1. (1998 Hunan Province Competition Question) Given $x, y \in (0, +\infty)$, and $\frac{19}{x}+\frac{98}{y}=1$, then what is the minimum value of $x+y$?
1. Let $\vec{a}=\left(\sqrt{\frac{19}{x}}, \sqrt{\frac{98}{y}}\right), \vec{b}=(\sqrt{x}, \sqrt{y})$,
$$
\begin{array}{l}
1=\frac{19}{x}+\frac{98}{y}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(\sqrt{19}+\sqrt{98})^{2}}{x+y}=\frac{117+4 \sqrt{38}}{x+y}, \\
x+y \geqslant 117+14 \sqrt{38},
\end{array}
$$

Therefore, the minimum value of $x+y$ is $117+14 \sqrt{38}$.