Volume 10 (2009), Issue 1, Article 29, 8 pp.
THE BEST CONSTANTS FOR A DOUBLE INEQUALITY IN A TRIANGLE YU-DONG WU, NU-CHUN HU, AND WEI-PING KUANG D EPARTMENT OF M ATHEMATICS Z HEJIANG X INCHANG H IGH S CHOOL S HAOXING 312500, Z HEJIANG P EOPLE ’ S R EPUBLIC OF C HINA .
[email protected] URL: http://www.staff.vu.edu.au/rgmia/members/Wu3.htm D EPARTMENT OF M ATHEMATICS Z HEJIANG N ORMAL U NIVERSITY J INHUA 321004, Z HEJIANG P EOPLE ’ S R EPUBLIC OF C HINA .
[email protected] D EPARTMENT OF M ATHEMATICS H UAIHUA U NIVERSITY H UAIHUA 418008, H UNAN P EOPLE ’ S R EPUBLIC OF C HINA .
[email protected]
Received 08 August, 2008; accepted 02 February, 2009 Communicated by S.S. Dragomir Dedicated to Professor Bi-Cheng Yang on the occasion of his 63rd birthday.
A BSTRACT. In this short note, by using some of Chen’s theorems and classic analysis, we obtain a double inequality for triangle and give a positive answer to a problem posed by Yang and Yin [6].
Key words and phrases: Inequality; Best Constant; Triangle. 2000 Mathematics Subject Classification. 51M16.
1. I NTRODUCTION AND M AIN R ESULTS For 4ABC, let a, b, c denote the side-lengths, A, B, C the angles, s the semi-perimeter, R the circumradius and r the inradius, respectively. The authors would like to thank Prof. Zhi-Hua Zhang and Dr. Zhi-Gang Wang for their careful reading and making some valuable suggestions in the preparation of this paper. 239-08
2
Y U -D ONG W U , N U -C HUN H U , AND W EI -P ING K UANG
In 1957, Kooistra (see [1]) built the following double inequality for any triangle: √ A B C 3 3 . (1.1) 2 < cos + cos + cos ≤ 2 2 2 2 In 2000, Yang and Yin [6] considered a new bound of inequality (1.1) and posed a problem as follows: Problem 1.1. Determine the best constant µ such that √ !µ 3 3 s 1−µ A B C (1.2) · ≤ cos + cos + cos 2 R 2 2 2 holds for any 4ABC. In this short note, we solve the above problem and obtain the following result. Theorem 1.1. Let λ ≥ λ0 = 1 and
√ 2 ln (2 − 2) + ln 2 µ ≤ µ0 = ≈ 0.7194536993. 4 ln 2 − 3 ln 3 Then the double inequality √ !µ √ !λ 3 3 s 1−µ A B C 3 3 s 1−λ (1.3) · ≤ cos + cos + cos ≤ · 2 R 2 2 2 2 R
holds for any 4ABC, while the constants λ0 and µ0 are both the best constant for inequality (1.3). Remark 1. When λ0 = 1, the right hand of inequality (1.3) is just the right hand of inequality (1.1). Remark 2. It is not difficult to demonstrate that: √ µ µ0 1−µ 1−µ0 0 1 0 3 3 s 2 s 1−µ · R < 2 0 < R < 2 0 3√ 3 , 2 √ µ0 323 ·
s 1−µ0 R
≥2 2
1 1−µ0
2 √
3 3
µ0 1−µ
0
≤
s R
≤
√ 3 3 2
.
2. P RELIMINARY R ESULTS In order to establish our main theorem, we shall require the following lemmas. Lemma 2.1 (see [3, 4, 5]). If the inequality s ≥ (>)f (R, r) holds for any isosceles triangle whose top-angle is greater than or equal to π3 , then the inequality s ≥ (>)f (R, r) holds for any triangle.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 29, 8 pp.
http://jipam.vu.edu.au/
T HE B EST C ONSTANTS FOR A D OUBLE I NEQUALITY IN A T RIANGLE
3
Lemma 2.2 (see [2, 3]). The homogeneous inequality s ≥ (>)f (R, r)
(2.1)
holds for any acute-angled triangle if and only if it holds for any acute isosceles triangle whose √ top-angle A ∈ π3 , π2 with 2r ≤ R < ( 2 + 1)r and any right-angled triangle with R ≥ √ ( 2 + 1)r. For the convenience of our readers, we give below the proof by Chen in [2, 3]. J Proof. Let O denote the circumcircle of 4ABC. Necessity is obvious from Lemma 2.1. Thus we only need to prove the sufficiency. It is well known that R ≥ 2r for any acute-angled triangle. So we consider the following two cases: √ (i) When 2r ≤ R < ( 2 + 1)r : In this case , we can construct an isosceles triangle J A1 B1 C1 whose circumcircle is also O and the top-angle of 4A1 B1 C1 (see Figure 2.1) is ! r 1 2r A1 = 2 arcsin 1+ 1− . 2 R
C
C1
I1
O
A
B1 I2
B A1 Figure 2.1:
It is easy to see that (see [4, 5]): R1 = R,
r1 = r,
s1 ≤ s
and
π π ≤ A1 < . 3 2
Thus we have (2.2)
s ≥ s1 ≥ f (R1 , r1 ) = f (R, r). because the inequality (2.1) holds for any acute isosceles triangle whose top-angle A ∈ π π , . 3 2
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 29, 8 pp.
http://jipam.vu.edu.au/
4
Y U -D ONG W U , N U -C HUN H U , AND W EI -P ING K UANG
√ (ii) When R ≥ ( 2 + 1)r: In this case we can construct a right-angled triangle A2 B2 C2 J whose inscribed circle is also I and the length of its hypotenuse is c2 = 2R (see Figure 2.2). This implies that
A2
C
A
O
I
C2
B2 B
Figure 2.2:
1 1 r2 = (a2 + b2 − c2 ) = r, R2 = c2 = R, 2 2 1 s2 = (a2 + b2 + c2 ) = 2R2 + r2 = 2R + r < s. 2 Thus we have the inequality (2.2) since the inequality (2.1) holds for any right-angled triangle. Lemma 2.3 (see [2, 3]). The homogeneous inequality (2.1) holds for any acute-angled triangle if and only if (2.3)
p
(1 − x)(3 +
x)3
2
≥ (>)f (2, 1 − x )
0≤x
)f (2, 1 − x2 )
√
2−1≤x
