Junior problems - Phần 2

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Junior problems - Phần 2 Junior problems J175. Let a, b ∈ (0, π ) such that sin2 a + cos 2b ≥ 1 1 sec a and sin2 b + cos 2a ≥ sec b. Prove that 2 2 2 1 cos6 a + cos6 b ≥ . 2 Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Prithwijit De, HBCSE, India We will use the following well-known trigonometric identities (a) sin2 x = 1 − cos2 x, (b) cos 2x = 2 cos2 x − 1, 1 (c) sec x = . cos x The inequalities can be written as 1 2 cos2 b cos a − cos3 a ≥ (1) 2 and 1 2 cos2 a cos b − cos3 b ≥ . (2) 2 π The signs of the inequalities are preserved because cos x is positive when x ∈ 0, . Now by 2 squaring both sides of (1) and (2) and adding them we get 1 cos6 a + cos6 b ≥ . 2 Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain; Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor a Vergata Roma, Italy; Tigran Hakobyan, Armenia. 1 Mathematical Reflections 6 (2010) J176. Solve in positive real numbers the system of equations x1 + x2 + · · · + xn = 1 1 1 1 1 3 x1 + x2 + · · · + xn + x1 x2 ···xn = n + 1. Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buzau, Romania Solution by Tigran Hakobyan, Armenia We have n2 1 1 1 = n2 + ··· + ≥ + x1 + x2 + · · · + xn x1 x2 xn and 1 1 = nn . ≥ x1 +x2 +···+xn n x1 x2 · · · xn n Thus, n3 + 1 ≥ nn + n2 which implies that n ≤ 2. If n = 1 we get a contradiction. For n = 2 we get x1 + x + 2 = 1 1 1 1 x1 + x2 + x1 x2 = 9 2, 1 , 3 , 2, 2 , 1 2 which is (n, x1 , x2 ) ∈ . 3 33 Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad P´blica u de Navarra, Spain; Perfetti Paolo, Dipartimento di Matematica, Universit` degli studi di Tor a Vergata Roma, Italy; Lorenzo Pascali, Universit` di Roma “La Sapienza”, Roma, Italy. a ...