Example 34 Let $n$ be a positive integer, $a, b$ be positive real numbers, and satisfy $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$ .
Since $a, b>0$, we have $a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1$, thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=$ $\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$, equality holds when $a=b=1$. Therefore, the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is 1.