证明:1<=5sin^4xcos^4x+5sin^2xcos^2x+1<=41/16

令t = (sinx cosx)² = (1/2 sin2x)² = 1/4 * sin²2x
由于sin2x∈[-1 , 1]，于是 t∈[0 , 1/4]
g(t) = f(x) = 5sin^4xcos^4x+5sin²xcos²x+1
= 5t² + 5t + 1
= 5(t + 0.5)² - 1/4
显然，g(t)在[0 , 1/4]上单调递增，因此
1=g(0)≤g(t)≤g(1/4)=41/16
即
1≤5sin^4xcos^4x+5sin²xcos²x+1≤41/16

解：令sin^2x=m,cos^2x=n.则m+n=1, 0<=m,n<=1.
于是sin^4x=4sin^2xcos^2x=4mn
5sin^4xcos^4x+5sin^2xcos^2x+1
=5*4mn(1-4mn)+5mn+1
令t=mn，则0<=t<=1/4
上式=-80t+25t+1
剩下的楼主应该自己可以解决了。