One, (20 points) Given $a^{2}+b^{2}=1$. For all real number pairs $(x, y)$ satisfying $x+y=1, x y \geqslant 0$, the inequality
$$
a y^{2}-x y+b x^{2} \geqslant 0
$$

always holds. When the product $a b$ takes the minimum value, find the values of $a$ and $b$.
Given $x+y=1, x y \geqslant 0$, we know that
$0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1$.
In equation (1), let $x=0, y=1$, we get $a \geqslant 0$; let $x=1, y=0$, we get $b \geqslant 0$.
Substituting $y=1-x$ into equation (1) yields
$$
a(1-x)^{2}-x(1-x)+b x^{2} \geqslant 0,
$$

which simplifies to $(1+a+b) x^{2}-(2 a+1) x+a \geqslant 0$.
It is easy to see that $1+a+b>0,0<\frac{2 a+1}{2(1+a+b)}<1$, so the graph (parabola) of the quadratic function $y=(1+a+b) x^{2}-(2 a+1) x+a$ opens upwards, and the x-coordinate of the vertex lies between 0 and 1.

By the problem's condition, inequality (2) holds for all real numbers $x$ satisfying $0 \leqslant x \leqslant 1$, so the discriminant
$$
\Delta=(2 a+1)^{2}-4(1+a+b) a \leqslant 0,
$$

which simplifies to $a b \geqslant \frac{1}{4}$.
From the system of equations
$$
\left\{\begin{array}{l}
a^{2}+b^{2}=1, \\
a b=\frac{1}{4},
\end{array}\right.
$$

eliminating $b$ yields $16 a^{4}-16 a^{2}+1=0$.
Thus, $a^{2}=\frac{2-\sqrt{3}}{4}$ or $a^{2}=\frac{2+\sqrt{3}}{4}$.
Since $a \geqslant 0$, we have
$$
a=\frac{\sqrt{6}-\sqrt{2}}{4} \text { or } a=\frac{\sqrt{6}+\sqrt{2}}{4} \text {. }
$$

Therefore, the solutions to the system of equations (3) are

Thus, there are two sets of values for $a$ and $b$ that satisfy the conditions, which are
$$
a=\frac{\sqrt{6}-\sqrt{2}}{4}, b=\frac{\sqrt{6}+\sqrt{2}}{4}
$$

and $a=\frac{\sqrt{6}+\sqrt{2}}{4}, b=\frac{\sqrt{6}-\sqrt{2}}{4}$.