AN INEQUALITY ON TERNARY QUADRATIC FORMS IN TRIANGLES

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1 AN INEQUALITY ON TERNARY QUADRATIC FORMS IN TRIANGLES NU-CHUN HU Department of Mathematics Zhejiang Normal University Jinhua 31004, Zhejiang People s Republic of China. nuchun@zjnu.cn vol. 10, iss. 1, art. 15, 009 Received: 07 May, 008 Accepted: 5 February, 009 Communicated by: S.S. Dragomir 000 AMS Sub. Class.: 6D15. Key words: Abstract: Positive semidefinite ternary quadratic form, arithmetic-mean geometric-mean inequality, Cauchy inequality, triangle. In this short note, we give a proof of a conjecture about ternary quadratic forms involving two triangles and several interesting applications. Page 1 of 13

2 1 Introduction 3 Preliminaries 4 3 Proof of the Main Theorem 9 4 Applications 11 vol. 10, iss. 1, art. 15, 009 Page of 13

3 1. Introduction In [3], Liu proved the following theorem. Theorem 1.1. For any ABC and real numbers x, y, z, the following inequality (1.1) x cos A + y cos B + z cos C yz sin A + zx sin B + xy sin C. In [6], Tao proved the following theorem. Theorem 1.. For any A 1 B 1 C 1, A B C, the following inequality (1.) cos A 1 cos A + cos B 1 cos B + cos C 1 cos C sin A 1 sin A + sin B 1 sin B + sin C 1 sin C. Then, in [4], Liu proposed the following conjecture. Conjecture 1.3. For any A 1 B 1 C 1, A B C and real numbers x, y, z, the following inequality (1.3) x cos A 1 cos A + y cos B 1 cos B + z cos C 1 cos C yz sin A 1 sin A + zx sin B 1 sin B + xy sin C 1 sin C. vol. 10, iss. 1, art. 15, 009 Page 3 of 13 In this paper, we give a proof of this conjecture and some interesting applications.

4 . Preliminaries For ABC, let a, b, c denote the side-lengths, A, B, C the angles, s the semiperimeter, S the area, R the circumradius and r the inradius, respectively. In addition we will customarily use the symbols (cyclic sum) and (cyclic product): 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.1 (see []). Let p i, q i (i = 1,, 3) be real numbers such that p i 0 (i = 1,, 3), 4p p 3 q 1, 4p 3 p 1 q, 4p 1 p q 3 and (.1) 4p 1 p p 3 p 1 q 1 + p q + p 3 q 3 + q 1 q q 3. Then the following inequality holds for any real numbers x, y, z, (.) p 1 x + p y + p 3 z q 1 yz + q zx + q 3 xy. Lemma.. For ABC, the following inequalities hold. (.3) (.4) cos B cos C sin A > sin A, cos C cos A sin B > sin B, cos A cos B 3 3 (.5) 4 sin C > sin C. Proof. We will only prove (.3) because (.4) and (.5) can be done similarly. Since vol. 10, iss. 1, art. 15, 009 Page 4 of 13 S = 1 bc sin A = s(s a)(s b)(s c)

5 and cos B s(s b) =, cos C s(s c) ca =, ab then it follows that cos B cos C sin A s(s b) s(s c) 3 3S ca ab b c 4s (s b)(s c) 7s (s a) (s b) (s c) a bc b 4 c 4 4 a 7(s a) (s b)(s c) b 3 c 3 (.6) 4b 3 c 3 7a (s a) (s b)(s c). On the other hand, by the arithmetic-mean geometric-mean inequality, we have the following inequality. 7a (s a) (s b)(s c) = a(s a) a(s a) (s b)(s c) [ a(s a) + 1 a(s a) + (s b)(s c) 3 ] 3 (b + c a) = 4 [bc < 4b 3 c 3. 4 Therefore the inequality (.6) holds, and hence (.3) ] 3 vol. 10, iss. 1, art. 15, 009 Page 5 of 13

