2. If positive real numbers $a, b$ satisfy $a+b=1$, then the minimum value of $\left(1+\frac{1}{a}\right)\left(1+\frac{8}{b}\right)$ is

保留源文本的换行和格式，翻译结果如下：

2. If positive real numbers $a, b$ satisfy $a+b=1$, then the minimum value of $\left(1+\frac{1}{a}\right)\left(1+\frac{8}{b}\right)$ is
2. 50 .

Substitute $a=1-b$, then we only need to consider the function defined on $(0,1)$:
$$
f(x)=\left(1+\frac{1}{x-1}\right)\left(1+\frac{8}{x}\right)=\frac{(2-x)(x+8)}{x(1-x)} .
$$

It is easy to see that $f^{\prime}(x)=\frac{(7 x-4)(4-x)}{x^{2}(1-x)^{2}}$, which shows that when $x \in\left(0, \frac{4}{7}\right)$, $f^{\prime}(x)0$, i.e., $f(x)$ is monotonically increasing on $\left(\frac{4}{7}, 1\right)$. Therefore, the minimum value of $f(x)$ is $f\left(\frac{4}{7}\right)=50$.