摘要：其中.证明:,(III)证明:.解: =3x2-2x+ = 3(x-)2+ >0 . ∴f(x)是R上的单调增函数．(II)∵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．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由知对一切n=1.2.-.都有xn<xn+1<x0<yn+1<yn．(III) = = yn2+xnyn+xn2-(yn+xn)+ ≤(yn+xn)2-(yn+xn)+

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