Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 4, Article 106, 2004
THE GENERALIZED SINE LAW AND SOME INEQUALITIES FOR SIMPLICES
SHIGUO YANG
D EPARTMENT OF M ATHEMATICS
A NHUI I NSTITUTE OF E DUCATION
H EFEI 230061
P EOPLE ’ S R EPUBLIC OF C HINA .
sxx@ahieedu.net.cn
Received 31 July, 2003; accepted 23 April, 2004
Communicated by A. Sofo
A BSTRACT. The sines of k-dimensional vertex angles of an n-simplex is defined and the law of
sines for k-dimensional vertex angles of an n-simplex is established. Using the generalized sine
law for n-simplex, we obtain some inequalities for the sines of k-dimensional vertex angles of an
n-simplex. Besides, we obtain inequalities for volumes of n-simplices. As corollaries, the generalizations to several dimensions of the Neuberg-Pedoe inequality and P. Chiakuei inequality,
and an inequality for pedal simplex are given.
Key words and phrases: Simplex, k-dimensional vertex angle, Volume, Inequality.
2000 Mathematics Subject Classification. 52A40.
1. I NTRODUCTION
The law of sines for triangles in E 2 has natural analogues in higher dimensions. In 1978, F.
Eriksson [1] defined the n-dimensional sines of the n-dimensional corners of an n-simplex in
n-dimensional Euclidean space E n and obtained the law of sines for the n-dimensional corners
of an n-simplex. In this paper, the sines of k-dimensional vertex angles of an n -simplex will
be defined, and the law of sines for k-dimensional vertex angles of an n-simplex will be established. Using the generalized sine law for simplices and a known inequality in [2], we get some
inequalities for the sines of k-dimensional vertex angles of an n-simplex.
Recently, Yang Lu and Zhang Jingzhong [2, 3], Yang Shiguo [4], Leng Gangson [5, 6] and
D. Veljan [7] and V. Volenec et al. [9] have obtained some important inequalities for volumes of
n-simplices. In this paper, some interesting new inequalities for volumes of n-simplices will be
established. As corollaries, we will obtain an inequality for pedal simplex and a generalization
to several dimensions of the Neuberg-Pedoe inequality, which differs from the results in [4], [5]
and [6].
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
106-03
2
S HIGUO YANG
2. T HE G ENERALIZED S INE L AW
FOR
S IMPLICES
Let Ai (i = 1, 2, . . . , n + 1) be the vertices of an n-dimensional simplex Ωn in the ndimensional Euclidean space E n , V the volume of the simplex Ωn and Fi (n − 1)-dimensional
content of Ωn . F. Eriksson defined the n-dimensional sines of the n-dimensional corners αi of
the n-simplex Ωn and obtained the law of sines for n-simplices as follows [1]
n+1
Q
(n − 1)!
Fj
Fi
j=1
(2.1)
=
(i = 1, 2, . . . , n + 1).
n sin α
(nV )n−1
i
In this paper, we will define the sines of the k-dimensional vertex angles of an n-dimensional
simplex and establish the law of sines for the k-dimensional vertex angles of an n-simplex.
Let Vi1 i2 ···ik be the (k − 1)-dimensional content of the (k − 1)-dimensional face Ai1 Ai2 · · · Aik
((k − 1)-simplex) of the simplex Ωn for k ∈ {2, 3, . . . , n + 1} and i1 , i2 , . . . , ik ∈ {1, 2, . . . , n +
−→
1}, O and R denote the circumcenter and circumradius of the simplex Ωn respectively. OAi =
Rui (i = 1, 2, . . . , n + 1), ui is the unit vector. The sines of the k-dimensional vertex angles of
the simplex Ωn are defined as follows.
Definition 2.1. Let αij denote the angle formed by the vectors ui and uj . The sine of a kdimensional vertex angle ϕi1 i2 ···ik of the simplex Ωn corresponding the (k − 1)-dimensional
face Ai1 Ai2 · · · Aik is defined as
1
sin ϕi1 i2 ···ik = (−Di1 i2 ···ik ) 2 ,
(2.2)
where
(2.3)
Di1 i2 ···ik
= 0 1
···
1
..
