J87. Prove that for any acute triangle ABC, the following inequality holds: 1 1 1 1 + + ≥ . −a2 + b2 + c2 a2 − b2 + c2 a2 + b2 − c2 2Rr Proposed by Mircea Becheanu, Bucharest, Romania First solution by Brian Bradie, VA, USA Using the Law of Cosines and the formula R= abc , 4rs we can rewrite the original inequality as b c a + + ≥ 4s = 2(a + b + c), cos α cos β cos γ (1) where α, β and γ are the acute angles in the triangle. Using the Law of Sines, we can write sin γ sin β c=a and b = a . sin α sin α Substituting into (1) yields tan α + tan β + tan γ ≥ 2(sin α + sin β + sin γ). (2) On (0, π2 ), tan x is convex and sin x is concave; it therefore follows from Jensen’s inequality that √ π α+β+γ = 3 tan = 3 3, and tan α + tan β + tan γ ≥ 3 tan 3 3 √ π 3 3 α+β+γ = 3 sin = . sin α + sin β + sin γ ≤ 3 sin 3 3 2 Hence, (2) holds with equality if and only if α = β = γ. Thus, the original inequality holds with equality if and only if the triangle is an equilateral triangle. Second solution by Mihai Miculita, Oradea, Romania Because 2Rr = 2 Sp · b2 abc 4S = abc 2p = abc a+b+c , the given inequality is equivalent to 1 1 abc 1 + 2 + 2 ≥ . (1) 2 2 2 2 2 2 +c −a a +c −b a +b −c a+b+c Mathematical Reflections 3 (2008) 5 Let us observe that since ABC is an acute triangle the following is true b2 + c2 − a2 > 0 ⇒ 2(b − c)2 (b2 + c2 − a2 ) ≥ 0 ⇔ (b − c)2 (2b2 + 2c2 − 2a2 ) ≥ 0 ⇔ (b − c)2 [(b + c)2 + (b − c)2 − 2a2 ] ≥ 0 ⇔ (b2 − c2 )2 + (b − c)4 − 2a2 (b − c)2 ≥ 0 ⇔ (b − c)4 − 2a2 (b − c)2 + a4 ≥ a4 − (b2 − c2 )2 ⇔ [a2 − (b − c)2 ]2 ≥ (a2 + b2 − c2 )(a2 + c2 − b2 ) p ⇔ a2 − (b − c)2 ≥ (a2 + b2 − c2 )(a2 + c2 − b2 ) p ⇔ (a + b − c)(a + c − b) ≥ (a2 + b2 − c2 )(a2 + c2 − b2 ). (2) Thus, using the AM-GM inequality and using the result in (2) we have that: 1 1 1 1 ≥p + 2 2 2 2 b2 + c2 − a2 a2 + c2 − b2 (b + c − a )(a2 + c2 − b2 ) 1 ≥ . (3) (b + c − a)(a + c − b) Summing up inequality (3) and the two obtained by a circular permutation of the letters we obtain 1 1 1 1 1 1 + + = + b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2 2 b2 + c2 − a2 a2 + c2 − b2 1 1 1 1 1 1 + + + + 2 b2 + c2 − a2 a2 + b2 − c2 2 a2 + c2 − b2 a2 + b2 − c2 1 1 + ≥ (b + c − a)(a + c − b) (b + c − a)(a + b − c) 1 + (a + c − b)(a + b − c) a+b+c = (b + c − a)(a + c − b)(a + b − c) 1 1 1 ⇒ 2 + 2 + 2 2 2 2 2 b +c −a a +c −b a + b2 − c2 a+b+c ≥ . (4) (b + c − a)(a + c − b)(a + b − c) It is known that p (b + c − a) + (a + c − b) (b + c − a)(a + c − b) ≤ = c. 2 Multiplying the above inequality with its respective ones obtained by circular permutation of letters we obtain (b + c − a)(a + c − b)(a + b − c) ≤ abc. (5) Mathematical Reflections 3 (2008) 6 Using (4) and (5) we readily obtain the desired inequality (1). Third solution by Ovidiu Furdui, Cluj, Romania We will use the following standard trigonometric formulae s = 4R cos B C A cos cos 2 2 2 and r = 4R sin A B C sin sin , 2 2 2 where s denotes the semiperimeter of triangle ABC. It is simply to check, by using the preceding formulas that 4sRr = abc. Let f : (0, π2 ) → R be the function defined by f (x) = cosx x . A calculation shows that x + x sin2 x + sin(2x) f ′′ (x) = > 0, cos3 x and hence, f is a convex function. Using the Law of Cosines combined with Jensen’s inequality for convex functions we get that X 1 1 1 1 X a 1 + + = = −a2 + b2 + c2 a2 − b2 + c2 a2 + b2 − c2 2bc cos A 2abc cos A cyclic ≥ 1 ·3· 2abc a+b+c 3 cos A+B+C 3 = cyclic 2s 1 = , abc 2Rr and the problem is solved. Fourth solution by Tarik Adnan Moon, Kushtia, Bangladesh X 1 1 ≥ 2 2 2 −a + b + c 2Rr cyc We know that, −a2 + b2 + c2 = 2bc · cos A. So,we need to prove that, X 1 1 ≥ 2bc · cos A 2Rr cyc Lemma 1: We know that, [ABC] = sr = abc 4R =⇒ 4sr = abc R After multiplying by 2abc we get, X cyc a abc 4sr ≥ = = 4s cos A Rr r By Cauchy-Schwarz inequaltiy we get, ! ! X a X ≥ a · cos A cos A cyc cyc Mathematical Reflections 3 (2008) X cyc !2 a = 4s2 ...(1) 7 Lemma 2: We know that, X cyc a · cos A ! = 2sr R So, it is left to prove that, X cyc a · cos A ! = 2sr ≤ s ⇔ R ≥ 2r R And we are done. Some words about the lemmas: Lemma 1: Straightforward, just need to use extended law of sines. Lemma 2:We know that,P a cos A = 2RQ sin A · cos A = R · sin 2A Then,we use the identity, sin 2A = 4 sin A Q and using the extended law of sines we obtain, 4R2 sin A = bc sin A = 2[ABC] P From these three we obtain, a cos A = 2[ABC] = 2sr R R Also solved by Andrea Munaro, Italy; Arkady Alt, San Jose, California, USA; Daniel Campos Salas, Costa Rica; Daniel Lasaosa, Universidad Publica de Navarra, Spain; G.R.A.20 Math Problems Group, Roma, Italy; Ivanov Andrey, Chisinau, Moldova; Athanasios Magkos, Kozani, Greece; Michel Bataille, France; Ricardo Barroso Campos, Spain; Roberto Bosch Cabrera, Cuba; Samin Riasat, Notre Dame College, Dhaka, Bangladesh; Vicente Vicario Garcia, Huelva, Spain. Mathematical Reflections 3 (2008) 8