\section*{Problem 4 - 211034}

Determine all real numbers \(x\) that satisfy the inequality:

\[
\frac{1-\sqrt{1-3 x^{2}}}{x}<1
\]
}

Obviously, \(x\) cannot be 0, as otherwise the fraction would not be defined. Similarly, \(|x| \leq \frac{1}{\sqrt{3}}\) must hold, because for \(|x|>\frac{1}{\sqrt{3}}\) immediately \(x^{2}>\frac{1}{3}\) and thus \(1-3 x^{2}<0\). The fraction therefore has the same sign as the denominator \(x\). If this is negative, the inequality is obviously satisfied, so we obtain a first solution interval, namely \(x \in\left[-\frac{1}{\sqrt{3}} ; 0\right)\).

On the other hand, if \(0<x\), the fraction is positive and subtracting 1 is equivalent to \(-\sqrt{1-3 x^{2}}<1-x\). Since \(\frac{1}{\sqrt{3}}(1-x)^{2}=1-2 x+x^{2}\) or \(0>-2 x+4 x^{2}=(-2 x)(1-2 x)\).

Since \(x>0\), \(-2 x<0\) and \(1-2 x>0\) or \(x<\frac{1}{2}\). Because \(\frac{1}{2}<\frac{1}{\sqrt{3}}\) holds, indeed all positive \(x\) with \(x<\frac{1}{2}\) satisfy the inequality (and the others do not).

In summary, we obtain that exactly those \(x\) from the set \(\left[-\frac{1}{\sqrt{3}} ; \frac{1}{2}\right) \backslash\{0\}\) satisfy the inequality.