Example 1 Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x$ $\geqslant 0$, $f(x)=x^{2}$. If for any $x \in[t, t+2]$, the inequality $f(x+t) \geqslant 2 f(x)$ always holds, then the range of the real number $t$ is ( ).
A. $[\sqrt{2},+\infty)$
B. $[2,+\infty)$
C. $(0,2]$
D. $[-\sqrt{2},-1] \cup[\sqrt{2}, 0]$
Solution: According to the problem, the function is increasing on $\mathbf{R}$, and the inequality $f(x+t) \geqslant 2 f(x)$ can be transformed into $f(x+t) \geqslant f(\sqrt{2} x)$,
so $x+t \geqslant \sqrt{2} x$, which means $t \geqslant(\sqrt{2}-1) x$.
Since the function $g(x)=(\sqrt{2}-1) x$ is increasing on $[t, t+2]$,

the maximum value of function $g(x)$ on $[t, t+2]$ is $(\sqrt{2}-1)(t+2)$.

From $t \geqslant(\sqrt{2}-1)(t+2)$, solving it yields $t \geqslant \sqrt{2}$, hence the answer is A.