1. Given vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $|\boldsymbol{a}+\boldsymbol{b}|=1$. Then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{b}$ is
$$
\begin{array}{l}
\text {-1. } \frac{1}{4} \\
a \cdot b=\frac{(a+b)^{2}}{4}-\frac{(a-b)^{2}}{4} \\
=\frac{1}{4}-\frac{(a-b)^{2}}{4} \leqslant \frac{1}{4}
\end{array}
$$

When $a$ and $b$ are equal and $|a|=\frac{1}{2}$, the equality holds.