sum of polynoms of given property

Solution 1:

you can let $$G(x)=f(x)e^{-x},-\infty<x<\infty$$ then we have $$G'(x)=f'(x)e^{-x}-f(x)e^{-x}=-P(x)e^{-x}\le 0$$ so this $G(x)$ is decreasing and $$\lim_{x\to+\infty}G(x)=\lim_{x\to+\infty}f(x)e^{-x}=0$$ $$f(x)e^{-x}=G(x)\ge \lim_{x\to+\infty}G(x)=0$$ so $$f(x)\ge 0$$

Solution 2:

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.