α
.
− 12 sin2 il2im
1
1 (l, m = 1, 2, . . . , k).
We will prove that
(2.4)
1
0 < (−Di1 i2 ···ik ) 2 ≤ 1.
If n = 2, the sine of the 2-dimensional vertex angle ϕij of the triangle A1 A2 A3 is the sine of
the angle formed by two edges Ak Ai and Ak Aj .
With the notation introduced above, we establish the law of sines for the k -dimensional
vertex angles of an n-simplex as follows.
Theorem 2.1. For an n-dimensional simplex Ωn in E n and k ∈ {2, 3, . . . , n + 1}, we have
(2.5)
Vi1 i2 ···ik
(2R)k−1
=
sin ϕi1 i2 ···ik
(k − 1)!
(1 ≤ i1 < i2 < · · · < ik ≤ n + 1).
Put ϕ12···i−1,i+1,...,n+1 = θi , V12···i−1,i+1,...,n+1 = Fi (i = 1, 2, . . . , n + 1), by Theorem 2.1 we
obtain the law of sines for the n-dimensional vertex angles of n-simplices as follows.
Corollary 2.2.
(2.6)
F1
F2
Fn+1
(2R)n−1
=
= ··· =
=
.
sin θ1
sin θ2
sin θn+1
(n − 1)!
If we take n = 2 in Theorem 2.1 or Corollary 2.2, we obtain the law of sines for a triangle
A1 A2 A3 in the form
a1
a2
a3
(2.7)
=
=
= 2R.
sin A1
sin A2
sin A3
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
T HE G ENERALIZED S INE L AW
AND
S OME I NEQUALITIES FOR S IMPLICES
3
Proof of Theorem 2.1. Let aij = |Ai Aj | (i, j = 1, 2, . . . , n + 1), then
αij
aij = 2R sin
,
2
(2.8)
sin2 ϕi1 i2 ···ik
0 1
···
1 1
= −Di1 i2 ···ik
..
1
2
.
− 8R2 ail im
1
0 1 ··· 1 1
k
2 −(k−1) = (−1) (8R )
· ..
a2il im
.
1
= − (l, m = 1, 2, . . . , k).
By the formula for the volume of a simplex, we have
(2.9)
sin2 ϕi1 i2 ···ik = −Di1 i2 ···ik
= (−1)k (8R2 )−(k−1) (−1)k 2k−1 (k − 1)!2 Vi21 i2 ···ik
=
(k − 1)!2 2
V
.
(2R)2(k−1) i1 i2 ···ik
From this equality (2.5) follows.
Now we prove that inequality (2.4) holds. When k ≥ 2, we have k ≤ 2k−1 . Using the
Voljan-Korchmaros inequality [3], we have
! k2
12
Y
1
k
(2.10)
Vi1 i2 ···ik ≤
ail im
.
(k − 1)! 2k−1
1≤l<m≤k
Equality holds if and only if the simplex Ai1 Ai2 · · · Aik is regular.
Combining inequality (2.10) with equality (2.5), we get
(2.11)
Vi1 i2 ···ik
(2R)k−1
≤
·
(k − 1)!
(2R)k−1
≤
·
(k − 1)!
(2R)k−1
≤
.
(k − 1)!
k
12
2k−1
k
αi i
sin l m
2
1≤l<m≤k
! k2
Y
12
2k−1
Using equality (2.5) and inequality (2.11), we get
1
0 < (−Di1 i2 ···ik ) 2 = sin ϕi1 i2 ···ik =
(k − 1)!
Vi i ···i ≤ 1.
(2R)k−1 1 2 k
For the sines of the k-dimensional vertex angles of an n-simplex, we obtain an inequality as
follows.
Theorem 2.3. Let ϕi1 i2 ···ik (1 ≤ i1 < i2 < · · · < ik ≤ n + 1) denote the k-dimensional vertex
angles of an n-simplex Ωn in E n , and λi > 0 (i = 1, 2, . . . , n + 1) be arbitrary real numbers,
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
4
S HIGUO YANG
then we have
X
(2.12)
2
λi1 λi2 · · · λik sin ϕi1 i2 ···ik
1≤i1 <i2 <···<ik ≤n+1
Pn+1 k
n! ·
i=1 λi
≤
.
(n − k + 1)!(k − 1)!(4n)k−1
Equality holds if λ1 = λ2 = · · · = λn+1 and the simplex Ωn is regular.
By taking λ1 = λ2 = · · · = λn+1 in the inequality (2.12), we get:
Corollary 2.4.
X
(2.13)
sin2 ϕi1 i2 ···ik ≤
1≤i1 <i2 <···<ik ≤n+1
n! · (n + 1)k
.
(n − k + 1)!(k − 1)!(4n)k−1
Equality holds if the simplex Ωn is regular.
To prove Theorem 2.3, we need a lemma as follows.
Lemma 2.5. Let Ωn be an n-simplex in E n , xi > 0 (i = 1, 2, . . . , n + 1) be real numbers, Vi1 i2 ···is+1 be the s-dimensional volume of the s-dimensional simplex Ai1 Ai2 · · · Ais+1 for
i1 , i2 , . . . , is+1 ∈ {1, 2, . . . , n + 1}. Put
X
Ms =
xi1 xi2 · · · xis+1 Vi21 i2 ···is+1 , M0 =
1≤i1 <i2 <···<ik ≤n+1
n+1
X
xi ,
i=1
then we have
(2.14)
Msl ≥
[(n − l)!(l!)3 ]s
(n! · M0 )l−s Mls (1 ≤ s < l ≤ n).
[(n − s)!(s!)3 ]l
Equality holds if and only if the intertial ellipsoid of the points A1 , A2 , . . . , An+1 with masses
x1 , x2 , . . . , xn+1 is a sphere.
For the proof of Lemma 2.5. the reader is referred to [2] or [9].
