An equivalent form of fundamental triangle inequality and its applicationsMihály Bencze , Hui-Hua Wu 1 2 and Shan-He Wu 2 1 Str. Harmanului 6, 505600 Sacele-Négyfalu, Jud. Brasov, Romania E-mail:
[email protected] 2 Department of Mathematics and Computer Science, Longyan University, Longyan Fujian 364012, PR China E-mail:
[email protected] Abstract: An equivalent form of the fundamental triangle inequality is given. The result is then used to obtain an improvement of the Leuenberger's inequality and a new proof of Garfunkel-Banko inequality. Keywords: fundamental triangle inequality; equivalent form; improvement; Garfunkel-Banko inequality; Leuenberger's inequality 2000 AMS Mathematics subject classi cation: 26D05, 26D15, 51M16 1 Introduction and main results In what follows, we denote by A, B, C the angles of triangle ABC , let a, b, c denote the lengths of its corresponding sides, and let s, R and r denote respectively the semi-perimeter, circumradius and inradius of a triangle. We will customarily use the symbol of cyclic sum such as f (a) = f (a) + f (b) + f (c), f (a, b) = f (a, b) + f (b, c) + f (c, a). The fundamental triangle inequality is one of the cornerstones of geometric inequalities for triangle. It reads as follows: 2R2 + 10Rr − r2 − 2(R − 2r) R2 − 2Rr s2 2R2 + 10Rr − r2 + 2(R − 2r) R2 − 2Rr. (1) The equality holds in left (or right) inequality of (1) if and only if the triangle is isosceles. As is well known, the inequality (1) is a necessary and su cient condition for the existence of a triangle with elements R, r and s. This classical inequality has many important applications in the theory of geometric inequality and has received a lot of attention. There exist a large number of papers that have been written about applying the inequality (1) to establish and prove the geometric inequalities for triangle, e.g., see [1-9] and the references cited in them. The objective of this paper is to give an equivalent form of the fundamental triangle inequality. As applications, we shall apply our results to a new proof of Garfunkel-Banko inequality and the improvement of the Leuenberger's inequality, it will be shown that our new inequality can e caciously reduce computational complexity in the proof of certain inequalities for triangle. We state the main result in the following theorem: Theorem 1. For any triangle ABC the following inequalities hold true: 1 Furthermore. R Let 1− 2r = 1 − δ. C be angles of an arbitrary triangle. 4 (2) δ =1− 1− 2r R . R (3) By the Euler's inequality R 2r (see [1]). 2 . 1 s2 1 16δ − 12δ 2 + 3δ 3 − δ 4 4 + 4δ − δ 3 − δ 4 .1 δ(4 − δ)3 4 where s2 R2 1 (2 − δ)(2 + δ)3 . R 0<δ 1. R 2 By applying identities (4) and (5) to inequality (3). then we have the inequality tan2 A 2 2 − 8 sin (6) A B C sin sin . Let A. we observe that 0 1− 2r < 1. 2 2 2 The equality holds in (6) if and only if the triangle ABC is equilateral. B. 3 Application to new proof of Garfunkel-Banko inequality Theorem 2. 