Example 7 Given $0<a<1$,
$$
\begin{array}{l}
f(x)=\log _{a}(x+1), \\
g(x)=2 \log _{a}(2 x+t),
\end{array}
$$
for $x \in[0,1]$, $f(x) \geqslant g(x)$ always holds. Find the range of the real number $t$.
$$
\begin{array}{l}
\text { Solution: From } f(x) \geqslant g(x) \text { we get } \\
\log _{a}(x+1) \geqslant 2 \log _{a}(2 x+t) \\
\Leftrightarrow\left\{\begin{array}{l}
x+1>0, \\
2 x+t>0, \\
x+1 \leqslant(2 x+t)^{2}
\end{array}\right. \\
\Leftrightarrow t \geqslant \sqrt{x+1}-2 x .
\end{array}
$$

Let $u=\sqrt{x+1}$, then $x=u^{2}-1$. Given $x \in$ $[0,1)$, we know $u \in[1, \sqrt{2})$,
$$
\begin{aligned}
\therefore & \sqrt{x+1}-2 x=u-2\left(u^{2}-1\right) \\
& =-2 u^{2}+u+2 \\
& =-2\left(u-\frac{1}{4}\right)^{2}+\frac{17}{8} .
\end{aligned}
$$

When $u=1$, $\sqrt{x+1}-2 x$ has a maximum value of 1.
Therefore, the range of $t$ is $t \geqslant 1$.