China Mathematical Competition

According to the given condition, we have $$f(x)=\{\begin{array}{ll}{x}^2& (x\ge 0),\\ -x^2& (x<0).\end{array}$$ 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 $[\sqrt{2},+\infty )$.