Example 9. Let $x>0$. Try to find the maximum value of $y=\sqrt{x}+\frac{1}{\sqrt{x}}-$ $\sqrt{x+\frac{1}{x}+1}$.
To construct the dual form of
$$
y=\sqrt{x}+\frac{1}{\sqrt{x}}-\sqrt{x+\frac{1}{x}+1}
$$

is
$$
u=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x+\frac{1}{x}+1},
$$

then \( y u=1, y=\frac{1}{u} \).
$$
\because u \geqslant 2+\sqrt{2+1}=2+\sqrt{3} \text{ (when } x=1
$$

is achieved),
$$
\therefore y=\frac{1}{u} \leqslant \frac{1}{2+\sqrt{3}}=2-\sqrt{3} \text{. (when } x=1
$$

is achieved)
Therefore, the maximum value of \( y \) is \( 2-\sqrt{3} \).