How to show that $f'(x)<2f(x)$ [closed]

From the assumptions we know that $f$, $f'$ and $f''$ are positive and increasing, so $$\lim_{x\to-\infty}f(x)\ge 0, \quad\text{and}\quad\lim_{x\to-\infty}f'(x)=\lim_{x\to-\infty} f''(x)=0.$$ It follows that $$f''(x)=\int_{-\infty}^xf'''(t) \, dt,$$ $$f'(x)=\int_{-\infty}^xf''(t) \, dt=\int_{-\infty}^x \left(\int_{-\infty}^tf'''(s) \, ds \right) \, dt=\int_{-\infty}^x(x-s)f'''(s) \, ds,\tag{1}$$ and $$f(x)\ge\int_{-\infty}^xf'(t) \, dt=\int_{-\infty}^x \left(\int_{-\infty}^t (t-s) f'''(s) \, ds \right) \, dt=\frac{1}{2}\int_{-\infty}^x(x-s)^2f'''(s)\,ds.\tag{2}$$ We will try to compare $f$ and $f'$ by comparing $x-s$ and $(x-s)^2$ in Equation (1) and (2). $(x-s)^2$ exceeds $x-s$ only when the latter exceeds some value. In order to use (2), the integration still needs to run to $x$ instead of somewhere to the left of $x$. We will subtract the "over-integration". We do this by adjusting the critical value $\lambda$ of $x-s$ vs $(x-s)^2$. In fact, it turns out we do not have to choose the optimal critical value for this particular problem, but getting the optimal critical value poses no harm.

For $\lambda>0$ to be determined below, as $x-s<\frac{(x-s)^2}{\lambda}$ for $s\in(-\infty,x-\lambda)$, we have $$\begin{eqnarray*} f'(x) &\le &\lambda^{-1}\int_{-\infty}^{x-\lambda}(x-s)^2f'''(s)ds+\int_{x-\lambda}^x(x-s)f'''(s)ds \tag{by (1)}\\ &\le & 2\lambda^{-1}f(x)+\int_{x-\lambda}^x [(x-s)-\lambda^{-1}(x-s)^2]f'''(s) \, ds\tag{by (2)}\\ &<& 2\lambda^{-1}f(x)+f(x)\int_{x-\lambda}^x[(x-s)-\lambda^{-1}(x-s)^2] \, ds\tag{$f'''(s)<f(x)$}\\ &=&\left(2\lambda^{-1}+\frac{\lambda^2}{6}\right)f(x). \end{eqnarray*}$$

Since $\min_{\lambda>0}(2\lambda^{-1}+\frac{\lambda^2}{6})=\frac{3}{\sqrt[3]{6}}<2$( for $\lambda=\sqrt[3]{6}$), the conclusion follows.