Journal of Inequalities in Pure and Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 155, 2006
A NOTE ON AN INEQUALITY FOR THE GAMMA FUNCTION
BAOGUO JIA
SCHOOL OFMATHEMATICS ANDSCIENTIFICCOMPUTER
ZHONGSHANUNIVERSITY
GUANGZHOU, CHINA, 510275.
Received 19 February, 2006; accepted 12 December, 2006 Communicated by A. Laforgia
ABSTRACT. By means of the convex properties of functionln Γ(x), we obtain a new proof of a generalization of a double inequality on the Euler gamma function, obtained by Jozsef Sándor.
Key words and phrases: Gamma function, Inequalities, Convex function.
2000 Mathematics Subject Classification. 33B15, 33C05.
The Euler gamma functionΓ(x)is defined forx >0by Γ(x) =
Z +∞
0
e−ttx−1dt.
Recently, by using a geometrical method, C. Alsina and M.S. Tomas [1] have proved the folowing double inequality:
Theorem 1. For allx∈[0,1]and all nonnegative integersn, one has 1
n! ≤ Γ(1 +x)n Γ(1 +nx) ≤1.
By using a representation theorem of the “digamma function” ΓΓ(x)0(x), J. Sándor [2] proved the following generalized result:
Theorem 2. For alla≥1and allx∈[0,1],one has 1
Γ(1 +a) ≤ Γ(1 +x)a Γ(1 +ax) ≤1.
In this paper, by means of the convex properties of functionln Γ(x), for0 < x < +∞, we will prove that
ISSN (electronic): 1443-5756 c
2006 Victoria University. All rights reserved.
This project was supported in part by the Foundations of the Natural Science Committee and Zhongshan University Advanced Research Centre, China.
047-06
2 B. JIA
Theorem 3. For alla≥1and allx >−1a, one has Γ(1 +x)a Γ(1 +ax) ≤1.
(i) For alla≥1and allx∈[0,1],one has 1
Γ(1 +a) ≤ Γ(1 +x)a Γ(1 +ax). (ii) For alla≥1and allx≥1, one has
1
Γ(1 +a) ≥ Γ(1 +x)a Γ(1 +ax). (iii) For alla∈[0,1]and allx∈[0,1], one has
1
Γ(1 +a) ≥ Γ(1 +x)a Γ(1 +ax). (iv) For alla∈[0,1]and allx≥1, one has
1
Γ(1 +a) ≤ Γ(1 +x)a Γ(1 +ax).
Our method is elementary. We only need the following simple lemma, see [3].
Lemma 4.
(a) Γ(x+ 1) =xΓ(x), for 0< x <+∞.
(b) Γ(n+ 1) =n!,for n = 1,2, . . .. (c) ln Γ(x)is convex on (0,+∞).
Proof of Theorem 3. Whena= 1, it is obvious.
Whena >1, by (c) of Lemma 4, we have Γ
u p + v
q
≤Γ(u)1pΓ(v)1q, wherep > 1, q >1,1p +1q = 1, u >0, v >0.
Letp=a, q = a−1a . Then Γ
1 au+
1− 1
av
≤Γ(u)a1Γ(v)1−1a, foru >0, v >0.
Letv = 1, u=ax+ 1. Note thatΓ(1) = 1, 1au+ (1− 1av) = x+ 1.
We obtain
Γ(x+ 1) ≥Γ(ax+ 1)a1, for x= u−1 a >−1
a.
Remark 5. Theorem 3 is a generalization of the right side inequality of Theorem 2.
Proof of Theorem 3.
(i) Let
f(x) = ln Γ(ax+ 1)−ln Γ(1 +a)−aln Γ(x+ 1).
SinceΓ(2) = 1, We havef(1) = 0.
f0(x) = a
Γ0(ax+ 1)
Γ(ax+ 1) − Γ0(x+ 1) Γ(x+ 1)
J. Inequal. Pure and Appl. Math., 7(5) Art. 155, 2006 http://jipam.vu.edu.au/
A NOTE ON ANINEQUALITY FOR THEGAMMAFUNCTION 3
Seth(t) = ln Γ(t).By (c) of the Lemma 4,ln Γ(x)is convex on(0,+∞). So(ln Γ(t))00 ≥ 0.
That is Γ0(t) Γ(t)
0
≥ 0. ThereforeΓ0(t) Γ(t)
is increasing. Becausea ≥ 1andx ∈ [0,1],one has ax+ 1 ≥x+ 1. So
Γ0(ax+ 1)
Γ(ax+ 1) ≥ Γ0(x+ 1) Γ(x+ 1)
Thusf0(x)≥0. In addition tof(1) = 0, we obtain thatf(x)≤0, fora ≥1andx∈[0,1].
So (i) is proved.
Note that
ax+ 1 ≥x+ 1, for a≥1 and x≥1;
ax+ 1≤x+ 1, for a ∈[0,1] and x∈[0,1];
ax+ 1 ≤x+ 1, for a∈[0,1] and x≥1.
So (ii), (iii), (iv) are obvious.
REFERENCES
[1] C. ALAINA AND M.S. TOMAS, A geometrical proof of a new inequality for the gamma func- tion, J. Ineq. Pure Appl. Math., 6(2) (2005), Art. 48. [ONLINE:http://jipam.vu.edu.au/
article.php?sid=517].
[2] J. SÁNDOR, A Note on certain inequalities for the Gamma function, J. Ineq. Pure Appl. Math., 6(3) (2005), Art. 61. [ONLINE:http://jipam.vu.edu.au/article.php?sid=534].
[3] W. RUDIN, Principle of Mathematical Analysis, New York: McGraw-Hill, 1976, p. 192–193.
J. Inequal. Pure and Appl. Math., 7(5) Art. 155, 2006 http://jipam.vu.edu.au/
