函数f(x)=(x+1)lnx-x+1.证明:(x-1)f(x)≥0.

当x≥1,f(x)=(x+1)lnx-x+1,f’(x)=(x+1)*1/x+lnx-1=1/x+1nx,
因为x≥1,则lnx≥0,1/x>0,所以f’(x)>0,所以f(x)在[1,+oo）上递增,
则f(x) ≥f(1)=0-1+1=0,又（x-1）≥0所以(x-1)f(x)≥0.
当1>x>0,f(x)=(x+1)lnx-x+1,f’(x)=(x+1)*1/x+lnx-1=1/x+1nx,f’’(x)=1/x-1/x^2=(x-1)/x^2,
因为1>x所以f’’(x)f’(1)=1>0,
则f(x)在（0,1）上递增,则f(x)