证明不等式《高等数学》

ln(1+0)=0,
假设x>0,prove x-ln(1+x)>0,let f(x)=x-ln(1+x),f(0) = 0,f'(x) = 1-[1/(1+x)] = x/(1+x)>0,so f(x) is increasing in [0,+正无穷),so f(x)>0,so x>ln(1+x)。
prove ln(1+x) - [x/(1+x)] >0,let g(x) = ln(1+x) - [x/(1+x)] = ln(1+x) - [1-1/(1+x))],
so g(0) = 0,g'(x) = 1/(1+x) -1/[(1+x)^2] = x/[(1+x)^2] >0,where x>0,
so g(x) is increasing in [0,+正无穷),so g(x)>0 for x>0,so ln(1+x) > x/(1+x)