\section*{Problem 1 - 111241}

All real numbers \(x\) are to be specified for which the expression

\[
\frac{2 x}{|x-3|-5}+\frac{1}{x+2}
\]

exists, and among these, all \(x\) are to be determined that satisfy the following inequality (2):

\[
\frac{2 x}{|x-3|-5}+\frac{1}{x+2} \geq 1
\]
}

In (1) all \(x\) can be inserted for which neither of the two denominators becomes 0, i.e., all \(x \in \mathbb{R} \backslash\{-2,8\}\). We distinguish the cases \(x \geq 3\) and \(x < 3\).

Case 1.1: For \(x \geq 3\), the denominator is positive, and inequality (2) is then equivalent to

\[
1-2 x \geq x+2 \Longleftrightarrow x \leq -\frac{1}{3}
\]

This case provides no solutions since \(x \geq 3\) and \(x \leq -\frac{1}{3}\) cannot be true simultaneously.

Case 1.2: For \(-2 < x < 3\), the denominator is positive, and inequality (2) is then equivalent to

\[
1-2 x \geq x+2 \Longleftrightarrow x \leq -\frac{1}{3}
\]

This case provides all \(x\) with \(-2 < x \leq -\frac{1}{3}\) as solutions of (2).

Case 2.2: For \(x < -2\), the denominator is negative, and the inequality sign reverses. Inequality (2) is then equivalent to \(1-2 x \leq x+2 \Longleftrightarrow x \geq -\frac{1}{3}\), which is not possible in this case.

In summary: Exactly all \(x \in \left(-2, -\frac{1}{3}\right] \cup (8, \infty)\) are solutions of (2).