10. (19th Austrian Mathematical Competition) In the region $\{(x, y) \mid x, y>0, x y=1\}$, find the range of the function
$$
f(x, y)=\frac{x+y}{[x][y]+[x]+[y]+1}
$$

where $[\alpha]$ denotes the integer part of $\alpha$.
By symmetry, let $x \geqslant 1$.
(1) When $x=1, y=1$, $f(1, 1)=\frac{1}{2}$.
(2) When $x>1$, let $x=n+\alpha$ $(n \geqslant 1, n \in \mathbb{N}, 0<\alpha<1)$. If $f(x, y)$ is an increasing function when $x>1$, then
$$
f(x, y) \in\left(\frac{n+1}{n+1}, \frac{n+1+\frac{1}{n+1}}{n+1}\right).
$$

Let $a_{n}=\frac{n+\frac{1}{n}}{n+1}, b_{1}=\frac{n+1+\frac{1}{n+1}}{n+1}$, then
$$
\begin{array}{l}
a_{n}-a_{n+1}=\frac{2-n}{n(n+1)(n+2)}, \\
b_{n}-b_{n+1}=\frac{2 n+3}{(n+1)^{2}(n+2)^{2}}>0.
\end{array}
$$

Therefore, when $n \geqslant 1$, we have
$$
\begin{array}{l}
a_{1}>a_{2}=a_{3}<a_{4}<a_{5}<\cdots, \\
b_{1}>b_{2}>b_{3}>b_{4}>b_{5}>\cdots.
\end{array}
$$

Thus, when $x>1$, the range of $f(x, y)$ is $\left(a_{2}, b_{1}\right)$, i.e., $\left(\frac{5}{6}, \frac{5}{4}\right)$.

Combining (1) and (2), the range of $f(x, y)$ is $\left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right)$.