Show that $\sup_{x \in [-a, a]}|f'(x)| \le \frac{\sup_{x \in [-a, a]}|f(x)|}{a} + \sup_{x \in [-a, a]}\frac{|f''(x)|(x^2 + a^2)}{2a}$

Question

Let $f(x)$ be twice differentiable on $[-a, a]$.

Show that:

$$\sup_{x \in [-a, a]}|f'(x)| \le \frac{\sup_{x \in [-a, a]}|f(x)|}{a} + \sup_{x \in [-a, a]}\frac{|f''(x)|(x^2 + a^2)}{2a}$$

I think it is somehow related to Taylor polynomial $x \to x_0$

$$ f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)(x - x_0)^2}{2} $$