China Mathematical Competition

According to the given condition, we have
$$f(x) = \begin{cases} x^2 & (x \ge 0), \\ -x^2 & (x < 0). \end{cases}$$
So $2f(x) = f(\sqrt{2}x)$. Therefore, the original inequality is equivalent to $f(x+a) \ge f(\sqrt{2}x)$.

As $f(x)$ is increasing over $\mathbb{R}$, then $x + a \ge \sqrt{2}x$, i.e.,
$$a \ge (\sqrt{2} - 1)x.$$
Furthermore, since $x \in [a, a+2]$, $(\sqrt{2}-1)x$ reaches $(\sqrt{2}-1)(a+2)$ the maximum value when $x = a+2$.
Therefore, $a \ge (\sqrt{2}-1)(a+2)$, from which we obtain $a \ge \sqrt{2}$, i.e., $a \in [\sqrt{2}, +\infty)$.

The answer is then $[\sqrt{2}, +\infty)$.