设f(x)在R上有3阶连续导数，且其直到3阶的导数恒取正值，若对任何x成立f'''(x)<=f(x)，试证明对任何x成立f'(x)<1.7f(x)

.....up不止为了10月

给一个简单的证明吧,这样的题目关键做辅助函数T(x)=∫[f'''(x)/f(x)]dx<=x=∫[1/f(x)]df''(x)=f''(x)/f(x) + ∫[f''(x)f'(x)/f²(x)]dx=f''(x)/f(x) + ∫[1/f²(x)]df²'(x)=f''(x)/f(x) +[f'(x)/f(x)]²*(1/2)+ (1/6)*∫[f'(x)/f(x)]³/dx>(1/6)*∫[f'(x)/f(x)]³/dx=(1/6))*∫[f'(x)/f(x)]³*b这里b在[0,x]区间==>x>=(1/6))*∫[f'(x)/f(x)]³*bf'(x)/f(x) < 6^(1/3)*(x/b)<=1.8

不好意思,式子比较复杂,最后那点写错了

还上一年级，晕！高一的都看不懂！

我觉得题目没意思......