Example 4 Let the constant $a$ be such that the inequality concerning $x$
$$
\frac{1}{1+\sqrt{x}} \geqslant a \sqrt{\frac{x}{x-1}}
$$

has non-zero real solutions. Find the maximum value of $a$.
Let $f(x)=\frac{1}{1+\sqrt{x}} \cdot \sqrt{\frac{x-1}{x}}=\frac{\sqrt{x-1}}{x+\sqrt{x}}$, then $a \leqslant \max _{x>1}\{f(x)\}$. Below we find $\max _{x>1}\{f(x)\}$.
$$
\begin{array}{l}
\because f^{\prime}(x) \\
=\frac{1}{2 \sqrt{x-1}(x+\sqrt{x})}-\frac{\sqrt{x-1}}{(x+\sqrt{x})^{2}}\left(1+\frac{1}{2 \sqrt{x}}\right) \\
=\frac{\sqrt{x}(x+\sqrt{x})-(x-1)(2 \sqrt{x}+1)}{2 \sqrt{x} \sqrt{x-1}(x+\sqrt{x})^{2}} \\
=-\frac{(\sqrt{x}+1)(x-\sqrt{x}-1)}{2 \sqrt{x} \sqrt{x-1}(x+\sqrt{x})^{2}}, \\
f^{\prime}(x)>0 \Leftrightarrow x-\sqrt{x}-1<0 \\
\Leftrightarrow x \in\left(1, \frac{3+\sqrt{5}}{2}\right),
\end{array}
$$
$\therefore$ When $x \in\left(1, \frac{3+\sqrt{5}}{2}\right)$, $f(x)$ is an increasing function; when $x \in\left(\frac{3+\sqrt{5}}{2},+\infty\right)$, $f(x)$ is a decreasing function.
$$
\begin{array}{l}
\therefore f(x)_{\max }=f\left(\frac{3+\sqrt{5}}{2}\right)=\frac{\sqrt{\frac{1+\sqrt{5}}{2}}}{2+\sqrt{5}} \\
=\sqrt{\frac{5 \sqrt{5}-11}{2}} .
\end{array}
$$

Thus, $a_{\text {max }}=\sqrt{\frac{5 \sqrt{5}-11}{2}}$.
Note: This problem is from "Mathematical Communications" Issue 9, 2001, "New Comprehensive Problems" Question 3. The original problem was to prove $a<\frac{1}{3}$. The problem provider pointed out in the "Problem Background" that "it is impossible to find the maximum possible precise value of $a$," but this problem provides the maximum precise value of $a$.