Sum of derivatives of a polynomial 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?
 A: 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, $\forall x$,
$$f(x) \ge \min(f) = f(z_0) = p(z_0) + f^\prime (z_0) = p(z_0) \ge 0 $$
A: Instead of $L_n := (1 + D + D^2 + \dots + D^{n})$, use $L_{\infty} := (1 + D + D^2 + \dots)$, which comes to the same thing for polynomials of degree $n$ or less. If we let
$$\sigma(x) = p(x)+p'(x)+p''(x)+\dots+p^{(n)}(x) = L_{\infty}(p)(x)$$
then we can sum the geometric progression in $D$ to get  
$$\sigma(x) = ((1 + D + D^2 + \dots)p)(x) = ((1-D)^{-1}p)(x)$$
Thus
$$p(x) = ((1-D)\sigma)(x)$$  
This might look like trickery, but if you check it against the original expression for $\sigma$ you can see that it works:  
$$((1-D)\sigma)(x) = p(x)+p'(x)+p''(x)+\dots+p^{(n)}(x) - 
(p'(x)+p''(x)+\dots+p^{(n)}(x))$$ 
$$= p(x)$$
Now define $\tau(x) = e^{-x}\sigma(x)$, so $\tau'(x) = e^{-x}(\sigma'(x) - \sigma(x)) = e^{-x}((D-1)\sigma)(x) = -e^{-x}p(x)$. By hypothesis, this is $ \le 0$ for all $x \in \mathbb R$. Also, $\sigma$ is a polynomial, so $\tau(x) \to 0$ as $x \to \infty$. Therefore $\tau(x) \ge 0$ for all $x$.  
Hence $\sigma(x) \ge 0$ for all $x$, which is what we wanted.
