11. Let $a, b$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-$ $a-2 b$ is $\qquad$.
11 .
$$
\begin{array}{l}
a^{2}+a b+b^{2}-a-2 b \\
=a^{2}+(b-1) a+b^{2}-2 b \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4} b^{2}-\frac{3}{2} b-\frac{1}{4} \\
=\left(a+\frac{b-1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 .
\end{array}
$$

When $a+\frac{b-1}{2}=0, b-1=0$,
i.e., $a=0, b=1$, the equality in the above inequality holds, so the minimum value sought is -1.