1. Let $n$ be a natural number, $a, b$ 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 $\qquad$.
$$
\begin{array}{l}
\text { 1. } \because a b>0 \\
\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, \quad a^{n} b^{n} \leqslant 1 . \\
\text { So } \quad \frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1 .
\end{array}
$$

When $a=b=1$, the above expression $=1$. Therefore, the minimum value is 1.