If $f(0)=f^{\prime}(0)=0$ and $|f''(x)|\leqslant |f(x)|+|f'(x)|$, then $f(x)\equiv 0$ Is there a simpler way to solve this problem with derivatives?

Let $f$ be second-order differentiable on $(-\infty,+\infty)$, and
$$f(0)=f^{\prime}(0)=0, \quad|f''(x)|\leqslant |f(x)|+|f'(x)|$$
Show that $$f(x)\equiv 0$$

I do as follows,for all $x\in\left[-\frac{1}{3},\frac{1}{3}\right]$,let$$M=\max\limits_{x\in\left[-\frac 1 3,\frac 1 3\right]}|f'(x)|$$Using Lagrange mean value theorem I got it$$\begin{align} 
|f'(x)| & = |f''(\eta_1)||x|\\ 
        & \leqslant \frac{|f(\eta_1)|}3+\frac{|f'(\eta_1)|}3 \\ 
        &  =\frac{|f'(\zeta _1)||\zeta_1|}3+\frac{|f'(\eta_1)} 3 \\ 
        & \leqslant \frac{2}3|f'(\xi_1)|, \text{ here }|f'(\xi_1)|=\max\{|f'(\zeta _1)|,|f'(\eta_1)|\} 
\end{align}$$So we have$$|f'(x)|\leqslant \frac{2^n}{3^n}|f'(\xi_n)| \leqslant \frac{2^n}{3^n}M\to 0(n\to\infty).$$
Hence $$|f'(x)|\equiv 0\equiv f'(x)\implies f(x)\equiv f(0)\equiv 0\text{ for }x\in \left[-\frac 1 3,\frac 1 3\right].$$Repeat the above steps and use induction complete the proof.But
I think my method is too complicated.
Do you have a simpler and refreshing proof? Thank you!
 A: Note that
\begin{align}
g(x):=&\ |f'(x)|+|f(x)| \\
=&\ \left|\int^x_0 f''(y)\ dy \right| +\left|\int^x_0f'(y)\ dy \right|\\
\le&\ \int^x_0 |f''(y)|\ dy + \int^x_0 |f'(y)|\ dy\\
\le&\ \int^x_0 2|f'(y)|+|f(y)|\ dy\\
\le&\ 2\int^x_0 g(y)\ dy.
\end{align}
If we call $G(x) = \int^x_0 g(y)\ dy$ then we have that
\begin{align}
G'(x) \le 2 G(x) \ \ \implies \ \ G'(x)-2G(x) \le 0.
\end{align}
Now, we will use some basic ODE trick, i.e. integrating factor method to get
\begin{align}
\frac{d}{dx}(e^{-2x}G(x))
=e^{-2x}G'(x)-2e^{-2x}G'(x) \le 0
\end{align}
which means
\begin{align}
e^{-2x}G(x)-G(0) = \int^x_0 \frac{d}{dy}(G(y)e^{-2y})\ dy \le 0.
\end{align}
Hence, we arrive at
\begin{align}
\int^x_0 |f'(y)|+|f(y)|\ dy =G(x) \le G(0)e^{2x} = 0.
\end{align}
So, $f(x) = 0$.
The ODE trick is also called Gronwall's inequality.
Edit: As mentioned by @MartinR, if $f''$ is not continuous then we need to be a little more careful.
Fix $x>0$. By the mean-value theorem, we have that
\begin{align}
|f(x)|+|f'(x)| =&\ |f(x)-f(0)|+|f'(x)-f'(0)|\\
\le&\ |f'(\xi_1)||x| + |f''(\xi_2)||x|
\end{align}
where $\xi_1, \xi_2 \in (0, x)$. Using the hypothesis, we have that
\begin{align}
|f(x)|+|f'(x)| \le 2\left(\sup_{\xi \in [0, x]}|f'(\xi)| +\sup_{\xi \in [0, x]}|f(\xi)|\right)|x|.
\end{align}
Choose $x>0$ such that $1-2|x|>0$ then it follows
\begin{align}
(1-2|x|)\left(\sup_{\xi \in [0, x]}|f'(\xi)| +\sup_{\xi \in [0, x]}|f(\xi)|\right)\le 0.
\end{align}
Hence it follows
\begin{align}
\sup_{\xi \in [0, x]}|f'(\xi)| +\sup_{\xi \in [0, x]}|f(\xi)| = 0
\end{align}
for the chosen $x$. Repeat the argument for the interval $[x, 2x]$ so on... This gives you the desired result.
