Let $f: \mathbb{R} \to \mathbb{R}$ be a $C^3$ function such that $f,f',f'',f'''>0$ and $f''' \le f.$ What is the smallest $c$ such that we can guarantee $f'<cf$? Since $f(x)=e^x$ works, we must have $c>1.$ On the other hand, I managed to show $c = 1.5^{1/3}$ works.
Proof: First, we note $\lim\limits_{x \to -\infty} f'(x) = \lim\limits_{x \to -\infty} f''(x) = 0.$ Now $f''' \le f \Rightarrow (f''^2)' = 2f''f''' \le 2f''f < 2f''f + 2f'^2 = (2ff')',$ so integrating yields $(f''^2)(x) - (f''^2)(x_0) < (2ff')(x)-2ff'(x_0)$ for any $x,x_0.$ Taking $x_0 \to -\infty$ gives us $f''^2 < 2ff'.$
We use $f''' \le f$ again, but multiply by $f'$ instead: $(f'f'')' = f''^2 + f'f''' \le f''^2 + f'f < 3ff' = (1.5f^2)'.$ Integrating, we get $(f'f'')(x) - (f'f'')(x_0) < (1.5f^2)(x) - (1.5f^2)(x_0) < (1.5f^2)(x).$ Take $x_0 \to -\infty$ again to get $f'f'' < 1.5f^2 \Rightarrow (\frac{1}{3}f'^3)' = f'^2f'' < 1.5f^2f' = (0.5f^3)'.$ Integrate and take $x_0 \to -\infty$ for the last time to get $\frac{1}{3}f'^3 < 0.5f^3 \Rightarrow f' < 1.5^{1/3} f.$
