1. Given $x, y \in R_{+}$, and $x+y<27$. Then the minimum value of $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x y}}+\frac{1}{\sqrt{27-x-y}}$ is
1. 1

Analysis: By Cauchy-Schwarz inequality:
$$
\begin{array}{l}
\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x y}}+\frac{1}{\sqrt{27-x-y}}=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{27-x-y}} \geq \frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{27-x-y}} \\
\geq \frac{9}{\sqrt{(x+y+27-x-y)(1+1+1)}}=1, \text { with equality if and only if } x=y=9
\end{array}
$$