4. 2. 13 * Given $x, y \in \mathbf{R}^{+}$, and $x+y=1$, find the minimum value of $\sqrt{x+\frac{1}{x}}+\sqrt{y+\frac{1}{y}}$.
Let $t=\sqrt{x+\frac{1}{x}}+\sqrt{y+\frac{1}{y}}$, then
$$
\begin{aligned}
t^{2}=\left(\sqrt{x+\frac{1}{x}}+\sqrt{y+\frac{1}{y}}\right)^{2} & =x+y+\frac{x+y}{x y}+2 \sqrt{\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)} \\
& =1+\frac{1}{x y}+2 \sqrt{x y+\frac{1}{x y}+\frac{y}{x}+\frac{x}{y}},
\end{aligned}
$$

Since $x y \leqslant \frac{1}{4}$, we have $x y+\frac{1}{x y} \geqslant \frac{1}{4}+4=\frac{17}{4}, \frac{y}{x}+\frac{x}{y} \geqslant 2$. Therefore, $t^{2} \geqslant 1+4+$
$$
2 \sqrt{\frac{17}{4}+2}=10.
$$

Thus, the desired minimum value is $\sqrt{10}$.