3. Find the numerical value of the expression

$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{2}{a b+1}
$$

if it is known that $a$ is not equal to $b$ 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

$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}=\frac{2}{a b+1}
$$

we get

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

expand all brackets in the numerator, combine like terms, we get the equality $(a-b)^{2}(a b-1)=0$. Since $a \neq b$, then $a b=1$. 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 $a$ is taken this way, and $b$ another way, then something will result" is rated 0 points.