4.1.13 * Given a real number $a$, solve the inequality $\frac{1}{1+\sqrt{1+x}}+\frac{1}{\sqrt{x+1}-1} \geqslant \sqrt{\frac{a}{x}}$.
When $a=0$, the inequality becomes $\frac{2}{x} \cdot \sqrt{x+1} \geqslant 0$, whose solution is $x=-1$ or $x>0$; when $a \neq 0$, the original inequality can be transformed into
$$
\frac{\sqrt{x+1}+1}{x}+\frac{\sqrt{x+1}-1}{x} \geqslant \sqrt{\frac{a}{x}} \Leftrightarrow \frac{2 \sqrt{x+1}}{x} \geqslant \sqrt{\frac{a}{x}} .
$$
(1) When $a>0$, the original inequality $\Leftrightarrow\left\{\begin{array}{l}x>0, \\ 2 \sqrt{x+1} \geqslant \sqrt{a x}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>0, \\ 4(x+1) \geqslant a x\end{array} \Leftrightarrow\right.\right.$ $\left\{\begin{array}{l}x>0, \\ (a-4) \cdot x \leqslant 4,\end{array}\right.$ thus when $0<a \leqslant 4$, the solution set is $x>0$; when $a>4$, the solution set of the inequality is $0<x \leqslant \frac{4}{a-4}$. When $a>4$, the solution set is $\left(0, \frac{4}{a-4}\right]$.