Volume 10 (2009), Issue 1, Article 15, 6 pp.
AN INEQUALITY ON TERNARY QUADRATIC FORMS IN TRIANGLES
NU-CHUN HU
D EPARTMENT OF M ATHEMATICS
Z HEJIANG N ORMAL U NIVERSITY
J INHUA 321004, Z HEJIANG
P EOPLE ’ S R EPUBLIC OF C HINA .
nuchun@zjnu.cn
Received 07 May, 2008; accepted 25 February, 2009
Communicated by S.S. Dragomir
A BSTRACT. In this short note, we give a proof of a conjecture about ternary quadratic forms
involving two triangles and several interesting applications.
Key words and phrases: Positive semidefinite ternary quadratic form, arithmetic-mean geometric-mean inequality, Cauchy
inequality, triangle.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In [3], Liu proved the following theorem.
Theorem 1.1. For any 4ABC and real numbers x, y, z, the following inequality holds.
A
B
C
(1.1)
x2 cos2 + y 2 cos2 + z 2 cos2 ≥ yz sin2 A + zx sin2 B + xy sin2 C.
2
2
2
In [6], Tao proved the following theorem.
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.
(1.3) x2 cos
A1
A2
B1
B2
C1
C2
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.
133-08
2
N.-C. H U
2. P RELIMINARIES
For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, s the semi-perimeter, S
the area, R the circumradius
and r the
P
Q inradius, respectively. In addition we will customarily
use the symbols (cyclic sum) and (cyclic product):
X
f (a) = f (a) + f (b) + f (c),
Y
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
4p1 p2 p3 ≥ p1 q12 + p2 q22 + p3 q32 + q1 q2 q3 .
(2.1)
Then the following inequality holds for any real numbers x, y, z,
p1 x2 + p2 y 2 + p3 z 2 ≥ q1 yz + q2 zx + q3 xy.
(2.2)
Lemma 2.2. For 4ABC, the following inequalities hold.
√
B
C
3 3 2
2 cos cos ≥
(2.3)
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
(2.5)
2 cos cos ≥
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
and
B
cos =
2
r
s(s − b)
,
ca
C
cos =
2
r
s(s − c)
,
ab
then it follows that
(2.6)
√
B
C
3 3 2
2 cos cos ≥
sin A
2
2
4 r
r
√
s(s − b) s(s − c)
3 3S 2
⇐⇒ 2
≥ 2 2
ca
ab
bc
4s2 (s − b)(s − c)
27s2 (s − a)2 (s − b)2 (s − c)2
⇐⇒
≥
a2 bc
b4 c 4
4
27(s − a)2 (s − b)(s − c)
⇐⇒ 2 ≥
a
b3 c 3
3 3
2
⇐⇒ 4b c ≥ 27a (s − a)2 (s − b)(s − c).
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 15, 6 pp.
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T ERNARY Q UADRATIC F ORMS IN T RIANGLES
3
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
3
(b + c − a)2
< 4b3 c3 .
= 4 bc −
4
Therefore the inequality (2.6) holds, and hence (2.3) holds.
Lemma 2.3. For 4ABC, the following equality holds.
X
sin4 A
(2R + 5r)s4 − 2(R + r)(16R + 5r)rs2 + (4R + r)3 r2
(2.7)
=
.
2R3 s2
cos2 B2 cos2 C2
Proof. By the familiar identity: a + b + c = 2s, ab + bc + ca = s2 + 4Rr + r2 , 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
− 13(a + b + c)2 (ab + bc + ca)2 − 7abc(a + b + c)3
+ 4(ab + bc + ca)3 + 19abc(ab + bc + ca)(a + b + c) − 6a2 b2 c2 ,
it follows that
X
a5 (b + c − a) = 4(2R + 5r)rs4 − 8(R + r)(16R + 5r)r2 s2 + 4(4R + r)3 r3 ,
and hence
X
X a 4 (b + c)2 − a2
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
Q
s
, it follows that
Thus, together with the familiar identity cos A2 = 4R
P
X
sin4 A cos2 A2
sin4 A
Q 2A
=
cos2 B2 cos2 C2
cos 2
P 4
sin A(1 + cos A)
Q
=
2 cos2 A2
sin4 A(1 + cos A) =
(2R + 5r)s4 − 2(R + r)(16R + 5r)rs2 + (4R + r)3 r2
.
