Journal of Inequalities in Pure and
Applied Mathematics
GEOMETRIC INEQUALITIES FOR A SIMPLEX
SHIGUO YANG
Department of Mathematics
Anhui Institute of Education
Hefei, 230061, P.R. China.
volume 6, issue 3, article 76,
2005.
Received 18 October, 2004;
accepted 11 March, 2005.
Communicated by: J. Sándor
EMail: sxx@ahieedu.net.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
096-04
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Abstract
In this paper, we study a problem of geometric inequalities for an n-simplex.
Some new geometric inequalities for a simplex are established. As special
cases, some known inequalities are deduced.
2000 Mathematics Subject Classification: 52A40, 51K16.
Key words: Simplex, Volume, Inradius, Circumradius, Inequality.
Geometric Inequalities for a
Simplex
The author would like to express his thanks to the editor for his kind help and invaluable suggestions in the formatting and writing of this paper.
Shiguo Yang
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Lemmas and Proofs of Theorems . . . . . . . . . . . . . . . . . . . . . . .
References
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3
4
7
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1.
Introduction
Let σn be an n-dimensional simplex in the n-dimensional Euclidean space E n ,
τ = {A0 , A1 , . . . , An } denote the vertex set of σn , V the volume of σn , R
and r the circumradius and inradius of σn , respectively. For i = 0, 1, . . . , n,
let ri be the radius of ith escribed sphere of σn , Fi the area of the ith face
fi = A0 · · · Ai−1 Ai+1 · · · An of σn . Let P be an arbitrary interior point of the
simplex σn , di the distance from the point P to the ith face fi of σn , hi the
altitude of σn from vertex Ai for i = 0, 1, . . . , n.
Let a0 , a1 and a2 denote the edge-lengths of triangle A0 A1 A2 (2-dimensional
simplex). An important inequality for a triangle was established by Janić (see
[1]) as follows:
(1.1)
a2
a2
a20
+ 1 + 2 ≥ 4.
r1 r 2 r2 r 0 r0 r1
Let P be an arbitrary interior point of the triangle A0 A1 A2 . Gerasimov (see [2])
obtained an inequality for the triangle A0 A1 A2 as follows:
(1.2)
d1 d2 d2 d0 d0 d1
1
+
+
≤ .
a1 a2 a2 a0 a0 a1
4
Geometric Inequalities for a
Simplex
Shiguo Yang
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2.
Main Results
We will extend inequalities (1.1) and (1.2) to an n-dimensional simplex. Our
main results are contained in the following theorem:
Theorem 2.1. For the n-dimensional simplex σn we have
(2.1)
n
X
i=0
n/(n−1)
2
Fi
(n − 1)n n3n /2(n−1)
≥ n
,
r0 · · · ri−1 ri+1 · · · rn
n (n + 1)(n−2)/2 (n!)n/(n−1)
Geometric Inequalities for a
Simplex
with equality iff the simplex σn is regular.
Shiguo Yang
By letting n = 2 in relation (2.1), inequality (1.1) is reobtained.
Theorem 2.2. Let P be an arbitrary interior point of the simplex σn , and let
θ ∈ (0, 1] be a real number. Then we have
(2.2)
n
X
i=0
(n!)2θ
(n + 1)(n−1)(1−θ) nn(3θ−1)
V n−2(n−1)θ ,
with equality iff the simplex σn is regular and the point P is the circumcenter of
σn .
If we take θ =
n
2(n−1)
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d0 · · · di−1 di+1 · · · dn
(F0 · · · Fi−1 Fi+1 · · · Fn )2θ−1
≤
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in inequality (2.2), we obtain the following corollary:
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Corollary 2.3. Let P be an arbitrary interior point of the simplex σn . Then we
have
(2.3)
n
X
i=0
d0 · · · di−1 di+1 · · · dn
(n!)n/(n−1)
≤
,
(F0 · · · Fi−1 Fi+1 · · · Fn )1/(n−1)
(n + 1)(n−2)/2 nn(n+2)/2(n−1)
with equality iff the simplex σn is regular and the point P is the circumcenter of
σn .
If n = 2 in inequality (2.3), then inequality (1.2) follows from inequality
(2.3).
By taking θ = 12 in inequality (2.2), we obtain a generalization of Gerber’s
inequality as follows:
Corollary 2.4. Let P be arbitrary interior point of the simplex σn . Then
(2.4)
n
X
Shiguo Yang
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d0 · · · di−1 di+1 · · · dn ≤
i=0
n!
