Let H(x) be polynomial where H(x)>=0 for all x in R. Prove that:
f(x)= H(x)+H'(x)+H"(x)+...>=0.
First, a few simple observations:
- $\displaystyle H$ must be a polynomial of even degree
- Taking derivatives enough times will eventually give zero
- The function $\displaystyle f(x)=H(x)+H'(x)+H''(x)+\ldots$ is a finite sum, and all of the derivatives are polynomials, so $\displaystyle f$ is a polynomial
- Since $\displaystyle H$ has even degree, $\displaystyle f$ has even degree too
Now, to show that $\displaystyle f(x)\geq 0$, we should just try to find its minimum. (And in this case, it is enough to do so. Be sure to justify this!) To do this, we set the derivative equal to zero:
$\displaystyle f'(x)=H'(x)+H''(x)+H'''(x)+\ldots=0$
and "solve". Suppose $\displaystyle x=c$ is a solution. We can put it back into $\displaystyle f$ to see what value we get:
$\displaystyle f(c)=H(c)+(H'(c)+H''(c)+\ldots)=H(c)+0=H(c)$
Since $\displaystyle H$ is positive for all $\displaystyle x$, we have that $\displaystyle f(c)=H(c)\geq 0$. Therefore, all minimum values of $\displaystyle f$ are greater than or equal to zero, so all of $\displaystyle f$ must be greater than or equal to zero. This proves it!
Your question basically boils down to showing that all positive even-degree polynomials have a minimum. I would probably try to show that if $\displaystyle x$ is high or low enough, the polynomial eventually becomes monotonic. (This is behaviour makes the "W"-shape where the tails of the polynomial run out to infinity.) Then I would cut the tails off. What would be left is some portion of the polynomial on a closed interval. Since polynomials are the nicest functions we have, the extreme value theorem applies, so it has to take a minimum on that interval (which then serves as a minimum for the whole thing).
Does this sketch work for you? If you have trouble writing it down formally, you can let us know and we can help you through it.