32. Suppose that $n$ is a positive integer, and $a, b$ are positive real numbers with $a+b=2$. Find the smallest possible value of
$$
\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}} \text {. }
$$
32. Ans: 1 .
Note that $a b \leq 1$ and $(a b)^{n} \leq 1$.
$$
\frac{1}{1+a^{n}}+\frac{1}{1+b^{n}}=\frac{1+a^{n}+b^{n}+1}{1+a^{n}+b^{n}+(a b)^{n}} \geq 1 .
$$

When $a=b=1$, we get 1 . Thus, the smallest value is 1 .