Let's determine all (real) numbers that satisfy the following inequality

$$
\frac{1}{\sqrt{1-x}}-\frac{1}{\sqrt{1+x}} \geq 1
$$
For the inequality to hold, it is necessary that the expressions under the square roots and the left side are positive, which means the first denominator must be less than the second, thus

$$
00$, hence $00$ inequality for $z$; by rearrangement,

$$
0 \geqq z^{2}+2 z-2=(z+1)^{2}-3=(z+1+\sqrt{3})(z+1-\sqrt{3}).
$$

The first factor on the right side is positive, so the second must be negative or equal to 0:

$$
\begin{gathered}
z+1-\sqrt{3}=\sqrt{1-x^{2}}+1-\sqrt{3} \leqq 0 \\
\sqrt{1-x^{2}} \leqq \sqrt{3}-1, \quad 1-x^{2} \leqq 4-2 \sqrt{3}, \quad x^{2} \geqq 2 \sqrt{3}-3 \\
x \geqq \sqrt{2 \sqrt{3}-3} \quad(=0.6812 \ldots)
\end{gathered}
$$

Accordingly, (1) holds if and only if $\sqrt{2 \sqrt{3}-3} \leqq x<1$.