2 4 R 4 After factoring out common factors. This completes the proof of Theorem 1. we obtain that 1 2 + 10 δ − δ 2 2 1 − δ − δ2 2 2 (5) 1 − 2 1 − 2 δ − δ2 2 2 (1 − δ) s2 R2 1 2 + 10 δ − δ 2 2 that is 1 − δ − δ2 2 1 + 2 1 − 2 δ − δ2 2 (1 − δ). 2 Proof of Theorem 1 We write the fundamental triangle inequality (1) as 2+ r2 2r 10r − 2 −2 1− R R R 1− 2r R s2 R2 2+ 10r r2 2r − 2 +2 1− R R R 1− 2r . the equality holds in left (or right) inequality of (2) if and only if the triangle is isosceles. the above inequality can be transformed into the desired inequality (2). the identity (4) is equivalent to the following identity: r 1 = δ − δ2 . (4) Also. 4 Application to the improvement of Leuenberger's inequality In 1960.02206078402 · · · . (9) By applying inequality (2) and identity (5) to inequality (9). Proof. (10) R a After three years. it is the root on the interval (0. 2 s2 B C r A . 15 ) of the following equation (13) 405k 5 + 6705k 4 + 129586k 3 + 1050976k 2 + 2795373k − 62181 = 0. it was rst proved by Banko in [11]. The Garfunkel-Banko inequality is proved. 3 . From the identities for triangle (see [2]): (4R + r)2 A = − 2. (11) 2(R + r) a Mitrinovi¢ et al [2. 5R + 12r + k0 (2r − R) a 1 where k0 = 0.Inequality (6) was proposed by Garfunkel as a conjecture in [10]. that is. sin sin sin = 2 2 2 4R it is easy to see that the Garfunkel-Banko inequality is equivalent to the following inequality: tan2 4− 2r R s2 r − 4+ 2 R R 2 (7) (8) 0. Leuenberger presented the following inequality concerning the sides and the circumradius of a triangle (see [1]) √ 3 1 . we give a simpli ed proof of Garfunkel-Banko inequality by means of the equivalent form of fundamental triangle inequality. Steining sharpened inequality (10) in the following form (see also [1]): √ 1 3 3 . a uni ed improvement of the inequalities (11) and (12) was given by Wu [12].173] showed another sharpened form of (10). it follows that 4− 2r R s2 r − 4+ R2 R 2 1 1 (4 − 2δ + δ 2 )(2 − δ)(2 + δ)3 − (4 − δ)2 (2 + δ)2 4 4 1 2 = − δ (2 + δ)2 (1 − δ)2 4 0. √ 1 11 3 . + (3 3 − 4)Rr (12) Recently. p. In this section. as follows 1 a 2R2 5R − r √ . which implies the required inequality (9). By using the identities for triangle (see [2]): ab = s2 + 4Rr + r2 . For any triangle (14) Proof. ABC the following inequality holds true: √ 25Rr − 2r2 1 . (15) 1 a 2 − 25Rr − 2r2 16R2 r2 = = (s2 + 4Rr + r2 )2 25Rr − 2r2 − . 1 1 δ(4 − δ)3 − 4 2 −δ 4 + 4δ 3 − 25 2 δ + 17δ 2 then.We show here a new improvement of the inequalities (11) and (12). from the monotonicity of the function f (x) = x2 − −δ 4 + 4δ 3 − we deduce that s4 25 − −δ 4 + 4δ 3 − δ 2 + 17δ R4 2 s2 1 + (4 − δ)2 (4 − δ 2 )2 δ 2 R2 16 1 2 1 25 1 δ (4 − δ)6 − δ(4 − δ)3 −δ 4 + 4δ 3 − δ 2 + 17δ + (4 − δ)2 (4 − δ 2 )2 δ 2 16 4 2 16 1 = (4 − δ)2 (1 − δ)2 (6 − δ)δ 3 8 0. Theorem 3. 4 . 2 4 4 25 2 2 δ 1 + 17δ x+ 16 (4−δ)2 (4−δ 2 )2 δ 2 . and the identity (5). 2 − 25Rr − 2r2 16R2 r2 0. we have abc = 4sRr. R2 16 (16) Note that the inequality (2): s2 R2 and 1 δ(4 − δ)3 4 = δ( 15 23 1 − δ) + δ 3 + δ 4 > 0. a 4Rr with equality holding if and only if the triangle ABC is equilateral. 