AN INEQUALITY ON TERNARY QUADRATIC
FORMS IN TRIANGLES
Ternary Quadratic Forms in
Triangles
NU-CHUN HU
Department of Mathematics
Zhejiang Normal University
Jinhua 321004, Zhejiang
People’s Republic of China.
Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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EMail: nuchun@zjnu.cn
Received:
07 May, 2008
Accepted:
25 February, 2009
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
26D15.
Key words:
Positive semidefinite ternary quadratic form, arithmetic-mean geometric-mean
inequality, Cauchy inequality, triangle.
Abstract:
In this short note, we give a proof of a conjecture about ternary quadratic forms
involving two triangles and several interesting applications.
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Contents
1
Introduction
3
2
Preliminaries
4
3
Proof of the Main Theorem
9
4
Applications
Ternary Quadratic Forms in
Triangles
11
Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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1.
Introduction
In [3], Liu proved the following theorem.
Theorem 1.1. For any 4ABC and real numbers x, y, z, the following inequality
holds.
(1.1)
x2 cos2
B
C
A
+ y 2 cos2 + z 2 cos2 ≥ yz sin2 A + zx sin2 B + xy sin2 C.
2
2
2
Ternary Quadratic Forms in
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Nu-Chun Hu
In [6], Tao proved the following theorem.
vol. 10, iss. 1, art. 15, 2009
Theorem 1.2. For any 4A1 B1 C1 , 4A2 B2 C2 , the following inequality holds.
(1.2)
cos
A1
A2
B1
B2
C1
C2
cos
+ cos
cos
+ cos
cos
2
2
2
2
2
2
≥ sin A1 sin A2 + sin B1 sin B2 + sin C1 sin C2 .
Then, in [4], Liu proposed the following conjecture.
Conjecture 1.3. For any 4A1 B1 C1 , 4A2 B2 C2 and real numbers x, y, z, the following inequality holds.
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A1
A2
B1
B2
C1
C2
(1.3) x2 cos
cos
+ y 2 cos
cos
+ z 2 cos
cos
2
2
2
2
2
2
≥ yz sin A1 sin A2 + zx sin B1 sin B2 + xy sin C1 sin C2 .
In this paper, we give a proof of this conjecture and some interesting applications.
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2.
Preliminaries
For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, s the semiperimeter, S the area, R the circumradius
P and r the inradius,
Q respectively. In addition
we will customarily use the symbols (cyclic sum) and (cyclic product):
X
Y
f (a) = f (a) + f (b) + f (c),
f (a) = f (a)f (b)f (c).
To prove the inequality (1.1), we need the following well-known proposition
about positive semidefinite quadratic forms.
Proposition 2.1 (see [2]). Let pi , qi (i = 1, 2, 3) be real numbers such that pi ≥ 0
(i = 1, 2, 3), 4p2 p3 ≥ q12 , 4p3 p1 ≥ q22 , 4p1 p2 ≥ q32 and
(2.1)
4p1 p2 p3 ≥ p1 q12 + p2 q22 + p3 q32 + q1 q2 q3 .
Then the following inequality holds for any real numbers x, y, z,
(2.2)
p1 x2 + p2 y 2 + p3 z 2 ≥ q1 yz + q2 zx + q3 xy.
Lemma 2.2. For 4ABC, the following inequalities hold.
√
B
C
3 3 2
(2.3)
2 cos cos ≥
sin A > sin2 A,
2
2
4
√
A
3 3 2
C
(2.4)
2 cos cos ≥
sin B > sin2 B,
2
2
4
√
A
B
3 3 2
2 cos cos ≥
(2.5)
sin C > sin2 C.
2
2
4
Proof. We will only prove (2.3) because (2.4) and (2.5) can be done similarly. Since
p
1
S = bc sin A = s(s − a)(s − b)(s − c)
2
Ternary Quadratic Forms in
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Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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and
r
r
B
s(s − b)
s(s − c)
C
cos =
,
cos =
,
2
ca
2
ab
then it follows that
√
C
3 3 2
B
2 cos cos ≥
sin A
2
2
4 r
r
√
3 3S 2
s(s − b) s(s − c)
≥ 2 2
⇐⇒ 2
bc
ab
ca
4s2 (s − b)(s − c)
27s2 (s − a)2 (s − b)2 (s − c)2
⇐⇒
≥
a2 bc
b4 c 4
27(s − a)2 (s − b)(s − c)
4
⇐⇒ 2 ≥
a
b3 c 3
3 3
2
(2.6)
⇐⇒ 4b c ≥ 27a (s − a)2 (s − b)(s − c).
On the other hand, by the arithmetic-mean geometric-mean inequality, we have the
following inequality.
27a2 (s − a)2 (s − b)(s − c)
1
1
= 108 · a(s − a) · a(s − a) · (s − b)(s − c)
2
2
1
3
a(s − a) + 12 a(s − a) + (s − b)(s − c)
2
≤ 108
3
2 3
(b + c − a)
= 4 bc −
< 4b3 c3 .
4
Therefore the inequality (2.6) holds, and hence (2.3) holds.
