 6 (2006), No. 1, Artical 29.
[25] A. W. Marshall, I. Olkin, and B. C. Arnold, Inequalities: Theory of Majorization and Its Applications, Second Edition, Springer, New York, 2011.
[26] K. S. Miller and S. G. Smako, Completely Monotonic Functions, Int. Transf. and Spec. Funct., 12(2001), No. 4, 389–402.
[27] D. S. Mitrinovi´c and P. M. Vasi´c, Analytic Inequalities, Springer, New York, 1970.
11 [28] K. Nantomah and M. M. Iddrisu, Extensions of Some Inequalities for the Gamma function, International Mathematical Forum, 9(2014), No. 5, 199–206.
[29] F. Qi and C. P. Chen , A Complete Monotonicity Property of the Gamma Function, J. Math Anal. Appl., 296 (2004), pp. 603–607.
[30] J. S´andor, A Note on Certain Inequalities for the Gamma Function, Journal of Inequalities in Pure and Applied Mathematics, 6 (2005), No. 3, Article 61.
[31] A. S. Shabani, Some Inequalities of the Gamma Function, Journal of Inequalities in Pure and Applied Mathematics, 8 (2007), No. 2, Article 49.
[32] N. M. Temme, Special Functions An Introduction to the Classical Functions of Mathematical Physics, John Wiley & Sons, New York, 1996.
[33] F. Tricomi and A. Erd´elyi, The Asymptotic Expansion of a Ratio of Gamma Functions, Pacific Journal of Mathematics, 1(1951), No. 1, 133–142.
[34] H. Vogt, A monotonicity Property of the Γ−Function, Journal of Inequalities in Pure and Applied Mathematics, 3 (2002), No. 5, Article 73.
[35] S. Zhimin, D. Xiangkai, and Y. Li, On Some New Inequalities for the Gamma Function, Octagon Mathematical Magazine, 17(2009), No. 1, 14–18.
12