If $\sum_{k=0}^2 c_kf^{(k)}(x) \ge 0$ show that $f(x)$ is a non-negative polynomial 
If for some polynomial $f(x)$ with real coefficients there exist $c_1,c_2 \in \mathbb{R}$, satisfying $c_1^2\ge 4c_2$, such that for all real values of $x$ the following inequality holds:
$$\sum_{k=0}^2 c_kf^{(k)}(x) \ge 0, \ \ (c_0=1)$$ then show that $f(x)$ is a non-negative polynomial. $f^{(k)}(x)$ stands for the $k^{th}$ derivative of $f$.

Assuming $f$ looks like $$f(x)=\sum_{i=0}^n a_ix^i$$ the given sum will become of the form
$$\sum_{k=0}^2 c_kf^{(k)}(x) = \sum_{i=0}^na_ix^i + c_1\sum_{i=0}^n ia_ix^{i-1} + c_2\sum_{i=0}^n i(i-1)a_ix^{i-2}=$$$$=a_0+c_1a_1+a_1x+\sum_{i=2}^n a_ix^{i-2}(x^2+c_1ix+c_2i(i-1)) \ge0$$ But this is no more simplified the problem. Any help is appreciated.
 A: We have that
$$
\left(1+c_1\frac{d}{dx}+c_2\frac{d^2}{dx^2}\right)f(x)\ge 0, \quad\text{for all $x\in\mathbb R$}.
$$
Since $c_1^2\ge 4c_2$, the polynomial $\xi^2+c_1\xi+c_2$ has real roots, say $\mu,\nu$, i.e.
$$
\xi^2+c_1\xi+c_2=(\xi-\mu)(\xi-\nu),
$$
and
$$
1+c_1\frac{d}{dx}+c_2\frac{d^2}{dx^2}
=\left(1-\mu\frac{d}{dx}\right)\left(1-\nu\frac{d}{dx}\right)
$$
Now it suffices to show to the following:
If $\mu\in \mathbb R$ and $f(x)-\mu f'(x)\ge 0$, for all $x$, then $f(x)\ge 0$, for all $x$.
If $\mu=0$, there is nothing to prove. Let $\mu> 0$. Clearly
$$
f(x)-\mu f'(x)\ge 0 \quad\Longrightarrow\quad 
\frac{1}{\mu}f(x)- f'(x)\ge 0
\quad\Longrightarrow\quad
\mathrm{e}^{-x/\mu}\big(f(x)/\mu-f'(x)\big)\ge 0 \quad\Longrightarrow\quad -\big(\mathrm{e}^{-x/\mu}f(x)\big)'\ge 0
\\ \quad\Longrightarrow\quad \int_{x_1}^{x_2}\big(\mathrm{e}^{-\mu x}f(x)\big)'\,dx\le 0 \quad\Longrightarrow\quad
\mathrm{e}^{-\mu x_2}f(x_2)\le\mathrm{e}^{-\mu x_1}f(x_1)
$$
for all $x_2\ge x_1$.
Since $\lim_{x_2\to\infty}\mathrm{e}^{-\mu x_2}f(x_2)=0$, we obtain from the above that
$$
0\le\mathrm{e}^{-\mu x_1}f(x_1)\quad\Longrightarrow\quad
0\le f(x_1)
$$
for all $x_1\in\mathbb R$.
If $\mu<0$, setting $\nu=-\mu>0,\,$ we have
$$
f(x)+\nu f'(x)\ge 0 \quad\Longrightarrow\quad 
\frac{1}{\nu}f(x)+ f'(x)\ge 0
\quad\Longrightarrow\quad
\mathrm{e}^{x/\nu}\big(f(x)/\nu+f'(x)\big)\ge 0 \quad\Longrightarrow\quad \big(\mathrm{e}^{x/\nu}f(x)\big)'\ge 0
\\ \quad\Longrightarrow\quad \int_{x_1}^{x_2}\big(\mathrm{e}^{\nu x}f(x)\big)'\,dx\ge 0 \quad\Longrightarrow\quad
\mathrm{e}^{\nu x_2}f(x_2)\ge\mathrm{e}^{\nu x_1}f(x_1)
$$
for all $x_2\ge x_1$.
Since $\lim_{x_1\to-\infty}\mathrm{e}^{\nu x_1}f(x_1)=0$, we obtain from the above that
$$
0\le\mathrm{e}^{\nu x_2}f(x_2)\quad\Longrightarrow\quad
0\le f(x_2)
$$
for all $x_2\in\mathbb R$.
A: Alternative solution:
Assume, for the sake of contradiction, that there exists $x_0$ such that $f(x_0) < 0$.
Clearly, $f(x)$ is of even degree with positive leading coefficient.
We claim that $c_2 > 0$. Suppose $c_2 \le 0$. Clearly, $f$ achieves its global minimum $f(x_1) < 0$ at some $x_1$ with $f'(x_1) = 0$ and $f''(x_1) \ge 0$
(necessary condition for local minimum) which contradicts $f(x_1) + c_1f'(x_1) + c_2f''(x_1) \ge 0$.
Since
$$(\mathrm{e}^{\frac{c_1}{2c_2}x}f)'' = \frac{1}{c_2}\mathrm{e}^{\frac{c_1}{2c_2}x}\left(\frac{c_1^2}{4c_2}f + c_1f' + c_2f''\right),$$
we have
$$\frac{1}{c_2}\mathrm{e}^{\frac{c_1}{2c_2}x}\left(f + c_1f' + c_2f''\right)
= (\mathrm{e}^{\frac{c_1}{2c_2}x}f)'' - \frac{c_1^2-4c_2}{4c_2^2}\mathrm{e}^{\frac{c_1}{2c_2}x}f .$$
Thus, we have
$$(\mathrm{e}^{\frac{c_1}{2c_2}x}f)'' - \frac{c_1^2-4c_2}{4c_2^2}\mathrm{e}^{\frac{c_1}{2c_2}x}f \ge 0, \ \forall x \in \mathbb{R}.\tag{1}$$
We split into two cases:

*

*$c_1 > 0$: Since $f$ is a polynomial of even degree with positive leading coefficient, noting also that $f(x_0) < 0$, there exists $x_2$ such that
$f(x_2) = 0$, and $f(x) > 0$ for all $x$ in $(-\infty, x_2)$.
Thus, from (1), $(\mathrm{e}^{\frac{c_1}{2c_2}x}f)'' \ge 0$ on $(-\infty, x_2)$, and thus
$\mathrm{e}^{\frac{c_1}{2c_2}x}f$ is convex on $(-\infty, x_2)$. However, this is impossible
since $\mathrm{e}^{\frac{c_1}{2c_2}x_2}f(x_2) = 0$, $\lim_{x\to - \infty} \mathrm{e}^{\frac{c_1}{2c_2}x}f = 0$,
and $\mathrm{e}^{\frac{c_1}{2c_2}x}f(x) > 0$ for all $x$ in $(-\infty, x_2)$.


*$c_1 < 0$: Since $f$ is a polynomial of even degree with positive leading coefficient, noting also that $f(x_0) < 0$, there exists $x_3$ such that
$f(x_3) = 0$, and $f(x) > 0$ for all $x$ in $(x_3, \infty)$. The rest is similar to 1).
(Q. E. D.)
