2. If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-$ $xy+y^{2}$ is $\qquad$
2. 1 .

Notice that
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also }|x \pm y| \leqslant|x|+|y| \leqslant 1 \text {, then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$

When $x$ and $y$ take 0 and 1 respectively, the equality holds.