3. Find all possible values of the expression

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

where $x$ and $y$ are any real numbers satisfying the condition $x+y \geq 1$.
3. To determine the range of values of a given expression, we will find for each $s \geq 1$ all numbers $p$ for which the system of equations

$$
x+y=s, \quad \frac{x+y}{x^{2}+y^{2}}=p
$$

has solutions in the set of real numbers (the assumption $s \geq 1$ implies that $p>0$).

From the first equation, we express $y=s-x$ and substitute it into the second equation. After simplification, we obtain a quadratic equation in the unknown $x$:

$$
2 p x^{2}-2 p s x+s(p s-1)=0 \text{. }
$$

This equation has real solutions if and only if its discriminant $D=4 p s(2-p s)$ is non-negative. Since both numbers $s$ and $p$ are positive, the inequality $D \geq 0$ holds if and only if $p s \leq 2$. This means that the system of equations has real solutions for every $p \leq \frac{2}{s} \leq 2$ and specifically for $s=1$ for every positive $p \leq 2$.

Thus, the expression in question takes all values from the interval $(0,2]$.

Another solution. From the Cauchy-Schwarz inequality $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right)$, which holds for any real numbers $x$ and $y$, we get the estimate

$$
0<\frac{x+y}{x^{2}+y^{2}} \leq \frac{x+y}{\frac{1}{2}(x+y)^{2}}=\frac{2}{x+y} \leq 2
$$

under the assumption $x+y \geq 1$. Conversely, for every $p \in(0,2]$, it suffices to set, for example, $x=y=\frac{1}{p}$. Then

$$
\frac{x+y}{x^{2}+y^{2}}=p \quad \text{ and } \quad x+y=\frac{2}{p} \geq 1.
$$

Thus, the set of values of the expression is the half-open interval $(0,2]$.

For a complete solution, 6 points are awarded. For proving that the values of the expression fall within the interval $(0,2]$, 4 points are awarded, and for proving that for every $p \in(0,2]$ there exist corresponding $x$ and $y$ such that the expression equals $p$, 2 points are awarded.