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
1
II. Fill-in-the-blank Questions
1. 【Analysis and Solution $\because a b>0$, $\therefore a b \leqslant\left(\frac{a+b}{2}\right)^{2}=1, a^{n} b^{n} \leqslant 1$. Therefore, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$. When $a=b=1$, the above expression $=1$, hence the minimum value is 1.