Proof of Theorem 2.3. By putting s = 1, l = k − 1 and xi = λi (i = 1, 2, . . . , n + 1) in the
inequality (2.14), we have
!k−1
!k−2
n+1
3
X
X
(n − k + 1)! · (k − 1)!
(2.15)
λi λj a2ij
≥
n! ·
λi
[(n − 1)!]k−1
i=1
1≤i<j≤n+1
X
×
λi1 λi2 · · · λik Vi21 i2 ···ik .
1≤i1 <i2 <···<ik ≤n+1
By Theorem 2.1, we have
Vi1 i2 ···ik =
(2.16)
(2R)k−1
sin ϕi1 i2 ···ik .
(k − 1)!
Using the known inequality [3]
(2.17)
X
λi λj a2ij ≤
n+1
X
!2
λi
R2 ,
i=1
1≤i<j≤n+1
Pn+1
with equality if and only if the point P = i=1 λi Ai is the circumcenter of simplex Ωn .
Combining (2.15) with (2.16) and (2.17), we obtain inequality (2.12). It is easy to see that
equality holds in (2.12) if λ1 = λ2 = · · · = λn+1 and simplex Ωn is regular.
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
T HE G ENERALIZED S INE L AW
AND
S OME I NEQUALITIES FOR S IMPLICES
5
3. S OME I NEQUALITIES FOR VOLUMES OF S IMPLICES
Let P be an arbitrary point inside the simplex Ωn and Bi the orthogonal projection of the point
P on the (n−1)-dimensional plane σi containing (n−1)-simplex fi = A1 · · · Ai−1 Ai+1 · · · An+1 .
Simplex Ωn = B1 B2 · · · Bn+1 is called the pedal simplex of the point P with respect to the simplex Ωn . Let ri = |P Bi | (i = 1, 2, . . . , n + 1), V be the volume of the pedal simplex Ωn , V (i)
and V (i) denote the volumes of two n-dimensional simplices Ωn (i) = A1 · · · Ai−1 P Ai+1 · · · An+1
and Ωn (i) = B1 · · · Bi−1 P Bi+1 · · · Bn+1 , respectively. Then we have an inequality for volumes
of just defined n-simplices as follows.
Theorem 3.1. Let P be an arbitrary point inside n-dimensional simplex Ωn and λi (i =
1, 2, . . . , n + 1) positive real numbers, then we have
" n
#n
n+1
X
1 X
λi V (i) V 1−n ,
(3.1)
λ1 · · · λi−1 λi+1 · · · λn+1 V (i) ≤ n
n
i=1
i=1
with equality if the simplex Ωn is regular, P is the circumcenter of Ωn and λ1 = λ2 = · · · =
λn+1 .
Now we state some applications of Theorem 3.1.
If taking λ1 = λ2 = · · · = λn+1 in inequality (3.1), we have
" n+1
#n
n+1
X
1 X
(3.2)
V (i) ≤ n
V (i) · V 1−n .
n
i=1
i=1
Since the point P is in the interior of the simplex Ωn , then
(3.3)
n+1
X
V (i) = V ,
n+1
X
V (i) = V.
i=1
i=1
Using (3.2) and (3.3) we obtain an inequality for the volume of the pedal simplex Ωn of the
point P with respect to the simplex Ωn as follows.
Corollary 3.2. Let P be an arbitrary point inside the n-simplex Ωn , then we have
1
(3.4)
V ≤ n V,
n
with equality if simplex Ωn is regular and P is the circumcenter of Ωn .
Corollary 3.3. Let P be an arbitrary point inside the n-simplex Ωn , then we have
(3.5)
n+1
X
V (i) · V (i) ≤
i=1
1
V 2,
(n + 1)nn
with equality if the simplex Ωn is regular and P is the circumcenter of Ωn .
Proof. Let λi = [V (i)]−1 (i = 1, 2, . . . , n + 1) in inequality (3.1); we get
n
n+1
n+1
X
Y
n+1
1−n
(3.6)
V (i) · V (i) ≤
V
V (j).
n
i=1
j=1
Using the arithmetic-geometric mean inequality and equality (3.3), we have
"
#n+1
n
n+1
n+1
X
X
n+1
1
1
V (i) · V (i) ≤
V 1−n
V (j)
=
V 2.
n
n
n + 1 j=1
(n + 1)n
i=1
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
6
S HIGUO YANG
It is easy to see that equality in (3.5) holds if the simplex Ωn is regular and the point P is the
circumcenter of Ωn .
−−→
Proof of Theorem 3.1. Let hi be the altitude of simplex Ωn from vertex Ai , P B i = ri ei , where
ei is the unit outer normal vector of the i-th side face fi = A1 · · · Ai−1 Ai+1 · · · An+1 of the
simplex Ωn , and n sin αk be the n-dimensional sine of the k-th corner αk of the simplex Ωn .
Wang and Yang [8] proved that
n
(3.7)
1
sin αn = [det(ei · ej )ij6=k ] 2
(k = 1, 2, . . . , n + 1).
By the formula for the volume of an n-simplex and (3.7), we have


