TITLE: Sum of derivatives of a polynomial
QUESTION [24 upvotes]: 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?

REPLY [21 votes]: I have to admit the following solution was proposed to me by a friend, I did not find it:
Let $$f(x) = p(x) + p^\prime(x) + ... + p^{(n)}(x) $$
Note that
$$ f^\prime = p^\prime + p^{\prime \prime} + ...+ p^{(n)} $$ 
that is,
$$ f = p + f^\prime$$
Clearly $n$ is even. Hence $f$ has even degree, too. 
This implies that $f$ attains it's absolute minimum (it's not a maximum, as $f$ behaves like $p$ at infinity) in some point $z_0$, hence 
$ f^\prime(z_0) = 0.$ Consequently, $\forall x$,
$$f(x) \ge \min(f) = f(z_0) = p(z_0) + f^\prime (z_0) = p(z_0) \ge 0 $$