6 Lemma.3. For ABC, the following equality (.7) sin 4 A cos B cos C = (R + 5r)s4 (R + r)(16r + 5r)rs + (4R + r) 3 r R 3 s. Proof. By the familiar identity: a + b + c = s, ab + bc + ca = s + 4Rr + r, abc = 4Rrs (see [5]) and the following identity a 5 (b + c a) = (a + b + c) 6 + 7(ab + bc + ca)(a + b + c) 4 vol. 10, iss. 1, art. 15, (a + b + c) (ab + bc + ca) 7abc(a + b + c) 3 + 4(ab + bc + ca) abc(ab + bc + ca)(a + b + c) 6a b c, it follows that a 5 (b + c a) = 4(R + 5r)rs 4 8(R + r)(16r + 5r)r s + 4(4R + r) 3 r 3, and hence sin 4 A(1 + cos A) = ( a ) 4 (b + c) a R bc = (a + b + c) a 5 (b + c a) 3R 4 abc = (R + 5r)s4 (R + r)(16r + 5r)rs + (4R + r) 3 r 16R 5. Page 6 of 13

7 Thus, together with the familiar identity cos A = s, it follows that 4R sin 4 A cos B cos C sin 4 A cos A = cos A = sin 4 A(1 + cos A) cos A Therefore the equality (.7) is proved. = (R + 5r)s4 (R + r)(16r + 5r)rs + (4R + r) 3 r R 3 s. Lemma.4. For ABC, the following inequality (.8) (R + 5r)s 4 + (R + 5r)(R + r)(r + r)s (4R + r) 3 r 0. Proof. First it is easy to verify that the inequality (.8) is just the following inequality. (.9) (R + 5r)[ s 4 + (4R + 0Rr r )s r(4r + r) 3 ] + r(14r + 31Rr 10r )(4R + 4Rr + 3r s ) Thus, together with the fundamental inequality + 4(R r)(4r 3 + 6R r + 3Rr 8r 3 ) 0. s 4 + (4R + 0Rr r )s r(4r + r) 3 0 vol. 10, iss. 1, art. 15, 009 Page 7 of 13 (see [5, page ]), Euler s inequality R r and Gerretsen s inequality s 4R + 4Rr + 3r (see [1, page 45]), it follows that the inequality (.9) holds, and hence (.8)

8 Lemma.5. For ABC, the following inequality (.10) sin 4 A cos B cos C + 64 sin A 4. Proof. By Lemma.3 and the familiar identity sin A = r, it follows that 4R (.11) sin 4 A cos B cos C + 64 sin A 4 (R+5r)s4 (R+r)(16R+5r)rs +(4R+r) 3 r R 3 s + 4r R 4 (R+5r)s4 +(R+5r)(R+r)(R+r)s (4R+r) 3 r R 3 s 0. Thus, by Lemma.4, it follows that the inequality (.11) holds, and hence (.10) vol. 10, iss. 1, art. 15, 009 Page 8 of 13

9 3. Proof of the Main Theorem Now we give the proof of inequality (1.1). Proof. First, it is easy to verify that (3.1) (3.) (3.3) cos A 1 cos A 0, cos B 1 cos B 0, cos C 1 cos C 0. Next, by Lemma., we have the following inequalities: (3.4) (3.5) (3.6) 4 cos B 1 cos B cos C 1 cos C sin A 1 sin A, 4 cos C 1 cos C cos A 1 cos A sin B 1 sin B, 4 cos A 1 cos A cos B 1 cos B sin C 1 sin C. Thus, in order that Proposition.1 is applicable, we have to show the following inequality. (3.7) 4 cos A 1 cos A cos A 1 sin A 1 cos A sin A + cos B 1 sin B 1 cos B sin B + cos C 1 sin C 1 cos C sin C + sin A 1 sin A. vol. 10, iss. 1, art. 15, 009 Page 9 of 13

