e^x-1>½x²+ln(x+1)f(x)=e^x-1-1/2*x^2-ln(x+1)f'(x)=e^x-x-1/(x+1)x>0, e^x>1+xx+1/(x+1)<1+x则 此时 f'(x)>0, 增函数又 lim(x-->0+)f(x)=x>0故有当x>0,e^x-1>½x²+ln(x+1)
