Example 1 Let positive real numbers $x, y$ satisfy $x y=1$, find the range of the function $f(x, y)=\frac{x+y}{[x][y]+[x]+[y]+1}$ (where $[x]$ denotes the greatest integer not exceeding $x$)
By symmetry, without loss of generality, assume $x \geqslant y$. When $x=1, y=1$, $f(1,1)=\frac{1}{2}$.
When $x>1$, let $x=n+\alpha$, where $n=[x]$, then $f(x, y)=\frac{n+\alpha+\frac{1}{n+\alpha}}{n+1}$. It is easy to see that $g(x)=x+\frac{1}{x}(x \geqslant 1)$ is increasing. Thus, $f(x, y) \in\left[\frac{n+\frac{1}{n}}{n+1}, \frac{n+1+\frac{1}{n+1}}{n+1}\right]$. Let $a_{n}=\frac{n+\frac{1}{n}}{n+1}=\frac{n^{2}+1}{n^{2}+n}$, $b_{n}=\frac{n+1+\frac{1}{n+1}}{n+1}=1+\frac{1}{(n+1)^{2}}$, then $a_{1}>a_{2}=a_{3}, a_{3}b_{2}>b_{3}>\cdots$. Therefore, when $x>1$, the range of $f(x, y)$ is $\left[a_{2}, b_{1}\right)$, i.e., $\left[\frac{5}{6}, \frac{5}{4}\right)$. In summary, the range of $f(x, y)$ is $\left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right)$.