2R3 s2
Therefore the equality (2.7) is proved.
=
Lemma 2.4. For 4ABC, the following inequality holds.
(2.8)
−(2R + 5r)s4 + 2(2R + 5r)(2R + r)(R + r)s2 − (4R + r)3 r2 ≥ 0.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 15, 6 pp.
http://jipam.vu.edu.au/
4
N.-C. H U
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 ]
+ 2r(14R2 + 31Rr − 10r2 )(4R2 + 4Rr + 3r2 − s2 )
+ 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.
Lemma 2.5. For 4ABC, the following inequality holds.
X
(2.10)
sin4 A
cos2 B2 cos2
C
2
+ 64
Proof. By Lemma 2.3 and the familiar identity
X
sin4 A
cos2 B2 cos2
C
2
+ 64
Y
sin2
Q
Y
sin2
sin A2 =
A
≤ 4.
2
r
,
4R
it follows that
A
≤4
2
(2R + 5r)s4 − 2(R + r)(16R + 5r)rs2 + (4R + r)3 r2 4r2
+ 2 ≤4
2R3 s2
R
4
2
−(2R + 5r)s + 2(2R + 5r)(2R + r)(R + r)s − (4R + r)3 r2
⇐⇒
≥ 0.
2R3 s2
⇐⇒
(2.11)
Thus, by Lemma 2.4, it follows that the inequality (2.11) holds, and hence (2.10) holds.
3. P ROOF OF THE M AIN T HEOREM
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
cos
cos
≥0,
2
2
C1
C2
cos
cos
≥0.
2
2
cos
(3.1)
(3.2)
(3.3)
Next, by Lemma 2.2, we have the following inequalities:
(3.4)
(3.5)
(3.6)
B1
B2
C1
C2
cos
· cos
cos
≥ sin2 A1 sin2 A2 ,
2
2
2
2
C1
C2
A1
A2
4 cos
cos
· cos
cos
≥ sin2 B1 sin2 B2 ,
2
2
2
2
A1
A2
B1
B2
4 cos
cos
· cos
cos
≥ sin2 C1 sin2 C2 .
2
2
2
2
4 cos
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 15, 6 pp.
http://jipam.vu.edu.au/
T ERNARY Q UADRATIC F ORMS IN T RIANGLES
5
Thus, in order that Proposition 2.1 is applicable, we have to show the following inequality.
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
C2
C1
sin2 C1 cos
sin2 C2 +
sin A1
sin A2 .
+ cos
2
2
However, in order to prove the inequality (3.7), we only need the following inequality.
(3.7) 4
(3.8)
Y
cos
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
In fact, by the Cauchy inequality and Lemma 2.5, we have that
"
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
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
"
#
X
Y
sin4 A2
A
2
×
+ 64
sin2
2
cos2 B22 cos2 C22
#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).
4. A PPLICATIONS
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
A2
B1
B2
C1
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 .
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 15, 6 pp.
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6
N.-C. H U
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 holds.
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 .
h2a cos
Corollary 4.5. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality holds.
R12 cos
A1
A2
B1
B2
C1
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 .
Corollary 4.6. For any 4ABC, 4A1 B1 C1 , 4A2 B2 C2 , the following inequality holds.
r12 cos
A1
A2
B1
B2
C1
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 .
R EFERENCES
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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, Middle-School Mathematics
(in Chinese), 5 (1996), 16–19.
[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.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 15, 6 pp.
http://jipam.vu.edu.au/