1. Given that $D$ is any point on side $A B$ of $\triangle A B C$ with area 1, $E$ is any point on side $A C$, and $F$ is any point on segment $D E$. Let $\frac{A D}{A B}=x, \frac{A E}{A C}=y, \frac{D F}{D E}=z$, and $y+z-x=\frac{1}{2}$. Try to find the maximum area of $\triangle B D F$. (2005 Hunan Province Mathematics Competition Problem)
1. Connect $B E$, then the area of $\triangle B D F$
$$S_{\triangle B D F}=z S_{\triangle B D E}=z(1-x) S_{\triangle A B D}=z(1-x) y S_{\triangle A B C}=z(1-x) y$$

By the AM-GM inequality, we get $z(1-x) y \leqslant\left(\frac{z+(1-x)+y}{3}\right)^{3}=\frac{1}{8}$. Equality holds if and only if $z=1-x=y, y+z-x=\frac{1}{2}$, i.e., $x=y=z=\frac{1}{2}$. Therefore, the maximum area of $\triangle B D F$ is $\frac{1}{8}$.