PUMaC 2009 · 代数（B 组） · 第 7 题
PUMaC 2009 — Algebra (Division B) — Problem 7
题目详情
- Find the maximal positive integer n , so that for any real number x we have sin x +cos x ≥ . n
解析
- Find the maximal positive integer n , so that for any real number x we have sin x +cos x ≥ . n 2 n 1 2 Solution. 8. For x = π we need ( − 1) ≥ , hence n is even. Since sin x + cos x = 1, and n n 2 × n/ 2 n 2 × n/ 2 we need to find the minimum of sin x + cos x = sin x + cos x , one would expect that the minimum occurs when sin x = cos x (in analogy with the AM-GM or the power mean √ 1 inequality). For x = π/ 4, we have sin x = cos x = 1 / 2 so we need 2 × ≥ 1 /n which n 2 2 implies n ≤ 8. This can be proved from the binomial formula or using calculus. The last step is to prove the inequality for n = 8. One way to do this is to use the power mean inequality: 8 2 8 2 sin x + cos x 1 sin x + cos x 1 8 2 ( ) ≥ ( ) 2 2 8 8 sin x +cos x 1 hence ≥ , which is the desired result. Using calculus, one can give a different 4 2 2 n n solution by finding the minimum of f ( x )=sin x + cos x .