n+1
1
1
1 Y  n
(3.8)
V (i) =
[det(rl rk el · ek )l,k6=i ] 2 =
rj  · sin αi .

n!
n!
j=1
j6=i
Using (3.8), (2.1) and nV = hi Fi , we get that
n+1
X
λ1 · · · λi−1 λi+1 · · · λn+1 V (i)
i=1
=
1
n!
n+1
X



n+1
Y
i=1
1
n!

λj rj  · n sin αi
j=1
j6=i

=

n+1 

X
n+1
Y
i=1
j=1
j6=i




= [(n!)2 · V ]−1




λj rj  (nV )n−1 (n − 1)! ·
n+1
X
i=1
n+1
Y
j=1
j6=i



n+1
Y
−1 



Fj 




λj rj hj  ,
j=1
j6=i
i.e.
(3.9)
(n!)2 V
n+1
X
λ1 · · · λi−1 λi+1 · · · λn+1 V (i) =
i=1
n+1
X
i=1

n+1
Y




λj rj hj  .
j=1
j6=i
Taking s = n − 1, l = n in inequality (2.14), we get


 n
! n+1 !n−1
n+1
n+1
n+1
Y
n3n X
X Y  2 
(3.10)
xj  Fi  ≥
x
xi
V 2(n−1) .


i
2
(n!)
i=1
i=1
i=1
j=1
j6=i
Let xi = (λi ri hi )−1 (i = 1, 2, . . . , n + 1) in inequality (3.10). Then we have


 
!n
n+1
n+1
n+1
X
n3n X Y
 
(3.11)
λi ri hi Fi2
≥
λj rj hj  Fi2  · V 2(n−1) .


