1. When $a=1, b=2$, we have
$$
27 \geqslant 4 k \Rightarrow k \leqslant \frac{27}{4} \text {. }
$$

By the AM-GM inequality, we get
$$
\begin{array}{l}
(a+b)(a b+1)(b+1) \\
=\left(a+\frac{b}{2}+\frac{b}{2}\right)\left(\frac{a b}{2}+\frac{a b}{2}+1\right)\left(\frac{b}{2}+\frac{b}{2}+1\right) \\
\geqslant 3 \sqrt[3]{a \times \frac{b}{2} \times \frac{b}{2}} \times 3 \sqrt[3]{\frac{a b}{2} \times \frac{a b}{2} \times 1} \times \\
3 \sqrt[3]{\frac{b}{2} \times \frac{b}{2} \times 1} \\
=\frac{27}{4} a b^{2} \text {. } \\
\end{array}
$$

In conclusion, the maximum value of $k$ is $\frac{27}{4}$.
1. Find the largest real number $k$, such that for any positive numbers $a, b$, we have
$$
(a+b)(a b+1)(b+1) \geqslant k a b^{2} .
$$