For which values of $x$ will

$$
\frac{x-1}{x+1}>\frac{x}{x-1}
$$
By moving the right-hand side of the inequality to the left and bringing the two fractions to a common denominator, we obtain the following new inequality:

$$
\frac{1-3 x}{x^{2}-1}>0
$$

Thus, the fraction on the left is positive. In this case, the numerator and the denominator are of the same sign, meaning they are either both positive or both negative. Let's consider each case in turn.

$$
\begin{aligned}
& 1^{0} \quad 1-3 x>0 \quad \text { and } \quad x^{2}-1>0 \\
& \text { then: } \\
& x1 \quad \text { or } \quad x\frac{1}{3} \quad \text { and } \quad-1<x<1
\end{aligned}
$$

The three inequalities hold simultaneously for values of $x$ such that $\frac{1}{3}<x<1$, meaning all real fractions greater than $\frac{1}{3}$ satisfy the inequality.

(Béla Grünhut, fóreálisk. VIII. o. t. Pécs).

The problem was also solved by: Berháth Friedmann, S. A. Ujhely; Aladár Visnya, Pécs.