Problem 11.4. For positive numbers $x$ and $y$, it is known that

$$
\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}=1
$$

What values can the product $x y$ take? List all possible options and prove that there are no others.
Answer: 1.

Solution. Note that for each positive $y$ the function

$$
f_{y}(x)=\frac{1}{1+x+x^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+x+y}
$$

strictly decreases on the ray $(0 ;+\infty)$, since the denominators of all three fractions increase. Therefore, the function $f_{y}$ takes each value no more than once. Moreover, it is not difficult to see that

$$
\begin{aligned}
f_{y}(1 / y) & =\frac{1}{1+1 / y+1 / y^{2}}+\frac{1}{1+y+y^{2}}+\frac{1}{1+1 / y+y}= \\
& =\frac{y^{2}}{1+y+y^{2}}+\frac{1}{1+y+y^{2}}+\frac{y}{1+y+y^{2}}=1
\end{aligned}
$$

from which we conclude that $x=1 / y$.

## Criteria

One of the largest suitable criteria is used:

7 6. Any correct solution to the problem.

3 6. The monotonicity of the left side with respect to $x$ or $y$ is noted.

3 6. It is proven that for any $y$ there is no more than one $x$ satisfying the equation.