7. 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 $\qquad$ .
$2=a+b \geqslant 2 \sqrt{a b} \Rightarrow a b \leqslant 1 \Rightarrow a^{n} b^{n} \leqslant 1$, equality holds when $a=b=1$.
Thus $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{2+a^{n}+b^{n}}{\left(1+a^{n}\right)\left(1+b^{n}\right)}=\frac{2+a^{n}+b^{n}}{1+a^{n}+b^{n}+a^{n} b^{n}} \geqslant 1$,
therefore, the minimum value of $\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}$ is 1.