PUMaC 2012 · 个人决赛（A 组） · 第 2 题

PUMaC 2012 — Individual Finals (Division A) — Problem 2

Let a, b, c be real numbers such that a + b + c = abc . Prove that + + ≥ . 2 2 2 a +1 b +1 c +1 4

解析

Let a, b, c be real numbers such that a + b + c = abc . Prove that + + ≥ . 2 2 2 a +1 b +1 c +1 4 Solution: As the condition and the inequality are invariant under the transformation ( a, b, c ) → ( − a, − b, − c ), [ ) π we may assume that at most one of a, b, c is negative, so WLOG let a, b ≥ 0. Let A, B ∈ 0 , 2 a + b be such that tan A = a and tan B = b , and let C = π − A − B . Then, c = = ab − 1 tan A +tan B − = − tan ( A + B ) = tan C . 1 − tan A tan B 1 1 1 1 1 1 2 2 2 We have + + = + + = cos A + cos B + cos C = 2 2 2 2 2 2 a +1 b +1 c +1 tan A +1 tan B +1 tan C +1 2 2 2 2 2 2 2 cos A + cos B + (sin A sin B − cos A cos B ) = cos A + cos B + (1 − cos A )(1 − cos B ) − 2 2 2 sin A sin B cos A cos B + cos A cos B = 1 − 2 cos A cos B (sin A sin B − cos A cos B ) = 1 − 2 cos A cos B cos C . [ ) 1 π We now show that cos A cos B cos C ≤ . Since A, B ∈ 0 , , cos A ≥ 0 and cos B ≥ 0, so 8 2 1 if cos C < 0 then cos A cos B cos C ≤ 0 ≤ . Otherwise we may assume cos C ≥ 0, so by 8 ( ) 3 cos A +cos B +cos C AM-GM we have cos A cos B cos C ≤ . Finally, since A, B, C ∈ [0 , π ] and 3 ( ) cos A +cos B +cos C A + B + C cos x is concave on this interval, we have by Jensen’s that ≤ cos = 3 3 ( ) π 1 1 1 1 1 cos = . Putting this together gives cos A cos B cos C ≤ , so + + = 2 2 2 3 2 8 a +1 b +1 c +1 3 1 − 2 cos A cos B cos C ≥ . 4