THE BEST CONSTANTS FOR A DOUBLE INEQUALITY IN A TRIANGLE YU-DONG WU Department of Mathematics Zhejiang Xinchang High School Shaoxing 1500, Zhejiang People s Republic of China. EMail: yudong.wu@yahoo.com.cn WEI-PING KUANG Department of Mathematics Huaihua University Huaihua 418008, Hunan People s Republic of China. EMail: sy78515@16.com Received: 08 August, 008 Accepted: 0 February, 009 Communicated by: S.S. Dragomir NU-CHUN HU Department of Mathematics Zhejiang Normal University Jinhua 1004, Zhejiang People s Republic of China. EMail: nuchun@zjnu.cn Page 1 of 16 000 AMS Sub. Class.: 51M16 Key words: Inequality; Best Constant;.
Abstract: 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]. Acknowledgment: Dedicatory: 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. Dedicated to Professor Bi-Cheng Yang on the occasion of his 6rd birthday. Page of 16
1 Introduction and Main Results 4 Preliminary Results 6 The Proof of Theorem 1.1 1 Page of 16
1. Introduction and Main Results For ABC, let a, b, c denote the side-lengths, A, B, C the angles, s the semiperimeter, R the circumradius and r the inradius, respectively. In 1957, Kooistra see [1] built the following double inequality for any triangle: 1.1 < cos A + cos B + cos C. In 000, Yang and Yin [6] considered a new bound of inequality 1.1 and posed a problem as follows: Problem 1. Determine the best constant µ such that µ s 1 µ A 1. cos R + cos B + cos C holds for any ABC. In this short note, we solve the above problem and obtain the following result. Theorem 1.1. Let and λ λ 0 = 1 µ µ 0 = ln + ln 0.71945699. 4 ln ln Then the double inequality µ s 1 µ A 1. cos R + cos B + cos C λ s 1 λ R Page 4 of 16
holds for any ABC, while the constants λ 0 and µ 0 are both the best constant for inequality 1.. Remark 1. When λ 0 = 1, the right hand of inequality 1. is just the right hand of inequality 1.1. Remark. It is not difficult to demonstrate that: µ0 s 1 µ0 R < 0 < s < 1 1 µ 0 R µ0 s 1 µ0 R 1 µ 0 1 µ 0 µ 0 1 µ 0, 1 µ 0 s R. Page 5 of 16
. Preliminary Results In order to establish our main theorem, we shall require the following lemmas. Lemma.1 see [, 4, 5]. If the inequality s >fr, r holds for any isosceles triangle whose top-angle is greater than or equal to π, then the inequality s > fr, r holds for any triangle. Lemma. see [, ]. The homogeneous inequality.1 s >fr, r holds for any acute-angled triangle if and only if it holds for any acute isosceles triangle whose top-angle A [ π, π with r R < + 1r and any rightangled triangle with R + 1r. For the convenience of our readers, we give below the proof by Chen in [, ]. Proof. Let O denote the circumcircle of ABC. Necessity is obvious from Lemma.1. Thus we only need to prove the sufficiency. It is well known that R r for any acute-angled triangle. So we consider the following two cases: i When r R < +1r : In this case, we can construct an isosceles triangle A 1 B 1 C 1 whose circumcircle is also O and the top-angle of A 1 B 1 C 1 see Figure 1 is A 1 = arcsin 1 1 + 1 r. R Page 6 of 16 It is easy to see that see [4, 5]: R 1 = R, r 1 = r, s 1 s and π A 1 < π.
C C 1 A I 1 O B 1 I B A 1 Thus we have Figure 1:. s s 1 fr 1, r 1 = fr, r. because the inequality.1 holds for any acute isosceles triangle whose topangle A [ π, π. ii When R + 1r: In this case we can construct a right-angled triangle A B C whose inscribed circle is also I and the length of its hypotenuse is c = R see Figure. This implies that Page 7 of 16
A C A I O C B B Figure : r = 1 a + b c = r, R = 1 c = R, s = 1 a + b + c = R + r = R + r < s. Thus we have the inequality. since the inequality.1 holds for any rightangled triangle. Page 8 of 16 Lemma. see [, ]. The homogeneous inequality.1 holds for any acuteangled triangle if and only if. 1 x + x >f, 1 x 0 x < 1,
and.4 5 x >f, 1 x 1 x < 1. Proof. Since the inequality.1 is homogeneous, we may assume R = without losing generality. i When r R < + 1r: By Lemma., we only need to consider the isosceles triangle whose top-angle A [ π, π. Let [ t = sin A 1,. Then we have see [4, 5].5 r = 4t1 t and s = 41 + t 1 t. Let x = t 1. Then the inequality.1 is just the inequality.. ii When R + 1r: We only need to consider a right-angled triangle. Let r = r R = 4t1 t 0, 1 t < 1. Thus we have Page 9 of 16.6 s = R + r = 4 + 4t1 t. Let x = t 1. Then the inequality.1 is just the inequality.4. This completes the proof Lemma..
