Three. (50 points)
The area of $\triangle ABC$ is 1, and there are two points $D, E$ inside it. Among the 10 triangles formed by any three points chosen from $D, E$, and $A, B, C$, the minimum area is $s$. Find the maximum value of $s$.

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Three, (50 points)
As shown in the figure, let $\frac{A M}{A B}=x, \frac{A N}{A C}=y, \frac{M D}{M N}=m, \frac{E N}{M N}=n$, then from $S_{\triangle A B D} \geqslant s, S_{\triangle A E C} \geqslant s, S_{\triangle A D E} \geqslant s, S_{\triangle B K C} \geqslant s$, we get
$$
\left\{\begin{array}{l}
y m \geqslant s \\
x n \geqslant s \\
x(1-y)(1-m-n) \geqslant s \\
y(1-\pi)(1-m-n) \geqslant s
\end{array}\right.
$$

From (1), (2) we get $m \geqslant \frac{s}{y}, n \geqslant \frac{s}{x}$, substituting into (3), (4), we get
$$
\left\{\begin{array}{l}
x(1-y)\left(1-\frac{s}{y}-\frac{s}{x}\right) \geqslant s \\
y(1-x)\left(1-\frac{s}{y}-\frac{s}{x}\right) \geqslant s
\end{array}\right.
$$

So $\left\{\begin{array}{l}\frac{x y(1-y)}{x+2 y-y(x+y)} \geqslant s \\ \frac{x y(1-x)}{2 x+y-x(x+y)} \geqslant s\end{array}\right.$
So $\frac{x y(1-y)+x y(1-x)}{[x+2 y-y(x+y)]+[2 x+y-x(x+y)]} \geqslant s$, i.e., $\frac{2 x y-x y(x+y)}{3(x+y)-(x+y)^{2}} \geqslant s$,
Using $x y \leqslant\left(\frac{x+y}{2}\right)^{2}$, we get
$\frac{(x+y)[2-(x+y)]}{4[3-(x+y)]} \geq s$.
Let $x+y=t$, then $\frac{t(2-t)}{4(3-t)} \geq s$,
By the discriminant method, we get $s \leqslant 1-\frac{\sqrt{3}}{2}$.
When $x=y=\frac{3-\sqrt{3}}{2}, m=n=\frac{3-\sqrt{3}}{6}$, $s$ can achieve the maximum value $1-\frac{\sqrt{3}}{2}$.