1.5.2 * Let real numbers $a, x, y$ satisfy the following conditions
$$
\left\{\begin{array}{l}
x+y=2 a-1, \\
x^{2}+y^{2}=a^{2}+2 a-3 .
\end{array}\right.
$$

Find the minimum value that the real number $xy$ can take.
By (1) $)^{2}-$ (2) we get
$$
x y=\frac{1}{2}\left((2 a-1)^{2}-\left(a^{2}+2 a-3\right)\right)=\frac{1}{2}\left(3 a^{2}-6 a+4\right) \text {. }
$$

Combining $x^{2}+y^{2} \geqslant 2 x y$, we know $a^{2}+2 a-3 \geqslant 3 a^{2}-6 a+4$, solving this yields $2-\frac{\sqrt{2}}{2} \leqslant a \leqslant 2+\frac{\sqrt{2}}{2}$. $x y=\frac{3}{2}(a-1)^{2}+\frac{1}{2} \geqslant \frac{3}{2}\left(2-\frac{\sqrt{2}}{2}-1\right)^{2}+\frac{1}{2}$, thus the minimum value of $x y$ is reached when $a=2-\frac{\sqrt{2}}{2}$, and the required minimum value is $\frac{11-6 \sqrt{2}}{4}$.