lnx>2(x-1)/(x+1)
另f(x)=lnx-2(x-1)/(x+1)
求导 f'(x)=1/x-[(x+1)-(x-1)]/(x+1)^2
=1/x-2/(x+1)^2
=[(x+1)^2-2x]/(x+1)^2
=(x^2+1)/(x+1)^2>0
所以f(x)在定义域单调递增
且当x=1时
f(x)=0
所以当x>1时,f(x)>0
所以lnx-2(x-1)/(x+1)>0,即lnx>2(x-1)/(x+1)