已知函数f(x)是在(0,+∞)上妹一点均可导函数,若xf'(x)>f(x)在x>0时恒成立求证:当x1>0,x2>0时,有f(x1+x2)>f(x1)+f(x2)

Question

已知函数f(x)是在(0,+∞)上妹一点均可导函数,若xf'(x)>f(x)在x>0时恒成立
求证:当x1>0,x2>0时,有f(x1+x2)>f(x1)+f(x2)

Answer 1

构建函数:[F(x)/x]’
[F(x)/x]’=[xF’(x)-f(x)]/x^2
[F(x)/x]在X>0上单增
F(x1+x1)/(x1+x2)>f(x1)/x1;
F(x1+x2)>[(x1+x2)*f(x1)/x1]` (1)
F(x1+x1)/(x1+x2)>f(x2)/x2
F(x1+x2)>[(x1+x2)* f(x2)/x2 ] (2)
(1) ,(2)相加整理就得出结论了!