8. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=x^{2}$. If for any $x \in[a, a+2]$, we have $f(x+a) \geqslant 2 f(x)$, then the range of the real number $a$ is . $\qquad$
8. $[\sqrt{2},+\infty)$.

From the problem, we know that $f(x)=\left\{\begin{array}{ll}x^{2}, & x \geqslant 0 \text {; } \\ -x^{2}, & x<0 .\end{array}\right.$
Thus, $2 f(x)=f(\sqrt{2} x)$.
Therefore, the original inequality is equivalent to
$$
f(x+a) \geqslant f(\sqrt{2} x) \text {. }
$$

By the monotonicity of $f(x)$, we know that $x+a \geqslant \sqrt{2} x$ holds for all $x \in[a, a+2]$, which means
$$
\begin{array}{l}
a \geqslant((\sqrt{2}-1) x)_{\max }=(\sqrt{2}-1)(a+2) \\
\Rightarrow a \geqslant \sqrt{2} .
\end{array}
$$