Prove: From the 2-variable mean inequality and the $n$-variable mean inequality, we get

that is,
$$\left(1+x^{2}\right)\left(1+y^{2}\right)=1+x^{2}+y^{2}+x^{2} y^{2} \geq 1+2 x y+x^{2} y^{2}=(1+x y)^{2}$$

that is,
$$\begin{array}{l} \left(1+x^{2}\right)\left(1+y^{2}\right) \geq(1+x y)^{2} \text {. } \\ x^{n-1} y+x y^{n-1} \leq \frac{1}{n}\left[(n-1) x^{n}+y^{n}\right]+\frac{1}{n}\left[x^{n}+(n-1) y^{n}\right]=x^{n}+y^{n}, \\ x^{n-1} y+x y^{n-1} \leq x^{n}+y^{n} \text {. } \end{array}$$

Thus, $\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}}=\frac{x^{n}\left(1+y^{2}\right)+y^{n}\left(1+x^{2}\right)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$
$$\begin{array}{l} =\frac{\left(x^{n}+y^{n}\right)+x y\left(x^{n-1} y+x y^{n-1}\right)}{\left(1+x^{2}\right)\left(1+y^{2}\right)} \\ \leq \frac{\left(x^{n}+y^{n}\right)+x y\left(x^{n}+y^{n}\right)}{(1+x y)^{2}} \\ =\frac{x^{n}+y^{n}}{1+x y} \end{array}$$

Therefore, we have $\quad \frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}} \leq \frac{x^{n}+y^{n}}{1+x y}$.