OCTOGON MATHEMATICAL MAGAZINE
Vol. 17, No.2, October 2009, pp 732-735
ISSN 1222-5657, ISBN 978-973-88255-5-0,
www.hetfalu.ro/octogon
732
New inequality in tetrahedron
Mihály Bencze26
ABSTRACT. In this paper we will prove a new inequality for the tetrahedron.
MAIN RESULTS
Theorem 1. (A refinement of Finsler-Hadwiger‘s inequality). In all triangle
ABC holds
√
¡
¢
6 (ab + bc + ca) − 5 a2 + b2 + c2 ≤ 4 3σ [ABC] ≤
¡
¢
≤ 2 (ab + bc + ca) − a2 + b2 + c2
Proof. Using the Schur‘s theorem for all x, y, z > 0 and t ∈ R we get
xt (x − y) (x − z) + y t (y − z) (y − x) + z t (z − x) (z − y) ≥ 0
If t = 2 and x = a, y = b, z = c, then we obtain
X
a4 + abc
X
a≥
X
¡
¢
ab a2 + b2
or
4
³X
ab
´2
+
³X
≥3
a2
´2
−4
³X
ab
´ ³X
´
a2 ≥
³X ´ Y
a
(b + c − a)
which is equivalent with
X
X
√
4 3σ [ABC] ≤ 2
ab −
a2
26
Received: 15.02.2007
2000 Mathematics Subject Classification. 26D05
Key words and phrases. Geometric inequalities, triangle inequalities, tetrahedron
inequalities.
New inequality in tetrahedron
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Using again the Schur‘s theorem we get
³ X
X ´ ³X ´
2
xy −
x2
x ≤ 9xyz
but 27xyz ≤ (
P
x)3 , therefore
2
X
xy −
X
r
2
3xyz
x ≤
³X ´
x
IfP
x + y = a,
Py + z =√b, z + x = c, then we obtain
6 ab − 5 a2 ≤ 4 3σ [ABC] .
Corollary 1.1. In all triangle ABC holds
4r2 + 16Rr − s2 ≤
√
3sr ≤ r (4R + r)
Proof. In Theorem 1 we use that
X
ab = s2 + r2 + 4Rr,
X
¡
¢
a2 = 2 s2 − r2 − 4Rr
and σ [ABC] = sr.
Theorem 2. Let ABCD be a tetrahedron with hA and mA the lengths of
the altitude and the median from vertex A to the opposite face BCD,
respectively, and V denote the volume of the tetrahedron. If
M = (BC − CD)2 + (BC − BD)2 + (BC − AC)2 + (BC − AB)2 +
+ (CD − BD)2 +(CD − AD)2 +(CD − AC)2 +(BD − AD)2 +(BD − AB)2 +
+ (AD − AC)2 + (AD − AB)2 + (AB − AC)2 ,
then
³X
hA
´ ³X
´ 128V
32r
M,
m2A ≥ √ +
9
3
where r denote the radius of insphere of tetrahedron.
P 2
P
Proof. We have
mA = 49 AB 2 . Denote SA the area of face BCD etc.,
and using the Theorem 1, we obtain
734
Mihály Bencze
√

2 + CD 2 + DB 2 ≥ 4S
BC
3 + (BC − CD)2 + (BC − DB)2 + (CD − DB)2
A


√

AC 2 + CD2 + DA2 ≥ 4SB √3 + (AC − CD)2 + (AC − DA)2 + (CD − DA)2

AB 2 + BD2 + DA2 ≥ 4SC√ 3 + (AB − BD)2 + (AB − DA)2 + (BD − DA)2


AB 2 + BC 2 + AC 2 ≥ 4SD 3 + (AB − BC)2 + (AB − AC)2 + (BC − AC)2
After addition we get
√ X
√ X
9X 2
AB 2 ≥ 4 3
SA + M or
SA + M or
mA ≥ 4 3
2
√
X
2
8 3X
2
SA + M.
mA ≥
9
9
P
Multiplying by
hA we obtain:
³X ´ ³X
´ 8√3 ³X ´ ³X ´ 2 ³X ´
2
hA
mA ≥
SA
hA + M
hA .
9
9
If we suppose that SA ≤ SB ≤ SC ≤ SD , then hA ≥ hB ≥ hC ≥ hD and from
Chebishev‘s inequality holds
³X ´ ³X ´
X
SA
hA ≥ 4
SA hA = 48V
P 1
P
1
Because
hA ≥ P161 = 16r, thus
hA = r , therefore
hA
³X ´ ³X
´ 8√3 ³X ´ ³X ´ 2 ³X ´
hA
m2A ≥
hA ≥
SA
hA + M
9
9
√
8 3
2
128V
32r
≥
· 48V + M · 16r = √ +
M
9
9
9
3
2
X
Equality holds if and only if AB = BC = CD = DA = BD = AC.
Open Question. Using the notation of Theorem 2, determine the best
constants c1 , c2 > 0 such that
³X ´ ³ X
´ 128V
128V
√ + c1 M R ≥
hA
m2A ≥ √ + c2 M r,
3
3
where R denote the radius of circumsphere of tetrahedron.
New inequality in tetrahedron
REFERENCE
[1] Octogon Mathematical Magazine (1993-2009)
Str. Hărmanului 6,
505600 Săcele-Négyfalu
Jud. Brasov, Romania
E-mail: [email protected]
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