Ternary Quadratic Forms in
Triangles
Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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Lemma 2.3. For 4ABC, the following equality holds.
(2.7)
X
sin4 A
cos2 B2 cos2 C2
(2R + 5r)s4 − 2(R + r)(16R + 5r)rs2 + (4R + r)3 r2
=
.
2R3 s2
2
2
Proof. By the familiar identity: a + b + c = 2s, ab + bc + ca = s + 4Rr + r ,
abc = 4Rrs (see [5]) and the following identity
X
a5 (b + c − a) = −(a + b + c)6 + 7(ab + bc + ca)(a + b + c)4
Ternary Quadratic Forms in
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Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
− 13(a + b + c)2 (ab + bc + ca)2 − 7abc(a + b + c)3
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+ 4(ab + bc + ca)3 + 19abc(ab + bc + ca)(a + b + c) − 6a2 b2 c2 ,
Contents
it follows that
X
a5 (b + c − a) = 4(2R + 5r)rs4 − 8(R + r)(16R + 5r)r2 s2 + 4(4R + r)3 r3 ,
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and hence
X
X a 4 (b + c)2 − a2
sin4 A(1 + cos A) =
2R P 2bc
(a + b + c) a5 (b + c − a)
=
32R4 abc
(2R + 5r)s4 − 2(R + r)(16R + 5r)rs2 + (4R + r)3 r2
.
=
16R5
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Q
s
Thus, together with the familiar identity cos A2 = 4R
, it follows that
P 4
X
sin A cos2 A2
sin4 A
Q 2A
=
cos2 B2 cos2 C2
cos 2
P 4
sin A(1 + cos A)
Q
=
2 cos2 A2
(2R + 5r)s4 − 2(R + r)(16R + 5r)rs2 + (4R + r)3 r2
=
.
2R3 s2
Ternary Quadratic Forms in
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Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
Therefore the equality (2.7) is proved.
Lemma 2.4. For 4ABC, the following inequality holds.
(2.8)
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−(2R + 5r)s4 + 2(2R + 5r)(2R + r)(R + r)s2 − (4R + r)3 r2 ≥ 0.
Proof. First it is easy to verify that the inequality (2.8) is just the following inequality.
(2.9) (2R + 5r)[−s4 + (4R2 + 20Rr − 2r2 )s2 − r(4R + r)3 ]
2
2
2
2
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2
+ 2r(14R + 31Rr − 10r )(4R + 4Rr + 3r − s )
+ 4(R − 2r)(4R3 + 6R2 r + 3Rr2 − 8r3 ) ≥ 0.
Thus, together with the fundamental inequality
−s4 + (4R2 + 20Rr − 2r2 )s2 − r(4R + r)3 ≥ 0
(see [5, page 2]), Euler’s inequality R ≥ 2r and Gerretsen’s inequality s2 ≤ 4R2 +
4Rr + 3r2 (see [1, page 45]), it follows that the inequality (2.9) holds, and hence
(2.8) holds.
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Lemma 2.5. For 4ABC, the following inequality holds.
X
(2.10)
sin4 A
cos2 B2 cos2
C
2
+ 64
Y
Proof. By Lemma 2.3 and the familiar identity
sin2
Q
A
≤ 4.
2
sin A2 =
r
,
4R
it follows that
Y
sin4 A
2 A
+
64
sin
≤4
B
C
2
cos2 2 cos2 2
(2R+5r)s4 −2(R+r)(16R+5r)rs2 +(4R+r)3 r2 4r2
+ 2 ≤4
⇐⇒
2R3 s2
R
−(2R+5r)s4 +2(2R+5r)(2R+r)(R+r)s2 −(4R+r)3 r2
⇐⇒
≥ 0.
2R3 s2
X
(2.11)
Thus, by Lemma 2.4, it follows that the inequality (2.11) holds, and hence (2.10)
holds.
Ternary Quadratic Forms in
Triangles
Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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3.
Proof of the Main Theorem
Now we give the proof of inequality (1.1).
Proof. First, it is easy to verify that
A1
A2
cos
≥0,
2
2
B1
B2
(3.2)
cos
cos
≥0,
2
2
C1
C2
(3.3)
cos
cos
≥0.
2
2
Next, by Lemma 2.2, we have the following inequalities:
cos
(3.1)
B1
B2
C1
C2
cos
· cos
cos
≥ sin2 A1 sin2 A2 ,
2
2
2
2
C1
C2
A1
A2
(3.5)
4 cos
cos
· cos
cos
≥ sin2 B1 sin2 B2 ,
2
2
2
2
A1
A2
B1
B2
(3.6)
4 cos
cos
· cos
cos
≥ sin2 C1 sin2 C2 .
2
2
2
2
Thus, in order that Proposition 2.1 is applicable, we have to show the following
inequality.
Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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4 cos
(3.4)
(3.7) 4
Ternary Quadratic Forms in
Triangles
A1 Y
A2
cos
2
2
A1
A2
B1
B2
≥ cos
sin2 A1 cos
sin2 A2 + cos
sin2 B1 cos
sin2 B2
2
2
2
2
Y
Y
C1
C2
sin2 C1 cos
sin2 C2 +
sin A1
sin A2 .