Lemma.4 [, 4, 5]. The homogeneous inequality.7 s <fr, r holds for any triangle if and only if it holds for any isosceles triangle whose topangle A 0, π ], or the following inequality holds.8 1 x + x <f, 1 x 1 < x 0. Lemma.5 see [, ]. The homogeneous inequality.7 holds for any acuteangled triangle if and only if it holds for any isosceles triangle whose top-angle A 0, π ], or the inequality.8 holds. Proof. As acute-angled triangles include all isosceles triangles whose top-angle is less than or equal to π, Lemma.5 straightforwardly follows from Lemma.4 and Lemma.1. Lemma.6. Define.9 G 1 x := ln 1 x + ln 1 + x ln 1 x + ln + x + ln 1 + x ln. Then G 1 is decreasing on 1, 1, and ln + ln.10 lim x G 1 x = 1 4 ln ln < G 1 x < 1 = lim x 1 + G 1x. Page 10 of 16 Proof. Let G 1 be the derivative of G 1. It is easy to see that.11 G 1x = 4xg 1 x ln 1 x+ ln +x+ ln 1+x ln 1 x +x
with.1 g 1 x := x 1 ln 1 x x+[ln +x ln ]+x+1 ln 1+x. Moreover, we know that.1 g 1x = ln 1 x ln + x + ln 1 + x + ln and.14 g 1x = 8x 1 x + x. Now we show that G 1 is decreasing on 1, 1. i It is easy to see that g 1x 0 when 1 < x 0, and g 1 is increasing on 1, 0]. Thus, g 1x g 10 = 0, and g 1 is decreasing on 1, 0]. Therefore, g 1 x g 1 0 = 0, and G 1x 0. This means that G 1 is decreasing on 1, 0]. ii It is easy to see that g 1x 0 when 0 x < 1, and g 1 is decreasing on [ 0, 1. Thus, g 1 x g 10 = 0, and g 1 is decreasing on [ 0, 1. Therefore, g 1 x g 1 0 = 0, and G 1x 0. This means that G 1 is decreasing on [ 0, 1. Combining i and ii, it follows that G 1 is decreasing on 1, 1 and.10 holds, and hence the proof is complete. Lemma.7. Define.15 G x := ln 1 x ln 5 x + ln 1 x ln. Page 11 of 16
Then G is increasing on [ 1, 1, and.16 G x G 1 = ln + ln. 4 ln ln Proof. Let G be the derivative of G. It is easy to see that.17 G x = and.18 h x = where 4xg x ln 5 x + ln 1 x ln 1 x 5 x, g x = xh x, 16x 1 x 5 x ; g x := 1 x [ln 1 x+ln 1 + x]+ x 5 ln 5 x + 5 x ln, and.19 h x = ln 1 x ln 5 x + ln. Thus it follows that h x < 0 when 1 x < 1, and h is decreasing on [ 1, 1, and h x h 1 = ln 7 4 + 4 < 0. Therefore g x > 0, and g is increasing on [ 1, 1, and g x g 1 = 6 + 1 ln 8 ln 8 ln + 1 > 0. Page 1 of 16 This means that G x > 0, and G is increasing on [ 1, 1, and.16 holds. The proof of Lemma.7 is thus completed.
. The Proof of Theorem 1.1 Proof. i The first inequality of 1. is equivalent to.1 cos A + cos B + cos C µ sin A + sin B + sin C 1 µ with application to the well known identity Taking sin A + sin B + sin C = s R. A π A, B π B and C π C, then inequality.1 is equivalent to µ. sin A + sin B + sin C sin A + sin B + sin C 1 µ for an acute-angled triangle ABC. By the well known identities sin A + sin B + sin C = 4 sin A sin B sin C, and sin A sin B sin C = rs R, the inequality. can be written as follows:. s R µ 1 µ rs s µ R R µ 1 µ r. R Page 1 of 16
Furthermore, by Lemma., the inequality. holds if and only if the following two inequalities µ 1 x + x µ 1.4 x 1 µ 0 x < 1 and.5 5 x µ µ 1 x 1 µ 1 x < 1 hold. In other words,.6 µ min 0 x<1 Gx where.7 Gx = { G1 x G x 0 x < 1, 1 x < 1, while G 1 x and G x are defined by.9 and.15 respectively. By Lemma.6 and Lemma.7, it follows that min Gx = G 1. 0 x<1 Thus the first inequality of 1. holds, and the best constant µ for inequality 1. is Page 14 of 16 µ 0 = ln + ln. 4 ln ln
ii By applying a similar method to i, it follows that the second inequality of 1. is equivalent to s λ λ 1 λ r.8. R R By Lemma.5, the inequality.8 holds if and only if the following inequality holds: λ 1 x + x λ.9 1 x 1 λ 1 < x 0, or equivalently,.10 λ sup G 1 x, 1<x 0 where G 1 x is given by.9. By Lemma.6, it follows that λ 1. Moreover, the second inequality of 1. holds when λ 0 = 1. Thus the second inequality of 1. holds and the best constant λ for inequality 1. is λ 0 = 1. The proof of Theorem 1.1 is hence completed. Page 15 of 16
References [1] O. BOTTEMA, R. Z. DJORDJEVIĆ, R.R. JANIĆ, D.S. MITRINOVIĆ, AND P.M. VASIĆ. Geometric Inequality. Wolters-Noordhoff Publishing, Groningen, The Netherlands, 1969. [] S.-L. CHEN, Inequalities involving R, r, s in acute-angled triangle, Geometric Inequalties in China, Jiangsu Educational Press, Nanjing 1996, No. 7-81. in Chinese [] S.-L. CHEN, The simplified method to prove inequalities in triangle, Studies of Inequalities, Tibet People s Press, Lhasa 000, 8. in Chinese [4] S.-L. CHEN, A new method to prove one kind of inequalities-equate substitution method, Fujian High-School Mathematics, No. 0-, 199. in Chinese [5] Y.-D. WU. The best constant for a geometric inequality, J. Inequal. Pure Appl. Math., 64 005, Art. 111. [ONLINE http://jipam.vu.edu.au/ article.php?sid=585]. [6] X.-Z. YANG AND H.-Y. YIN, The comprehensive investigations of trigonometric inequalities for half-angles of triangle in China, Studies of Inequalities, Tibet People s Press, Lhasa 000, No.1 174. in Chinese Page 16 of 16