Example 12 Let $x, y, z$ be real numbers greater than -1. Find the minimum value of
$$\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}}$$
Given $x, y, z > -1$, then $\frac{1+x^{2}}{1+y+z^{2}}, \frac{1+y^{2}}{1+z+x^{2}}, \frac{1+z^{2}}{1+x+y^{2}}$ have positive numerators and denominators, so
$$\begin{aligned}
& \frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \\
\geqslant & \frac{1+x^{2}}{1+z^{2}+\frac{1+y^{2}}{2}}+\frac{1+y^{2}}{1+x^{2}+\frac{1+z^{2}}{2}}+\frac{1+z^{2}}{1+y^{2}+\frac{1+x^{2}}{2}} \\
= & \frac{2 a}{2 c+b}+\frac{2 b}{2 a+c}+\frac{2 c}{2 b+a}
\end{aligned}$$

where $a=\frac{1+x^{2}}{2}, b=\frac{1+y^{2}}{2}, c=\frac{1+z^{2}}{2}$.
By the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
& \frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \\
\geqslant & \frac{(a+b+c)^{2}}{a(b+2 c)+b(c+2 a)+c(a+2 b)} \\
= & \frac{3(a b+b c+a c)+\frac{1}{2}\left[(b-c)^{2}+(c-a)^{2}+(a-b)^{2}\right]}{3(a b+b c+a c)} \\
\geqslant & 1
\end{aligned}$$

and the minimum value is achieved when $a=b=c=1$.
Therefore, the minimum value sought is 2.