8. Given $a, b \in \mathbf{R}^{+}, a+b=1, a b+\frac{1}{a b}$ the minimum value is $\qquad$ $-$
8. $\frac{17}{4}$.
$t=a b \leqslant\left(\frac{a+b}{2}\right)^{2}=\frac{1}{4}$. Also, $t+\frac{1}{t}$ is monotonically decreasing on $\left(0, \frac{1}{4}\right]$, so when $t=a b=\frac{1}{4}$, there is a maximum value.