10 However, in order to prove the inequality (3.7), we only need the following inequality. (3.8) sin A 1 sin A + sin B 1 cos B 1 cos C 1 cos B cos C cos C 1 cos A 1 + sin C 1 cos A 1 cos B 1 sin C + 8 cos A cos B sin B cos C cos A sin A 1 8 sin A 4. In fact, by the Cauchy inequality and Lemma.5, we have that [ sin A 1 sin A + sin B 1 sin B cos B 1 cos C 1 cos B cos C cos C 1 cos A 1 cos C cos A sin C 1 sin C + cos A 1 cos B 1 cos A cos B [ ] sin 4 A 1 cos B 1 cos C sin A 1 [ sin 4 A cos B cos C A sin sin A 8 sin A Therefore the inequality (3.8) holds, and hence (3.7) Thus, together with inequality (3.4) (3.7), Proposition.1 is applicable to complete the proof of (1.1). ] ] vol. 10, iss. 1, art. 15, 009 Page 10 of 13

11 4. Applications Let P be a point in the ABC. Recall that A, B, C denote the angles, a, b, c the lengths of sides, w a, w b, w c the lengths of interior angular bisectors, m a, m b, m c the lengths of medians, h a, h b, h c the lengths of altitudes, R 1, R, R 3 the distances of P to vertices A, B, C, r 1, r, r 3 the distances of P to the sidelines BC, CA, AB. Corollary 4.1. For any ABC, A 1 B 1 C 1, A B C, the following inequality a cos A 1 cos A + b cos B 1 cos B + c cos C 1 cos C bc sin A 1 sin A + ca sin B 1 sin B + ab sin C 1 sin C. Corollary 4.. For any ABC, A 1 B 1 C 1, A B C, the following inequality wa cos A 1 cos A + w b cos B 1 cos B + w c cos C 1 cos C w b w c sin A 1 sin A + w c w a sin B 1 sin B + w a w b sin C 1 sin C. Corollary 4.3. For any ABC, A 1 B 1 C 1, A B C, the following inequality m a cos A 1 cos A + m b cos B 1 cos B + m c cos C 1 cos C m b m c sin A 1 sin A + m c m a sin B 1 sin B + m a m b sin C 1 sin C. vol. 10, iss. 1, art. 15, 009 Page 11 of 13 Corollary 4.4. For any ABC, A 1 B 1 C 1, A B C, the following inequality

12 h a cos A 1 cos A + h b cos B 1 cos B + h c cos C 1 cos C h b h c sin A 1 sin A + h c h a sin B 1 sin B + h a h b sin C 1 sin C. Corollary 4.5. For any ABC, A 1 B 1 C 1, A B C, the following inequality R1 cos A 1 cos A + R cos B 1 cos B + R 3 cos C 1 cos C R R 3 sin A 1 sin A + R 3 R 1 sin B 1 sin B + R 1 R sin C 1 sin C. Corollary 4.6. For any ABC, A 1 B 1 C 1, A B C, the following inequality r 1 cos A 1 cos A + r cos B 1 cos B + r 3 cos C 1 cos C r r 3 sin A 1 sin A + r 3 r 1 sin B 1 sin B + r 1 r sin C 1 sin C. vol. 10, iss. 1, art. 15, 009 Page 1 of 13

13 References [1] O. BOTTEMA, R.Ž. DJORDJEVIĆ, R.R. JANIĆ, D.S. MITRINOVIĆ AND P.M. VASIĆ, Geometric Inequalities, Wolters-Noordhoff Publishing, Groningen, [] S.J. LEON, Linear Algebra with Applications, Prentice Hall, New Jersey, 005. [3] J. LIU, Two results about ternary quadratic form and their applications, Middle- School Mathematics (in Chinese), 5 (1996), [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), 8. Kluwer Academic Publishers Group, Dordrecht, [6] C.G. TAO, Proof of a conjecture relating two triangle, Middle-School Mathematics (in Chinese), (004), vol. 10, iss. 1, art. 15, 009 Page 13 of 13