7. 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 $\qquad$
Solution: $[\sqrt{2},+\infty)$.
From the problem, we know that $f(x)=\left\{\begin{array}{l}x^{2}(x \geqslant 0), \\ -x^{2}(x<0),\end{array}\right.$, then $2 f(x)=f(\sqrt{2} x)$.
Therefore, the original inequality is equivalent to $f(x+a) \geqslant f(\sqrt{2} x)$.
Since $f(x)$ is an increasing function on $\mathbf{R}$, we have $x+a \geqslant \sqrt{2} x$, which means $a \geqslant(\sqrt{2}-1) x$.
Also, $x \in[a, a+2]$. Therefore, when $x=a+2$, $(\sqrt{2}-1) x$ reaches its maximum value of $(\sqrt{2}-1)(a+2)$.
Thus, $a \geqslant(\sqrt{2}-1)(a+2)$, solving this gives $a \geqslant \sqrt{2}$.
Hence, the range of values for $a$ is $[\sqrt{2},+\infty)$.