Polynomial or Exponential PROBLEM:
Let $f(x)$ be a polynomial function.
It is known that for every $x$:
$$ f'(x) \leq f(x) $$
Prove/disprove:
For every $x$:
$$ f(x) \geq 0 $$
MY INTUITION:
Suppose by contradiction that $f(z)<0$ for some $z$. Then $f'(z)<0$ too, so $f$ must go down and down to $-\infty$.
Moreover, the rate of decrease must by at least exponential because:
$$ |f'(x) \geq f(x)| $$
and we know that equality holds for the exponential function.
This contradict the fact that $f(x)$ is polynomial.
MY QUESTIONS: Is my intuition true? If so, how to formalize it? If not, then what is the correct answer?

Your intuition seems quite good. Here are a couple notes on some ways to make it rigorous: 
1) Since $f$ is a polynomial, $\displaystyle \lim_{x \to \pm \infty} |f'(x)|/|f(x)| = 0$ (unless $f \equiv 0$, but in that case you're already done.)
2) If for some $x_0$, $f(x_0)<0$, then since $f$ is continuous, there is a certain interval $(a,b)$ containing $x_0$ such that $f(x) < 0$ for all $x \in (a,b)$. This means that $f'(x)<0$ for every $x \in (a,b)$ implying that $f$ is strictly decreasing on $(a,b)$. In particular, you can deduce that $f(b) < f(x_0)<0$. Now, let $$A = \sup\{ y \geq b \mid f(x)<0 \text{ for every } x \in (a,y) \}$$ If $A \neq \infty$ then by the same work as before you get that $f(A) < f(x_0) < 0$, implying there is some $\delta > 0$ such that for all $x \in (a, A+\delta)$, $f(x)<0$ contradicting the choice of $A$, hence $A = \infty$.
3) From the previous mark, if $f(x_0)<0$, then $\forall x \geq x_0$, $f(x) < 0$ and hence $f'(x) \leq f(x) < 0$. Now, as you pointed out, this means $\forall x \geq x_0$, $|f'(x)|/|f(x)| \geq 1$. But this contradicts the first point.
