Let $p(x)$ be a polynomial of degree $n$ satisfying $p(x)\geq 0$ for all $x$. That is, for all $x$, $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \geq 0$, $a_n\neq 0$.
Show that p(x)+p'(x)+p''(x)+\cdots+p^{(n)}(x)\geq 0 for all $x$ where $p^{(i)}(x)$ is the $i^\text{th}$ derivative.
My interest: I know that, we can rewrite the LDE as follows
p(x)+p'(x)+p''(x)+\cdots+p^{(n)}(x) = Lp(x)
where $L := I + D + D^2 + \cdots + D^{(n)}$. Can we say anything about a linear operator of this kind so that it does not change the sign of the input it takes? I can try to solve the question by writing for all the derivatives and factoring them using $p(x)$, but I think there should be a clever way of showing this by the properties of $L$. Can I figure out a solution by just looking at $L$ and the sign of $p(x)$ as in the question? Where should I look for that?