Example 8 Suppose for all real numbers $x$, the inequality
$$
\begin{array}{l}
x^{2} \log _{2} \frac{4(a+1)}{a}+2 x \log _{2} \frac{2 a}{a+1} \\
+\log _{2} \frac{(a+1)^{2}}{4 a^{2}}>0
\end{array}
$$

always holds. Find the range of real numbers $a$.
Strategy: Treat $x$ as the main variable and $a$ as the main parameter, transforming the inequality into $\log _{2} \frac{a+1}{2 a}>-\frac{3 x^{2}}{x^{2}-2 x+2}$. Since $-\frac{3 x^{2}}{x^{2}-2 x+2}=-\frac{3 x^{2}}{(x-1)^{2}+1} \leqslant 0$, it follows that $\log _{2} \frac{a+1}{2 a}>0$. Solving this inequality will give the range of $a$.
Solution: The inequality can be rewritten as
$$
\begin{array}{l}
3 x^{2}+\left(x^{2}-2 x+2\right) \log _{2} \frac{a+1}{2 a}>0 . \\
\because x^{2}-2 x+2=(x-1)^{2}+1>0, \\
\therefore \log _{2} \frac{a+1}{2 a}>-\frac{3 x^{2}}{(x-1)^{2}+1} . \\
\text { For } x \in \mathbf{R}, -\frac{3 x^{2}}{(x-1)^{2}+1} \leqslant 0, \\
\therefore \log _{2} \frac{a+1}{2 a}>0 \Leftrightarrow \frac{a+1}{2 a}>1 .
\end{array}
$$

Solving this gives $0<a<1$.