Let's calculate the value of the expression

$$
\frac{1}{1+x+x y}+\frac{1}{1+y+y z}+\frac{1}{1+z+z x}
$$

given that $x y z=1$.
Let's extend the second term by $x$, and the third term by $x y$. Thus, in the denominator, we can use the assumption in the term of at least the third degree:

$$
\begin{gathered}
\frac{1}{1+y+y z}=\frac{x}{x+x y+x y z}=\frac{x}{x+x y+1} \\
\frac{1}{1+z+z x}=\frac{x y}{x y+x y z+x^{2} y z}=\frac{x y}{x y+1+x}
\end{gathered}
$$

We immediately see that in this way, the three denominators are equal. The sum of the three numerators is also the same, so the value of the expression is $1$, provided, of course, that the denominator is not 0.

The extension is allowed because, according to the assumption, neither $x$ nor $x y$ can be 0.

Orava Sándor (Balassagyarmat, Szántó Kovács J. g I. o. t.)