Proving that $f'(x)\le f(x), \forall x\in \mathbb{R}$

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 $f(x) \geq 0$.

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.