\section*{Problem 4 - 201024}

Determine all real numbers \(z\) with \(0<z<1\) that, when added to their reciprocal, yield at least 4!
}

Since \(z>0\), it can be transformed as follows

\[
z+\frac{1}{z} \geq 4 \quad \Rightarrow \quad z^{2}+1 \geq 4 z \quad \Rightarrow \quad z^{2}-4 z+1 \geq 0
\]

The quadratic equation \(z^{2}-4 z+1=\) has the solutions

\[
0<z_{1}=2-\sqrt{3}<1 \quad \text { and } \quad 1<z_{2}=2+\sqrt{3}
\]

A quadratic function \(y=z^{2}-4 z+1=(z-2)^{2}-3\) has its roots at \(z_{1}\) and \(z_{2}\) and is an upward-opening parabola. Therefore, the function has negative function values in the interval \(\left(z_{1} ; 1\right)\).

The solutions sought in the problem statement are thus all real numbers \(0<z \leq 2-\sqrt{3}\).