答案:
(1)∵f'(x)=3x2-2x+=3(x-)2+>0,
∴f(x)是R上的单调增函数.
(2)∵0<x0<,即x1<x0<y1.又f(x)是增函数,
∴f(x1)<f(x0)<f(y1).即x2<x0<y2.
又x2=f(x1)=f(0)=>0=x1,y2=f(y1)=f()=<=y1,
综上,x1<x2<x0<y2<y1.
用数学归纳法证明如下:
①当n=1时,上面已证明成立.
②假设当n=k(k≥1)时有xk<xk+1<x0<yk+1<yk.
当n=k+1时,
由f(x)是单调增函数,有f(xk)<f(xk+1)<f(x0)<f(yk+1)<f(yk),
∴xk+1<xk+2<x0<yk+2<yk+1
由①②知对一切n=1,2,都有xn<xn+1<x0<yn+1<yn.
(3)==yn2+xnyn+xn2-(yn+xn)+≤(yn+xn)2-(yn+xn)+
=[(yn+xn)-]2+.
由(Ⅱ)知0<yn+xn<1.
∴-<yn+xn-<,
∴<()2+=