思路：x>0时,(x+1)ln(x+1/x)>1可以变形：(x+1)ln(x+1/x)>1ln(x+1/x)＞1/(x+1)ln[(x+1)/x]＞1/(x+1)-ln[(x+1)/x]＜-1/(x+1)ln{[(x+1)/x]^-1}=ln[x/(x+1)]=ln[1-1/(x+1)]＜-1/(x+1)ln[1-1/(x+1)]+1/(x+1)＜0∴只要证明...