Three, (18 points) Given that the line $l$ passes through the point $P(2,1)$, intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and $P A = P B$.
(1) Find the function expression of the line $l$;
(2) Let $\odot Q$ be the incircle of the right triangle $\triangle A O B$, which is tangent to $O A$, $O B$, and $A B$ at points $D$, $E$, and $F$ respectively. Prove that the lengths of $A D$ and $B E$ are the two roots of the equation $x^{2} - 2 \sqrt{5} x + 4 = 0$.
(1) As shown in Figure 6, establish a coordinate system. Construct $\mathrm{Rt} \triangle A O B$ according to the problem. Draw $P H \perp O A$, with the foot of the perpendicular at $H$.

Since $P A = P B$, it follows that $O H = H A$.
Therefore, $A(4,0)$.

Let the function of line $l$ be $y = k x + b$.
Since points $A(4,0)$ and $P(2,1)$ lie on line $l$, we have:
Thus, the function of line $l$ is $y = -\frac{1}{2} x + 2$.
(2) From (1), in $\mathrm{Rt} \triangle A O B$, $A O = 4$, $B O = 2$, and $A B = 2 \sqrt{5}$.
Since $\odot Q$ is the incircle of $\mathrm{Rt} \triangle A O B$, we have:
$A D = A F$, $B E = B F$, and $O D = O E$.
Therefore, $A D + B E = A F + B F = A B = 2 \sqrt{5}$.
Observing, in a right triangle, the relationship between the inradius $r$ and the three sides is:
$$
\begin{array}{l}
O D = \frac{A O + B O - A B}{2} \text{ (i.e., } r = \frac{a + b - c}{2} \text{) } \\
= \frac{4 + 2 - 2 \sqrt{5}}{2} = 3 - \sqrt{5}.
\end{array}
$$

Thus, $A D = A O - O D = 4 - (3 - \sqrt{5}) = 1 + \sqrt{5}$,
$$
B E = B O - O E = 2 - (3 - \sqrt{5}) = \sqrt{5} - 1.
$$

Therefore, $A D \cdot B E = (\sqrt{5} + 1)(\sqrt{5} - 1) = 4$.
By the relationship between the roots and coefficients of a quadratic equation, $A D$ and $B E$ are the two roots of the equation $x^{2} - 2 \sqrt{5} x + 4 = 0$.