if $f(x)\geq0$ for all $x$, then $f+f'+f''+f'''+ \cdots + f^n \geq0$ Suppose $f$ is a polynomial of degree $n$, and $f\geq0$ for all $x$, prove that $f+f'+f''+f'''+\cdots +f^n\geq0$. (so just functions of the form $ax^n+bx^{n-1}+.....c$. We are not going to include irrational functions, exponential, logs, etc.)
So for this problem, I looked at a few functions and graphed them to see why this statement would be true. The first thing I found out is that the polynomial's highest degree must be even, otherwise there would be some $x$ where $f(x)$ is negative. Now this implies the odd numbered derivatives will have an odd number for the term with the highest degree.
Having figured that out, I was wondering whether the sum of every successive would change the function so that $f+f'$ has negative values, and $f+f'+f''$ would have positive values and so on, and this does turn out to be true for the ones I've tried. If this is true in general, that would mean that $f+f'$ has some negative values, $f+f'+f''$ has not negative values, and $f+f'+f''+f'''+....f^{n-1}$ has negative values until we get to $f+f'+f''+f'''+....f^{n-1}+f^n$ which has no negative values.
Ok but having figured this out, I'm not sure as to how to proceed with the proofs because the above assumptions I've made are not lemmas or proofs that I was given.
 A: It is known that a polynomial that is nonnegative on all reals must be the sum of squares of two polynomials. As a result, it suffices to prove the claim when the given polynomial $f$ can be written as $g^2$ for some polynomial $g$. We may show that
$$\frac{d}{d^k}g^2=\sum_{i=0}^k\binom ki g^{(i)}g^{(k-i)}$$
by induction on $k$, and so, if $g$ has degree $m=n/2$, the sum we desire is
$$\sum_{k=0}^{2m}\sum_{i=0}^k\binom kig^{(i)}g^{(k-i)}.$$
As $g^{(i)}=0$ for $i>m$, this sum can be rewritten as
$$\sum_{j_1=0}^m\sum_{j_2=0}^m\binom{j_1+j_2}{j_1}g^{(j_1)}g^{(j_2)}.$$
For a given $x$, write $\frac{1}{j!}g^{(j)}(x)=x_j$, so $x_0=x$. We will show that, for any reals $x_0,\dots,x_m$,
$$\sum_{j_1=0}^m\sum_{j_2=0}^m(j_1+j_2)!x_{j_1}x_{j_2}\geq 0,$$
which will finish the proof. We note that we have
$$k!=\int_0^\infty e^{-t}t^kdt,$$
so our sum is
$$\int_0^\infty \sum_{j_1=0}^m\sum_{j_2=0}^mx_{j_1}x_{j_2}t^{j_1+j_2}e^{-t}dt=\int_0^\infty e^{-t}\left(\sum_{j=0}^m x_jt^j\right)^2dt;$$
for each $t$, the integrand is nonnegative, and so the whole integral is nonnegative, finishing the proof.
A: Let $g(x) = \sum\limits_{k=0}^n f^{(k)}(x)$. Notice
$$(e^{-x} g)' = e^{-x}(g'-g)
= e^{-x}\sum_{k=0}^n (f^{(k+1)} - f^{(k)})
= e^{-x}(f^{(n+1)} - f)$$
Since $f$ is a non-negative polynomial of degree $n$, we get
$$f^{(n+1)}\equiv 0 \quad\implies\quad (e^{-x}g)' = -e^{-x} f \le 0\quad\forall x \in \mathbb{R}$$
This means the function $e^{-x}g(x)$ is non-increasing.
Since $f(x)$ is non-negative for all $x$, $n$ has to be an even integer and the leading coefficient of $f(x)$ is positive. It is easy to see $g(x)$ is
a polynomial with same degree $n$ and leading coefficient as $f(x)$. This means for sufficiently positive $y$, we will have $g(y) > 0$.
For any $z \in \mathbb{R}$, take a $y > z$ large enough to make $g(y) > 0$.
Using the fact $e^{-x} g(x)$ is non-increasing, we obtain
$$e^{-z}g(z) \ge e^{-y}g(y) > 0\quad\implies\quad g(z) > 0$$
This is slightly stronger than we want to show. Namely, $g(x)$ is not only non-negative for all $x$, it is positive for all $x$.
A: Since $g(x)$ has positive leading coefficient, $\exists$ $g(a) = \min g(x)$: $$g(x) \geqslant g(a)=g'(a)+f(a)=0+f(a) \geqslant 0$$
