Assume that:$f(0) \geq 0$ and $f^\prime(x) \geq f(x)$ show $f(x)\geq 0 \forall x \in (0,\infty)$

$f$ satisfies the ODE $$u'(x) = u(x) + g(x)$$ with initial conditions $u(0) = f(0)$, and where $g(x) = f'(x)-f(x) \geq 0.$

This ODE has a solution $$u(x) = f(0)e^x + e^x\int_0^x e^{-y}g(y)\,dy$$ which is clearly positive everywhere. Uniqueness isn't immediate since $g(x)$ is not necessarily continuous, but since $0$ is the unique solution to the ODE $$(f-u)' = (f-u)$$ with initial conditions $(f-u)(0) = 0$, it follows that $f=u$.

$$f^\prime(x) \geq f(x) \Longleftrightarrow e^{-x}(f^\prime(x) - f(x))\ge 0 \Longleftrightarrow \frac{d}{dx}[e^{-x}f(x)]\ge 0$$

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