Romanian Mathematical Olympiad

a) For $a = 0$ the result is obvious. For $a \ne 0$, consider the differentiable function $g : \mathbb{R} \to \mathbb{R}$ defined by $g(x) = F(x) \cdot e^{\frac{x}{a}}$. We have
$$g'(x) = f(x) \cdot e^{\frac{x}{a}} + \frac{1}{a} \cdot F(x) \cdot e^{\frac{x}{a}} = \frac{1}{a} \cdot e^{\frac{x}{a}} \cdot \left( F(x) + a \cdot f(x) \right).$$
For $a > 0$ it is obvious that $g'(x) \ge 0$, for $x \in \mathbb{R}$, implying that $g$ is non-decreasing. As $\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} F(x) \cdot e^{\frac{x}{a}} = 0$, we get $g(x) \ge 0$, for any $x \in \mathbb{R}$, thus $F(x) = g(x) \cdot e^{-\frac{x}{a}} \ge 0$, for $x \in \mathbb{R}$.
If $a < 0$, $g'(x) \le 0$, for $x \in \mathbb{R}$, implying that $g$ is non-increasing. As $\lim_{x \to \infty} g(x) = \lim_{x \to -\infty} F(x) \cdot e^{\frac{x}{a}} = 0$, we obtain $g(x) \ge 0$, for $x \in \mathbb{R}$. It follows that $F(x) = g(x) \cdot e^{-\frac{x}{a}} \ge 0$, for any $x \in \mathbb{R}$.

b) Let $\mathcal{P} = \{f \mid f : \mathbb{R} \to \mathbb{R}, f \text{ is a polynomial function}\}$. For any real $a$ consider the function $T_a : \mathcal{P} \to \mathcal{P}$, defined by $T_a(f) = f + a \cdot f'$, i.e., $T_a(f)(x) = f(x) + a \cdot f'(x)$, for any $x \in \mathbb{R}$. As for any $f \in \mathcal{P}$ and any $\alpha \in \mathbb{R}$, we have $\lim_{|x|\to\infty} \frac{f(x)}{e^{|\alpha \cdot x|}} = 0$, by a), for real $a$, we have the implication
$$f \in \mathcal{P} : T_a(f)(x) \ge 0, \forall x \in \mathbb{R} \implies f(x) \ge 0, \forall x \in \mathbb{R}.$$
Consider the roots $r_1, r_2, \dots, r_n$ of $g$, and $s_k = -r_k$ for $k \in \{1, 2, \dots, n\}$ their opposites. Then $g = (X - r_1)(X - r_2) \dots (X - r_n) = (X + s_1)(X + s_2) \dots (X + s_n)$.
By Vieta, we obtain the coefficients of $g$:
$$a_k = \sum_{1 \le i_1 < i_2 < \dots < i_k \le n} s_{i_1} s_{i_2} \dots s_{i_k}, \quad \text{for any } k \in \{1, 2, \dots, n\}.$$
For $m \in \{1, 2, \dots, n\}$ and any $f \in \mathcal{P}$, we get
$$(T_{s_m} \circ \dots \circ T_{s_2} \circ T_{s_1})(f) = f + \left(\sum_{i=1}^m s_i\right) f' + \left(\sum_{1 \le i_1 < i_2 \le m} s_{i_1} s_{i_2}\right) f'' + \dots \\ \dots + \left(\sum_{1 \le i_1 < i_2 < \dots < i_k \le m} s_{i_1} s_{i_2} \dots s_{i_k}\right) f^{(k)} + \dots + (s_1 s_2 \dots s_m) f^{(m)}. \quad (Q(m))$$
If for $m \in \{1, 2, \dots, n-1\}$, $Q(m)$ is supposed to be true, we have
$$\begin{align*} (T_{s_{m+1}} \circ T_m \circ \dots T_{s_2} \circ T_{s_1})(f) &= T_{s_{m+1}}(T_m \circ \dots \circ T_{s_1})(f) = \\ &= T_{s_{m+1}} \left( \sum_{J \subseteq \{1, 2, \dots, m\}} \left( \prod_{j \in J} s_j \right) f^{|J|} \right) = \\ &= \left( \sum_{J \subseteq \{1, 2, \dots, m\}} \left( \prod_{j \in J} s_j \right) f^{|J|} \right) + \\ &\quad + s_{m+1} \cdot \left( \sum_{J \subseteq \{1, 2, \dots, m\}} \left( \prod_{j \in J} s_j \right) f^{|J|} \right)' \\ &= \sum_{J_1 \subseteq \{1, 2, \dots, m, m+1\}} \left( \prod_{j \in J_1} s_j \right) f^{|J_1|} \end{align*}$$
so $Q(m+1)$ is also true.

From $Q(n)$, we have
$$(T_{s_n} \circ \dots \circ T_{s_2} \circ T_{s_1}) (f) = f + a_1 \cdot f' + a_2 \cdot f'' + \dots + a_n \cdot f^{(n)}.$$
By the hypothesis, we have
$$(T_{s_n} \circ \dots \circ T_{s_2} \circ T_{s_1}) (f)(x) = f(x) + a_1 \cdot f'(x) + a_2 \cdot f''(x) + \dots + a_n \cdot f^{(n)}(x) \geq 0,$$
for all $x \in \mathbb{R}$.
Successively, applying a), for $m \in \{1, 2, \dots, n\}$, we get
$$(T_{s_m} \circ \dots \circ T_{s_2} \circ T_{s_1}) (f)(x) \geq 0, \quad \text{for any } x \in \mathbb{R}.$$
In particular $f(x) \geq 0$, for any $x \in \mathbb{R}$.