6. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=x^{2}$. If for any $x \in[a, a+2]$, the inequality $f(x+a) \geqslant 2 f(x)$ always holds, then the range of the real number $a$ is $\qquad$ .
Given $f(x)=\left\{\begin{array}{ll}x^{2}, & x \geqslant 0 \\ -x^{2}, & x<0\end{array}\right.$, then $f(x)$ is a monotonic function on $\mathbf{R}$ and $f(\sqrt{2} x)=2 f(x)$.
Thus, $f(x+a) \geqslant 2 f(x) \Rightarrow x+a \geqslant \sqrt{2} x \Rightarrow(\sqrt{2}-1) x-a \leqslant 0, x \in[a, a+2]$ $\Rightarrow\left\{\begin{array}{l}(\sqrt{2}-1) a-a \leqslant 0, \\ (\sqrt{2}-1)(a+2)-a \leqslant 0\end{array} \Rightarrow a \geqslant \sqrt{2}\right.$. Therefore, the range of $a$ is $[\sqrt{2},+\infty)$.