+ cos
2
2
Y
cos
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However, in order to prove the inequality (3.7), we only need the following inequality.
(3.8)
sin2 A1
sin2 A2
sin2 B1
sin2 B2
·
+
·
cos B21 cos C21 cos B22 cos C22
cos C21 cos A21 cos C22 cos A22
Y
Y
sin2 C1
sin2 C2
A1
A2
+
·
+
8
sin
·
8
sin
≤ 4.
A1
B1
A2
B2
2
2
cos 2 cos 2 cos 2 cos 2
Nu-Chun Hu
In fact, by the Cauchy inequality and Lemma 2.5, we have that
"
sin2 A2
sin2 B1
sin2 B2
sin2 A1
·
+
·
cos B21 cos C21 cos B22 cos C22
cos C21 cos A21 cos C22 cos A22
Y
Y
sin2 C1
sin2 C2
A1
A2
+
·
+
8
sin
·
8
sin
A1
B1
A2
B2
2
2
cos 2 cos 2 cos 2 cos 2
"
#
X
Y
sin4 A1
A1
≤
+
64
sin2
B1
C1
2
2
2
cos 2 cos 2
"
#
4
X
Y
sin A2
A2
×
+ 64
sin2
B2
C2
2
2
2
cos 2 cos 2
Ternary Quadratic Forms in
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vol. 10, iss. 1, art. 15, 2009
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#2
≤ 16
Therefore the inequality (3.8) holds, and hence (3.7) holds. Thus, together with
inequality (3.4)–(3.7), Proposition 2.1 is applicable to complete the proof of (1.1).
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4.
Applications
Let P be a point in the 4ABC. Recall that A, B, C denote the angles, a, b, c the
lengths of sides, wa , wb , wc the lengths of interior angular bisectors, ma , mb , mc the
lengths of medians, ha , hb , hc the lengths of altitudes, R1 , R2 , R3 the distances of P
to vertices A, B, C, r1 , r2 , r3 the distances of P to the sidelines BC, CA, AB.
Corollary 4.1. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality
holds.
a2 cos
A1
B1
C1
A2
B2
C2
cos
+ b2 cos
cos
+ c2 cos
cos
2
2
2
2
2
2
≥ bc sin A1 sin A2 + ca sin B1 sin B2 + ab sin C1 sin C2 .
Corollary 4.2. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality
holds.
wa2 cos
A1
A2
B1
B2
C1
C2
cos
+ wb2 cos
cos
+ wc2 cos
cos
2
2
2
2
2
2
≥ wb wc sin A1 sin A2 + wc wa sin B1 sin B2 + wa wb sin C1 sin C2 .
Corollary 4.3. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality
holds.
m2a cos
A1
A2
B1
B2
C1
C2
cos
+ m2b cos
cos
+ m2c cos
cos
2
2
2
2
2
2
≥ mb mc sin A1 sin A2 + mc ma sin B1 sin B2 + ma mb sin C1 sin C2 .
Corollary 4.4. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality
Ternary Quadratic Forms in
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Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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holds.
h2a cos
A1
A2
B1
B2
C1
C2
cos
+ h2b cos
cos
+ h2c cos
cos
2
2
2
2
2
2
≥ hb hc sin A1 sin A2 + hc ha sin B1 sin B2 + ha hb sin C1 sin C2 .
Corollary 4.5. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality
holds.
R12
A2
B1
B2
C1
C2
A1
cos
cos
+ R22 cos
cos
+ R32 cos
cos
2
2
2
2
2
2
≥ R2 R3 sin A1 sin A2 + R3 R1 sin B1 sin B2 + R1 R2 sin C1 sin C2 .
Corollary 4.6. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality
holds.
r12 cos
B1
C1
A1
A2
B2
C2
cos
+ r22 cos
cos
+ r32 cos
cos
2
2
2
2
2
2
≥ r2 r3 sin A1 sin A2 + r3 r1 sin B1 sin B2 + r1 r2 sin C1 sin C2 .
Ternary Quadratic Forms in
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Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
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References
[1] O. BOTTEMA, R.Ž. DJORDJEVIĆ, R.R. JANIĆ, D.S. MITRINOVIĆ AND
P.M. VASIĆ, Geometric Inequalities, Wolters-Noordhoff Publishing, Groningen, 1969.
[2] S.J. LEON, Linear Algebra with Applications, Prentice Hall, New Jersey, 2005.
[3] J. LIU, Two results about ternary quadratic form and their applications, MiddleSchool Mathematics (in Chinese), 5 (1996), 16–19.
Ternary Quadratic Forms in
Triangles
Nu-Chun Hu
vol. 10, iss. 1, art. 15, 2009
[4] J. LIU, Inequalities involving nine sine (in Chinese), preprint.
[5] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND V. VOLENEC, Recent Advances in
Geometric Inequalities, Mathematics and its Applications (East European Series), 28. Kluwer Academic Publishers Group, Dordrecht, 1989.
[6] C.G. TAO, Proof of a conjecture relating two triangle, Middle-School Mathematics (in Chinese), 2 (2004), 43–43.
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