(n +
1)(n−1)/2 nn/2
V,
with equality iff the simplex σn is regular.
Using inequality (2.4) and the arithmetic-geometric mean inequality we get
Gerber’s inequality [3] as follows:
(2.5)
Geometric Inequalities for a
Simplex
n
Y
i=0
di ≤
(n!)(n+1)/n
V (n+1)/n .
n(n+1)/2 (n + 1)1/2n
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Theorem 2.5. Let P be an arbitrary interior point of the simplex σn . Then we
have
(2.6)
n
X
i=0
1
r
≥ (n + 1)nn+1 · n+1 ,
d0 · · · di−1 di+1 · · · dn
R
with equality iff the simplex σn is regular and the point P is the circumcenter of
σn .
If the point P is the incenter I of the simplex σn , i.e. di = r(i = 0, 1, . . . , n),
then the following n-dimensional Euler inequality stated in [4] is obtained from
(2.6):
(2.7)
R ≥ nr.
Geometric Inequalities for a
Simplex
Shiguo Yang
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3.
Lemmas and Proofs of Theorems
To prove the theorems stated above, we need some lemmas as follows.
Let mi (i = 0, 1, . . . , n) be positive numbers, Vi0 i1 ···ik denote the k-dimensional
volume of the k-dimensional simplex Ai0 Ai1 · · · Aik for Ai0 , Ai1 , . . . , Aik ∈ τ .
Put
X
Mk =
mi0 mi1 · · · mik Vi20 i1 ···ik ,
(1 ≤ k ≤ n),
M0 =
0≤i0 <i1 <···<ik ≤n
n
X
Geometric Inequalities for a
Simplex
mi .
Shiguo Yang
i=0
Lemma 3.1. For positive numbers mi (i = 0, 1, . . . , n) and the n-dimensional
simplex σn , we have
(3.1)
Mkl
[(n − l)!(l!)3 ]k
≥
(n! · M0 )l−k Mlk ,
[(n − k)!(k!)3 ]l
Contents
(1 ≤ k < l ≤ n),
with equality iff the simplex σn is regular and m0 = m1 = · · · = mn .
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Lemma 3.2.
(3.2)
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n
Y
i=0
!
Fi
n
n2 −1
≥
1
(n + 1)1/2
3n
n
n!2
1
2(n−1)
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with equality iff the simplex σn is regular.
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For the proof of Lemmas 3.1 and 3.2, the reader is referred to [5] or [1].
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Lemma 3.3.
(3.3)
n
X
h0 · · · hi−1 hi+1 · · · hn
i=0
r0 · · · ri−1 ri+1 · · · rn
≥ (n + 1)(n1 )n ,
with equality iff the simplex σn is regular.
For the proof of Lemma 3.3, see [5].
Lemma 3.4.
Geometric Inequalities for a
Simplex
nn/2 (n + 1)(n+1)/2 n
V ≥
r ,
n!
(3.4)
with equality iff the simplex σn is regular.
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This is also known, see [5] or [1].
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Proof of Theorem 2.1. Without loss of generality, let F0 ≤ F1 ≤ · · · ≤ Fn . By
the known formula ([1])
(3.5)
nV
,
j=0 Fj − 2Fi
ri = Pn
(i = 0, 1, . . . , n),
j=1
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1
1
1
≤ Q
≤ ··· ≤ Q
.
Fj rj
Fj rj
Fj rj
j=0
j6=i
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it follows that r0 ≤ r1 ≤ · · · ≤ rn and
Q
Shiguo Yang
j=0
j6=n
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Using the Chebyshev inequality, we have
! n
n
n
n/(n−1)
X
Y
X F 1/(n−1)
Fi
i
Q
Q
(3.6)
=
Fi
r
Fj rj
j
i=0
i=0
i=0
j=0
j6=i
j=0
j6=i

n
Y
1
≥
n+1
!
Fi
i=0
n
X
i=0
!
1/(n−1)
Fi

n
X
1 

.