2
(n!) i=1
i=1
j=1
j6=i
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
T HE G ENERALIZED S INE L AW
AND
S OME I NEQUALITIES FOR S IMPLICES
7
Using inequality (3.11) and ri Fi = nV (i), hi Fi = nV , we get


#n
" n+1
n+1
n+1
n
X
X Y
n

2(n−1)
λi V (i) ≥
(3.12)
Vn
V
λj rj hj  .

2
(n!)
i=1
i=1
j=1
j6=i
Substituting equality (3.9) into inequality (3.12) we get inequality (3.1). It is easy to prove that
equality in (3.1) holds if simplex Ωn is regular, P is the circumcenter of Ωn and λ1 = λ2 =
· · · = λn+1 . Theorem 3.1 is proved.
Finally, we shall establish some inequalities for volumes of two n-simplices. As corollaries, the generalizations to several dimensions of the Neuberg-Pedoe inequality and P.Chiakui
inequality will be given.
Let ai (i = 1, 2, 3) denote the sides of the triangle A1 A2 A3 with area ∆, and a0i (i = 1, 2, 3)
denote the sides of the triangle A01 A02 A03 with area ∆0 , then
!
3
3
X
X
2
2
(3.13)
a2i
a0j − 2 (a0i ) ≥ 16∆∆0 ,
i=1
j=1
with equality if and only if ∆A1 A2 A3 is similar to ∆A01 A02 A03 .
Inequality (3.13) is the well-known Neuberg-Pedoe inequality.
In 1984, P. Chiakui [9] proved the following sharpening of the Neuberg-Pedoe inequality:
!
3
3
X
X
2
2
(3.14)
a2i
a0j − 2 (a0i )
i=1
j=1
≥8
(a01 )2 + (a02 )2 + (a03 )2 2
a21 + a22 + a23
0 2
∆
+
2
2
2 (∆ )
2
2
2
0
0
0
a1 + a2 + a3
(a1 ) + (a2 ) + (a3 )
!
,
with equality if and only if ∆A1 A2 A3 is similar to ∆A01 A02 A03 .
Recently, Leng Gangson [5] has extended inequality (3.14) to the edge lengths and volumes
of two n-simplices. In this paper, we shall extend inequality (3.14) to the volumes of two nsimplices and the contents of their side faces. As a corollary, we get a generalization to several
dimensions of the Neuberg-Pedoe inequality. Let Ai (i = 1, 2, . . . , n + 1) be the vertices of
n-simplex Ωn in E n , V the volume of the simplex Ωn and Fi (n − 1)- dimensional content of
the (n − 1)-dimensional face fi = A1 · · · Ai−1 Ai+1 · · · An+1 of Ωn . For two n-simplices Ωn and
Ω0n and real numbers α, β ∈ (0, 1], we put
1
n+1
n+1
X
X
n3
n+1 n
α
0 β
(3.15)
σn (α) =
Fi , σn (β) =
(Fi ) , bn =
.
n+1
n!2
i=1
i=1
We obtain an inequality for volumes of two n-simplices as follows.
Theorem 3.4. For any two n-dimensional simplices Ωn and Ω0n and two arbitrary real numbers
α, β ∈ (0, 1], we have
!
n+1
n+1
X
X
β
β
(3.16)
Fiα
Fj0 − 2 (Fi0 )
i=1
j=1
(n − 1)2 α σn (β) 2(n−1)α/n
β σn (α)
0 2(n−1)β/n
≥
bn
V
+ bn
(V )
.
2
σn (α)
σn (β)
Equality holds if and only if simplices Ωn and Ω0n are regular.
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
8
S HIGUO YANG
Using inequality (3.16) and the arithmetic-geometric mean inequality, we get the following
corollary.
Corollary 3.5. For any two n-dimensional simplices Ωn and Ω0n and two arbitrary real numbers
α, β ∈ (0, 1], we have
!
n+1
n+1
X
X
β
β
β
(3.17)
Fiα
Fj0 − 2 (Fi0 )
≥ b(α+β)/2
(n2 − 1)(V α (V 0 ) )(n−1)/n .
n
i=1
j=1
Equality holds if and only if simplices Ωn and Ω0n are regular.
If we let α = β in Corollary 3.5, we get Leng Gangson’s inequality [6] as follows. For any
θ ∈ (0, 1] we have
!
n+1
n+1
X
X
θ
θ
(3.18)
Fiθ
Fj0 − 2 (Fi0 ) ≥ bθn (n2 − 1)(V V 0 )(n−1)θ/n ,
i=1
j=1
with equality if and only if simplices Ωn and Ω0n are regular.
