6. Given point $A(2,2), P(x, y)$, and $x, y$ satisfy
$$
\left\{\begin{array}{l}
0<x, y \leqslant 2, \\
x+y \geqslant 2, \\
\frac{1}{x}+\frac{1}{y} \geqslant 2 .
\end{array}\right.
$$

Then the range of $|P A|$ is $\qquad$
6. $\left[\frac{\sqrt{7}}{2}, 2\right)$.

Since $x, y > 0$, simplifying $\frac{1}{x} + \frac{1}{y} \geqslant 2$ yields $\left(x - \frac{1}{2}\right)\left(y - \frac{1}{2}\right) \leqslant \frac{1}{4}$.

In the Cartesian coordinate system $x O y$, the plane region under the constraint conditions is the shaded part shown in Figure 7, excluding points $M(2, 0)$ and $N(0, 2)$.

Clearly, $|P A|$ achieves its minimum value when $P$ is on the curve $C$:
$$
\left(x - \frac{1}{2}\right)\left(y - \frac{1}{2}\right) = \frac{1}{4}
$$

Below, we find the minimum value of $|P A|$.
$$
\begin{array}{l}
|P A|^2 = (x - 2)^2 + (y - 2)^2 \\
= \left(x - \frac{1}{2} - \frac{3}{2}\right)^2 + \left(y - \frac{1}{2} - \frac{3}{2}\right)^2 \\
= \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 - 3\left(x - \frac{1}{2}\right) - 3\left(y - \frac{1}{2}\right) + \frac{9}{2} \\
= \left[\left(x - \frac{1}{2}\right) + \left(y - \frac{1}{2}\right)\right]^2 - 3\left[\left(x - \frac{1}{2}\right) + \left(y - \frac{1}{2}\right)\right] + 4 \\
= \left[\left(x - \frac{1}{2}\right) + \left(y - \frac{1}{2}\right) - \frac{3}{2}\right]^2 + \frac{7}{4} \\
\geqslant \frac{7}{4}.
\end{array}
$$

The equality holds if and only if $\left(x - \frac{1}{2}\right) + \left(y - \frac{1}{2}\right) = \frac{3}{2}$. Solving the system of equations $\left\{\begin{array}{l}\left(x - \frac{1}{2}\right) + \left(y - \frac{1}{2}\right) = \frac{3}{2}, \\ \left(x - \frac{1}{2}\right)\left(y - \frac{1}{2}\right) = \frac{1}{4} .\end{array}\right.$
yields $\left\{\begin{array}{l}x = \frac{5 + \sqrt{5}}{4}, \\ y = \frac{5 - \sqrt{5}}{4}\end{array}\right.$ or $\left\{\begin{array}{l}x = \frac{5 - \sqrt{5}}{4}, \\ y = \frac{5 + \sqrt{5}}{4} .\end{array}\right.$
Upon verification, both solutions satisfy the conditions.
Therefore, $|P A|_{\min} = \frac{\sqrt{7}}{2}$.
Clearly, $|P A|$ achieves its maximum value of 2 when $P$ is at point $M$ or $N$. However, points $M$ and $N$ are not within the region, so the range of $|P A|$ is $\left[\frac{\sqrt{7}}{2}, 2\right)$.