Q


F
r
j j
i=0
j=0
j6=i
Substituting Fj = nV
(j = 0, 1, . . . , n) into the right side of inequality (3.6)
hj
and using the arithmetic-geometric mean inequality we get
! n
!
n
n
n/(n−1)
X
Y
X 1/(n−1)
Fi
1
Q
(3.7)
Fi
Fi
≥
n
+
1
r
j
i=0
i=0
i=0
j=0
j6=i
≥
n
Q
n
1 X h0 · · · hi−1 hi+1 · · · hn
×
(nV )n i=0 r0 · · · ri−1 ri+1 · · · rn
2n2
(n −1)
Fi
i=0
(nV )n
n
X
h0 · · · hi−1 hi+1 · · · hn
i=0
r0 · · · ri−1 ri+1 · · · rn
Geometric Inequalities for a
Simplex
Shiguo Yang
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By inequalities (3.7), (3.2) and (3.3) we obtain relation (2.1). It is easy to see
that equality in (2.1) holds iff the simplex σn is regular. The proof of Theorem
2.1 is thus complete.
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Proof of Theorem 2.2. Taking k = n − 1, l = n in inequality (3.1), we can
write
!n
n
X
m0 · · · mi−1 mi+1 · · · mn Fi2
(3.8)
i=0
n
X
n3n
≥ 2
n!
i=0
!
mi
n
Y
!n−1
mi
V 2(n−1) .
i=0
λi Fi−2
By putting m0 · · · mi−1 mi+1 · · · mn =
(i = 0, 1, . . . , n) in equality (3.8),
we get
!n n
!
! n
!
n
n
2(n−1)
Y
Y
X F2
(nV
)
1X
i
λi
Fi2 ≥
(3.9)
λi
.
n i=0
(n − 1)!2
λi
i=0
i=0
i=0
We now prove that the following inequality (3.10) is valid for any number θ ∈
(0, 1]:
n
(3.10)
1X
λi
n i=0
≥
!n
n
Y
i=0
λi
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n
X
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Fi2θ
!n
Geometric Inequalities for a
Simplex
n
X
F 2θ
i
i=0
λi
!
(n + 1)2(n−1)θ (nV )2(n−1)θ
·
.
nn(1−θ)
(n − 1)!2θ
When θ = 1, inequalities (3.10) and (3.9) are the same, so inequality (3.10) is
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valid for θ = 1. For θ ∈ (0, 1), using inequality (3.9) we have
!n n
n
Y
1X
(3.11)
λi
Fi2θ
n i=0
i=0
!n n
#θ "
!n #1−θ
"
n
n
Y
X
1
1X
=
λi
Fi2 ·
λi
n i=0
n i=0
i=0
"
! n
!#θ "
!n #1−θ
n
n
X F2
1X
(nV )2(n−1) Y
i
≥
λi
·
λi
.
(n − 1)!2
λi
n i=0
i=0
i=0
Geometric Inequalities for a
Simplex
Shiguo Yang
By Maclaurin’s inequality ([1]) we have
n
1 X
λ0 · · · λi−1 λi+1 · · · λn
n + 1 i=0
! n1
n
1 X
λi ,
≤
n + 1 i=0
i.e.
n
1X
λi
n i=0
(3.12)
!n
(n + 1)n−1
≥
nn
From (3.11) and (3.12) we can write
n
(3.13)
1X
λi
n i=0
!n
n
Y
i=0
n
Y
i=0
!
λi
n
X
1
λ
i=0 i
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Fi2θ
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≥
n
X
!"
λi
i=0
1
n 2θ θ
X
F
i
λθi
i=0
#θ "
#1−θ
1
1−θ
n X
1
·
λ1−θ
i
i=0
θ
n−1 1−θ
(n + 1)
(nV )2(n−1)
×
.
nn
(n − 1)!2
By Hölder’s inequality ([1]) we have
"
(3.14)
1
n 2θ θ
X
F
i=0
i
λθi
#θ "
1
1−θ
n X
1
·
λ1−θ
i
i=0
#1−θ
≥
n
X
F 2θ
i
i=0
λi
Geometric Inequalities for a
Simplex
.
Shiguo Yang
Using (3.13) and (3.14) we get relation (3.9).
PnTaking λi = di Fi (i = 0, 1, . . . , n) in equality (3.9) and noting the fact that
i=0 di Fi = nV , we get inequality (2.2). It is easy to prove that equality in
(2.2) holds iff the simplex σn is regular and the point P is the circumcenter of
σn . The proof of Theorem 2.2 is thus complete.
Proof of Theorem 2.5. Inequality (3.9) can be written also as
n
(3.15)
n3n 2(n−1) X
V
λ0 · · · λi−1 λi+1 · · · λn Fi2 ≤
n!2
i=0
n
X
i=0
!n
λi
n
Y
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Fi2 .
