Proving that $f'(x)\le f(x), \forall x\in \mathbb{R}$ Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that


*

*$0 \le f''(x) \le f(x)$

*$f'(x)\ge 0, \forall x\in \mathbb{R}$.


Thus, prove that $f'(x)\le f(x), \forall x\in \mathbb{R}$.    
I couldn't make much progress, but I observed that $f$ is increasing and convex while $f'$ is increasing from the hypothesis.     
Then I tried to evaluate the derivative of $h:\mathbb{R} \to \mathbb{R}$ with
$$ h(x)=f(x)-f'(x)$$
and I got that 
$$h'(x)=f'(x)-f''(x), \forall x\in \mathbb{R}$$
Now, I would be done if I could prove that $h'(x)\ge 0$ for all $x\in \mathbb{R}$. But I'm not sure yet if this approach is really that straightforward.
I also tried to use that $f$ is convex, but to no avail.     
EDIT: Maybe that we should somehow use the fact that a differentiable function $g:\mathbb{R}\to\mathbb{R}$ is convex if and only if $g(x)\ge g(y)+g'(y)(x-y),\  \forall x,y\in \mathbb{R}$ 
 A: With
$0 \le f''(x) \le f(x), \; \forall x \in \Bbb R, \tag 1$
and
$f'(x) \ge 0, \; \forall x \in \Bbb R, \tag 2$
we have
$0 \le f'(x) f''(x) \le f(x) f'(x), \; \forall x \in \Bbb R, \tag 3$
or
$0 \le \dfrac{1}{2} ((f'(x))^2)' \le \dfrac{1}{2} (f^2(x))', \tag 4$
or
$0 \le ((f'(x))^2)' \le  (f^2(x))', \tag 5$
we integrate this 'twixt some arbitrary $L \in \Bbb R$ and $x \in \Bbb R$ to obtain
$(f'(x))^2 - (f'(L))^2 = \displaystyle \int_L^x ((f'(s))^2)'  \; ds \le \int_L^x (f^2(s))' \; ds = f^2(x) - f^2(L). \tag 6$
Now in light of (1), 
$f(x) \ge 0, \forall x \in \Bbb R, \tag 7$
so if we define
$\alpha = \inf \{ f(x), \; x \in \Bbb R \}, \tag 8$
then
$\alpha \ge 0 \tag 9$
and, by virtue of (2), $f(x)$ is monotonically decreasing with decreasing $x$; together these imply that
$\displaystyle \lim_{x \to -\infty} f(x) = \alpha.  \tag{10}$
We next consider $f'(x)$ as $x \to -\infty$; again in accord with (1) we see that $f'(x)$ is monotonically decreasing with decreasing $x$, and by (2) it too is bounded below by $0$. I claim that in fact
$\displaystyle \lim_{x \to -\infty} f'(x) = 0; \tag{11}$
for if not, setting
$\beta = \inf \{f'(x), x \in \Bbb R\} > 0, \tag{12}$
then we may assert that
$f'(x) \ge \beta, \; \forall x \in \Bbb R; \tag{13}$
then picking 
$x_0, x_1 \in \Bbb R, \; x_0 < x_1, \tag{14}$
it follows that
$f(x_1) - f(x_0) = \displaystyle \int_{x_0}^{x_1} f'(s) \; ds \ge \int_{x_0}^{x_1} \beta \; ds = \beta(x_1 - x_0), \tag{15}$
whence
$f(x_1) - f(x_0)  \ge \beta(x_1 - x_0), \tag{16}$
or
$f(x_0) - f(x_1)  \le -\beta(x_1 - x_0), \tag{16}$
that is,
$f(x_0) \le  f(x_1)  - \beta(x_1 - x_0); \tag{17}$
but it is easily seen that this implies that 
$f(x_0) \to -\infty \; \text{as} \; x_0 \to -\infty, \tag{18}$
which contradicts (1).  Thus
$\beta = 0 \tag{19}$
and
$f'(x) \to 0 \; \text{as} \; x \to \infty. \tag{20}$
Returning now to (6), we have
$(f'(x))^2 - (f'(L))^2 \le f^2(x) - f^2(L),  \tag {21}$
and letting
$L \to -\infty \tag{22}$
we reach
$(f'(x))^2 \le f^2(x) - \alpha^2 \le f^2(x),  \tag {23}$
and since both
$f(x), f'(x) \ge 0, \forall x \in \Bbb R, \tag{24}$
we may at last conclude that
$f'(x) \le f(x), \; \forall x \in \Bbb R, \tag{25}$
$OE\Delta$.
Finally, note that in light of (10) and (11) we have
$\displaystyle \lim_{x \to -\infty} (f(x) - f'(x))$ $= \lim_{x \to -\infty} f(x) - \lim_{x \to -\infty}f'(x) = \alpha - 0 \ge 0, \tag{26}$
as our OP requested be proved in a comment to Clement Yung's answer.
A: For a non-negative, increasing, convex function, we have $\lim_{x \to -\infty}(f(x) - f'(x)) \geq 0$ exists. Consider:
$$
g(x) = (f(x) - f'(x))e^x
$$
Differentiating yields:
$$
g'(x) = (f(x) - f''(x))e^x \geq 0
$$
We have that $\lim_{x \to -\infty} g(x) \geq 0$. Since $g'(x) \geq 0$, $g$ is increasing so this implies $g(x) \geq 0$ $\forall x \in \mathbb{R}$. Since $e^x > 0$, we have $f(x) \geq f'(x)$.

EDIT: To prove that $\lim_{x \to -\infty}(f(x) - f'(x)) \geq 0$, we shall show that $\lim_{x \to -\infty} f(x) \geq 0$ and $\lim_{x \to -\infty} f'(x) = 0$.
$\lim_{x \to -\infty} f(x)$ exists because of monotone convergence theorem. Take any $(x_n)_{n \in \mathbb{Z}^+}$ such that $x_n \to -\infty$, and we may assume WLOG that $x_n$ is monotone (as otherwise we can remove the terms which violate monotonicity, and the asymptotic behavior remains unchanged). Since $x_n \geq 0$ $\forall n$, we have that $x_n$ converges. Since $f(x) \geq 0$, the limit must also be non-negative.
Similarly, we can prove that $\lim_{x \to -\infty} f'(x) \geq 0$ as we have $f'(x) \geq 0$ and $f''(x) \geq 0$ so $f'$ is a monotonically increasing function bounded from below. We show further that $\lim_{x \to -\infty} f'(x) = 0$ by contradiction, in which if $\lim_{x \to -\infty} f'(x) = \epsilon > 0$, then for some $M > 0$ we have $f'(x) > \frac{\epsilon}{2}$ for all $x < -M$. This would violate the assumption that $f(x) \geq 0$.
