函数f(x)=[(sinx)^2+1/2010(sinx)^2][(cosx)^2+1/2010(cosx)^2]的最小值是

设(sinx)^2=a,(cosx)^2=b,则a+b=1,a≥0,b≥0.f(x)=[(sinx)^2+1/2010(sinx)^2][(cosx)^2+1/2010(cosx)^2]=(a+1/(2010a))( b+1/(2010b))=ab+a/(2010b)+ b/(2010a)+ 1/(2010^2ab)=ab+(a^2+b^2)/(2010ab) + 1/(2010^2ab)...