求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值

题目

求函数F(x)=[in²X+1/(2010sin²X)][os²x+1/(2010cos²X)]的最小值

答案

令a=(sinx)^2,b=(cosx)^2
则F(x)=(a+1/2010a)(b+1/2010b)
=ab+a/2010b+b/2010a+1/(2010^2*ab)
=ab+[(a^2+b^2)/2010ab]+1/(2010^2*ab)
=ab+[(a+b)^2-2ab]/2010ab+1/(2010^2*ab)
=ab+(1-2ab)/2010ab+1/(2010^2*ab)
=ab+1/2010ab-1/1005+1/(2010^2*ab)
=ab+(2011/2010^2ab)-1/1005
>=2根号(2011/2010^2)-1/1005
=2根号[（2011）/2010] -1/1005