1. If real numbers $x, y$ satisfy $x^{2}+y^{2}+x y=1$, then the maximum value of $x+y$ is $\qquad$
$1=(x+y)^{2}-x y \geqslant(x+y)^{2}-\left(\frac{x+y}{2}\right)^{2}=\frac{3}{4}(x+y)^{2} \Rightarrow x+y \leqslant \frac{2 \sqrt{3}}{3}$, equality holds when $x=y=\frac{\sqrt{3}}{3}$. Therefore, the maximum value of $x+y$ is $\frac{2 \sqrt{3}}{3}$.