TITLE: sum of polynoms of given property
QUESTION [5 upvotes]: I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.
I have to prove that:

$f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.

I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.

REPLY [3 votes]: Because $f$ is a nonzero polynomial, it has only finitely many zeros $x_1< x_2 < \dots < x_r$. Moreover, $f(x) \underset{x \to \pm \infty}{\sim} P(x)$ so $P(x) \geq 0$ on $(- \infty,x_1]$ and $[x_r,+ \infty)$. Notice also that the sign of $f$ is constant on $[x_i,x_{i+1}]$.
Let $1 \leq i \leq r-1$. Because $f(x_i)=f(x_{i+1})=0$ there exists $y_i \in (x_i,x_{i+1})$ such that $f'(y_i)=0$ by mean value theorem. But $f(y_i)=P(y_i)+f'(y_i) \geq 0$. Therefore, $f$ is non-negative on $[x_i,x_{i+1}]$.
Finally, you deduce that $f$ is non-negative.