Example 5 Let $x, y \in (0, +\infty)$, and $\frac{19}{x} + \frac{98}{y} = 1$. Find the minimum value of $x + y$.
(1998, Hunan Province High School Mathematics Competition)
Solution: Since $x, y \in \mathbf{R}_{+}$, we have
$$
1=\frac{19}{x}+\frac{98}{y} \geqslant \frac{(\sqrt{19}+\sqrt{98})^{2}}{x+y} \text {, }
$$

which means $x+y \geqslant(\sqrt{19}+\sqrt{98})^{2}=117+14 \sqrt{38}$.
Equality holds if and only if $\frac{\sqrt{19}}{x}=\frac{\sqrt{98}}{y}$, i.e.,
$$
x=\sqrt{19}(\sqrt{19}+\sqrt{98}), y=\sqrt{98}(\sqrt{19}+
$$
$\sqrt{98})$.
$$
\text { Therefore, }(x+y)_{\min }=117+14 \sqrt{38} \text {. }
$$