\section*{Problem 1 - 021231}

For which values of \(x\) does the following inequality hold

\[
\frac{1}{\sqrt{1+x}}-\frac{1}{\sqrt{1-x}} \geq 1
\]
}

First, it is important to note that the range of \(x\) is already restricted by the formula as follows: \(-1<x<1\). Now, we seek the boundary points of the inequality, i.e., we first calculate:

\[
\begin{array}{r}
\frac{1}{\sqrt{1+x}}-\frac{1}{\sqrt{1-x}}=1 \\
\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x^{2}}}=1 \\
\frac{1-x+1+x+2 \sqrt{1-x^{2}}}{1-x^{2}}=1 \\
\frac{2-2 \sqrt{1-x^{2}}}{1-x^{2}}=1 \\
2-2 \sqrt{1-x^{2}}=1-x^{2} \\
0=1-x^{2}-2+2 \sqrt{1-x^{2}}
\end{array}
\]

Now, we substitute with \(z=\sqrt{1-x^{2}}\).

\[
0=z^{2}+2 z-2 \Rightarrow z_{1,2}=-1 \pm \sqrt{3}
\]

Using the substitution, \(x\) can be determined: \(x=\sqrt{1-z^{2}}\).

For \(z_{2}=-1-\sqrt{3}\), no \(x\) can be calculated. For \(z_{1}=-1+\sqrt{3}\), we get \(x \approx \pm 0.6813\). To clarify where the actual boundary lies (squaring can introduce extraneous solutions) and how the curve behaves, we substitute values from each interval into the equation.

\[
\begin{array}{rrr}
x=-0.7: & 1.059 \geq 1 & \text { (true) } \\
x=0: & 0 \geq 1 \quad & \text { (false) } \\
x=0.7: & -1.059 \geq 1 & \text { (false) }
\end{array}
\]

Thus, the inequality holds for \(-1<x \leq -0.6813\).