3. Find the numerical value of the expression

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

if it is known that $x$ is not equal to $y$ and the sum of the first two terms is equal to the third.
Answer: 2.

Solution. Let's bring the condition to a common denominator

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

we get

$$
\frac{\left(x^{2}+y^{2}+2\right)(x y+1)-2\left(x^{2}+1\right)\left(y^{2}+1\right)}{\left(x^{2}+1\right)\left(y^{2}+1\right)(x y+1)}=0
$$

expand all brackets in the numerator, combine like terms, we get the equality $(x-y)^{2}(x y-1)=0$. Since $x \neq y$, then $x y=1$. From this, the third term equals 1, and then the entire sum equals 2.

Criteria. Full solution - 7 points. Correctly brought to a common denominator, like terms correctly combined, but without further progress - 1 point. Reasoning of the type "if $x$ is taken this way, and $y$ another way, then something will result" is rated 0 points.