lnx>2(x-1)/(x+1)
另f(x)=lnx-2(x-1)/(x+1)
求導 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)在定義域單調(diào)遞增
且當x=1時
f(x)=0
所以當x>1時,f(x)>0
所以lnx-2(x-1)/(x+1)>0,即lnx>2(x-1)/(x+1)