To prove Theorem 3.4, we need some lemmas as follows.
Lemma 3.6. For an n-simplex Ωn and arbitrary number α ∈ (0, 1], we have
Qn+1 2α
3n α
1
n
i=1 Fi
V 2(n−1)α ,
(3.19)
Pn+1 2α ≥
(n−1)α+1
2
(n
+
1)
(n!)
i=1 Fi
with equality if and only if simplex Ωn is regular.
Proof. If taking l = n, s = n − 1 and xi = Fi2 (i = 1, 2, . . . , n + 1) in inequality (2.14), we
get an inequality as follows
n+1
n+1
X
(n + 1)n (n!)2 Y 2
2(n−1)
Fi ≥ V
Fi2 ,
n3n
i=1
i=1
or
(3.20)
n+1
(n + 1)nα (n!)2α Y 2α
Fi ≥ V 2(n−1)α
n3nα
i=1
n+1
X
!α
Fi2
.
i=1
It is easy to prove that equality in (3.20) holds if and only if simplex Ωn is regular. From inequality (3.20) we know that inequality (3.19) holds for α = 1. For α ∈ (0, 1), using inequality
(3.20) and the well-known inequality
! α1
n+1
n+1
X
X
1
(3.21)
Fi2 ≥ (n + 1)
Fi2α
,
n
+
1
i=1
i=1
we get inequality (3.19). It is easy to see that equality in (3.19) holds if and only if the simplex
Ωn is regular.
Lemma 3.7. For an n-simplex Ωn (n ≥ 3) and an arbitrary number α ∈ (0, 1], we have
!2
n+1
n+1
X
X
α
(3.22)
Fi
−2
Fi2α ≥ bαn (n2 − 1)V 2(n−1)α/n ,
i=1
i=1
with equality if and only if the simplex Ωn is regular.
For the proof of Lemma 3.7, the reader is referred to [6].
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
http://jipam.vu.edu.au/
T HE G ENERALIZED S INE L AW
AND
S OME I NEQUALITIES FOR S IMPLICES
9
Lemma 3.8. Let ai (i = 1, 2, 3) and ∆ denote the sides and the area of the triangle (A1 A2 A3 ),
respectively. For arbitrary number α ∈ (0, 1], denote by ∆α the area of the triangle (A1 A2 A3 )α
with sides aαi (i = 1, 2, 3), then the following inequality holds
α
3 16 2
2
(3.23)
∆α ≥
∆
.
16 3
For α 6= 1, equality holds if and only if a1 = a2 = a3 .
For the proof of Lemma 3.8, the reader is referred to [9].
Lemma 3.9. Let numbers xi > 0, yi > 0 (i = 1, 2, . . . , n + 1), σn =
n+1
P
xi , σn0 =
i=1
(3.24)
σn σn0 − 2
n+1
X
"
xi yi ≥
i=1
σn0
1
2 σn
σn2 − 2
n+1
X
!
x2i
+
i=1
σn
σn0
2
(σn0 ) − 2
n+1
P
yi , then
i=1
n+1
X
!#
yi2
,
i=1
with equality if and only if
y1
y2
yn+1
=
= ··· =
.
x1
x2
xn+1
Proof. Inequality (3.24) is
n+1
n+1
n+1
X
σn0 X 2 σn X 2
xi + 0
yi ≥ 2
xi yi .
σn i=1
σn i=1
i=1
(3.25)
Now we prove that inequality (3.25) holds. Using the arithmetic-geometric mean inequality, we
have
σn0 2 σn 2
x + y ≥ 2xi yi
(i = 1, 2, . . . , n + 1).
σn i σn0 i
Adding up those n + 1 inequalities, we get inequality (3.25). Equality in (3.25) holds if and
0
only if σσnn x2i = σσn0 yi2 (i = 1, 2, . . . , n + 1), i.e.
n
y1
y2
yn+1
σ0
=
= ··· =
= n.
x1
x2
xn+1
σn
Proof of Theorem 3.4. For n = 2, consider two triangles (A1 A2 A3 )α and (A01 A02 A03 )β . Using
inequality (3.14) and Lemma 3.8, we have
!
3
3
X
X
1 α σ20 (β) α
β
β
β σ2 (α)
α
0
0
0 β
(3.26)
ai
aj − 2 (ai )
≥
b
∆ + b2 0
(∆ ) .
2 2 σ2 (α)
σ2 (β)
i=1
j=1
Equality in (3.26) holds if and only if a1 = a2 = a3 and a01 = a02 = a03 . Hence, inequality (3.16)
holds for n = 2.
For n ≥ 3, taking xi = Fiα , yi = (Fi0 )β (i = 1, 2, . . . , n + 1) in inequality (3.24), we get
!
n+1
n+1
X
X
β
β
(3.27)
Fiα
Fj0 − 2 (Fi0 )
i=1
j=1
=
n+1
X
!
Fiα
i=1
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
n+1
X
i=1
!
β
(Fi0 )
−2
n+1
X
β
Fiα (Fi0 )
i=1
http://jipam.vu.edu.au/
10
S HIGUO YANG



