Four. (20 points)
Determine all possible values of the parameter $a$ such that for all $x, y$ we have $2 a x^{2}+2 a y^{2}+4 a x y-2 x y-y^{2}-2 x+1 \geqslant 0$.


The original inequality is equivalent to:
$$
2 a(x+y)^{2} \geqslant(x+y)^{2}-(x-1)^{2} .
$$

If $x=-y$, then (1) becomes: $0 \geqslant-(x-y)^{2}$
Then (1) holds for any $a$.
If $x \neq -y$, from (1) we get $a \geqslant \frac{1}{2}-\frac{(x-1)^{2}}{2(x+y)^{2}}$.
Taking $x=1$, we get $a \geqslant \frac{1}{2}$.
Conversely, if $a \geq \frac{1}{2}$, then for all $x \neq -y$, we have
$$
a \geqslant \frac{1}{2}-\frac{(x-1)^{2}}{2(x+y)^{2}}
$$

Thus, (1) holds in this case.
In summary, $a \geqslant \frac{1}{2}$.