One, (20 points) Given $a^{2}+b^{2}=1$. For all real numbers $x$ satisfying $0 \leqslant x \leqslant 1$, the inequality
$$
\begin{array}{l}
a(1-x)(1-x-a x)-b x(b-x-b x) \\
\geqslant 0
\end{array}
$$

always holds. When the product $a b$ takes the minimum value, find the values of $a$ and $b$.
I. Organize equation (1) and substitute $a^{2}+b^{2}=1$ to get
$$
(1+a+b) x^{2}-(2 a+1) x+a \geqslant 0 \text {. }
$$

In equation (2), let $x=0$, we get $a \geqslant 0$; let $x=1$, we get $b \geqslant 0$.

It is easy to see that $1+a+b>0,0<\frac{2 a+1}{2(1+a+b)}<1$. Therefore, 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 is between 0 and 1.

According to the problem, equation (2) holds for all real numbers $x$ that satisfy $0 \leqslant x \leqslant 1$. Therefore,
$$
\Delta=(2 a+1)^{2}-4(1+a+b) a \leqslant 0,
$$

which means $a b \geqslant \frac{1}{4}$.
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 $a=\frac{\sqrt{6}-\sqrt{2}}{4}$ or $a=\frac{\sqrt{6}+\sqrt{2}}{4}$.
Therefore, the solutions to the system of equations (3) are
Thus, the minimum value of $a b$ is $\frac{1}{4}$, at which point, $a$ and $b$ have two sets of values, which are
$$
a=\frac{\sqrt{6}-\sqrt{2}}{4}, b=\frac{\sqrt{6}+\sqrt{2}}{4}
$$

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