3. Given $x, y \geq 1, x+y=8$, then the maximum value of $\left|\sqrt{x-\frac{1}{y}}+\sqrt{y-\frac{1}{x}}\right|$ is
Answer: $\sqrt{15}$
Difficulty: Easy
Examining: Monotonic function, Mean inequality
$$
\begin{array}{l}
\text { Analysis: Let } z=\left|\sqrt{x-\frac{1}{y}}+\sqrt{y-\frac{1}{x}}\right|, \\
z^{2}=x+y-\frac{x+y}{x y}+2 \sqrt{x y+\frac{1}{x y}-\frac{(x+y)^{2}-2 x y}{x y}} \\
=8-\frac{8}{x y}+2 \sqrt{x y-\frac{63}{x y}+2}
\end{array}
$$

Notice that $z^{2}$ is monotonically increasing with respect to $x y$, and $x y \leq\left(\frac{x+y}{2}\right)^{2}=16$. Therefore, $z \leq \sqrt{15}$, and when $x=y=4$, $\left|\sqrt{x-\frac{1}{y}}+\sqrt{y-\frac{1}{x}}\right|=\sqrt{15}$. Hence, the maximum value of $\left|\sqrt{x-\frac{1}{y}}+\sqrt{y-\frac{1}{x}}\right|$ is $\sqrt{15}$.