!2
n+1
n+1

0
X
X
1 σn (β) 
≥
Fiα − 2
Fi2α 
2  σn (α)
i=1
i=1


!2
n+1
n+1

X
X
σn (α) 
2β 
0 β
.
+ 0
(Fi )
−2
Fi

σn (β)
i=1
i=1
Using inequality (3.27) and Lemma 3.7, we get
!
n+1
n+1
X
X
n2 − 1 α σn0 (β) 2(n−1)α/n
β
α
0
β σn (α) 2(n−1)β/n
Fi
Fj
≥
bn
V
+ bn 0
V
.
2
σ
(α)
σ
(β)
n
n
i=1
i=1
Hence, inequality (3.16) is true for n ≥ 3. For n ≥ 3, it is easy to see that equality in (3.16)
holds if and only if two simplices Ωn and σn0 are regular. Theorem 3.4 is proved.
R EFERENCES
[1] F. ERIKSSON, The law of sines for tetrahedra and n-simplics, Geometriae Dedicata, 7 (1978),
71–80.
[2] JINGZHONG ZHANG AND LU YANG, A class of geometric inequalities concerning the masspoints system, J. China. Univ. Sci. Techol., 11(2) (1981), 1–8.
[3] JINGZHONG ZHANG AND LU YANG, A generalization to several dimensions of the NeubergPedoe inequality, Bull. Austral. Math. Soc., 27 (1983), 203–214.
[4] SHIGUO YANG, Three geometric inequalities for a simplex, Geometriae Dedicata, 57 (1995),
105–110.
[5] GANGSON LENG, Inequalities for edge-lengths and volumes of two simplices, Geometriae Dedicata, 68 (1997), 43–48.
[6] GANGSON LENG, Some generalizations to several dimensions of the Pedoe inequality with applications, Acta Mathematica Sinica, 40 (1997), 14–21.
[7] D. VELJAN, The sine theorem and inequalities for volumes of simplices and determinants, Lin.
Alg. Appl., 219 (1995), 79–91.
[8] SHIGUO YANG AND JIA WANG, An inequality for n-dimensional sines of vertex angles of a
simplex with some aplications, Journal of Geometry, 54 (1995), 198–202.
[9] D.S. MITRINOVIĆ, J.E. PEČARIĆ
Dordrecht, Boston, London, 1989.
J. Inequal. Pure and Appl. Math., 5(4) Art. 106, 2004
AND
V. VOLENEC, Recent Geometric Inequalities, Kluwer,
http://jipam.vu.edu.au/