In $\triangle ABC$, the area is $1, DE \parallel AB$, connect $BD$, let the largest area among $\triangle DCE$, $\triangle ABD$, and $\triangle BDE$ be $y$. Find the minimum value of $y$.

The area of $\triangle ABC$ is $1, DE \parallel AB$, connect $BD$, and let the largest area among $\triangle DCE$, $\triangle ABD$, and $\triangle BDE$ be $y$. Find the minimum value of $y$.
Let $\frac{C D}{A C}=x$. Since $\triangle C D E \sim \triangle A B C$, we have $S_{\triangle C D E}=x^{2}$, $\frac{A D}{A C}=1-x$. Therefore, $S_{\triangle A B D}=1-x$, $S_{\triangle B D E}=1-x^{2}-(1-x)=x-x^{2}=x(1-x)$, where $0<x<1$.

From $x(1-x)<1-x$, we know that $S_{\triangle B D E}$ cannot be the largest. When $S_{\triangle C D E}=S_{\triangle A B D}$, it is always possible to move $D E$ to make one of the triangles have a larger area. Therefore, the maximum area can only be achieved when the other two triangles have equal areas (i.e., both are at their maximum values). Thus, we have
$y=1-x=x^{2}, x=\frac{-1+\sqrt{5}}{2}$ (the negative value is discarded).
$$
\therefore y=1-x=\frac{3-\sqrt{5}}{2} \text {. }
$$

Therefore, the minimum value of $y$ is $\frac{3-\sqrt{5}}{2}$.
Note: Using a similar idea, the maximum value of the smallest area among the three can also be found to be $\frac{1}{4}$.
(Guo Wangming, Xiechang Middle School, Xianyang City, Hubei Province, 433017)