1. If real numbers $x, y$ satisfy $x^{2}+y^{2}+x y=1$, then the maximum value of $x+y$ is $\qquad$ .
$-1 . \frac{2 \sqrt{3}}{3}$.
From the condition, we know
$$
\begin{array}{l}
(x+y)^{2}=x y+1 \leqslant\left(\frac{x+y}{2}\right)^{2}+1 \\
\Rightarrow(x+y)^{2} \leqslant \frac{4}{3} \\
\Rightarrow-\frac{2 \sqrt{3}}{3} \leqslant x+y \leqslant \frac{2 \sqrt{3}}{3} .
\end{array}
$$

Thus, the maximum value of $x+y$ is $\frac{2 \sqrt{3}}{3}$.