Let $f$ be twice differentiable such that $f(0)=0,\ f'(0)>0$ and $f''\geq f$. Prove $f>0$ on $(0,\infty)$
My attempt- Expanding $f$'s taylor polynomial we get $$\frac{f(x)}{x}=\frac{f'(0)x+R_1(x)}{x}=f'(0)+\frac{R_1(x)}{x}$$ Because $R_1(x)/x\rightarrow0$ then theres a neighborhood $(0,\delta)$ in which $ \left |\frac{R_1(x)}{x} \right |<\frac{|f'(0)|}{2}$ and hence $f$ is positive.
Now we assume there's a point $x_0>\delta$ such that $f(x_0)=0$ (wlog we take the "first one" after $\delta$). $f$ has a maxima on $[0,x_0]$ achieved at $x_{\max}$. We get $f(x_{\max})>0>f''(x_\max)$ by contradiction.
Is my proof correct? and if so, would you change anything about it? Thanks in advance.