If $f(x) \geq 0$, then $g(x) =f(x)+f'(x)+f''(x)+...+f^r(x) \geq 0$

Question

I saw the other questions regarding this problem. I am trying to approach it a bit differently.

We can write $g(x)$ as $(D^r+D^{r-1}+D^{r-2}+...+D+1)f=g(x) \implies \frac{(D^{r+1}-1)}{(D-1)}f(x)=g(x) \implies f(x) = (1+D^{r+1}+...)[g(x)-g'(x)]=g(x)-g'(x) \geq 0$.

[$r$ =$deg f(x)$, $f(x)$ is a polynomial, greater or equal to zero for all $x$. ]

Now, $1 \geq g'(x)/g(x) \implies e^x \geq \ln{g(x)}$, which is defined for all $x$. Now, $g(x)$ has to be $>0$ $\forall x$, for that to occur.

Can someone take it from here?