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Journal of Inequalities in Pure and
Applied Mathematics
A GEOMETRICAL PROOF OF A NEW INEQUALITY FOR THE
GAMMA FUNCTION
volume 6, issue 2, article 48,
2005.
C. ALSINA AND M.S. TOMÁS
Secció de Matemàtiques
ETSAB. Univ. Politècnica Catalunya
Diagonal 649, 08028 Barcelona, Spain.
Received 07 March, 2005;
accepted 19 March, 2005.
Communicated by: Th.M. Rassias
EMail: claudio.alsina@upc.es
EMail: maria.santos.tomas@upc.es
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
085-05
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Abstract
Using the inclusions between the unit balls for the p-norms, we obtain a new
inequality for the gamma function.
2000 Mathematics Subject Classification: 33B15, 33C05, 26D07.
Key words: Gamma function, Hypergeometric function, Inequalities.
The authors thank Prof. Hans Heinrich Kairies for his interesting remarks concerning
this paper.
A Geometrical Proof of a New
Inequality for the Gamma
Function
C. Alsina and M.S. Tomás
Since the gamma function
Z
Γ(x) =
Title Page
∞
−t x−1
e t
dt,
x>0
0
is one of the most important functions in Mathematics, there exists an extensive
literature on its inequalities (see [1], [2]).
Our aim here is to present and prove the inequalities
1
Γ(1 + x)n
≤
≤ 1 x ∈ [0, 1],
n!
Γ(1 + nx)
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n ∈ N.
As we will show the above inequalities follow immediately from a key geometrical argument. From now on for any r > 0, p, n ≥ 1 we will consider the
notation:
n,r
Dk·k
= {(x1 , . . . , xn ) ∈ Rn /k(x1 , . . . , xn )kp < r}
p
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J. Ineq. Pure and Appl. Math. 6(2) Art. 48, 2005
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for the n-ball of radius r for the p-norm k(x1 , . . . , xn )kp = (|x1 |p + · · · + |xn |p )1/p .
To this end, we need to prove the following:
Lemma 1. For all n in N, p ≥ 1 and r > 0 we have:
n
Γ 1 + p1
n,r
rn .
= 2n (1)
Volume Dk·k
p
n
Γ 1+ p
1,r
Proof. For n = 1, Dk·k
is the interval (−r, r), whose measure is 2r, i.e.,
p
1
p
A Geometrical Proof of a New
Inequality for the Gamma
Function
Γ 1+
r
2r = 2 1
Γ 1+ p
C. Alsina and M.S. Tomás
and (1) holds. By induction, let us assume that (1) holds for n−1. Then we note
that |x1 |p + · · · + |xn |p < rp is equivalent to |x1 |p + · · · + |xn−1 |p < rp − |xn |p
and by virtue of the induction hypothesis we have
Z
n,r
dx1 . . . dxn
Volume Dk·kp =
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n,r
Dk·k
p
Z
=2
0
r
n−1,(r p −|xn |p )1/p
p
Dk·k
dx1 . . . dxn−1
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J. Ineq. Pure and Appl. Math. 6(2) Art. 48, 2005
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n−1
Γ 1 + p1
n−1
(rp − xpn ) p dxn
=2
2n−1 0
Γ 1 + n−1
p
n−1
Z 1
Γ 1 + p1
n−1
n
n
r
(1 − z p ) p dz,
=2
0
Γ 1 + n−1
p
Z
r
where z = xn /r.
If we consider F (a, b, c, z) the first hypergeometric function (see [3]), then
Z
n−1
1 n
1 n−1
p p
,−
,1 + ,z
(1 − z ) dz = zF
p
p
p
and by well-known properties of the hypergeometric function we deduce:
n−1
Γ 1 + p1
1 n−1
1
n,r
n
n
r F
Volume Dk·kp = 2
,−
,1 + ,1
p
p
p
Γ 1 + n−1
p
n−1
Γ 1 + p1
Γ 1 + p1 Γ 1 +
rn
= 2n n−1
Γ 1+ p
Γ 1 + np
n
1
Γ 1+ p
rn .
= 2n Γ 1 + np
n−1
p
A Geometrical Proof of a New
Inequality for the Gamma
Function
C. Alsina and M.S. Tomás
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Therefore we have
Theorem 2. For all n ∈ N and x in (0, 1) we have
1
Γ(1 + x)n
≤
≤ 1.
n!
Γ(1 + nx)
Proof. For all n in N and p ≥ 1, from the inclusions
n,1
n,1
n,1
Dk·k
⊆ Dk·k
⊆ Dk·k
,
p
∞
1
A Geometrical Proof of a New
Inequality for the Gamma
Function
we deduce
Volume
n,1
n,1
n,1
≤
Volume
D
≤
Volume
D
Dk·k
k·kp
k·k∞ ,
1
C. Alsina and M.S. Tomás
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so by Lemma 1:
n
Γ 1 + p1
Γ(2)n
n
≤ 2n
2
≤2
n
Γ(n + 1)
Γ 1+
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n
p
and with 1/p = x, bearing in mind that Γ(2) = 1, Γ(n + 1) = n!,
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1
Γ(1 + x)n
≤
≤ 1.
n!
Γ(1 + nx)
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n
From this it follows immediately that the function Γ(1 + x) /Γ(1 + nx) is
strictly decreasing on (0, 1].
J. Ineq. Pure and Appl. Math. 6(2) Art. 48, 2005
http://jipam.vu.edu.au
References
[1] H. ALZER, On some inequalities for the gamma and psi functions, Math of
Comp., 66(217) (1997), 373–389.
[2] H. ALZER, Sharp bounds for the ratio of q−Gamma functions, Math
Nachr., 222 (2001), 5–14.
[3] E.W. WEISSTEIN, Hypergeometric function, [ONLINE: http://
mathworld.wolfram.com/HypergeometricFunction.html].
A Geometrical Proof of a New
Inequality for the Gamma
Function
C. Alsina and M.S. Tomás
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