11.1. Find the value of the expression $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{2}{1+x y}$, given that $x \neq y$ and the sum of the first two terms of the expression is equal to the third.
Answer: 2.

Solution. Let's write the condition of the equality of the sum of the first two terms to the third one as: $\frac{1}{1+x^{2}}-\frac{1}{1+x y}=\frac{1}{1+x y}-\frac{1}{1+y^{2}} \quad$ and bring it to a common denominator:
$\frac{x(y-x)}{(1+x^{2})(1+x y)}=\frac{y(y-x)}{(1+y^{2})(1+x y)}$. Given $x \neq y$, we cancel $x-y$ and $1+x y$: $x(1+y^{2})=y(1+x^{2})$. The latter is equivalent to $(x-y)(x y-1)=0$, and again we cancel $x-y$, obtaining $x y=1$ and the desired expression equals $\frac{2}{1+x y} \cdot 2=2$.

Grading criteria. Any suitable $x \neq y$ guessed and the answer: 1 point.