TITLE: Assume that:$f(0) \geq 0$ and $f^\prime(x) \geq f(x)$ show $f(x)\geq 0 \forall x \in (0,\infty)$
QUESTION [7 upvotes]: $f$  is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$ and $f(0) \geq 0$ and $f^\prime(x) \geq f(x)$
to show $f(x)\geq 0 \forall x \in (0,\infty)$  
my answer: if $\exists x_0 \in (0,\infty)$
s.t.  $f(x_0)<0$ then $\frac{f(x_0)-f(0)}{x_0} = f^\prime(a_1) \leq 0$
now $a_1 \in (0,x_0)  $
$\frac{f(a_1)-f(0)}{a_1}=f^\prime(a_2) \leq 0$  
continuing this way we get a sequence $(a_i) \rightarrow 0$ s.t.$f^\prime(a_i)\leq 0 \leq f(a_i)$ which is a contradiction
is this correct or not?

REPLY [2 votes]: $$f^\prime(x) \geq f(x) \Longleftrightarrow e^{-x}(f^\prime(x) - f(x))\ge 0 \Longleftrightarrow \frac{d}{dx}[e^{-x}f(x)]\ge 0$$
This implies after integration that: $$f(x)-f(0)= \int_0^x  \frac{d}{dt}[e^{-t}f(t)]dt  \ge 0$$
Hence $$f(x)\ge f(0)\ge 0$$