答：a=1,f(x)=alnx+2/(x+1)f(x)=lnx+2/(x+1),x>=1求导：f'(x)=1/x-2/(x+1)²f'(x)=(x²+2x+1-2x) / [x(x+1)²]f'(x)=(x²+1) / [x(x+1)²]>0恒成立所以：f(x)是单调递增函数所以：f(x)>=f(1)所...