2 R2 r 2 16s 16R2 r2 1 6 s2 r 2 16R s4 − R4 17r 4r2 − 2 R R s2 + R2 4r r2 + 2 R R 2 = 1 s4 25 − −δ 4 + 4δ 3 − δ 2 + 17δ 16R6 s2 r2 R4 2 s2 1 + (4 − δ)2 (4 − δ 2 )2 δ 2 . which is stated in the Theorem 3 below. The above inequality with the identity (16) lead us to that 1 a The Theorem 3 is thus proved. S. pp. 1969. References [1] O.022(2r − R) 5R + 12r + k0 (2r − R) 4Rr 4Rr As a further improvement of the inequality (14). it is worth noticing that the inequality (13) and the inequality (14) are incomparable in general. r = 3 − 3 . R. then R = 1. Geometric Inequalities. R (19) Acknowledgments. Math. J. China. Kuang. Jinan. Addenda to the Monograph: Recent Advances in Geometric Inequalities (I). Netherlands. third ed. 2003 (in Chinese). J. Kluwer Academic Publishers. Mitrinovi¢ and P. Pe£ari¢ and V. Groningen. China. Zhang. [6] Sh. < − − 5R + 12r + 0. Determine the best constant k for which the inequality below holds 1 a 1 4Rr 25Rr − 2r2 + k 1 − 2r R r3 . [2] D. Shandong Science and Technology Press. Djordjevi¢. 5 . S. Bottema. Open Problem.116 (electronic). E. Lasha. J. The research was supported by the program for new century excellent talents in the newly-built universities of Fujian province of China. √ √ 25Rr − 2r2 5R − r 3 1 √ . we propose the following signi cant problem.Remark.11934 · · · > 0. c = 1. [4] J. which can be observed by the following fact. M. Dordrecht. Liu. Wu and Zh. V. Applied Inequalities. Vasi¢. b = 2.-H. Mitrinovi¢. then R = 1. Wolters-Noordho. (17) a 4Rr R 2R2 + (3 3 − 4)Rr √ √ √ 1 25Rr − 2r2 3 3 3 . Mitrinovi¢.-H. Jani¢.. Recent Advances in Geometric Inequalities. A class of inequalities related to the angle bisectors and the sides of a triangle. 1989.023(2r − R) √ √ √ Letting a = 2. r = 2 − 1. Inequal. Chen. b = 1. BOTTEMA. C. S. c = 2. (11) and (12) because from the Euler's inequality R 2r it is easy to verify that the following inequalities hold for any triangle. E.00183 · · · < 0. [5] B. 7 (3) (2006). (18) a 4Rr 2(R + r) R In addition. 4Rr 5R + 12r + k0 (2r − R) 4Rr 5R + 12r + 0. Z. D. What We See. Pe£ari¢. R. Volenec. [3] D. Pure Appl. R. Journal of Ningbo University. Tibet People's Publishing House. Article 108. 2004 (in Chinese). direct calculating gives √ √ √ √ 11 3 11 3 25Rr − 2r2 25Rr − 2r2 = −0. Volenec and J. direct calculating gives 2 √ √ √ √ 25Rr − 2r2 11 3 25Rr − 2r2 11 3 − > − = 0. The inequality (14) is stronger than the inequalities (10). √ √ Letting a = 3. 4(2)(1991). Q. . [10] J. 6 (3) (2003). 9 (1983). au/v6n3. The best constant for a geometric inequality. Pure Appl.edu. 10 (1984). Crux Math.-H. Rep. Math.-L.117 (electronic). Wu and Zh.[7] Sh. Geometric Inequalities in China. Coll. 168. Wu. Certain inequalities for the elements R.vu. China. Garfunkel. Jiangsu.-H. Problem 825. D. 79. [8] Sh. Lasha. Tibet People's Publishing House.6 (4) (2005).html]. RGMIA Res. [12] Y. Chen. [9] Sh. Solution of Problem 825. Banko. Crux Math. Some strengthened results on Gerretsen's inequality.. Chen. J. Article 111. Article 16 [Available online at URL: http://rgmia. 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