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0
σn0
A00 A01
· · · A0n ,
Fi0
Let V denote the volume of the n-dimensional simplex
=
being the area of the ith face fi0 of σn0 . By Cauchy’s inequality and inequality
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(3.15), we have
(3.16)
n
n3n n−1 0 n−1 X
V
(V )
λ0 · · · λi−1 λi+1 · · · λn Fi Fi0
2
n!
i=0
# 12
"
n
3n
X
n
V 2(n−1)
λ0 · · · λi−1 λi+1 · · · λn Fi2
≤
2
n!
i=0
# 21
"
n
3n
X
n
2(n−1)
2
×
(V 0 )
λ0 · · · λi−1 λi+1 · · · λn (Fi0 )
2
n!
!n n i=0
! n
!
n
X
Y
Y
0
≤
λi
Fi
Fi .
i=0
i=0
i=0
V = (n + 1)
1/2
n!2
n3n
Shiguo Yang
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If we suppose that σn0 is a regular simplex with F00 = F10 = · · · = Fn0 = 1. then
0
Geometric Inequalities for a
Simplex
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2(n−1)
,
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so inequality (3.16) becomes
Close
(3.17)
(n + 1)
(n−1)/2 3n/2
n
n!
V n−1
n
X
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λ0 · · · λi−1 λi+1 · · · λn Fi
i=0
≤
n
X
i=0
!n
λi
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n
Y
i=0
Fi .
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By letting λ0 = λ1 = · · · = λn = 1 in inequality (3.17), we get
1
n3n/2(n−1)
≥ 1/(n−1)
V
n!
(n + 1)(n+1)/2(n−1)
(3.18)
×
n
X
1
F0 · · · Fi−1 Fi+1 · · · Fn
i=0
Now by Cauchy’s inequality we have
! n
n
X
X
d0 · · · di−1 di+1 · · · dn
i=0
i=0
1
d0 · · · di−1 di+1 · · · dn
1
! (n−1)
.
Geometric Inequalities for a
Simplex
!
≥ (n + 1)2 ,
Shiguo Yang
i.e.
n
X
(3.19)
i=0
1
≥ P
n
d0 · · · di−1 di+1 · · · dn
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(n + 1)2
.
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d0 · · · di−1 di+1 · · · dn
JJ
J
i=0
Using (3.19), (2.4) and (3.18), we get
(3.20)
n
X
i=0
≥
≥
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1
d0 · · · di−1 di+1 · · · dn
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(n + 1)(n+3)/2 nn/2 1
·
n!
V
Quit
(n + 1)
Close
(n2 +n−4)/2(n−1) n2 /2(n−1)
n
(n − 1)!n/(n−1)
n
X
i=0
1
! (n−1)
1
Qn
Fi
i=0
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1
(n−1)
Fi
.
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By inequality (3.20), formula
Pn
i=0
n
Y
Fi =
nV
r
and the known inequality ([1]):
2
(n + 1)(n −1)/2
2
Fi ≤ n+1 (n2 −3n−4)/2 Rn −1 ,
n! n
i=0
(3.21)
we get
(3.22)
n
X
i=0
1
d0 · · · di−1 di+1 · · · dn
≥ (n + 1)
(n−3)/2(n−1) (2n2 −n−2)/2(n−1)
n
Geometric Inequalities for a
Simplex
· n!
1/(n−1)
1
(n−1)
V
1
· n+1 .
r
R
Relations (3.22) and (3.4) imply inequality (2.6). It is easy to prove that equality
in (2.6) holds iff the simplex σn is regular and the point P is the circumcenter
of σn . The proof of Theorem 2.5 is thus complete.
Shiguo Yang
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References
[1] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND V. VOLENEC, Recent Advances
in Geometric Inequalities, Kluwer Acad. Publ., Dordrecht, Boston, London,
1989, 434–547.
[2] Ju.I. GERASIMOV, Problem 848, Mat. v Škole, 4 (1971), 86.
[3] L. GERBER, The orthecentric simplex as an extreme simplex, Pacific. J.
Math., 56 (1975), 97–111.
[4] M.S. KLAMKIN, Problem 85–26, SIAM. Rev., 27(4) (1985), 576.
Geometric Inequalities for a
Simplex
Shiguo Yang
[5] J.Zh. ZHANG AND L. YANG, A class of geometric inequalities concerning
the masspoint system, J. China Univ. Sci. Technol., 11(2) (1981), 1–8.
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