令x1>x2>=1 则f(x1)-f(x2) =-x1+2x1+x2-2x2 =(x2-x1)(x2+x1)-2(x2-x1) =(x2-x1)(x2+x1-2) x1>x2 x2-x1<0 x1>1,x2>=1 x2+x1-2>0 所以(x2-x1)(x2+x1-2)<0 即x1>x2>=1时，f(x1)
