TITLE: $p(x)\geq 0 \forall x\Rightarrow p(x)+p'(x)+p''(x)+...+p^{(n)}(x)\geq 0$
QUESTION [6 upvotes]: $p(x)\geq 0 \forall x \in \mathbb{R} \Rightarrow p(x)+p'(x)+p''(x)+...+p^{(n)}(x)\geq 0$, where p(x) is a polynomial of degree n.
I showed:
$a_{n}+...+a_{0}\geq 0$,
$p(x)+p'(x)+p''(x)+...+p^{(n)}(x)=\sum_{m=0}^{n}\sum_{k=m}^{n}a_{k}\frac{n!}{(n-k)!}p(x)^{(k-m)}=\sum_{k=0}^{n}\sum_{m=0}^{k}a_{k}\frac{n!}{(n-k)!}p(x)^{(k-m)}$

REPLY [4 votes]: An another proof,
$Q=P+P'+...+P^{(n)}$
Notice 
$$
P=Q-Q'
$$
Thus 
$$
\forall x\in \mathbb{R}, Q(x)\geq Q'(x).
$$
Let
$$
 \forall x \in \mathbb{R}, \varphi(x)= e^{-x}Q(x)
$$
So,
$$
\varphi'(x)=e^{-x}(Q'(x)-Q(x))\leq 0
$$
Therefore $\varphi$ is decreasing on $\mathbb{R}$. She tends to $0$ when $x \longrightarrow +\infty$, so $Q(x)\geq 0$