1. Let $n$ be a natural number, $a, b$ be positive real numbers, and satisfy the condition $a+b=2$, then the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is $\qquad$.
1. To find the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$, we can first consider special cases,

When $a=1, b=1, n=1$, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=1$.
When $a=\frac{1}{2}, b=\frac{3}{2}, n=1$, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{2}{3}+\frac{2}{5}$.
From this, we can conjecture that $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}} \geqslant 1$. We will now prove this:
Since $a>0, b>0$, we have $\frac{a+b}{2} \geqslant \sqrt{a b}$.
Substituting $a+b=2$, we get $\sqrt{a b} \leqslant 1$, which means $a b \leqslant 1, a^{n} b^{n} \leqslant 1$.
Thus, $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+1+b^{n}}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant \frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+1}=1$. The equality holds if and only if $a=b=1$. Therefore, the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is 1.